Regularised integrals, sums and traces.

An analytic point of view

(Preliminary draft)

Sylvie PAYCHA

July 15, 2010

Dedicated to my late uncle Francois Paycha †1. May 20081

Introduction

Regularisation techniques, implemented in quantum field theory, number theory andgeometry, might seem very arbitrary and uncanonical at first glance. They neverthe-less conceal canonical concepts, such as canonical integrals, sums and traces, which wewant to bring to the forefront in these lectures. Cut-off and dimensional regularisationare prototypes of regularisation techniques used in quantum field theory2. But we alsohave in mind Riesz3 and Hadamard finite parts4 methods used in number theory5.Regularisation techniques also comprise zeta regularisation used in physics in the formof zeta determinants to compute effective actions6, or in geometry7 and particularlyin the context of infinite dimensional manifolds8 and index theory as a substitute forthe equivalent heat-kernel methods 9.Regularised integrals, discrete sums and traces obtained by means of a regularisationprocedure present many discrepancies responsible for various anomalies10. In contrast,the underlying canonical integrals, discrete sums and traces are well-behaved. Canon-ical integrals are indeed covariant, translation invariant, and obey Stokes’ property,

1who did not live long enough to see this manuscript finished. He himself was a doctor with noserious acquaintance with mathematics but he showed unlimited curiosity for scientific research of allsorts.

2Just to quote a few books amongst the vast literature on the subject, see e.g. [Col], [CMa],[D], [Sm1],[Sm2] as well as more specific references in the context of renormalisation such as[Et],[CMa],[He],[HV],[Sp], [Z]

3also called modified dimensional regularisation.4which amounts to cut-off regularisation.5See e.g. [Ca] for an introductory presentation.6starting with pioneering work by Hawkins [Haw], see other applications in [El], [EORZ] and

further developments in string theory, see e.g. [D] and [AJPS] for a mathematical presentation.7with the work of Ray and Singer [RaSi] on analytic torsion where the zeta determinant was first

introduced in mathematics8e.g. for the geometry of loop groups, see [Fr]9starting with pioneering work by Atiyah and Singer [APS1, APS2, APS3] and later by Quillen

[Q], Bismut and Freed [BF], see also more recent work by Scott [Sc1] in the context of the familyindex theorem.

10For a treatment of anomalies in physics (see e.g. [D] and [N] for a mathematical presentation)from the point of view of discrepancies also called trace anomalies, see e.g. [CDP], [Mi]

1

canonical discrete sums are Zd-translation invariant, and canonical traces vanish oncommutators.Since we focus on ultraviolet divergences, namely divergences for large values of themomentum, it seems reasonable to pick out a specific class of functions whose control-lable behaviour in the large will enable them to be integrated and summed up usingappropriate methods. It turns out that functions of the form σs(ξ) = (1+|ξ|2)−

s2 which

arise in Feynman integrals for s = 2, functions of the form τs(ξ) = |ξ|−s χ(ξ) where χis a smooth cut-off function that gets rid of infrared divergences, which arise in num-ber theory for negative integer values of s, and operators of the form As = (∆ + 1)−

s2

(whose symbol is σs) for a generalised Laplacian ∆ and some integer s, which arisein infinite dimensional geometry and index theory for integer values of s, are all ofpseudodifferential nature. Classical and more generally, log-polyhomogeneous pseu-dodifferential symbols and operators form a natural class to consider in the frameworkof regularisation.The symbols and operators of the type mentioned above that one encounters typi-cally have integer order (here −s), a feature which is the main source of anomalies inphysics and the cause of many a discrepancy. These obstacles disappear when workingwith non-integer order symbols and operators, for which integrals, sums and tracesare canonically defined. The basic idea behind dimensional, Riesz or zeta regularisa-tion is to embed integer order symbols σ or operators A inside holomorphic familiesof symbols σ(z) or operators A(z) so as to perturb the order of the symbol or theoperator away from integers. In the examples mentioned above, natural holomorphicextensions are σs(z) = (1 + |ξ|2)−

s−z2 , τs(z) = |ξ|−(s+z) χ(ξ) and As(z) = (∆ + 1)

s−z2

which coincide with the original symbols σs, τs and operator As at z = 0.Away from integer order valued symbols (resp. operators) ordinary manipulations canbe carried out on integrals and sums (resp. traces) which legitimise physicists’ heuris-tic computations. Borrowing the physicists’ metaphorical language, this amounts to(holomorphically) embedding the integer11 dimensional world into a complex dimen-sional one where the objects are canonically defined away from integer dimensions.The problem remains to get back to integer dimensions or integer orders by means ofregularised evaluators at z = 0 which pick up a finite part in a Laurent expansion.The freedom of choice left at this stage is responsible for the one-parameter renormal-isation group which plays a central role in quantum field theory.In these lectures, we hope to modestly help clarify some aspects of this vast picturein setting some of these heuristic considerations on firm mathematical ground by pro-viding analytic tools to describe regularisation techniques in a common framework,whether those used in physics, number theory or geometry. The focus is set on the un-derlying canonical integral, discrete sum and trace which are characterised by naturalproperties such as Stokes’ and translation invariance or cyclicity. Various anomaliesare investigated, which all turn out to be local insofar as they can be expressed interms of the noncommutative residue, another central figure in these lectures.We do not claim to present breakthrough results but rather a unified outlook onknown results scattered in the physics and mathematics literature with pedestrianproofs, which we try to make accessible to the nonspecialist. We hope in this way toopen new perspectives on and further expand openings to concepts such as regularisedintegrals, sums and traces. Far from being exhaustive, these lectures leave out variousimportant regularisation techniques such as Epstein-Glaser [EG], Pauli-Villars [PV]and lattice regularisation techniques, as well as other regularisation artefacts such as

11which is 4 for usual space-time.

2

b-integrals [Mel] and relative determinants [Mu] used to investigate the geometry ofcertain manifolds with boundaries. We also leave aside the realm of noncommutativegeometry where zeta-type regularisation procedures are extended to abstract pseu-dodifferential calculus as well as the ambitious renormalisation issue, which would beneeded to make sense of multiple divergent integrals such as multiloop Feynman di-agrams in physics, multiple sums such as multiple zeta values in number theory andthe number of lattice points on convex cones. Here we only tackle simple integralsand discrete sums.These lectures, which are essentially self-contained, are based on joint work withvarious collaborators, among whom Dominique Manchon, Jouko Mickelsson, StevenRosenberg, Simon Scott and former PhD students Alexander Cardona, CatherineDucourtioux, Jean-Pierre Magnot, Carolina Neira and Marie-Francoise Ouedraogowhom I would like to thank most warmly. I am also grateful to many students in France(Clermont-Ferrand), Burkina Faso (Ouagadougou), Germany (Gottingen, Hannoverand Regensburg), Colombia (Bogota and Villa de Leyva) and Lebanon (Beyrouth),who attended my various courses (for some lecture notes and review articles see [Pa1],[Pa2], [Pa3]) on regularisation techniques which triggered this manuscript, for theyindirectly contributed to improving this presentation.Last but not least, I am very grateful to Rita Paycha who helped me improve theEnglish of this text by her careful reading and to Arthur Greenspoon for his valuablehelp in editing these notes.

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The lectures are organised around seven chapters, the first of which serves as a prepa-ration for the subsequent ones.

1. The Gamma function extended to non-positive integer points

1.1 The Gamma function; first properties

1.2 Riesz regularisation

1.3 Hadamard’s “finite part” method

1.4 The extended Gamma function and homogeneous distributions; discrepan-cies

2. The canonical integral and the noncommutative residue on symbols

2.1 Classical and log-polyhomogeneous symbols with constant coefficients onRd

2.2 The noncommutative residue

2.3 Closed linear forms on symbol valued forms

2.4 Uniqueness of the canonical integral on admissible sets

2.5 Cut-off (or Hadmard finite parts) regularised integral

2.6 Uniqueness of the canonical integral on non-integer order symbols

2.7 Discrepancies of the cut-off integral

2.8 Uniqueness of the noncommutative residue on classical symbols

2.9 Translation invariance versus Stokes’ property

3. The noncommutative residue as a complex residue

3.1 Regularised evaluators

3.2 Holomorphic families of symbols and meromorphic extensions of integrals

3.3 Extensions to log-polyhomogeneous symbols

3.4 Dimensional versus cut-off regularised integrals

3.5 Discrepancies of regularised integrals

4. The canonical sum on non-integer order classical symbols

4.1 The Euler-Maclaurin formula

4.2 Cut-off discrete sums on Zd subordinated to convex polytopes

4.3 Zd- translation invariant linear forms on symbols

4.4 A characterisation of the noncommutative residue as a Zd-translaiton in-variant linear form

4.5 Regularised discrete sums on symbols

4.6 The (Hurwitz) zeta function and zeta functions associated with quadraticforms

5. The canonical trace and the noncommutative residue on operators

5.1 From pseudodifferential symbols to pseudodifferential operators

5.2 Basic properties of pseudodifferential operators

4

5.3 Pseudodifferential operators on manifolds

5.4 From closed linear forms on symbols to traces on operators

5.5 First characterisation of the noncommutative residue

6. The noncommutative residue on operators as a complex residue andzeta regularisation

6.1 Complex powers

6.2 A fundamental formula

6.3 Zeta regularised traces

6.4 Logarithms of admissible operators

6.5 Weighted traces of differentiable families

6.6 Discrepancies

6.7 Regularised traces; locality versus non locality

6.8 The index as a superresidue

6.9 Uniquenesss of the canonical trace and the noncommutative residue onoperators

7. Multiplicative and conformal anomaly of regularised determinants

7.1 Weighted and zeta determinants

7.2 Multiplicative anomaly of weighted determinants

7.3 Multiplicative anomaly of the zeta determinant

7.4 Conformally covariant operators

7.5 Conformal anomalies

7.6 Conformal anomaly of the zeta determinant

5

1 The Gamma function extended to non-positive in-teger points

This first chapter addresses the issue of how to extend the Gamma function to non-positive integer points and serves as a preparation for the more general issue as howto make sense of certain types of divergent integrals. Whereas here the divergenceis at zero, later in the notes the divergences will take place at infinity but the wayone cures these divergences is similar. The Gamma function offers a good toy modelto compare regularisation methods mentioned in the introduction. Extending theGamma function to non-positive integers arises as an instance of the more generalproblem of extending homogeneous distributions at negative integers. We show thatRiesz and Hadamard’s “finite part” regularisation methods lead to the same extendedhomogeneous distributions (see Theorem 1) and hence to the same extended Gammafunction, a feature which arises again further in these notes. We discuss discrepanciesinduced by the regularisation procedure, which are a first hint to further obstructionswe will encounter while working with regularised integrals.Extending the Gamma function to the whole complex plane is indeed related to theproblem of extending homogeneous distributions

Fa : φ 7→∫ ∞

0

xa φ(x) dx

considered by Hadamard and Riesz (see e.g. [Sch] Chapter II) to all complex values awhere φ is a Schwartz function on R+. One wants to assign to the poles of Γ a finitevalue, and hence to assign a finite value F−k(φ) to negative integers −k.Let

S(R+) = f ∈ C∞(R+),∀α, β ∈ N,∃Cα,β , s.t. |xα∂βf(x)| ≤ Cα,β ∀x ∈ R+

denote the space of Schwartz functions on R+, i.e. smooth functions f on R+ whosederivatives ∂βf(x) go faster to zero as x tends to infinity, faster than any inverse powerx−β .The map φ : x 7→ e−x defines a Schwartz function on R+ such that all its derivativesare also Schwartz functions on R+. The integral

Γ(b) :=∫ ∞

0

xb−1 φ(x) dx =∫ ∞

0

xb−1 e−x dx

which holds for Re(b) > 0 is called the Gamma function.

1.1 The Gamma function; first properties

Let a be a complex number the real part of which satisfies Re(a) > −1. After inte-grating by parts k times we get:

Fa(φ) =(−1)k

(a+ 1) · · · (a+ k)Fa+k(φ(k)). (1)

Since φ(k) is a Schwartz function the expression on the r.h.s. is well defined provideda+k has its real part larger than −1. The expression (−1)k

(a+1)···(a+k)Fa+k(φ(k)) therefore

6

defines an extension of Fa to the half plane Re(a) > −k−1. Given a complex numbera, there is a positive integer k such that Re(a) > −k − 1 and we set

Fa(φ) :=(−1)k

(a+ 1) · · · (a+ k)Fa+k(φ(k)). (2)

Exercise 1 Show that this definition does not depend on the choice of k > −Re(a)−1by checking that Fa(φ) = (−1)k+l

(a+1)···(a+k+l) Fa+k+l(φ(k+l)) for any positive integer l.

Equation (2) therefore provides an extension of Fa(φ) to all complex values a /∈ −N.In the case φ(x) = e−x considered here, this provides an extension of the Gammafunction defined on the half plane Re(b) > −k:

Γ(b) :=1

b(b+ 1) · · · (b+ k − 1)Γ(b+ k), (3)

to the whole complex plane outsidenon positive integers b ∈ −N ∪ 0.By (2) the map a 7→ Fa(φ) is a meromorphic function on the half plane Re(a) > −1−kwith simple poles in −1,−2, · · · ,−k. It follows that the map a 7→ Fa(φ) extendsto a meromorphic function a 7→ Fa(φ) on the plane with poles at negative integers−k ∈ −N given by:

Resa=−kFa(φ) = lima→−k

(a+ k) Fa(φ) =(−1)k

(−k + 1) · · · (−1)

∫ ∞0

φ(k)(x) dx =φ(k−1)(0)(k − 1)!

,

(4)where for a meromorphic function f with simple pole at z0 we have set

Resz=z0f := limz→z0

((z − z0)f(z)) .

Exercise 2 Show that Γ has poles at integers −k ∈ −N ∪ 0 given by

Resb=−kΓ(b) =(−1)k

k!. (5)

From now on we use the same notation Γ for the extension Γ. By (3) we have thefollowing recursive formula:

Γ(b+ k) = b(b+ 1) · · · (b+ k − 1) Γ(b) if Re(b) > 0. (6)

ClearlyΓ(k) = (k − 1)! ∀k ∈ N.

The following straightforward statements are useful for forthcoming applications.

Proposition 1 1. The Gamma function is differentiable at any positive integer kand:

Γ′(1) = −γ; Γ′(k) =

k−1∑j=1

1j− γ

(k − 1)! ∀k ∈ N− 1,

whereγ := −

∫ ∞0

log x e−x dx

is the Euler constant.

7

2. The inverse of the Gamma function 1Γ(z) defined on the half plane Re(z) > 0

extends to a holomorphic map at z = 0 and

1Γ(z)

= z + γz2 + o(z2). (7)

In particular,(

1Γ

)′ (0) = 1.

Proof:

1. By formula (6) the derivative of Γ at 1 reads:

Γ′ (1) = ∂zΓ(1 + z)|z=0

=∫ ∞

0

log x e−x dx

= −γ.

The derivative at k ∈ N− 1 reads:

Γ′ (k) = limz→0

Γ (k + z)− Γ (k)z

= limz→0

(k + z − 1) · · · (z + 1) · Γ (z + 1)− (k − 1)!z

= (k − 1)!

k−1∑j=1

1j− γ

= Γ(k)

k−1∑j=1

1j− γ

,

so that for k ≥ 2Γ′(k)Γ(k)

=k−1∑j=1

1j− γ. (8)

2. By (6) we have Γ(z) = Γ(z+k)z(z+1)···(z+k−1) so that in particular

1Γ(z)

=z

Γ(z + 1)=

z

Γ(1) + Γ′(1)z + o(z)= z + γz2 + o(z2).

tu

Exercise 3 Show thatΓ(z) ∼0

1z− γ. (9)

Hint: Formula (9) follows from (3) setting b = z, k = 1 or equivalently from (7)

The following proposition gives the asymptotics at negative integers.

Proposition 2 At negative integers −k ∈ −N we have:

Γ(−k + z) ∼0(−1)k

k!

1z− γ +

k∑j=1

1j

. (10)

8

Proof: By formula (3) we have:

Γ(−k + z) =Γ(z)

(z − k) · · · (z − 1)

∼0

1z − γ

(z − k) · · · (z − 1)

∼0(−1)k

k!

(1z− γ)(

1 +z

k

)· · · (1 + z),

from which (10) follows. tu

Remark 1 Formula (10) yields back the residue at z = −k given by formula (5).

The convolution of the distributions Fa and Fb gives:

Fa−1 ∗ Fb−1(t) =∫ ∞

0

ra−1(r − t)b−1dr

= tb+a−1

∫ ∞0

ua−1(u− 1)b−1du

= B(a, b)Fa+b−1(t).

Here

B(a, b) :=∫ 1

0

ta−1(1−t)b−1dt = 2∫ π

2

0

(sin t)2a−1(cos t)2b−1dt =∫ ∞

0

ta−1(1+t)−(a+b)dt

is the Beta function. Applying Fa ∗ Fb to the Schwartz function φ : r 7→ e−r yields

Fa ? Fb(φ) =∫ ∞

0

∫ ∞0

ra−1 (r − t)b−1e−t dr dt

=∫ ∞

0

∫ ∞0

ra−1 sb−1e−(s+r) dr ds

= Γ(a) Γ(b)

from which we deduce that

B(a, b) =Γ(a)Γ(b)Γ(a+ b)

.

1.2 Riesz regularisation

We now want to make sense of F−k(φ) in spite of the fact that a = −k arises as apole in (2). Following Riesz (see e.g. [Sch]), we pick the constant term at z = 0 in theLaurent expansion of the map z 7→ F−k+z(φ).

Proposition 3 Given a Schwartz function φ on R+ and a complex number a, themap z 7→ Fa+z(φ) is meromorphic on the plane with simple poles in −a − N. Theconstant term in the Laurent expansion:∫ ∞,Riesz

0

xa φ(x) dx := limz→0

(Fa+z(φ)− 1

zResz=0Fa+z(φ)

)(11)

coincides with the ordinary integral∫∞

0xa φ(x) dx whenever Re(a) > −1.

9

1. If a /∈ −N, then∫ ∞,Riesz

0

xa φ(x) dx = Fa(φ) =(−1)k

(a+ 1) · · · (a+ k)Fa+k(φ(k)),

where k is any integer such that Re(a+ k) > −1.

2. If a = −k ∈ −N, then∫ ∞,Riesz

0

x−k φ(x) dx =φ(k−1)(0)(k − 1)!

k−1∑j=1

1j− 1

(k − 1)!

∫ ∞0

log xφ(k)(x)dx,

setting the sum over j equal to zero if k = 1.

Proof: The case a /∈ −N follows from the previous discussion. We therefore provethe proposition when a = −k for some integer k ∈ N.

1. Let us start with k = 1. Integrating by parts for Re(z) > −1, we have:

F−1+z(φ) = −1z

∫ ∞0

xzφ′(x)dx

which yieldsResz=0F−1+z(φ) = φ(0).

Moreover we have

limz→0

(F−1+z(φ)− 1

zResz=0F−1+z(φ)

)= lim

z→0

(−1z

∫ ∞0

xzφ′(x)dx− φ(0)z

)= lim

z→0

(−∫ ∞

0

xz − 1z

φ′(x) dx)

= −∫ ∞

0

log xφ′(x)dx.

2. When k > 1, integrating by parts k times we get for Re(z) > −1:

F−k+z(φ) =(−1)k

(−k + 1 + z) · · · (z − 1)z

∫ ∞0

xzφ(k)(x)dx.

10

From (4) we deduce that

limz→0

(F−k+z(φ)− 1

zResz=0F−k+z(φ)

)= lim

z→0

((−1)k

(−k + 1 + z) · · · (z − 1)z

∫ ∞0

xzφ(k)(x)dx− 1z

φ(k−1)(0)(k − 1)!

)= lim

z→0

((−1)k

(−k + z + 1) · · · (z − 1)

∫ ∞0

xz − 1z

φ(k)(x)dx

+1z

[(−1)k

(−k + z + 1) · · · (z − 1)

∫ ∞0

φ(k)(x)dx− φ(k−1)(0)(k − 1)!

])

= − 1(k − 1)!

∫ ∞0

log xφ(k)(x)dx+ limz→0

φ(k−1)(0)z

k−1∏j=1

1j − z

− 1(k − 1)!

= − 1

(k − 1)!

∫ ∞0

log xφ(k)(x)dx+ limz→0

(φ(k−1)(0)

[Ψk(z)−Ψk(0)

z

])= − 1

(k − 1)!

∫ ∞0

log xφ(k)(x)dx+ Ψ′k(0)φ(k−1)(0)

= − 1(k − 1)!

∫ ∞0

log xφ(k)(x)dx+φ(k−1)(0)(k − 1)!

k−1∑j=1

1j,

where we have set Ψk(z) :=∏k−1j=1

1j−z and used (5).

tu

Applying the previous proposition to φ(x) = e−x leads to the following Riesz ex-tension of the Gamma function.Given a complex number b, the map z 7→ Γ(b+ z) is meromorphic on the plane withsimple poles in −b − N ∪ 0. The finite part of the integral Γ(b + z) defined by theconstant term in the Laurent expansion:

ΓRiesz(b) := limz→0

(Γ(b+ z)− 1

zResz=0Γ(b+ z)

)(12)

coincides with the ordinary Gamma function Γ(b) whenever Re(b) > 0.

Exercise 4 Show that

1. if b /∈ −N ∪ 0, then

ΓRiesz(b) =Γ(b+ k)

b (b+ 1) · · · (b+ k − 1)

for any k such that Re(b) + k > 0.

2. Show that at non-positive integers b = −k we have

ΓRiesz(0) = −γ; ΓRiesz(−k) =(−1)k

k!

k∑j=1

1j− γ

if k > 0.

11

1.3 Hadamard’s “finite part” method12

As before, φ denotes a Schwartz function on R+. Clearly, for Re(a) > −1,

Fa(φ) :=∫ ∞

0

xa φ(x) dx = limε→0

∫ ∞ε

xa φ(x) dx.

In particular, for Re(b) > 0, we have

Γ(b) = limε→0

∫ ∞ε

xb−1 e−x dx.

Following Hadamard (see e.g. [Sch]) we extend Fa(φ) to all complex values of a bypicking the constant term denoted by

∫∞,Had

0xa φ(x) dx in the asymptotic expansion

of the map ε 7→∫∞εxa φ(x) dx as ε→ 0.

Applying this to φ(x) = e−x yields an alternative extension

ΓHad(b) :=∫ ∞,Had

0

xb−1 e−x dx

of the Gamma function to the whole complex plane.

Let us introduce some notation. A[ε] denotes the algebra of analytic functions inthe variable ε and for any complex number α and any positive integer l we set

A[ε] εα := f(ε) εα, f ∈ A[ε]; A[ε] logl ε := f(ε) logl ε, f ∈ A[ε]. (13)

We first need a technical lemma.

Lemma 1 Let φ denote a Schwartz function on R+. The map ε 7→∫∞ε

log xφ(x) dxlies in A[ε]

⊕A[ε] log ε and limε→0

∫∞ε

log xφ(x) dx =∫∞

0log xφ(x) dx.

In particular, the map ε 7→∫∞ε

log x e−x dx lies in A[ε]⊕A[ε] log ε and we have

limε→0

∫ ∞ε

log x e−x dx = −γ.

Proof: The second part of the lemma clearly follows from the first one applied toφ(x) = e−x. We therefore prove the first part of the lemma. We have∫ ∞ε

log xφ(x) dx =∫ ∞

0

log xφ(x) dx−∫ ε

0

log xφ(x) dx

=∫ ∞

0

log xφ(x) dx− ε∫ 1

0

log(ε x)φ(ε x) dx

=∫ ∞

0

log xφ(x) dx− ε∫ 1

0

log xφ(ε x) dx− ε log ε∫ 1

0

φ(ε x) dx.

A Taylor expansion of φ at 0 shows that ε 7→∫ 1

0φ(εx) dx lies in A[ε]. So does the map∫ 1

0log xφ(εx) dx lie in A[ε]; indeed

∫ 1

0log xφ(εx) dx = −ε

∫ 1

0(x log x − x)φ′(ε x) dx −

12I thank Bing Zhang for interesting discussions and comments concerning this type of regularisa-tion procedure.

12

φ(ε) and a Taylor expansion of φ and φ′ at 0 provides the required asymptotic expan-sion. Hence the map ε 7→

∫∞ε

log xφ(x) dx lies in A[ε]⊕A[ε] log ε and we have

limε→0

∫ ∞ε

log xφ(x) dx =∫ ∞

0

log xφ(x) dx.

tu

Proposition 4 Let φ denote a Schwartz function on R+.

1. If Re(a) > −1, the map ε 7→∫∞εxa φ(x) dx lies in C

⊕A[ε]εa+1 and we have

limε→0

∫ ∞ε

xa φ(x) dx =∫ ∞

0

xa φ(x) dx.

2. If a /∈ −N, the map ε 7→∫∞εxa φ(x) dx lies in C

⊕A[ε] εa+1 with constant term

in the asymptotic expansion given by∫ ∞,Had

0

xa φ(x) dx =(−1)k

(a+ 1) · · · (a+ k)

∫ ∞0

xa+k φ(k)(x) dx

for any integer k such that a+ k > −1.

3. If a = −k ∈ −N, the map ε 7→∫∞εxa φ(x) dx lies in A[ε]ε−k+1

⊕A[ε] log ε with

constant term in the asymptotic expansion given by∫ ∞,Had

0

x−k φ(x) dx = − 1(k − 1)!

∫ ∞0

log xφ(k)(x) dx+φ(k−1)(0)(k − 1)!

k−1∑j=1

1j,

where the sum is set to zero if k = 1.

Proof:

1. If Re (a) > −1 we have∫ ∞ε

xa φ(x) dx =∫ ∞

0

xa φ(x) dx−∫ ε

0

xa φ(x) dx

=∫ ∞

0

xa φ(x) dx− εa+1

∫ 1

0

xa φ(ε x) dx.

The map ε 7→∫ 1

0xa φ(ε x) dx defines an analytic function in ε so that ε 7→∫∞

εxa φ(x) dx lies in C

⊕A[ε]εa+1 and limε→0

∫∞εxa φ(x) dx =

∫∞0xa φ(x)dx.

2. If a /∈ −N, we proceed by induction on k with Re(a) ∈] − (k + 1),−k] usingintegration by parts to show that

∫∞εxa φ(x) dx lies in C

⊕A[ε]εa+1. The step

k = 0 holds by the previous item. One integration by parts provides the induc-tion step. Indeed, provided the map ε 7→

∫∞εxa+1 φ(x) dx lies in C

⊕A[ε]εa+2

then the map ε 7→∫∞εxaφ(x) dx lies in C

⊕A[ε]εa+1 since∫ ∞

ε

xa φ(x) dx = −∫ ∞ε

xa+1

a+ 1φ′(x) dx+

[xa+1

a+ 1φ(x)

]∞ε

= − 1a+ 1

∫ ∞ε

xa+1 φ′(x) dx− εa+1

a+ 1φ(ε).

13

For Re(a) ∈]− (k + 1),−k], using a Taylor expansion of φ(l) at 0 we get∫ ∞ε

xa φ(x) dx

=(−1)k

(a+ 1) · · · (a+ k)

∫ ∞ε

xa+k φ(k)(x) dx+ (−1)kεa+k

(a+ 1) · · · (a+ k)φ(k−1)(ε) + · · ·

+ (−1)jεa+j

(a+ 1) · · · (a+ j)φ(j−1)(ε) + · · · − εa+1

a+ 1φ(ε)

=(−1)k

(a+ 1) · · · (a+ k)

∫ ∞ε

xa+k φ(k)(x) dx

+k∑j=1

(−1)j∞∑ij=0

εa+j+ij

(a+ 1) · · · (a+ j) ij !φ(j−1+ij)(0).

The constant term reads∫ ∞,Had

0

xa φ(x) dx =(−1)k

(a+ 1) · · · (a+ k)

∫ ∞0

xa+k φ(k)(x) dx

since the remaining terms do not contribute to the constant term as a is not anegative integer

3. If a = −1 then,∫ ∞ε

x−1 φ(x) dx = −∫ ∞ε

log xφ′(x) dx+ [log xφ(x)]∞ε

= −∫ ∞ε

log xφ′(x) dx− log ε φ(ε)

which lies in A[ε]⊕A[ε] log ε by Lemma 1 and has finite part at zero given by∫ ∞,Had

0

x−1 φ(x) dx = −∫ ∞

0

log xφ′(x) dx.

4. If a = −k for some integer k > 1, then by induction on k we show that the mapε 7→

∫∞εx−k φ(x) dx lies in A[ε]ε−k+1

⊕A[ε] log ε. The previous step gives the

statement for k = 1. Using integration by parts we easily prove the inductionstep: ∫ ∞

ε

x−k φ(x) dx = −∫ ∞ε

x−k+1

−k + 1φ′(x) dx+

[x−k+1

−k + 1φ(x)

]∞ε

=1

k − 1

∫ ∞ε

x−k+1 φ′(x) dx+ε−k+1

k − 1φ(ε).

14

Iterating the integration by parts procedure gives∫ ∞ε

x−k φ(x) dx =1

(k − 1)!

∫ ∞ε

x−1 φ(k−1)(x) dx+ε−k+1

k − 1φ(ε) + · · ·

+ε−k+j

(k − 1) · · · (k − j)φ(j−1)(ε) + · · ·+ ε−1

(k − 1)!φ(k−2)(ε)

= − 1(k − 1)!

∫ ∞ε

log xφ(k)(x) dx− log ε(k − 1)!

φ(k−1)(ε)

+ε−k+1

k − 1φ(ε) + · · ·+ ε−k+j

(k − 1) · · · (k − j)φ(j−1)(ε)

+ · · ·+ ε−1

(k − 1)!φ(k−2)(ε), (14)

which after a Taylor expansion around ε = 0 of φ and its derivatives, gives riseto the following finite part∫ ∞,Had

0

x−k φ(x) dx = − 1(k − 1)!

∫ ∞0

log xφ(k)(x) dx+φ(k−1)(0)

(k − 1)!(k − 1)+ · · ·

+φ(k−1)(0)

(k − 1) · · · (k − j) (k − j)!+ · · ·+ φ(k−1)(0)

(k − 1)!

= − 1(k − 1)!

∫ ∞0

log xφ(k)(x) dx+φ(k−1)(0)(k − 1)!

k−1∑j=1

1k − j

.

tu

Applying Proposition 4 to φ(x) = e−x leads to the following extension of the Gammafunction:

ΓHad(b) := fpε=0

∫ ∞ε

xb−1 e−x dx.

Exercise 5 1. Show that if Re(b) > 0, the map ε 7→∫∞εxb−1 e−x dx lies in A[ε].

The corresponding asymptotic expansion has constant term given by

ΓHad(b) = limε→0

∫ ∞ε

xb−1 e−x dx = Γ(b).

2. Show that for b /∈ Z≤0, the map ε 7→∫∞εxb−1 e−x dx lies in C

⊕A[ε] εb. The

corresponding asymptotic expansion has constant term given by

ΓHad(b) =Γ(b+ k)

b(b+ 1) · · · (b+ k − 1).

3. Show that if b = −k ∈ −N ∪ 0, the map ε 7→∫∞εx−k−1 e−x dx lies in

A[ε]ε−k+1⊕A[ε] log ε. The corresonding asymptotic expansion has constant

term given by

ΓHad(0) = −γ if k = 0

ΓHad(−k) =(−1)k

k!

k∑j=1

1j− γ

,15

where the sum over j is set to zero when k = 1.

1.4 The extended Gamma function and homogeneous distri-butions; discrepancies

The following theorem sums up the results of the two previous paragraphs.

Theorem 1 The Riesz and Hadamard regularisation methods yield the same extendeddistribution Fa for any complex value a, which when applied to a Schwartz function φreads:

Fa(φ) :=∫ ∞,Riesz

0

xa φ(x) dx =∫ ∞,Had

0

xa φ(x) dx (15)

that coincides with the ordinary integral∫∞

0xa φ(x) dx whenever Re(a) > −1.

1. If a /∈ −N, then

Fa(φ) =(−1)k

(a+ 1) · · · (a+ k)Fa+k(φ(k)),

where k is any integer such that Re(a+ k) > −1.

2. Furthermore, for a positive integer k

F−k(φ) =φ(k−1)(0)(k − 1)!

k−1∑j=1

1j− 1

(k − 1)!

∫ ∞0

log xφ(k)(x)dx,

setting the sum over j equal to zero if k = 1.

This applied to the Schwartz function φ(x) = e−x confirms the results of Exercises5 and 4, which show that Riesz and Hadamard finite part regularisations lead to thesame extended Gamma function Γ(−k) at non-positive integers:

Γ(0) := ΓHad(0) = ΓRiesz(0) = −γ if k = 0

Γ(−k) := ΓHad(−k) = ΓRiesz(−k) =(−1)k

k!

k∑j=1

1j− γ

if k > 0.

Extending a homogeneous distribution Fa → Fa to negative integers unfortunatelyhas a cost; we lose various properties along the way.

1.4.1 Loss of homogeneity

Set for t > 0, φt = t−1 φ(t−1·). Then,

Fa(φt) = Fa(φt) = ta Fa(φ) if Re(a) > −1

i.e. Fa is a homogeneous distribution of degree a. As a result, since φ(k)t =

t−k−1φ(k)(t−1·), for a /∈ −N we have

Fa(φt) :=(−1)

(a+ 1) · · · (a+ k)Fa+k

(φ

(k)t

)= taFa(φt)

16

i.e. Fa is still a homogeneous distribution. However, for a = −k with k ∈ N we have

Fa(φt) =1

(k − 1)!

t−k k∑j=1

φ(k−1)(0)j

− t−k−1

∫ ∞0

log(x)φ(k)(t−1 x) dx

for

=1

(k − 1)!

t−k k∑j=1

φ(k−1)(0)j

− t−k∫ ∞

0

log(tx)φ(k)(x) dx

= t−k

[Fa(φt) +

φ(k−1)(0) log t(k − 1)!

]= t−k

[Fa(φt) + Resa=−kFa(φ) log t

],

so that unless φ(k−1)(0) vanishes, the extended distribution is no longer homogeneous.Thus a discrepancy arises with the loss of homogeneity of the extended homogeneousdistribution at negative integers.

Exercise 6 Show that for any complex value a:

Fa(φt) = ta(Fa(φt) + Rez=0Fa+z(φ) log t

).

Consequently, the discrepancy ta Rez=0Fa+z(φ) log t vanishes whenever the residueResz=0Fa+z(φ) vanishes.

1.4.2 The extended Gamma function: obstruction to the functional equa-tion

The ordinary Gamma function Γ obeys the following property outside the poles b in−N ∪ 0:

Γ(b) = (b− 1) Γ(b− 1) (16)

but the extended Gamma function does not anymore at the poles for b = −k in−N ∪ 0, since

Γ(−k) 6= (−k − 1) Γ(−k − 1).

A discrepancy therefore arises since property (16) breaks down at non-positive integers.

Exercise 7 Show that for any complex value b:

Γ(b) = (b− 1)(

Γ(b− 1) + Resz=0Γ(b+ z)).

Consequently, the discrepancy (b − 1) Resz=0Γ(b + z) vanishes whenever the residueResz=0Γ(b+ z) vanishes.

17

2 The canonical integral and noncommutative residueon symbols

In order to control divergences for large values of the momentum variable, it is usefulto restrict the class of functions under consideration. We therefore focus on pseu-dodifferential symbols, and specifically on classical and log-polyhomogeneous symbolswith constant coefficients, whose behaviour at infinity is polyhomogeneous and log-polyhomogeneous. We introduce two well-known useful linear forms on certain classesof symbols, the noncommutative residue on the algebra of classical symbols and thecanonical integral on non-integer order log-polyhomogeneous symbols.On the one hand, we characterise (see Theorem 2) the canonical integral as the unique(modulo a multiplicative factor) linear form on certain classes of symbols (includingthe set of non-integer order symbols) which vanishes on partial derivatives (Stokes’property). Any other linear extension of the ordinary integral to non-integer ordersymbols coincides with the canonical integral (see Theorem 3).On the other hand, we measure (see Theorem 6) in terms of a noncommutative residue,the obstruction that prevents the cut-off regularised integral obtained by Hadamardfinite parts from obeying Stokes’ property on the whole algebra of classical symbols.Finally we characterise the noncommutative residue as the unique ( modulo a mul-tiplicative factor) linear form on the whole algebra of classical symbols that obeysStokes’ property (see Theorem 4) or equivalently, that is translation invariant underthe action of Rd (see Theorem 5).

2.1 Classical and log-polyhomogeneous symbols with constantcoefficients on Rd

Pseudodifferential symbols fall into different classes according to their asymptoticbehaviour at infinity. We only provide a few definitions and refer the reader to [Sh,Ta, Tr2] for further details on classical pseudodifferential symbols.Given a real number r, let us denote by Srcc(Rd) the set of smooth complex valuedfunctions on Rd called symbols with constant coefficients, such that for anymultiindex β ∈ Nd there is a constant Cβ satisfying the following requirement:

|∂βξ σ(ξ)| ≤ Cβ |〈ξ〉r−|β| ∀ξ ∈ Rd, (17)

where we have set 〈ξ〉 =√

1 + |ξ|2 with | · | the Euclidean norm of ξ.

Remark 2 The presence of 〈ξ〉 takes care of infrared divergences, namely divergencesfor small values of |ξ|.

Let Scc(Rd) =⋃r∈R Srcc(Rd). We call smoothing any symbol in the set

S−∞cc (Rd) =⋂a∈CSacc(Rd).

Exercise 8 Show that S−∞cc (Rd) = S(Rd), where S(Rd) is the algebra of Schwartzfunctions on Rd.

“Equality modulo smoothing symbols” defined by σ ∼ σ′ ⇐⇒ σ − σ′ ∈ S−∞cc (Rd)yields an equivalence relation in Srcc(Rd) for any real number a.

18

For symbols σ ∈ Srcc(Rs), σk ∈ Srkcc (Rd), k ∈ N0 where rk decreases as k tends toinfinity and r0 = r, we further set(

σ ∼∑k∈N0

σk

)

⇐⇒

∀α ∈ R, ∃K(α) ∈ N, s.t. K ≥ K(α)⇒ σ −∑k≤K

σk ∈ Sαcc(Rd)

.

We single out classical and log-polyhomogeneous symbols with constant coefficients.

Definition 1 Let a be a complex number.

1. A symbol σ ∈ Scc(Rd) is classical (or polyhomogeneous) of order a if:

σ(ξ) ∼∞∑j=0

χ(ξ)σa−j(ξ) (18)

where

• χ is any smooth function on Rd such that χ vanishes in a small neighborhoodof 0 and is identically one outside the unit ball,

• σa−j ∈ C∞(Rd) is positively homogeneous of order a− j, i.e.

σa−j(tξ) = ta−jσa−j(ξ)

for any t > 0 and any ξ ∈ Rd.

2. A symbol σ ∈ S(Rd) is log-polyhomogeneous of log-type k for some non-negative integer k if

σ(ξ) ∼k∑l=0

∞∑j=0

χ(ξ)σa−j,l(ξ) logl |ξ| (19)

with σa−j,l, l = 0, · · · , k positively homogeneous functions of order a− j.

Remark 3 These definitions are independent of the choice of cut-off function χ withthe above mentioned properties. Indeed, if χ′ is another cut-off function χ with thesame properties then (χ− χ′)σa−j and (χ− χ′)σa−j,l are smoothing symbols for anyindices j and l.

Exercise 9 Show that the map ξ 7→ 1|ξ|2+1 defines a classical symbol of order −2. For

what values of d does its asymptotic expansion contain a positively homogeneous termof degree −d?

Exercise 10 Show that ξ 7→ log(|ξ|2 + 1) is log-polyhomogeneous of log-type 1 andorder zero.

Exercise 11 More generally, show that any smooth rational function is a classicalsymbol and that any k-th logarithmic power thereof is a log-polyhomogeneous symbolof log-type k.

19

Let CSa,kcc (Rd) ⊂ Scc(Rd) denote the class of log-polyhomogeneous symbols of ordera and log-type k and let us set CSa,∗cc (Rd) :=

⋃k∈N0

CSa,kcc (Rd). Then CSacc(Rd) :=CSa,0cc (Rd) corresponds to the set of classical symbols of order a.

Exercise 12 Show that CSacc(Rd) ⊂ SRe(a)cc (Rd) for any complex number a.

Exercise 13 Show that CSZcc(Rd) := ∪a∈Z ∈ CSacc(Rd) equipped with the ordinary

product of functions is an algebra.

Exercise 14 Show that the intersection⋂a∈C CS

acc(Rd) coincides with the algebra

S−∞cc (Rd) of smoothing symbols.

Definition 2 The component σL := σa ∈ C∞(Rd) of highest homogeneity degree of aclassical symbol σ in CSacc(Rd) is called the leading symbol of the symbol.

Exercise 15 Check that

σa(ξ) = limt→∞

t−aσ(tξ) ∀ξ ∈ Rd.

The ordinary product of functions sends CSacc(Rd)×CSbcc(Rd) to CSa+bcc (Rd) provided

b− a ∈ Z; letCScc(Rd) = 〈

⋃a∈C

CSacc(Rd)〉 (20)

denote the algebra generated by all classical symbols with constant coefficients on Rdequipped with the ordinary product of functions.

Exercise 16 Show that CScc(Rd)∩L1(Rd) = CS<−dcc (Rd), where CS<pcc (Rd) :=⋃

Re(a)<p CSacc(Rd)

stands for the set of classical symbols whose order has real part smaller than p.

Let us denote by

CS /∈Zcc (Rd) :=

⋃a∈/∈Z

CSacc(Rd) and CS /∈Z,kcc (Rd) :=

⋃a∈/∈Z

CSa,kcc (Rd) (21)

for any non-negative integer k. In spite of the fact that it does not form an algebra,the set CS /∈Z,k

cc (Rd) of non-integer order symbols will play an important role in thefollowing.

Let us finally equip the set CSacc(Rd) of classical symbols of order a with a Frechetstructure with the help of the following semi-norms labelled by multiindices β andintegers j ≥ 0, N (see [Ho]):

supx∈Rd(1 + |x|)−Re(a)+|β| ‖∂βxσ(x)‖

supx∈Rd(1 + |x|)−Re(a)+N+|β|‖∂βx

σ − N−1∑j=0

χ(x)σa−j

(x)‖

sup|x|=1‖∂βxσa−j(x)‖, (22)

where χ is any smooth function which vanishes in a neighborhood of zero and isidentically one outside the unit ball. We shall refer to this topology as the topologyon symbols of constant order.In this topology, we have

σ = limN→∞

N−1∑j=0

χσa−j

which justifies the notation σ ∼∑∞j=0 χσa−j .

20

2.2 The noncommutative residue

The noncommutative residue, which is a central character in these lectures, was origi-nally introduced by Adler [A] and Manin [Man] in the one dimensional case and laterextended to all dimensions by Wodzicki in [Wo1] (see also [Wo2] and [Ka] for a re-view). It was simultaneously discovered by Guillemin [Gu2] and an alternative proofwas recently given by Ponge in [Po2].To describe the residue we need the expression of the measure on the unit sphereSd−1 = ξ ∈ Rd,

∑di=1 ξ

2i = 1 induced by the canonical volume form on Rd. The

Liouville (or radial) field

X(ξ) =d∑i=1

ξi ∂i (23)

on Rd transforms the canonical volume form ω(ξ) = dξ1 ∧ · · · ∧ dξd on Rd to a (d− 1)-form

dµS(ξ) := iXω(ξ) =d∑j=1

(−1)j−1 ξj dξ1 ∧ · · · ∧ dξj ∧ · · · ∧ dξd (24)

which defines a volume form on Sd−1. It is covariant as a result of the covarianceproperty of the canonical volume form on Rd. The noncommutative residue picksout the −d-th degree homogeneous part of the symbol; a = −d corresponds to thethreshold value below which the integral

∫Rd |ξ|

a dξ diverges.

Definition 3 The noncommutative residue is a linear form on CScc(Rd) defined by

res(σ) :=1

(2π)d

∫Sd−1

σ−d(ξ) dµS(ξ) =∫Sd−1

σ−d(ξ) dSξ, (25)

where we have set dSξ = dµS(ξ)(2π)d

.

The noncommutative residue is a singular linear form, i.e. it vanishes on smoothingsymbols.

Exercise 17 Show the following covariance property of the residue:

|det(C)| res(σ C) = res(σ) ∀C ∈ Gld(R).

Exercise 18 1. (Euler’s theorem)For any positively homogeneous function f of degree a on Rd − 0,

d∑i=1

ξi∂if(ξ) = a f(ξ).

Hint:Apply the radial field (23) to a homogeneous function f of degree a.

2. Any positively homogeneous function f on Rd − 0 of degree a 6= −d is afinite sum of partial derivatives, i.e. there exist positively homogeneous functionsfi, i = 1, · · · , d such that

f(ξ) =d∑i=1

∂ifi(ξ). (26)

21

3. Show that this extends to a homogeneous function of degree −d with vanishingresidue (see e.g.[FGLS]).Hint: Observe that the Laplacian in polar coordinates (r, ω) ∈ R+

0 × Sd−1 reads∆ = −

∑di=1 ∂

2i = −r1−d∂r(rd−1∂r) + r−2∆Sd−1 , where ∆Sd−1 is the Laplacian

on the unit sphere and show the following equivalence:

f ∈ Ker(res)⇐⇒ ∃g ∈ C∞(Sd−1), f∣∣Sd−1= ∆Sd−1g.

Setting F (r ω) := g(ω) r2−d and fi := ∂iF , which is positively homogeneous ofdegree −d+ 1, check that f =

∑di=1 ∂ifi.

This leads to the following result which is reminiscent of the characterisation of topdegree exact forms by the vanishing of their integral over Rd.

Proposition 5 Any symbol σ ∈ CScc(Rd) with vanishing residue is, modulo somesmoothing symbol, a finite sum of partial derivatives, i.e. there exist symbols τi ∈CScc(Rd), i = 1, · · · , d such that

σ ∼d∑i=1

∂iτi. (27)

If σ has order a, the symbols τi can be chosen of order a+ 1.

Proof: We write σ ∼∑∞j=0 χσa−j with σa−j ∈ C∞(Rd−0) positively homogeneous

of degree a− j. Since res(σa−j) = 0, by Exercise 18 there are positively homogeneousfunctions τi,a−j+1 of degree a − j + 1 such that

∑di=1 ∂iτi,a−j+1 = σa−j . Let τi ∼∑∞

j=1 χ τi,a−j+1; then

σ ∼d∑i=0

∞∑j=0

χ∂iτi,a−j+1 ∼d∑i=1

∂iτi. (28)

Indeed, ∂iχ has compact support so that the difference σ −∑ni=1 ∂iτi is smoothing.

Since the τi are by construction of order a + 1, statement (27) of the propositionfollows. tuRemark 4 The symbols τi are not unique since are determined modulo smoothingsymbols and constants.

2.3 Closed linear forms on symbol valued forms

Symbol valued forms are defined in a similar manner to ordinary differential forms.Let S be a subset of CScc(Rd) and let

ΩkS :=

∑|I|=k

αI(ξ) dξI , αI ∈ S

,

the set of S-valued degree k forms on Rd. Provided S is stable under partial differ-entiation, exterior differentiation on forms extends to S- valued forms (see (5.14) in[LP]):

d : ΩkS → Ωk+1S

α(ξ) dξi1 ∧ · · · ∧ dξik 7→n∑i=1

∂iα(ξ) dξi ∧ dξi1 ∧ · · · ∧ dξik .

22

We call a symbol valued form α closed if dα = 0 and exact if α = d β where β is asymbol valued form; this gives rise to the following cohomology groups:

HkS := α ∈ ΩkS, dα = 0 / d β, β ∈ Ωk−1S.

Definition 4 A linear form λ on S extends to S-valued forms by:

λ : Ω•S → Cα(x) dxi1 ∧ · · · ∧ dxik 7→ δk−dλ(α) if i1 < · · · < ik.

This extension is closed if it vanishes on exact forms, in which case it induces a linearform

λ : H•S → C.

Let us recall Stokes’ theorem (see e.g. [BG]): if M is an oriented d-dimensionalmanifold with boundary ∂M equipped with the induced orientation; then for any(d − 1)-form α on M with compact support,

∫Mdα =

∫∂M

α. In particular, if M isboundaryless,

∫Mdα = 0.

We shall use the following consequence of Stokes’ theorem: let M be a compactoriented d-dimensional Riemannian manifold with boundary ∂M equipped with theinduced metric, with outward pointing unit normal vector field ν on ∂M , then for anysmooth vector field X on M ,∫

M

divX d vol =∫∂M

〈X, ν〉 d σ, (29)

where divX :=∑di=1 ∂iXi is the divergence of X, dvol is the volume measure on M

and d σ the induced measure on the boundary.If M is a submanifold of Rd equipped with the measure induced by the canonicalmeasure on Rd, when applying (29) to the vector field X = f ei for some i in 1, · · · , dwith e1, · · · , ed is the orthonormal basis in Rd and f a smooth function on M ,equation (29) reads ∫

M

∂if d vol =∫∂M

f 〈ei, ν〉 d σ. (30)

Exercise 19 Show that integration along Rd vanishes on partial derivatives of smooth-ing symbols.

Motivated by this example, we set the following definition.

Definition 5 A linear form on a subspace S of CScc(Rd) stable under partial differ-entiation fulfils Stokes’ property if

λ(∂iσ) = 0 ∀i ∈ 1, · · · , d, ∀σ ∈ S.

Closed linear forms on ΩS are in one to one correspondence with linear forms λ on Swith Stokes’ property since

d(σ dξ1 ∧ · · · ∧ dξi ∧ · · · ∧ dξd

)= (−1)i ∂iσ dξ1 ∧ · · · ∧ dξd

=⇒ λ(d(σ dξ1 ∧ · · · ∧ dξi ∧ · · · ∧ dξd

))= (−1)i λ(∂iσ).

By abuse of terminology, we shall also say that λ is closed.

23

Exercise 20 Show that the map res introduced in Definition 3 is closed on CScc(Rd).

The following easy integration by parts result will be useful in what follows.

Exercise 21 Assuming that S is stable under partial differentiation and multiplica-tion by polynomials, show that for any closed linear form λ : S → C, we have

|β| < |α| =⇒ λ(ξ 7→ ξβ∂αξ σ(ξ)) = 0 ∀σ ∈ S. (31)

Hint: Use an inductive proof by repeated integrations by parts.

2.4 Uniqueness of the canonical integral on admissible sets

We saw that the residue defines a closed and covariant linear form on CScc(R), avery useful property which enables integration by parts and change of variables to becarried out. It nevertheless presents an important drawback, namely that it vanisheson L1-symbols. Instead, we now consider an extension of the ordinary integration mapon L1-symbols, called the canonical integral. The price to pay for this is that it doesnot live on the whole algebra of symbols.In order to characterise the canonical integral, we first classify closed linear formson smoothing symbols on Rd with constant coefficients starting with the followingtechnical but useful exercise. Let l be a non-negative integer.

Exercise 22 Show that

σ ∈ CSa,lcc (Rk+1) =⇒(

(ξ, t) 7→∫ t

−∞σ(ξ, u) du

)∈ CSa+1,l

cc (Rk+1) (32)

for any complex number a with real part smaller than −1 and deduce that

σ ∈ S−∞cc (Rk+1) =⇒(

(ξ, t) 7→∫ t

−∞σ(ξ, u) du

)∈ S−∞cc (Rk+1). (33)

This leads us to the following lemma.

Lemma 2 Let α ∈ ΩdCSa,lcc (Rd) for some complex number a with real part smallerthan −d. Then∫

Rdα = 0⇐⇒

(∃β ∈ Ωd−1CSa+1,l

cc (Rd), such that α = dβ).

Proof: The implication from right to left follows from Stokes’ theorem, by which∫Rdd β = lim

R→∞

∫B(0,R)

dβ = limR→∞

∫S(0,R)

β.

Since β lies in Ωd−1CSa+1,lcc (Rd), it is a symbol valued form of order smaller than

−d+ 1 and hence bounded by 〈ξ〉−d+1−ε for some positive ε.Since limR→∞

∫S(0,R)

〈ξ〉−d+1−εdξ = 0, it follows that∫Rdd β = lim

R→∞

∫S(0,R)

β = 0.

For the implication from right to left, setting α = α(t, ξ) dt∧ dξ1 ∧ · · · ∧ dξd, we know

24

from (32) that β(t, ξ) =∫ t−∞ α(u, ξ) du lies in CSa+1,l(Rd) so that β = β(t, ξ) dξ1 ∧

· · · ∧ dξd satisfies the requirement since

dβ = ∂tβ(t, ξ) dt ∧ dξ1 ∧ · · · ∧ dξd = α(t, ξ) dt ∧ dξ1 ∧ · · · ∧ dξd.

tuIn view of the next proposition which characterises linear forms on smoothing symbolsthat fulfil Stokes’ property, we introduce a smoothing symbol τ on Rd such that∫

Rd τ(ξ) dξ = 1 and set ρ := τ(ξ) dξ1 ∧ · · · ∧ dξd.

Proposition 6 1. Let a be a complex number with real part smaller than −d.

α ∈ ΩdCSa,lcc (Rd)⇐⇒ ∃β ∈ Ωd−1CSa+1,lcc (Rd), α = d β + ρ

∫Rdα

2. Linear forms on a subset S of CS<−dcc (Rd) stable under partial differentiation,which vanish on partial derivatives, are proportional to the ordinary integral.

3. Linear forms on S−∞cc (Rd) with Stokes’ property are proportional to the ordinaryintegral.

Proof:

1. This follows from Lemma 2 applied to α− ρ∫

Rd α instead of α.

2. By the first part of the proposition that

α ∈ ΩdS ⇐⇒ ∃β ∈ Ωd−1CScc(Rd), α = d β + ρ

∫Rdα.

A linear form λ on S which vanishes on derivatives induces a linear form λ whichvanishes on exact forms dβ ∈ ΩS so that

λ(α) = λ(d β) + λ(ρ)∫

Rdα = c

∫Rdα,

where we have set c = λ(ρ).

3. The proof is similar to the second item. By Exercise 14, we have S−∞cc (Rd) =⋂Re(a)<−d CS

acc(Rd). It then follows from the first part of the proposition that

α ∈ ΩdS−∞cc (Rd)⇐⇒ ∃β ∈ Ωd−1S−∞cc (Rd), α = d β + ρ

∫Rdα.

A linear form λ on S−∞cc (Rd) which satisfies Stokes’ property induces a closedlinear form λ on ΩS−∞cc (Rd) so that

λ(α) = λ(d β) + λ(ρ)∫

Rdα = c

∫Rdα,

where we have set c = λ(ρ). tu

We now wish to extend the ordinary integral on smoothing symbols to linearforms on a class of admissible subsets of the kernel of the residue.

25

Definition 6 We call a subset S of CScc(Rd) such that

S−∞cc (Rd) ⊂ S ⊂ Ker(res)

admissible if

(a) it is stable under partial differentiation

σ ∈ S =⇒ ∂iσ ∈ S ∀i ∈ 1, · · · , d,

(b) it is stable under taking “primitives”, by which me mean that there aresymbols τi as in (28)

σ ∼d∑i=1

∂iτi

which lie in S together with their homogeneous components, i.e.

τi ∼∞∑j=0

τi,a−j χ =⇒ τi,a−j χ ∈ S ∀j ∈ Z+.

Remark 5 Since S might not contain the constants, we do not require that any“primitive” τi lies in S.

The kernel Ker(res) of the noncommutative residue, which is stable under partialdifferentiation, does not satisfy the second requirement, since the derivatives of ahomogeneous function τ of degree −d with non-vanishing residue have vanishingresidue. There are nevertheless various admissible subsets of the kernel of theresidue, such as the class of non-integer order symbols or of odd-class symbolsin odd dimensions.

Exercise 23 Check that the set CS /∈Zcc (Rd) of non-integer order symbols is ad-

missible. Observe that it does not contain the constants.

Exercise 24 Check that when d is odd, the set

CSoddcc (Rd) := σ ∈ CSZ

cc(Rd), σa−j(− ξ) = (−1)a−jσa−j(ξ) ∀j ∈ Z+, ∀ξ ∈ Rd(34)

of odd-class symbols is admissible.

The following theorem classifies linear forms with Stokes’ property on a class ofadmissible sets. The full cohomology of such sets can be computed, see [L2].

Theorem 2 Let S be a subset of CScc(Rd) which is admissible and satisfies thefollowing inclusions:

S−∞cc (Rd) ⊂ S ⊂ Ker(res).

The ordinary integration map∫

Rd uniquely extends to a linear form on S withStokes’ property which we call canonical integral and denote by −

∫Rd .

Any other linear form on S with Stokes’ property is proportional to −∫

Rd .

26

Proof: Since the linear form λ on S satisfies Stokes’ property, it induces aclosed linear form λ on ΩS so that, with the notations of Proposition 6, for anyα ∈ Ω

(S ∩ L1(Rd)

)λ(α) = λ(d β) + λ(ρ)

∫Rdα = c

∫Rdα.

where we have set c = λ(ρ). Hence λ restricts on L1 symbols to a linear formproportional to the ordinary integration map:

∃c ∈ C, λS∩L1(Rd) = c

∫Rd.

We now want to extend this linear form on L1 symbols to the whole set S. By(18) a symbol σ ∈ S can be written:

σ(ξ) =N−1∑j=0

σa−j(ξ)χ(ξ) + σχ(N)(ξ) ∀ξ ∈ Rd (35)

where χ is a smooth function which vanishes in a neighborhood of 0 and isidentically one outside the unit ball and where σχ(N) is a symbol whose orderhas real part smaller than −d. Since λ is determined on the L1-part σχ(N) of thesymbol, by linearity it suffices to determine λ on a finite number of expressionsof the type σa−j χ involving positively homogeneous components σa−j .We therefore need to extend the integral to a linear form λ with Stokes’ propertyon expressions of the type f χ with f a positively homogeneous function inKer(res). By Exercise 18

res(f) = 0 =⇒ f =d∑i=1

∂ifi

for some positively homogeneous functions f1, · · · , fd. The set S being admis-sible, the fi’s can be chosen such that fi χ lies in S; we are therefore left todefine λ(∂ifi χ) for any homogeneous function fi such that fi χ lies in S. Sinceit should satisfy Stokes’ property, the linear form λ on ∂ifi χ reads:

λ(∂ifi χ) = −λ(fi ∂iχ) = −c∫

Rdfi ∂iχ. (36)

The second equality follows from the fact that ∂iχ is smoothing and the fact thatthe linear form λ is proportional to ordinary integration on smoothing symbolsas a result of Proposition 6. To make sure that equation (36) defines λ onS ∩ Ker(res) consistently, we observe that this definition is independent of thechoice of primitive fi + c of gi := ∂ifi. Indeed, applying (29) to X = χ ei wehave ∫

B(0,1)

∂iχdξ =∫Sd−1

χ 〈ei, ν〉 dµS =∫Sd−1〈ei, ν〉 dµS = 0,

since the outward pointing normal vector ν to Sd−1 points in opposite directionsat diametrically situated points of the sphere.

27

Applying these constructions to each homogeneous function σa−j with Re(a)−j ≥ −d defines λ on S by

λ(σ) := −d∑i=1

N−1∑j=0

∫Rdτa−j,i ∂iχ+

∫Rdσχ(N).

This determines λ uniquely and the fact that λ satisfies Stokes’ property followsfrom its very construction.

tu

Exercise 25 With the notations of Theorem 2, if moreover S is invariant under theaction of GLd(R), i.e. if:

σ ∈ S =⇒ σ C ∈ S ∀C ∈ GLd(R),

show that −∫

Rd is covariant, i.e.:

|detC| −∫

Rdσ C = −

∫Rdσ ∀C ∈ GLd(R). (37)

Hint: It suffices to check the result on symbols f χ ∈ S with f positively homogeneous,for which the proof easily follows from integration by parts and the covariance of theordinary integration writing f =

∑di=1 ∂ifi.

Applying Theorem 2 to the admissible set of non-integer order classical symbols leadsto the following statement.

Corollary 1 Ordinary integration uniquely extends to a linear form −∫

Rd on the subsetCS /∈Z

cc (Rd) which satisfies Stokes’ property. Any other linear form on CS /∈Zcc (Rd) which

satisfies Stokes’ property is proportional to the canonical integral −∫

Rd .

Hence, the canonical integral is covariant and satisfies Stokes’ property, as a conse-quence of which changes of variables and integration by parts turn out to be legitimatefor the canonical integral on non-integer order symbols. This justifies the use of thesetechniques by physicists who work in non-integer dimensions (which, as we shall seelater in the notes corresponds to non-integer order integrands) when implementing thedimensional regularisation procedure.

Exercise 26 Let V be a finite dimensional complex linear space. Show that any linearform on CS /∈Z

cc (Rd)⊗End(V ) which satisfies Stokes’ property and vanishes on expres-sions of the form σ ⊗ [u, v] for any two endomorphisms u and v of V is proportionalto the canonical integral −

∫Rd tr.

Hint: By a density argument, one only needs to check that a linear form λ onCS /∈Z

cc (Rd) ⊗ End(V ) with Stokes’ property which vanishes on elements of the formσ ⊗ [u, v] reads λ(σ ⊗ u) = c tr(u) −

∫Rd σ for some constant c. To prove this fact, fix

a symbol σ in S and show that λ induces a linear form λσ : u 7→ λ(σ ⊗ u) which isproportional to the matrix trace tr:

∃cσ ∈ C, λσ = cσtr(u).

28

2.5 Cut-off (or Hadamard finite parts) regularised integral

The (momentum) cut-off integral widely used in the physics literature and in themathematics literature under the name “Hadamard finite parts”, provides a realisationof the canonical integral on non-integer order symbols. The obstruction which preventsit from extending to a linear form with Stokes’ property on the whole algebra ofsymbols is measured in terms of a noncommutative residue. Let us start with asomewhat technical exercise.

Exercise 27 Given a symbol σ ∈ CSacc(Rd) for some complex number a, the mapR 7→

∫B(0,R)

σ(ξ) d ξ has an asymptotic expansion as R → ∞ of the form (with thenotations of (18)):∫

B(0,R)

σ(ξ) d ξ ∼R→∞ α0(σ) +∞∑

j=0,a−j+d6=0

Ra−j+d

a− j + d

∫|ξ|=1

σa−j(ξ) dS ξ

+ logR∫Sd−1

σ−d((ξ) dSξ (38)

for some scalar α0(σ) corresponding to the finite part as R→∞.Hint: Use (18) to write σ =

∑N−1j=0 σa−j(ξ)χ(ξ) + σχ(N) with the order of σχ(N) de-

creasing with N and split the integral accordingly:∫B(0,R)

σ(ξ) d ξ

=∫B(0,1)

σ(ξ)χ(ξ) d ξ +N−1∑j=0

∫B(0,R)−B(0,1)

σa−j(ξ) d ξ +∫B(0,R)−B(0,1)

σχ(N)(ξ) d ξ.

The result then follows from the homogeneity of σa−j.

Definition 7 We call (momentum) cut-off (see e.g. [ZJ] for a physicist’s pointof view) (also called Hadamard finite part method in mathematics 13, see also [Ge])the regularised integral of a symbol σ in CSacc(Rd) for some complex number a, theconstant term in the asymptotic expansion of the map R 7→

∫B(0,R)

σ(ξ) d ξ as R tendsto ∞:

CScc(Rd) → R

σ 7→ −∫

Rdσ(ξ) dξ := fpR→∞

∫B(0,R)

σ(ξ) dξ,

so that

−∫

Rdσ(ξ) dξ :=

N−1∑j=0

∫B(0,1)

σa−j(ξ)χ(ξ) d ξ +∫

Rdσχ(N)(ξ) dξ

−N−1∑

a−j+d 6=0, j=0

1a− j + d

(∫Sd−1

σa−j(ξ) dsξ). (39)

where N > Re(a) + d is chosen arbitrarily large.

13see e.g. [Sch] Example 2 chapter II

29

Note that the cut-off integral extends the ordinary integral since

−∫

Rdσ(ξ) dξ =

∫Rdσ(ξ) d ξ ∀σ ∈ CS<−dcc (Rd).

Exercise 28 Show that

−∫

Rdσ(ξ) dξ =

∫B(0,1)

σ(ξ) dξ +∫

Rd−B(0,1)

σ(ξ)−Na−1∑j=0

σa−j(ξ)

dξ

−N−1∑

a−j+d6=0, j=0

1a− j + d

(∫Sd−1

σa−j(ξ) dsξ)

where N is any positive integer chosen large enough for σχ(N) to be in L1(Rd). Recoverfrom there the fact that −

∫Rd σ(ξ) dξ =

∫Rd σ(ξ) d ξ for any σ in CS<−dcc (Rd).

Remark 6 Replacing dξ (resp. dSξ) by the normalised measure dξ = dξ(2π)d

(resp.

dSξ = dSξ(2π)d

), the asymptotic expansion (38) reads:∫B(0,R)

σ(ξ) d ξ ∼R→∞ −∫

Rdσ(ξ) d ξ (40)

+∞∑

j=0,a−j+d6=0

Ra−j+d

a− j + d

∫Sd−1

σa−j(ξ) dS ξ + logR res(σ).

Exercise 29 Show that the cut-off regularised integral vanishes on polynomials.Hint: We need to show that −

∫Rd ξ

α dξ = 0 for any multiindex α ∈ Nd0.

Exercise 30 Show that −∫

R41

|ξ|2+1 dξ = 0.

Exercise 31 Show that −∫

R41

(|ξ|2+1)2+k dξ = π2

2 k(k+1) for positive integers k whereaswhen the order of the symbol is minus the dimension14,we have −

∫R4

1(|ξ|2+1)2 dξ = −2π2.

The previous constructions extend to log-polyhomogeneous symbols.

Proposition 7 Let σ be a symbol in CSa,kcc (Rd) for some complex number a and somenon-negative integer k.

1. The map R →∫B(0,R)

σ(x, ξ) dξ has an asymptotic expansion as R → ∞ indecreasing powers of R with coefficients given by polynomial expressions of degreek + 1 in logarithmic powers of R.

2. The constant term in this expansion, called the cut-offf regularised integral14this corresponds to a logarithmic divergence in the physics terminology

30

of σ, reads:

−∫

Rdσ(ξ) dξ := fpR→∞

∫B(0,R)

σ(ξ) dξ

=n−1∑j=0

∫B(0,1)

χ(ξ)σa−j(ξ) dξ +∫

Rdσχ(N)(ξ) dξ

+N−1∑j=0

k∑l=0

(−1)l+1l!(a− j + d)l+1

∫Sd−1

σa−j,l(ξ) dξ

=∫|ξ|≤1

σ(ξ) dξ +∫|ξ|≥1

σ(ξ)−n−1∑j=0

σa−j(ξ)

dξ

+N−1∑j=0

k∑l=0

(−1)l+1l!(a− j + d)l+1

∫Sd−1

σa−j,l(ξ) dSξ. (41)

which yields back (39) when k = 0. Here as before, we have set dξ = dξ(2π)d

.

Proof: The proof is derived using (19) by which we have∫B(0,R)

σ(ξ) dξ =k∑l=0

N−1∑j=0

∫B(0,R)

χ(ξ)σa−j,l(ξ) logl |ξ| dξ +∫B(0,R)

σχ(N)(ξ) dξ

with N chosen large enough so that σχ(N) lies in L1(Rd). tu

Remark 7 In practice we consider the normalised cut-off integral denoted by −∫

Rd σ:

−∫

Rdσ(ξ) dξ =

1(2π)d

−∫

Rdσ(ξ) dξ.

We take the following definition from Kontsevich and Zagier [KZ].

Definition 8 A period is a complex number, whose real and imaginary parts arevalues of absolutely convergent integrals of rational functions with rational coefficientsover a domain in Rd given by polynomial (in)equalities with rational coefficients.

Remark 8 As pointed out by Bogner and Weinzierl [BW] logarithms of rationalfunctions with rational coefficients are also allowed in integrands since log f(ξ) =∫ 1

0f(ξ)−1

(f(ξ)−1)t+1 dt.

Example 1 1. Any algebraic number is a period. More generally, irrational al-gebraic numbers and functions are expressible as integrals of rational functionsover rational domains, so that in the above definition, “rational” can be replacedby “algebraic”.

2. The irrational number π =∫ ∫

x2+y2≤1dxdy, which is transcendental by Linde-

mann’s Theorem, is a period.

31

Thus, the algebra P formed by periods is strictly larger than the field Q of algebraicnumbers. Clearly, an absolutely convergent integral over Rd of a rational functionwith rational coefficients or a product of such a function with powers of the logarithmthereof, is a period. The subsequent corollary tells us this holds true for the finite partof non-convergent such integrals. The following exercise serves as a useful preliminarytechnical result.

Exercise 32 1. Show that if the symbol σ is a rational function with rational co-efficients, then so are its homogeneous components σa−j.Hint: This can be shown by induction on j starting from the fact that for z = 0we have σa(ξ) = limt→∞ (t−a σ(t ξ)) where a is the order of σ.

2. Show that this extends to homogeneous components σa−j,l of powers of logarithmsof σ.Hint: Combine an induction on j with an induction on m = k − l where k isthe logarithmic degree of σ starting from the fact that

σa,k(ξ) = limt→∞

(1

ta logk tσ(t ξ)

).

We now restrict to rational functions with rational coefficients (resp. products of suchfunctions with powers of the logarithm thereof), which we know to be classical (resp.log-polyhomogeneous) symbols by Exercise 11.

Corollary 2 Let σ be a smooth rational function on Rd with rational coefficients ora product of such a function with a power of the logarithm thereof.Then the cut-off regularised integral −

∫Rd σ(ξ)dξ is a period.

Proof: By (41) the cut-off integral −∫

Rd σ(ξ) dξ is a sum of terms∫|ξ|≤1

σ(ξ) dξ,∫|ξ|≥1

(σ(ξ)−

∑N−1j=0 σa−j(ξ)

)dξ and (−1)l+1l!

(a−j+d)l+1

∫|ξ|=1

σa−j,l(ξ) dSξ all of which areperiods since the domains of integration are given by polynomial inequalities withrational coefficients and the integrands are by assumption rational functions withrational coefficients or products of such functions with powers of the logarithm thereof.tu

2.6 Uniqueness of the canonical integral on non-integer ordersymbols

Two features characterise the canonical integral, the fact that it extends the ordinaryintegral beyond L1-symbols and the fact that it obeys Stokes’ property. We beginwith a description of all possible linear extensions of the ordinary integration map.

Proposition 8 Linear extensions of the ordinary integration map on CScc(Rd) ∩L1(Rd) are of the form:

λ(σ) = −∫

Rdσ +

∑j∈Z+

λj(σ−d+j) (42)

for some linear forms λj , j ∈ Z+ on C∞(Rd − 0), with λj identically zero forsufficiently large j.

32

Proof: Let us set λ := λ−−∫

Rd , which by assumption vanishes on CS<−dcc (Rd). Withthe notations of (35), let us write a symbol σ :=

∑N−1j=0 χσa−j + σχ(N) with N chosen

large enough for σχ(N) to be in CS<−dcc (Rd). Then

λ(σ) =N−1∑j=0

λ(χσa−j) + λ(σχ(N)) =N−1∑j=0

λ(χσa−j).

On the one hand, λ(χσ−d−j) vanishes for any positive integer j since it vanishes onCS<−dcc (Rd). On the other hand, λ(χσa−j) is independent of χ; indeed λ(χ1 σa−j)−λ(χ2 σa−j) = λ((χ1 − χ2)σa−j) = 0 for two smooth cut-off functions χ1 and χ2 since(χ1−χ2)σa−j is a smoothing symbol. The expression λ(σ) therefore only depends ona finite number of homogeneous components σ−d+j , j ∈ Z+.This combined with the linearity of λ implies the existence of linear forms λj , j ∈ Z+

on C∞(Rd − 0) that are identically zero for j sufficiently large such that

λ(σ) =∑j∈Z+

λj(σ−d+j).

tuThe following straightforward consequence provides another justification for the ter-minology “canonical integration”. It also justifies in hindsight the fact that physicistsoperate with divergent integrals as they would with convergent ones, once they havecomplexified the dimension, which means they are actually operating away from inte-ger order operators.

Theorem 3 The ordinary integration map on CScc(Rd)∩L1(Rd) uniquely extends toa linear map on CS /∈Z

cc (Rd), which therefore coincides with the canonical integrationmap −

∫Rd and obeys Stokes’ property.

Proof: By (42), if σ ∈ CS /∈Zcc (Rd) then σ−d+j = 0 for any non-negative integer j

so that λ(σ) = −∫

Rd σ. Linear extensions of the ordinary integration map thereforecoincide with the canonical integral on CS /∈Z

cc (Rd). tu

2.7 Uniqueness of the noncommutative residue on classicalsymbols

We show that (Proposition 9) the noncommutative residue is the unique (modulo amultiplicative constant) singular linear form on CScc(Rd) which obeys Stokes’ prop-erty. We then drop the “singularity” assumption and prove in Theorem 4 the unique-ness (modulo a multiplicative constant) of the noncommutative residue as a (nonnecessarily singular) linear form with Stokes’ property from the result of Proposition9.

Proposition 9 The noncommutative residue is the unique (modulo a multiplicativeconstant) singular linear form on CScc(Rd) which satisfies Stokes’ property.

Proof: Let λ be such a linear form. Given any smooth function χ which is zero in aneighborhood of 0 and identically one outside the unit ball, the symbol χ(ξ) |ξ|−d hasnon-vanishing residue; we normalise it to get a symbol ρ with residue equal to one.

33

Given a symbol σ ∈ CScc(Rd), σ− res(σ)ρ has vanishing residue so that by (27) thereexist symbols τi ∈ CScc(Rd), i = 1, · · · , d such that

σ − res(σ) ρ ∼d∑i=1

∂iτi. (43)

Thus,

λ(σ)− res(σ)λ(ρ) = λ(σ − res(σ) ρ) =d∑i=1

λ(∂lτi) = 0

which implies that λ is proportional to the residue. tu

Theorem 4 The noncommutative residue is the unique (modulo a multiplicative fac-tor) linear form on CScc(Rd) that satisfies Stokes’ property.

Proof: A linear form λ on CScc(Rd) that satisfies Stokes’ property restricts to oneon CS<−dcc (Rd) = CScc(Rd)∩L1(Rd) with the same property. By Proposition 6, suchlinear forms are proportional to the usual integral so that

∃c ∈ C, λ∣∣CScc(Rd)∩L1(Rd)= c

∫Rd.

Let us consider the linear form λ := λ − c −∫

Rd , which is singular since it vanishes onCS<−dcc (Rd).Let us introduce the symbols τ(ξ) = χ(ξ) |ξ|−d where χ is any smooth function whichvanishes in a small neighborhood of zero and is one outside the unit ball, and the homo-geneous functions τi(ξ) = ξi |ξ|−d which satisfy the following identity;

∑di=1 ∂iτi = 0.

Since λ vanishes on smoothing symbols and λ satisfies Stokes’ property, for any smoothcut-off function χ we have

0 =d∑i=1

λ (∂i(χ τi)) since λ is singular

=d∑i=1

λ(∂i(χ τi))− cd∑i=1

−∫

Rd∂i(χ τi)

= −cd∑i=1

−∫

Rd∂i(χ τi) by Stokes′ property

= −cd∑i=1

res(ξ 7→ χ(ξ) τi(ξ) ξi |ξ|−2)

= −c Vol(Sd−1)(2π)d

by Theorem 6. Hence c = 0 and λ = λ. Thus the linear form λ is both singularand satisfies Stokes’ property. From Proposition 9 we know it is proportional to thenoncommutative residue, which ends the proof of the theorem. tu

Exercise 33 Let V be a finite dimensional complex linear space. Show that any linearform on CScc(Rd)⊗ End(V ) which satisfies Stokes’ property and vanishes on expres-sions of the form σ ⊗ [u, v] for any two endomorphisms u and v of V is proportionalto the noncommutative residue res tr.Hint: See Exercise 26.

34

2.8 Translation invariance versus Stokes’ property

The integration map∫

Rd is known to be the only (modulo a multiplicative constant)translation invariant linear form on L1-functions. We characterise the noncommuta-tive residue as the unique (up to a multiplicative factor) translation invariant linearform on CScc(Rd).The following proposition shows that Rd acts via translations on the algebra CScc(Rd)of classical symbols.

Proposition 10 Given a symbol σ in CSacc(Rd), for any η ∈ Rd, the translated symbolt∗ησ := σ(·+ η) lies in CSacc(Rd) and

t∗ησ(ξ) ∼∞∑|α|=0

∂αξ σ(ξ)ηα

α!

as a symbol in ξ.

Proof:

1. We first check that t∗ησ is a symbol by showing that for any η in Rd and for anymultiindex α, there is a constant Cα(η) such that∣∣∂αξ σ(ξ + η)

∣∣ ≤ Cα(η)〈ξ〉Re(a)−|α| ∀ξ ∈ Rd. (44)

Since σ is a symbol of order a, we know there exists a constant Cα such that∣∣∂αξ σ(ξ + η)∣∣ ≤ Cα〈ξ + η〉Re(a)−|α| ∀ξ ∈ Rd.

Since lim|ξ|→∞〈ξ+η〉〈ξ〉 = 1, there are constants C ′(η) and C ′′(η) such that

C ′(η)〈ξ〉 ≤ 〈ξ + η〉 ≤ C ′′(η)〈ξ〉,

hence the existence of a constant Dα(η) such that

〈ξ + η〉Re(a)−|α| ≤ Dα(η) 〈ξ〉Re(a)−|α|.

The constant Cα(η) := Dα(η)Cα thus satisfies (44).

2. Let us now show that t∗ησ is classical. Since σ is classical, by (18) we have

σ(ξ + η) ∼∞∑j=0

σa−j(ξ + η)χ(ξ + η),

where as before χ is a smooth function which vanishes in a neighborhood ofzero and is identically one outside the unit ball. Since ξ 7→ χ(ξ + η)− χ(ξ) hascompact support, it is smoothing and

σ(ξ + η) ∼∞∑j=0

σa−j(ξ + η)χ(ξ). (45)

35

Let us now investigate each term t∗ησa−j for any non-negative integer j. TheTaylor expansion at η = 0 reads:

t∗ησa−j(ξ) =∑

|β|≤N−1

∂βξ σa−j(ξ)ηβ

β!

+∑|β|=N

(∫ 1

0

(1− u)N−1∂βξ σa−j((1− u)ξ + uη)du)ηβ

β!. (46)

Using the mean value theorem for integrals applied to the continuous functionu 7→ (1 − u)N−1

∣∣∣∂βξ σa−j((1− u)ξ + uη)∣∣∣ on the compact interval [0, 1], we can

bound the remainder term from above by∣∣∣∣∫ 1

0

(1− u)N−1∂βξ σa−j((1− u)ξ + uη)∣∣∣∣ du ≤ ∣∣∣∂βξ σa−j((1− u0)ξ + u0η)

∣∣∣≤ C〈(1− u0)ξ + u0η〉Re(a)−j−|β|

for some u0 ∈]0, 1[ and some constant C depending on u0. Differentiating theremainder term α times with respect to ξ changes β to α + β and the aboveestimate therefore shows that the remainder term defines a symbol of ordera−N . Hence, we can write

t∗ησ(ξ) ∼∞∑j=0

(t∗ησ)a−j(ξ)χ(ξ), where (t∗ησ)a−j(ξ) :=∑

j+|β|=i

∂βξ σa−j(ξ)ηβ

β!

is positively homogeneous of degree a− j. Inserting this in (45) yields

t∗ησ(ξ) ∼∞∑i=0

∑j+|β|=i

∂βξ σa−j(ξ)χ(ξ)ηβ

β!∼

∞∑|α|=0

∂αξ σ(ξ)ηα

α!.

tuLet S be a subset of CScc(Rd) stable under Rd- translations, i.e.:

σ ∈ S =⇒ t?ησ ∈ S ∀ξ ∈ Rd.

Definition 9 A linear form λ : S → C is said to be Rd -translation invariantwhenever it is invariant under the action of the translation group:

t∗ηλ = λ ∀η ∈ Rd (resp. ∀η ∈ Zd),

where we have sett?ηλ(σ) := λ(t?ησ) ∀σ ∈ CScc(Rd).

The following proposition relates Rd- translation invariance and Stokes’ property.

Proposition 11 A translation invariant linear form λ on S satisfies Stokes’ property.Conversely, a linear form which vanishes on partial derivatives and on L1-symbols, istranslation invariant.

36

Proof: Let η ∈ Rd. Clearly, t∗ηλ = λ implies that λ∂iλ = limu→0t∗u ei

λ−λu = 0 where

ei denotes the i-th vector in the canonical orthonormal basis of Rd. Conversely, givena symbol σ in CScc(Rd), from the proof of Proposition 10 we know that there is aninteger K such that RηK(σ) := t∗ησ −

∑K−1|α|=0 ∂

βξ σ

ηβ

β! lies in L1(Rd). Thus,

λ(t∗ησ − σ

)=

K−1∑|α|=1

λ(∂βξ σ)ηβ

β!+ λ(RηK(σ)) = 0

under the assumptions of the proposition. tu

Exercise 34 Deduce that the noncommutative residue is Rd-translation invariant.

The following result which is a straightforward consequence of Theorem 4 characterisestranslation invariant linear forms on CScc(Rd).

Theorem 5 Any translation invariant linear form on CScc(Rd) is proportional to thenoncommutative residue.

2.9 Discrepancies of the cut-off integral

We describe various discrepancies of the cut-off integral on integer order symbols, allof which can be measured in terms of the noncommutative residue.

2.9.1 Obstruction to rescaling invariance

Proposition 12 Under a rescaling B(0, R)→ B(0, λR) for some positive real num-ber λ, the cut-off integral of a symbol σ ∈ CS /∈Z

cc (Rd) is modified by a term proportionalto the noncommutative residue:

fpR→∞

∫B(0,λR)

σ(ξ) d ξ = −∫

Rdσ(ξ) dξ + log λ res(σ).

It remains unchaged for a non-integer order symbol.

Proof: This easily follows from the fact that the only term logR whose rescalinglog λR might add contributions to the finite part arises in the asymptotic expansion(40) with a factor res(σ). This factor vanishes when the order is non-integer. tu

2.9.2 Obstruction to covariance

The subsequent proposition generalises Proposition 12, which corresponds to the caseA = λ Id.

Proposition 13 [L1] Under the action of an invertible matrix A ∈ GLd(R) the cut-off regularised integral transforms as follows:

|detA| −∫

Rdσ(Aξ) d ξ = −

∫Rdσ(ξ) dξ −

∫Sd−1

σ−d(ξ) log |A−1ξ| dSξ.

It behaves covariantly for a non-integer order symbol:

|detA| −∫

Rdσ(Aξ) d ξ = −

∫Rdσ(ξ) dξ ∀σ ∈ CS /∈Z

cc (Rd),

which confirms the result of Exercise 25.

37

Proof: In view of the asymptotic expansion (18) and the covariance property ofthe usual integral, it is sufficient to prove the result for a finite number of positivelyhomogeneous components of the symbol σa−j . Let f be positively homogeneous ofdegree a outside the unit ball. We want to compare |detA| −

∫Rd f(Aξ) d ξ with (see

(39))

−∫

Rdf(ξ) dξ :=

∫B(0,1)

f(ξ) d ξ − 1− δa+d

a+ d

∫Sd−1

f(ξ) dsξ.

We have:

|detA| −∫

Rdf(Aξ) d ξ

= |detA| fpR→∞∫B(0,R)

f(Aξ) d ξ

= fpR→∞

∫AB(0,R)

f(ξ) d ξ

=∫B(0,1)

f(ξ) d ξ + fpR→∞

∫AB(0,R)−B(0,1)

f(ξ) d ξ

=∫B(0,1)

f(ξ) d ξ +∫Sd−1

f(ξ)

(fpR→∞

∫ R |A−1ξ|−1

1

ra+d−1 dr

)dSξ

=∫B(0,1)

f(ξ) d ξ +∫Sd−1

f(ξ) fpR→∞

(Ra+d

(|A−1ξ|−1

)a+d − 1a+ d

(1− δa+d) +(logR− log |A−1ξ|

)δa+d

)dSξ

=∫B(0,1)

f(ξ) d ξ − 1− δa+d

a+ d

∫Sd−1

f(ξ)− δa+d

∫Sd−1

f(ξ) log |A−1ξ| dSξ

= −∫

Rdf(ξ) d ξ − δa+d

∫Sd−1

f(ξ) log |A−1ξ| dSξ.

2.9.3 Obstruction to Stokes’ property

The cut-off regularised integral, which extends the ordinary integral to a linear formon the algebra CScc(Rd) of classical symbols unfortunately does not fulfil Stokes’property. The obstruction to its closedness is measured in terms of a noncommutativeresidue, which vanishes on non integer order symbols. Related results can be found in[MMP].

Theorem 6 Given a symbol σ in CS∗,∗cc (Rd) we have

−∫

Rd∂iσ(ξ) dξ = res0(ξ 7→ χ(ξ)σ(ξ) ξi |ξ|−2), (47)

independently of the choice of smooth cut-off function χ which vanishes in a neighbor-hood of zero and is one outside the unit ball.

38

Proof: By (29) for any index i in 1, · · · , d and any symbol σ ∈ CSa,kcc (Rd) for somenon-negative integer k, we have

−∫

Rd∂iσ(ξ) dξ = fpR→∞

(∫B(0,R)

∂iσ(ξ) dξ

)

= fpR→∞

(Rd∫B(0,1)

∂iσ(Rξ)

)dξ

= fpR→∞

(Rd−1

∫B(0,1)

∂i(σ(Rξ)) dξ

)

= fpR→∞

(Rd−1

∫Sd−1

σ(Rξ) 〈ei, ν〉 dSξ).

Here as before dSξ = 1(2π)d

dSξ, is the induced measure on the unit sphere Sd−1 and

η = ( ξ1|ξ| , · · · ,ξd|ξ| ) the outward pointing normal vector at ξ. Hence, by the asymptotic

expansion (19)

−∫

Rd∂iσ(ξ) dξ =

k∑l=0

N−1∑j=0

fpR→∞Rd−1

(∫Sd−1

σa−j,l(Rξ) 〈ei, ν〉 dSξ logl |Rξ|)

+ fpR→∞

(Rd−1

∫Sd−1

σχ(N)(Rξ) 〈ei, ν〉 dSξ)

=k∑l=0

N−1∑j=0

fpR→∞

(Ra−j+d−1 loglR

∫Sd−1

σa−j,l(ξ) 〈ei, ν〉 dSξ)

= δa−j+d−1

∫Sd−1

σa−j,0(ξ) 〈ei, ν〉 dSξ

=∫Sd−1

σ−d+1,0(ξ) ξi dSξ

= res0(ξ 7→ χ(ξ)σ(ξ) ξi |ξ|−2),

independently of the choice of smooth cut-off function χ which vanishes in a neigh-borhood of zero and is one outside the unit ball.tu

The following corollary confirms the uniqueness result of Theorem 2 on non-integerorder symbols.

Corollary 3 The cut-off regularised integral satisfies Stokes’ property on CS /∈Zcc (Rd),

where it coincides with the canonical integral, but does not satisfies Stokes’ propertyon CScc(Rd), neither does it on Ker(res).

Proof: By Theorem 6 we have −∫

Rd ∂iσ(ξ) dξ = 0 if the order of σ is non-integer whichshows that the cut-off regularised integral satisfies Stokes’ property on non-integerorder symbols. By the uniqueness of the canonical integral non-integer order symbols,the cut-off integral coincides with the canonical integral on non-integer order symbols.To show it does not satisfy Stokes’ property on the whole algeba CScc(Rd) take σi(ξ) =χ(ξ) ξi

|ξ|d where, as before χ is a smooth cut-off function which is 1 outside the unit

39

ball and vanishes in a neighborhood of zero. Then

res

(ξ 7→

d∑i=1

χ(ξ)σi(ξ) ξi|ξ|−2

)= res

(ξ 7→ χ(ξ) |ξ|−d

)=

Vol(Sd−1)(2π)d

6= 0.

Hence, −∫

Rd

(∑di=1 ∂iσ

i(ξ))dξ 6= 0 so that the cut-off integral does not satisfy Stokes’

property.Since res(σi) = 0 whereas −

∫Rd ∂iσ

i 6= 0, the cut-off regularised integral does not satisfyStokes’ property on Ker(res). tu

2.9.4 Obstruction to translation invariance

Theorem 7 Under the action of a translation tη by a η ∈ Rd, the cut-off regularisedintegral transforms as follows. For any σ ∼

∑∞j=0 σa−j χ ∈ CScc(Rd) we have:

−∫

Rdt∗ησ(ξ) d ξ = −

∫Rdσ(ξ) dξ+

∑0≤j≤Re(a)+d

∑|α|=a−j+d

(∫B(0,1)

∂α (σa−j(ξ)χ) (ξ) dξ

)ηα

α!

,where a is the order of σ. The sum over α vanishes if σ ∈ CS<−dcc (Rd) or if a isnon-integer so that the cut-off integral is translation invariant on L1-symbols and onnon-integer order symbols:

−∫

Rdt∗ησ(ξ) d ξ = −

∫Rdσ(ξ) dξ ∀σ ∈ CS /∈Z

cc (Rd) ∪ CS<−dcc (Rd).

Proof : In view of the asymptotic expansion (18) and the covariance property of theusual integral, it is sufficient to prove the result for a finite number 0 ≤ j ≤ Re(a) +d of positively homogeneous components of the symbol σa−j . Let f be positivelyhomogeneous of degree a outside the unit ball. For large enough R we have:

−∫

Rdt∗ηf(ξ) d ξ −−

∫Rdf(ξ) dξ

= fpR→∞

(−∫t−η∗B(0,R)

f(ξ) dξ −−∫B(0,R)

f(ξ) dξ

)

= fpR→∞

(∫t−η∗B(0,R)B(0,R)

f(ξ) dξ −∫B(0,R)t−η∗B(0,R)

f(ξ) dξ

)

= fpR→∞Ra+d

(∫t−R−1η∗B(0,1)B(0,1)

f(ξ) dξ −∫B(0,1)t−R−1η∗B(0,1)

f(ξ) dξ

)

= fpR→∞Ra+d

(∫t−R−1η∗B(0,1)B(0,1)

f(ξ) dξ −∫B(0,1)t−R−1η∗B(0,1)

f(ξ) dξ

)

= fpR→∞Ra+d

∫B(0,1)

(f(ξ +R−1η)− f(ξ)

)dξ

=∑α

fpR→∞

(Ra+d−|α|

∫B(0,1)

∂αf(ξ) dξ

)ηα

α!

=∑

|α|=a+d

(∫B(0,1)

∂αf(ξ) dξ

)ηα

α!,

40

where we have set AB := A− (A∩B) for two sets A and B. Replacing f by σa−j χyields the result of the theorem. tu

41

3 The noncommutative residue as a complex residue

Evaluating at zero meromorphic functions presenting a pole at that point is a commontask among physicists. The minimal subtraction scheme used by physicists is one wayto extract a finite part at zero which coincides with ordinary evaluation at zero in theabsence of poles. It yields a regularised evaluator at zero which extends ordinary eval-uation at that point. We classify regularised evaluators (Theorem 8), which we thenapply to meromorphic extensions of integrals of classical symbols (Theorem 9) andlog-polyhomogeneous symbols (Theorem 10) of the ordinary integral on L1-symbols.Theorems 9 and 10 are based on joint work with Simon Scott [PS]. These results arethen applied to compare dimensional regularisation with cut-off regularisation (seeProposition 14).

3.1 Regularised evaluators

“Evaluating” meromorphic functions in one variable at zero with poles at that pointrequires a regularisation procedure. We describe all possible linear extensions of theordinary evaluation at zero to germs of meromorphic functions at zero.Let Merk0(C) be the set of germs of meromorphic functions at zero 15 with poles atzero of order no larger than k, and let Hol0(C) be the set of germs of holomorphicfunctions at zero.Inspired by Speer [Sp] we introduce the notion of regularised evaluator.

Definition 10 We call regularised evaluator at zero any linear form

λ : Mer0(C)→ C

which extends the following evaluator at zero on holomorphic functions at zero:

ev0 : Hol0(C) → Cf 7→ f(0).

One way of building a regularised evaluator at zero is to compose the evaluator withan appropriate Rota-Baxter operator.

Definition 11 A linear operator R : A → A on an (not necessarily associative)algebra A over a field F is a called Rota-Baxter16 of weight λ ∈ F if it satisfies thefollowing Rota-Baxter relation:

R(a)R(b) = R(R(a) b) +R(aR(b)) + λR(ab).

Remark 9 If λ 6= 0, replacing R by λ−1R gives rise to a Rota-Baxter operator ofweight 1.

If f(z) =∑∞i=k

aizi we set Resj0(f) := a−j , called the j-th residue of f at zero. Let

Mer0(C) := ∪∞k=0Merk0(C). The projection map

R+ : Mer0(C) → Hol0(C)

f 7→

z 7→ f(z)−k∑j=1

Resj0(f)zj

if f ∈ Merk0(C)

15i.e. equivalence classes of meromorphic functions defined on a neighborhood of zero for theequivalence relation f ∼ g if f and g coincide on some open neighborhood of zero.

16The concept of Rota-Baxter operator was actually introduced by Glen Baxter in [Ba] and furtherdeveloped by Gian-Carlo Rota in [Rot].

42

corresponds to what physicists call minimal subtraction scheme. Whereas R+(f)corresponds to the holomorphic part, R−(f) = (1−R+)(f) corresponds to the “polepart”.The contribution R+(fg) to the holomorphic part of the product of two meromorphicfunctions f and g differs from the product R+(f)R+(g) of the holomorphic parts of fand g by contributions of the poles through R−(f) and R−(g):

R+(f g) = R+(f)R+(g) +R+(f R−(g)) +R+(g R−(f)). (48)

Exercise 35 The maps R+ and R− are both Rota-Baxter maps of weight −1 onMer0(C), i.e.

R+(a)R+(b) = R+(R+(a) b) +R+(aR+(b))− R+(ab)

and similarly for R−.

Combining the evaluation at zero with the map R+ provides a first regularised evalu-ator on Merk0(C) at zero: The map

evreg0 : Merk0(C) → C

f 7→ ev0 R+(f), (49)

is a linear form that extends the ordinary evaluator ev0(f) = f(0) defined on thespace Hol0(C) of holomorphic functions at zero. The following result provides a clas-sification of regularised evaluators. The parameter µ that arises here is related to therenormalisation group parameter in quantum field theory.

Theorem 8 Regularised evaluators at zero on Merk0(C) are of the form:

λ0 = evreg0 +

k∑j=1

µj Resj0 (50)

for some constants µ1, · · · , µk.In particular, regularised evaluators at zero on Mer1

0(C) are of the form:

λ0 = evreg0 + µRes0. (51)

for some constant µ.

Proof: A linear form λ0 which extends ev0 coincides with ev0 on the range of R+

and therefore fulfils the following identity; λ0 R+ = ev0 R+ = evreg0 . Thus, for any

f ∈ Merk0(C),

λ0(f) = λ0 (R+(f)) + λ(R−(f))

= evreg0 +

k∑j=1

cj Resj0(f)

where we have set cj := λ(z−j). tu

43

3.2 Holomorphic families of symbols and meromorphic exten-sions of integrals

The noncommutative residue deserves its name as it can be realised as a complexresidue of the cut-off integral of a holomorphic family of symbols. The notion ofholomorphic family of classical pseudodifferential symbols was first introduced byGuillemin in [Gu3] and extensively used by Kontsevich and Vishik in [KV]. Theidea is to embed a symbol σ in a family z 7→ σ(z) depending holomorphically on acomplex parameter z.

A function f : Ω→ E on complex domain Ω with values in a topological vector space Eis holomorphic at z0 ∈ Ω if there is a vector f ′(z0) in E such that

∣∣∣ f(z)−f(z0)z−z0 − f ′(z0)

∣∣∣tends to zero as z tends to z0; it is holomorphic on Ω if this holds at each point z0

in Ω. Known results for Banach space valued holomorphic functions (see e.g. [Hil]Chapter 8) generalise to (sequentially) complete Hausdorff locally convex topologicalvector space valued functions17, i.e. to E-valued functions with the topology of thecomplete Hausdorff space E defined by a family of semi-norms | · |α, α ∈ A. Inductivelimits of Frechet spaces (known as LF spaces) of interest to us fall in this class ofspaces.In particular, holomorphicity implies analyticity by means of the Cauchy formula (see[Hil] Theorem 8.1.1 in the Banach case), which for a holomorphic function f at z0 ∈ Ωreads:

f(z0) =1

2iπ

∫|ζ−z0|=r

f(ζ)z − ζ

dζ,

where r is a positive number such that the disk centered at z0 of radius r is containedin Ω. One first observes that convergence as z tends to z0 holds uniformly on compactsubsets of Ω in a neighborhood of z0 so that f : Ω → E is uniformly complex-differentiable on compact subsets in a neighborhood of z0. By induction one showsthat it is infinitely (uniformly) complex-differentiable (see e.g. [Hil] Theorem 8.1.5 inthe Banach case) on (compact subsets of) Ω in a neighborhood of z0 with derivativegiven by

f (k)(z0) =k!

2iπ

∫|z−z0|=r

f(ζ)(z − ζ)k+1

dζ ∀ k ∈ N.

It follows that (see [Hil] Theorem 8.1.6 in the Banach case)

‖f (k)(z0)‖αk!

≤ Max|z−z0|=r‖f(z)‖αrk

∀k ∈ N ∀α ∈ A. (52)

17I thank Jean-Louis Ducourtioux for pointing this out to me; one can actually prove that weakholomorphicity implies analyticity.

44

For any complex number z such that |z − z0| < r we write

f(z) =1

2iπ

∫|u−z0|=r

f(u) (u− z)−1 du

=1

2iπ

∫|u−z0|=r

f(u)1− z−z0

u−z0(u− z0)−1 du

=1

2iπ

∞∑k=0

(z − z0)k∫|u−z0|=r

f(u) (u− z0)−(k+1) du since∣∣∣∣ z − z0

u− z0

∣∣∣∣ < 1

=∞∑k=0

f (k)(z0)(z − z0)k

k!. (53)

By (52) this series converges uniformly on any disk |z − z0| < r′ < r so that f is(uniformly) analytic in (compact subsets) of a neighborhood of z0. Note that theradius of convergence is independent of α.This applies to the space E := S(U) (or more generally to E := S(U) ⊗ End(V )where V is some finite dimensional vector space) of all symbols on an open subset Uof Rn considered here, seen as the inductive limit of the Frechet spaces Fν := S≤ν(U)of symbols whose order has real part non larger than ν ∈ R. The Frechet structureon Fν is given by the following semi-norms labelled by multiindices α, β and positiveintegers i (see [Ho]):

supx∈Ki,ξ∈Rn(1 + |ξ|)−ν+|β| ‖∂αx ∂βξ σ(x, ξ)‖,

where Ki, i ∈ N is a countable sequence of compact sets covering U .

Definition 12 A family σ(z) of classical symbols on Rd parametrised by a domain Ωis holomorphic at point z0 ∈ Ω if:

1. σ(z) is holomorphic at z0 as a function of z with values in C∞(Rd) and

σ(z) ∼∑j≥0

σα(z)−j(z) ∈ CSα(z)cc (Rd), (54)

where the function α : Ω→ C is holomorphic at z0;

2. for any integer N ≥ 1 the remainder

σ(N)(z) := σ(z)−N−1∑j=0

σα(z)−j(z)

is holomorphic at z0 as a function of z with values in C∞(Rd) with kth z-derivative

σ(k)(N)(z) := ∂kz (σ(N)(z)) (55)

a symbol on Rd of order α(z)−N + ε for any ε > 0 locally uniformly in z, i.e.the k-th derivative ∂kzσ(N)(z) satisfies a uniform estimate (17) in z on compactsubsets in Ω.

If σ(z) is holomorphic at every point z0 ∈ Ω, it is called a holomorphic family ofsymbols parametrised by Ω.

45

Remark 10 We shall also use finite sums of holomorphic families of symbols, inwhich case the notion of holomorphic order no longer makes sense; we then need tohandle each holomorphic symbol in the sum separately.

Differentiation in the parameter z does not modify the order and leads to the followinguseful relation (see Lemma 1.16 in [PS]).

Exercise 36 Let z 7→ σ(z) be a holomorphic family of classical symbols in CScc(Rd)of holomorphic order α(z) parametrised by Ω. Show that

∂z(σα(z)−j(z)

)∣∣Sd−1

=(

(∂zσ)α(z)−j (z))∣∣

Sd−1

and deduce the following identity:

∂z

∫Sd−1

σα(z)−j(z) =∫Sd−1

(∂zσ)α(z)−j (z).

The following theorem recalls a result of Kontsevich and Vishik [KV] which relatesthe complex residue of the cut-off integral of a holomorphic family of symbols to thenoncommutative residue of the symbols and yields back a result derived in [PS] whichdescribes the finite part of the cut-off integral of a holomorphic family of symbols.

Theorem 9 Let z 7→ σ(z) be a holomorphic family of classical symbols in CScc(Rd)of holomorphic order α(z) parametrised by Ω.

1. The map

z 7→ −∫

Rdσ(z)(ξ)dξ

is meromorphic with simple poles in Ω ∩ α−1 (Z ∩ [−d,+∞[).

2. [KV] Provided α′(z0) 6= 0, then the complex residue at z0 ∈ Ω∩α−1 (Z ∩ [−d,+∞[)is given by:

Resz0

(−∫

Rdσ(z)(ξ)dξ

)= − 1

α′(z0)res(σ(z0)). (56)

Consequently, the finite part evaluated at z0 ∈ Ω using any regularised evaluatorλz0(f(z)) = λ (f(z0 + ·)) at z0 differs from the ordinary finite part at z0 evaluatedusing evreg

z0 by:

λz0

(−∫

Rdσ(z)(ξ)dξ

)− evreg

z0

(−∫

Rdσ(z)(ξ)dξ

)= µ res(σ(z0))

for some complex constant µ.

3. [PS] Provided α(z) is a non-constant affine function of z, the finite part at z0 ∈ Ωdiffers from the cut-off regularised integral −

∫Rd σ(z0)(ξ)dξ by

evregz0

(−∫

Rdσ(z)(ξ)dξ

)−−∫

Rdσ(z0)(ξ)dξ = − 1

α′(z0)res(σ′(z0)), (57)

where we have extended the noncommutative residue to the possibly non-classicalsymbol τ(z0)(ξ) = σ′(z0)(ξ)18 by implementing the same formula as in Definition3

res(τ(z0)) :=∫Sd−1

τ−d(z0)(ξ) dSξ.

18The asymptotic expansion of τ(z0)(ξ) as |ξ| → ∞ might present logarithmic terms in log |ξ|,which vanish on the unit sphere and therefore do not explicitly arise in the following definition.

46

Remark 11 Formula (57) formally follows from (56) applied to the family τ(z) =σ(z)−σ(z0)

z−z0 since τ(z0) = σ′(z0). However, the proof is not quite so straightforwardsince τ(z) is not a holomorphic family of classical symbols (outside z0) but only alinear combination of such a holomorphic family σ(z) and a constant symbol σ(z0).

Proof:

1. Using (18) we can write σ(z)(ξ) =∑N−1j=0 χ(ξ)σa−j(z)(ξ) + σ(N)(z)(ξ) with N

chosen large enough for σ(N)(z) to lie in L1(Rd) in a small neighborhood of somepoint z0. Thus the cut-off regularised integral which reads:

−∫

Rdσ(z)(ξ) dξ =

∫Rdσ(N)(z)(ξ) dξ (58)

+N−1∑j=0

∫B(0,1)

χ(ξ)σα(z)−j(z)(ξ) dξ

−N−1∑

j 6=α(z)−d

1α(z)− j + d

∫Sd−1

σa−j(z)(ω) dSω,

where B(0, 1) stands for the unit ball, provides a meromorphic extension withsimple pole at z0 whenever there is some j0 ∈ Z+ such that α(z0) = j0 − d i.e.whenever α(z0) ∈ [−d,+∞[∩Z.

2. Inserting the Taylor expansion at order zero of σα(z)−j0

σα(z)−j0(z) = σ−d(z0) +O(z − z0)

and the Taylor expansion at order one of α in (58)

α(z) = α(z0) + α′(z0)(z − z0) + o(z − z0)

we deduce that

Resz0 −∫

Rdσ(z)(ξ)dξ = −Resz0

(1

α(z)− α(z0)

∫Sd−1

σα(z)−j0(z)(ω) dSω)

= Resz0

(1

α′(z0)(z − z0)

∫Sd−1

σα(z)−j0(z)(ω) dSω)

= − 1α′(z0)

∫Sd−1

σ−d(z)(ω) dSω

= − 1α′(z0)

res0(σ(z0)),

The rest of the statement then follows from (50).

3. By Exercise 36, the Taylor expansion at order one of σα(z)−j0 reads:

σα(z)−j0(z) = σ−d(z0) +(σα(z)−j0

)′ (z0)(z − z0) + o(z − z0)= σ−d(z0) + (σ′(z0))−d (z − z0) + o(z − z0)= σ−d(z0) + σ′−d(z0)(z − z0) + o(z − z0)

47

Inserting this in (58) yields

evregz0 −∫

Rdσ(z)(ξ) dξ −−

∫Rdσ(z0)(ξ) dξ

= −evregz0

(1

α(z)− α(z0)

∫Sd−1

σα(z)−j0(z)(ω) dSω)

= − 1α′(z0)

∫Sd−1

σ′−d(z0)(ω) dSω

= − 1α′(z0)

res0(σ′(z0))

since α(z) is an affine function of z.

tu

Exercise 37 Recover from (56) combined with covariance, translation invariance andStokes’ property of the cut-off integral on non-integer order symbols, the fact that thenonmmutative residue fulfills these properties on the whole algebra CScc(Rd).Hint: Observe that for any matrix A ∈ GLd(R) we have

|detA| res(σA) = −α′(z0) |detA|Res0

(−∫

Rdσ(z) Adξ

)= −α′(z0) Res0

(−∫

Rdσ(z)dξ

)= res(σ),

where we have set σ := σ(0). This gives back the result of Exercise 17. A similar proofleads to translation invariance and Stokes’ property.

Exercise 38 Check that for any classical symbol σ with non-integer order a and forany holomorphic perturbation σ(z) (so that σ(0) = σ) of order −qz + a with q 6= 0such that σ(0) = σ the map z 7→ −

∫Rd σ(z) is holomorphic at z = 0 and we have

limz→0−∫

Rdσ(z) = −

∫Rdσ = ev0

(∫Rdσ(z)

)∀σ ∈ CS /∈Z

cc (Rd). (59)

Exercise 39 Let σ(z) be a holomorphic family such that σ(0) is a smooth rationalfunction on Rd with rational coefficients and σ′(0) a product of such a function witha power of the logarithm of a function of the same type. Then the regularised integralevreg

0 −∫

Rd σ(z) is a period.Hint: This follows from combining (57) with z0 = 0 and Corollary 2.

3.3 Extension to log-polyhomogeneous symbols

Theorem 9 generalises to log-polyhomogeneous symbols. We shall need the follow-ing generalisation of Exercise 36, which together with the theorem that follows, wasderived in unpublished joint work with Simon Scott. Whereas differentiation withrespect to z preserves the order, it increases the logarithmic type by one unit.

Lemma 3 For a non-negative integer k, let

σ(z)(ξ) ∼∞∑j=0

σα(z)−j(ξ) χ(ξ)

48

in CSα(z),kcc (Rd) be a holomorphic family of log-polyhomogeneous symbols with holo-

morphic order α(z). For any non-negative integer i, ∂iσ(z) lies in CSα(z),k+icc (Rd)

and for any 1 ≤ l ≤ k and |ξ| = 1 we have

∂z(σα(z)−j, l(z)(ξ)) = σ′α(z)−j, l(z)(ξ)− α′(z)σ(z)α(z)−j, l−1(ξ)

∂z(σα(z)−j, 0(z)(ξ)) = σ′α(z)−j, 0(z)(ξ),

σ′α(z)−j, k+1(z)(ξ) = α′(z)σα(z)−j, k(z)(ξ), (60)

where σ′α(z)−j,l stands for (∂zσ)α(z)−j,l. Moreover

∂iz

(k∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

σ(z)α(z)−j, l(ξ) dSξ

)

=k+j∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

(∂izσ(z)

)α(z)−j, l (ξ) dSξ. (61)

Proof: To simplify the presentation, we only prove the case i = 1; the case i > 1 canbe proved by induction along the same line of reasoning.On the one hand (60) follows from the relation:

∂z(σ(z)α(z)−j(ξ)

)=

k∑l=0

∂z

[σ(z)α(z)−j,l(ξ) logl |ξ|

]=

k∑l=0

∂z

[|ξ|α(z)−jσ(z)α(z)−j,l(ξ |ξ|−1) logl |ξ|

]= α′(z)

k∑l=0

σ(z)α(z)−j,l(ξ) logl+1 |ξ|

+k∑l=0

|ξ|α(z)−j∂z[σ(z)α(z)−j,l(ξ |ξ|−1)

]logl |ξ|,

which for |ξ| = 1 yields for 1 ≤ l ≤ k

∂z(σ(z)α(z)−j,l(ξ)

)= α′(z)σ(z)α(z)−j,l−1(ξ) + ∂z

(σ(z)α(z)−j,l(ξ)

)and the required identities for l = 0 and l = k + 1.On the other hand we have

∂z

(k∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

σ(z)α(z)−j, l(ξ) dSξ

)

= α′(z)k∑l=0

(−1)l+2(l + 1)!(α(z)− j + d)l+2

∫Sd−1

σ(z)α(z)−j, l(ξ) dSξ

+k∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

∂z(σ(z)α(z)−j, l(ξ)

)dSξ. (62)

49

generalises to log-polyhomogeneous symbols by (60). Hence, the last expression inequation (62) reads

k∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

∂z(σ(z)α(z)−j, l(ξ)

)dSξ

= − 1α(z)− j + d

∫Sd−1

σ′α(z)−j, 0(z)(ξ) dSξ

+k∑l=1

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

σ′α(z)−j, l(z)(ξ) dSξ

− α′(z)k∑l=1

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

σα(z)−j, l−1(z)(ξ) dSξ

=k+1∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

σ′α(z)−j, l(z)(ξ) dSξ

+(−1)k+1(k + 1)!

(α(z)− j + d)k+2

∫Sd−1

σ′α(z)−j, k+1(z)(ξ) dSξ

− α′(z)k−1∑l=0

(−1)l+2(l + 1)!(α(z)− j + d)l+2

∫Sd−1

σα(z)−j, l(z)(ξ) dSξ

=k+1∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

σ′α(z)−j, l(z)(ξ) dSξ

+ α′(z)(−1)k+1(k + 1)!

(α(z)− j + d)k+2

∫Sd−1

σα(z)−j, k(z)(ξ) dSξ

+ α′(z)k−1∑l=0

(−1)l+1(l + 1)!(α(z)− j + d)l+2

∫Sd−1

σα(z)−j, l(z)(ξ) dSξ

=k+1∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

σ′α(z)−j, l(z)(ξ) dSξ

+ α′(z)k∑l=0

(−1)l+1(l + 1)!(α(z)− j + d)l+2

∫Sd−1

σα(z)−j, l(z)(ξ) dSξ,

where we have used the fact that σ′α(z)−j,k+1(ξ) = α′(z)σα(z)−j,k(ξ).

Adding this to the first term∑kl=0

(−1)l+2(l+1)!α′(z)(α(z)−j+d)l+2

∫Sd−1

σ(z)α(z)−j, l(ξ)

α(z)−j+d dSξ in (62)yields

∂z

(k∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

σ(z)α(z)−j, l(ξ) dSξ

)

=k+1∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

σ′α(z)−j, l(z)(ξ) dSξ,

which gives (61) when j = 1. tuThe following theorem, based on unpublished joint work with Simon Scott, generalises

50

results derived by Kontsevich and Vishik for classical operators and later by Lesch tolog-polyhomogeneous symbols in [L1]. The case k = 0 yields back Theorem 9.

Theorem 10 Let for any non-negative integer k, z 7→ σ(z) ∈ CSα(z),kcc (Rd) be a

holomorphic family of log-polyhomogeneous symbols on a domain Ω ⊂ C with non-constant affine order α(z) = a− q z.

1. [L1] The map

z 7→ −∫

Rdσ(z)(ξ)dξ

is meromorphic with poles at points zj ∈ Ω∩]−∞, a+dq ] ∩ Z of order no larger

than k + 1.For any σ ∈ CS∗,kcc (Rd) and any holomorphic family σ(z) ∈ CS∗,kcc (Rd)∗,k withorder α(z) = −qz + a such that σ(0) = a and q 6= 0, we have

Resjz0

(−∫

Rdσ(z)(ξ) dξ

)=

k∑l=j−1

l!(j + l + 1)! ql+1

resl(σ(l+1−j)(z0)) (63)

so that the following expression defined in terms of the complex residue of orderk + 1 at 0:

resk(σ) :=∫Sd−1

σ−d,k(ξ) dSξ =qk+1

k!Resk+1

0 −∫

Rdσ(z)(ξ) dξ, (64)

is independent of the family σ(z).Consequently, the finite part evaluated at z0 ∈ Ω using any regularised evaluatorλz0(f(z)) = λ (f(z0 + ·)) at z0 differs from the ordinary finite part at z0 evaluatedusing evreg

z0 by:

λz0

(−∫

Rdσ(z)(ξ)dξ

)− evreg

z0

(−∫

Rdσ(z)(ξ)dξ

)=

k∑j=0

µj resj(σ(k+1−j)(z0))

for some constants µj , j = 0, · · · , k.

2. (joint work with Simon Scott) Furthermore, we have

evregz0 −∫

Rdσ(z)(ξ) dξ = −

∫Rdσ(z0)(ξ) dξ +

k∑l=0

1(l + 1) ql+1

resl(σ(l+1)(z0)

)(65)

where we have setresl(σ) :=

∫Sd−1

σ−d,l(ξ) dSξ. (66)

Moreover, for positive integers r, the coefficient of (z−z0)r

r! in the Laurent expan-sion of −

∫Rd σ(z)(ξ) dξ at z0 reads

sr(σ) = −∫

Rdσ(r)(z0)(ξ) dξ +

k∑l=0

l! r!(r + l + 1)! ql+1

resl(σ(r+l+1)(z0)

).

In particular, if σ lies in CScc(Rd) we have:

sr(σ) = −∫

Rdσ(r)(z0)(ξ) dξ +

1(r + 1) q

res(σ(r+1)(z0)

). (67)

51

Proof: Let z0 ∈ Ω and j0 ∈ N∪0 be the integer such that α(z0) +d− j0 = 0. Withthe notation of (41) applied to σ(z) we choose N large enough so that σ(N)(z) liesin L1(Rd) in a neighborhood of z0. From Proposition 7 we know that the followingidentity holds in a neighborhood of z0:

−∫

Rdσ(z)(ξ) dξ

=N−1∑j=0

∫B(0,1)

χ(ξ)σα(z)−j(ξ) dξ +∫

Rdσ(N)(z)(ξ) dξ

+N−1∑j=0

k∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

σα(z)−j,l(z)(ξ) dSξ

=N−1∑j=0

∫B(0,1)

χ(ξ)σα(z)−j(z)(ξ) dξ +∫

Rd−1σ(N)(z)(ξ) dξ

+N−1∑i 6=j0

k∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

σα(z)−j,l(z)(ξ) dSξ

+k∑l=0

(−1)l+1l!(α(z)− α(z0))l+1

∫Sd−1

σα(z)−j0,l(z)(ξ) dSξ. (68)

1. Since σ(z) is a holomorphic family of polyhomogeneous symbols, there is a powerseries expansion

σα(z)−j(z) =∞∑r=0

σ(r)α(z0)−j(z0)

(z − z0)r

r!. (69)

The first line of (68) gives rise to a holomorphic expression. When j 6= j0, theexpression α(z)− j + d does not vanish so that the second line of equation (68)is holomorphic as a function of z.On the other hand, the third line in (68) yields a meromorphic function withpoles at points z0 such that α(z0) ∈ Z. Since the order function z 7→ α(z) =−qz + a is affine, using a Taylor expansion

σα(z)−j,l(z) =∞∑r=0

σ(r)α(z0)−j,l(z0)

(z − z0)r

r!

we havek∑l=0

(−1)l+1l!(α(z)− α(z0))l+1

∫Sd−1

σα(z)−j0,l(z)(ξ) dSξ (70)

=k∑l=0

R+l+1∑r=0

[(−1)l+1l!

(α′(z0))l+1

∫Sd−1

σ(r)−d,l(z0)(ξ) dSξ

](z − z0)r−l−1

r!

+ O((z − z0)R)

Thus

Resjz0

(−∫

Rdσ(z) dξ

)=

k∑l=j−1

l!(l + 1− j)! ql+1

∫Sd−1

σ(l+1−j)−d;l (z0) dSξ.

52

When j = k + 1 this yields

Resk+1z0

(−∫

Rdσ(z) dξ

)=

k!ql+1

∫Sd−1

σ−d;k(ξ) (z0) dSξ.

The rest of the statement then follows from (51).

2. The first line of (68) gives rise to the following Taylor expansion:

N−1∑j=0

∫B(0,1)

χ(ξ)σα(z)−j(z)(ξ) dξ +∫

Rdσ(N)(z)(ξ) dξ

=N−1∑j=0

∫B(0,1)

χ(ξ)σα(z0)−j(z)(ξ) dξ +∫

Rdσ(N)(z0)(ξ) dξ

+R∑r=1

N−1∑j=0

∫B(0,1)

χ(ξ)σ(r)α(z0)−j(z)(ξ) dξ

(z − z0)r

r!

+R∑r=1

(∫Rdσ

(r)(N)(z0)(ξ) dξ

)(z − z0)r

r!

+ o((z − z0)R

)= −

∫Rdσ(z0)(ξ) dξ

+R∑r=1

(−∫

Rdσ(r)(z)(ξ) dξ

(z − z0)r

r!

)+ o

((z − z0)R

), (71)

where we have used (41) applied to σ(z) and σ(r)(z). On the other hand, itfollows from Lemma 3 that

k∑l=0

(−1)l+1l!(α(z)− j + d)l+1

∫Sd−1

σα(z)−j,l(z)(ξ) dSξ

=k∑l=0

R∑r=0

∂rz

((−1)l+1l!

(α(z)− j + d)l+1

∫Sd−1

σα(z)−j,l(z)(ξ)dSξ)|z=z0

(z − z0)r

r!

+ o((z − z0)R

)=

R∑r=0

[k+r∑l=0

(−1)l+1l!(α(z0)− j + d)l+1

∫Sd−1

σ(r)α(z0)−j,l(z0)(ξ) dSξ

](z − z0)r

r!

+ o((z − z0)R

).

(72)

Inserting (70) which we split in three parts corresponding to r ≤ l, r = l+ 1, r >

53

l + 1, (71), (72) back into (68) we find:

−∫

Rdσ(z)(x, ξ)dξ

=k∑l=0

l∑r=0

[(−1)l+1l!

(α′(z0))l+1

∫Sd−1

σ(r)−d,l(z0)(ξ) dSξ

](z − z0)r−l−1

r!dSξ

+ −∫

Rdσ(z0)(ξ) dξ

+k∑l=0

(−1)l+1

(l + 1) (α′(z0))l+1

∫Sd−1

σ(l+1)−d,l (z0)(x, ξ) dSξ

+R∑r=1

−∫T∗xU

σ(r)(z0)(z − z0)r

r!

+R∑r=0

[k∑l=0

(−1)l+1l!

(α′(z0))l+1

∫Sd−1 σ

(r+l+1)−d,l (z0)(ξ) dSξ(r + l + 1)!

](z − z0)r

+ o((z − z0)R

]=

k∑l=0

l∑r=0

[(−1)l+1l!

(α′(z0))l+1resl

(σ(r)(z0)

)] (z − z0)r−l−1

r!

+ −∫

Rdσ(z0)(ξ) dξ

+k∑l=0

(−1)l+1

(l + 1) (α′(z0))l+1resl

(σ

(l+1)−d,l (z0)

)(z − z0)r

+R∑r=1

−∫

Rd−1σ(r)(z0)

(z − z0)r

r!

+R∑r=0

k∑l=0

l!(r + l + 1)! ql+1

resl(σ(r+l+1)(z0)

)(z − z0)r

+ o((z − z0)R

).

In particular, for non-negative integers r, the coefficient of (z−z0)r

r! in the mero-morphic expansion of −

∫Rd σ(z)(ξ) dξ at z0 reads

sr(σ) = −∫

Rdσ(r)(z0)(ξ) dξ +

k∑l=0

l! r!(r + l + 1)! ql+1

resl(σ(r+l+1)(z0)

).

Relabelling the terms and combining these results yields the second part of thetheorem.

tu

Exercise 40 Let σ be a symbol in CS∗,∗cc (Rd). Show that if σ(z)(ξ) = σ(ξ) |ξ|−z forany |ξ| ≥ 1, then evreg

0 −∫

Rd σ(z)(ξ) dξ = −∫

Rd σ(ξ) dξ.

54

Corollary 4 Let for any non-negative integer k, σ(z) ∈ CSα(z),kcc (Rd) be a holomor-

phic family of log-polyhomogeneous symbols with non-constant affine order in z with zrunning in a complex domain Ω such that at a given complex number z0 and for anynon-negative integer r, the maps ξ 7→ σ(r)(z0)(ξ) are rational functions on Rd withrational coefficients or logarithms thereof.Then the coefficients of the Laurent expansion of the map z 7→ −

∫Rd σ(z)(ξ)dξ at z0 are

periods.

Proof: Recall from (63) that the coefficients in the Laurent expansion read for 1 ≤j ≤ k + 1

Resjz0

(−∫

Rdσ(z)(ξ) dξ

)=

k∑l=j−1

l!(j + l + 1)! ql+1

resl(σ(l+1−j)(z0)

)and that for any non-negative integer r, the ceofficient of (z−z0)r

r! reads

Resz0

(−∫Rd σ(z)(ξ) dξ(z − z0)r+1

)= −∫

Rdσ(r)(z0)(ξ) dξ+

k∑l=0

l! r!(r + l + 1)! ql+1

resl(σ(r+l+1)(z0)

).

Under the assumptions of the corollary, expressions of the form

resl(σ(i)) =∫Sd−1

σ(i)−d;l(z0) dSξ =

∫|ξ|≤1

σ(i)−d;l(z0) dSξ

are periods. The statement then follows from the fact that by Corollary 2, cut-offintegrals −

∫Rd σ

(i)(z0)(ξ) dξ are also periods. tu

3.4 Dimensional versus cut-off regularised integrals

Dimensional regularisation, often used in the physics literature (see e.g.[Col], [Et],[HV], [N]) is a prototype of holomorphic regularisation which we view here as a mod-ified Riesz regularisation. In contrast, physicists tend to view Riesz regularisation asa modified dimensional regularisation and compare it with cut-off regularisation.Let Hol0

(CScc(Rd)

)(resp. Hol0

(CS∗,∗cc (Rd)

)) denote the space of holomorphic germs

at zero of classical (resp. log-polyhomogeneous) symbols, with a holomorphic familyof symbols defined as in Definition 12.

Definition 13 A holomorphic regularisation procedure on CScc(Rd) (resp. onCS∗,∗(Rd)) is a linear embedding:

R : CScc(Rd) → Hol0(CScc(Rd)

)σ 7→ (z 7→ σ(z))

(resp. filtered embeddings Rk : CS∗,kcc (Rd) → Hol0(CS∗,kcc (Rd)

)for any non-negative

integer k such that Rk+1|CS∗,kcc (Rd) = Rk) which sends a symbol σ to a holomorphicgerm of symbols σ(z) with non-constant affine order α(z) in z and such that σ(0) = σ.

Example 2 Riesz regularisations19 are holomorphic regularisations of the type

R(σ)(z)(x) = σ(ξ) |ξ|−z ∀|ξ| ≥ 1 (73)19also called modified dimensional regularisation in the physics literature, see e.g. [Sch] Example

3 chapter II for a mathematical presentation.

55

and we shall also consider the slightly more general holomorphic regularisations (whichas we shall see below include dimensional regularisation) of the type

R(σ)(z)(ξ) = H(z)σ(ξ) |ξ|−z ∀|ξ| ≥ 1. (74)

with H holomorphic such that H(0) = 1.

Exercise 41 1. Show that the holomorphic regularisation

R(σ)(z)(ξ) = σ(ξ) (1− χ(ξ)) + χ(ξ)σ(ξ) |ξ|−z,

where χ is a smooth cut-off function which vanishes in a small neighborhood ofzero and is identically one outside the unit ball, is a Risez regularisation.

2. Show that the holomorphic regularisation

R(σ)(z)(ξ) = σ(ξ) 〈ξ〉−z,

where we have set 〈ξ〉 =√

1 + |ξ|2 is asymptotically of Riesz type, namely:

R(σ)(z)(ξ) ' σ(ξ) |ξ|−z.

3. Show thatR(σ)(z)(ξ) = σ(ξ) (1 + q(ξ))−

z2 , (75)

where q is a positive definite quadratic form, is a holomorphic regularisationscheme.

Remark 12 In the above examples σ 7→ σ(z) is not an algebra morphism sinceσ(z) τ(z) 6= (σ τ) (z).

The following exercise is reminiscent of the fact that Feynman integrals regularisedvia Riesz (or modified dimensional) regularisation have Laurent expansions whosecoefficients are periods [BW].

Exercise 42 Let σ(ξ) be a symbol in CSa,∗cc (Rd) for some complex number a such thatthe maps ξ 7→ σ(ξ) and ξ 7→ σa−j(ξ) are rational functions on Rd with rational coef-ficients or logarithms thereof. Show that for a holomorphic regularisation of the typeσ(z)(ξ) = σ(ξ) (1 + q(ξ))−

z2 , with q a positive definite quadratic form with rational

coefficients, the coefficients of the Laurent expansion −∫

Rd σ(z)(ξ) dξ at zero are alsoperiods.Hint: Use Corollary 4.

By Theorem 9, to a holomorphic regularisation R : σ 7→ σ(z) we assign a meromorphicmap z 7→ −

∫Rd R(σ)(z) with a simple pole at z = 0. Combining it with the regularised

evaluator evreg0 at zero defined in (49) which amounts to taking the finite part at zero,

we build a linear form defined as follows.

Definition 14 Given a holomorphic regularisation R : σ 7→ σ(z) on CS∗,∗cc (Rd), wecall R-regularised integral of σ the linear form

−∫ R

Rd: CScc(Rd) → C

σ 7→ evreg0 −

∫RdR(σ).

56

Exercise 43 Show that Riesz regularisation coincides with cut-off regularisation:

−∫ R

Rdσ = −

∫Rdσ ∀σ ∈ CS∗,∗cc (Rd).

Hint: Note that σ(r)(0)(ξ)|Sd−1 = 0 for any positive integer r.

Exercise 44 1. Show that

−∫ R

Rdσ(ξ) dξ = lim

z→0−∫

Rdσ(z)(ξ) dξ =

∫Rdσ(ξ) dξ ∀σ ∈ CS<−d,∗cc (Rd).

2. Show that

−∫ R

Rdσ(ξ) dξ = lim

z→0−∫

Rdσ(z)(ξ) dξ = −

∫Rdσ(ξ) dξ ∀σ ∈ CS /∈Z∗

cc (Rd),

which confirms the fact that by Theorem 3, any linear extension of the ordi-nary integration map to non-integer order symbols coincides with the canonicalintegral.

We need a formula for the volume of the sphere in terms of the Gamma function.

Exercise 45 Show that the volume of the unit sphere Sd−1 in Rd is given by:

Vol(Sd−1) =2π

d2

Γ(d2

) (76)

Hint: Use the fact that the Gaussian integral,∫Rde−|x|

2dx = π

d2 (77)

satisfies the following equality(∫Re−x

2dx

)2

= Vol(Sd−1) ·∫ ∞

0

e−r2rd−1dr,

and then consider the even d = 2k and odd d = 2k + 1 dimensional cases. The resultthen follows using the duplication formula 20

Γ(z) Γ(z +

12

)= 21−2z

√π Γ(2z)

applied to z = k.

Definition 15 Let H(z) := Vol(Sd−z−1)Vol(Sd−1)

where we have set Vol(Sd−z−1) := 2πd−z

2

Γ( d−z2 )which, at z = 0, corresponds to the volume of the unit sphere in d dimensions. Let Rbe a regularisation of the type described in Example (74),

R(σ)(z)(ξ) = (1− χ(ξ))σ(ξ) + χ(ξ)H(z)σ(ξ) |ξ|−z

20I thank Stefan Weinzierl for pointing out this formula to me.

57

where χ is any smooth cut-off function which is identically one outside the unit balland vanishes in a neighborhood of 0.For any symbol σ ∈ CScc(Rd) we call

−∫ dim.reg

Rdσ(ξ) dξ := −

∫ RRd

σ(ξ) dξ

the dimensional regularised integral of σ.

The terminology “dimensional regularisation” is justified by the following exercisewhich shows how on radial symbols, dimensional regularisation amounts to “complex-ifying” the dimension d→ d− z.

Exercise 46 Show that for any radial symbol σ(ξ) = f(|ξ|) ∈ CScc(Rd) we have

−∫ dim.reg

Rdσ(ξ) dξ = evreg

0

(Vol(Sd−z−1)

(2π)d−∫ ∞

0

f(r) rd−z dr). (78)

The following proposition comparesR-regularised integrals with the cut-off regularisedintegral.

Proposition 14 Let σ be a symbol in CScc(Rd) of order a.

1. For a holomorphic regularisation R that sends σ to a symbol σ(z) of non-constantaffine order α(z) such that σ(0) = σ, we have

−∫ R

Rdσ(ξ)dξ −−

∫Rdσ(ξ)dξ = − 1

α′(0)res(σ′(0)). (79)

2. In particular, if R is of type (74), i.e. if R(σ)(z)(ξ) = H(z)σ(ξ) |ξ|−z if |ξ| ≥1, then

−∫ R

Rdσ(ξ)dξ −−

∫Rdσ(ξ)dξ = H ′(0) res(σ).

3. Riesz regularised integrals coincide with cut-off regularised integrals.

Proof: Let us set σ(z) := R(σ)(z).

1. By (57) at z0 = 0 we have

−∫ R

Rdσ(ξ)dξ −−

∫Rdσ(ξ)dξ = evreg

0 −∫

Rdσ(z)(ξ)dξ −−

∫Rdσ(ξ)dξ

= − 1α′(0)

res(σ′(0)). (80)

2. If R(σ)(z)(ξ) = H(z)σ(ξ) |ξ|−z if |ξ| ≥ 1 is of type (74), then α(z) = −z and

σ′(z)(ξ) = H ′(z)σ(ξ) |ξ|−z − z H(z)σ(ξ) log |ξ| ξ|−z if |ξ| ≥ 1

so that res(σ′(0)) = H ′(0) res(σ) and −∫R

Rd σ(ξ)dξ −−∫

Rd σ(ξ)dξ = H ′(0) res(σ).

3. Setting H ≡ 1 yields the result for Riesz regularisation.

58

tu

The following result yields back Corollary 2 when R is Riesz’s regularisation.

Proposition 15 Let σ be a smooth rational function on Rd with rational coefficientsor a product of such a function with a power of the logarithm thereof. Let R : σ 7→ σ(z)be a holomorphic regularisation order α(z) affine in z with rational coefficients, suchthat σ′(0) is a product of a smooth rational function on Rd with rational coefficientswith a power of the logarithm of such a function. Then −

∫RRd σ(x) dξ is a period.

Proof: Let us set σ(z) := R(σ)(z). The result easily follows from

−∫ R

Rdσ(ξ)dξ = −

∫Rdσ(ξ)dξ − 1

α′(0)res(σ′(0))

combined with the result of Corollary 2, which tells us that −∫

Rd σ(ξ)dξ is a periodtogether with the fact that res(σ′(0)) =

∫|ξ|=1

σ′−n(0)(x) d− Sξ is a period as a resultof the assumptions on σ′(0). tu

Exercise 47 Let P be a polynomial in d variables and let P (z)(ξ) := P (ξ) 〈ξ〉z be aholomorphic regularisation where as before we have set 〈ξ〉 :=

√1 + |ξ|2. Show that

the map z 7→ −∫

Rd P (z) is holomorphic at zero and that

limz→0−∫

RdP (z) = −

∫RdP = 0. (81)

Hint: Use (29) for the last identity.

Exercise 48 Show that in even dimensions d = 2k, the dimensional regularised in-tegral defined by (78) of a radial symbol σ(ξ) = f(|ξ|) relates to its cut-off regularisedintegral by

−∫ dim.reg

Rdσ(ξ)dξ −−

∫Rdσ(ξ)dξ = −

log π + γ −k−1∑j=1

1j

res(σ).

Hint: Here H(z) :=(

2πd−z

2

Γ( d−z2 )

)(2π

d2

Γ( d2 )

)−1

so that H ′(0) = − log π+ Γ′(k)Γ(k) = − log π−

γ +∑k−1j=1

1j , where γ is the Euler constant.

Exercise 49 Show that when d = 2k = 4 and σ(ξ) = 1|ξ|2+1 we have

−∫ dim.reg

R4

1|ξ|2 + 1

dξ = −∫

R4

1|ξ|2 + 1

dξ +log π + γ −

∑k−1j=1

1j

8π2=

log π + γ − 18π2

.

Hint: Use the fact that −∫

R41

|ξ|2+1 dξ vanishes by Exercise 30.

In the above examples the map σ 7→ res(σ′(0)) arising in (79) is proportional to theresidue. This is a special instance of a more general feature common to regularisationsof the type R(σ)(z) ∼ σ τ(z). The following results confirm those of Proposition 8.

59

Corollary 5 Given a holomorphic regularisation R : σ 7→ σ(z) such that R(σ)(z) ∼σ τ(z) there is a finite number of smooth functions φi, i = 0, · · · , k on Sd−1 with k anon-negative integer depending on the order of σ such that

−∫ R

Rdσ(ξ)dξ −−

∫Rdσ(ξ)dξ =

k∑j=1

∫Sd−1

φj(ξ)σ−d+j(ξ) dSξ. (82)

Proof: Since τ(0) = Id has order zero, so has the derivative τ ′(0) order zero. Thusby Exercise 36 we have for |ξ| = 1

σ′−d(0)(ξ) = (τ ′(0)(ξ)σ(ξ))−d =∑j∈Z+

τ ′−j(0)(ξ)σ−d+j(ξ),

where the sum is finite since j is bounded from above by d + ord(σ). It follows thatres(σ′(0)) =

∑j∈Z+

∫Sd−1 φj(ξ)σ−d+j(ξ) dSξ where we have set φj = τ ′−j(0). tu

Exercise 50 Show that for R(σ)(z)(ξ) = σ(ξ) 〈ξ〉−z we have

−∫ R

Rdσ(ξ)dξ −−

∫Rdσ(ξ)dξ =

(−1)k

2

∑k∈Z+

∫Sd−1

σ−d+2k+2(ξ) dSξ.

Use Exercise 18 to prove that

−∫ R

Rdσ(ξ)dξ −−

∫Rdσ(ξ)dξ =

12

∫Sd−1

σ−d(ξ) dSξ.

Hint: When k 6= 0 write σ−d+k =∑dj=1 ∂jfj for some homogeneous functions fj and

use Stokes’ theorem.

3.5 Discrepancies of regularised integrals

We end this chapter with three applications of Formula (57) by which we measurediscrepancies of the regularised integral −

∫RRd in terms of noncommutative residues in-

volving the derivative at zero of the holomorphic symbol obtained from the originalsymbol via the regularisation R.

Theorem 11 Given a holomorphic regularisation R : σ 7→ σ(z) on CScc(Rd) withσ(z) of non constant affine order α(z), then for any on σ ∈ CScc(Rd)

−∫ R

Rdt∗ησ −−

∫ RRd

σ = − 1α′(0)

res((t∗ησ)′(0))− t∗η(σ′(0))

)∀η ∈ Rd,

−∫ R

Rd∂iσ = − 1

α′(0)res ((∂iσ)′(0)− ∂i(σ′(0))) ∀i ∈ 1, · · · , d,

and

|det(A)| −∫ R

Rdσ A−−

∫ RRd

σ = −|det(A)|α′(0)

res ((σ A)′(0)− σ′(0) A) ∀A ∈ GLd(R).,

60

Proof: We first observe that by translation invariance of the canonical integral on noninteger order symbols (see Theorem 7), we have the following identity of meromorphicfunctions:

−∫

Rdt∗η(σ(z)) = −

∫Rdσ(z) ∀η ∈ Rd,

from which we deduce that

−∫ R

Rdt∗ησ −−

∫ RRd

σ = fpz=0

(−∫

Rd(t∗ησ)(z)−−

∫Rdσ(z)

)= fpz=0

(−∫

Rd(t∗ησ)(z)−−

∫Rdt∗η(σ(z))

)= − 1

α′(0)res((t∗ησ)′(0))− t∗η(σ′(0))

).

Similarly, we observe that Stokes’ property of the canonical integral on non integerorder symbols (see Corollary 3) implies the following identity of meromorphic func-tions:

−∫

Rd∂i(σ(z)) = 0,

from which we deduce that

−∫ R

Rd∂iσ = fpz=0

(−∫

Rd(∂iσ)(z)−−

∫Rd∂i(σ(z))

)= − 1

α′(0)res ((∂iσ)′(0))− ∂i(σ′(0))) .

using (57) applied to the holomorphic families (∂iσ)(z) and ∂i(σ)(z), which coincideat z = 0.Along the same line of proof, by the covariance of the canonical integral on non integerorder symbols (see Proposition 13) we write

|det(A)| −∫

Rdσ(z) A = −

∫Rdσ(z) ∀A ∈ GLd(R),

from which we deduce that

|det(A)| −∫ R

Rdσ A−−

∫ RRd

σ = |det(A)| fpz=0

(−∫

Rd(σ A)(z)

)−−∫

Rdσ(z)

= |det(A)| fpz=0

(−∫

Rdσ(z) A−−

∫Rdσ(z) A

)= −|det(A)|

α′(0)res ((σ A)′(0)− σ′(0) A) ,

using (57) applied to the holomorphic families σ(z) A and (σ A)(z), which coincideat z = 0. tu

It follows from Theorem 11 that any regularised integral −∫R

Rd restricted to non-integerorder symbols has the required properties such as translation invariance and Stokes’property. This was to be expected since we know by Exercise 44 that −

∫RRd coincides on

CS /∈Zcc (Rd) with the canonical integral, which indeed fulfills these properties.

61

Corollary 6 Given a holomorphic regularisation R : σ 7→ σ(z) on CScc(Rd), theregularised integral −

∫RRd is translation invariant, covariant and fulfills Stokes’ property

on non integer order symbols. Indeed, for any symbol σ ∈ CS /∈Zcc (Rd) we have:

−∫ R

Rdt∗ησ = −

∫ RRd

σ ∀η ∈ Rd,

−∫ R

Rd∂iσ = 0 ∀i ∈ 1, · · · , d,

and

|det(A)| −∫ R

Rdσ A = −

∫ RRd

σ ∀A ∈ GLd(R).

Proof: This follows from Theorem 11 using the fact that the residue vanishes on non-integer order symbols and that the derivative σ′(0) at zero of a holomorphic familyσ(z) has same order as the original symbol σ = σ(0). tu

Exercise 51 Using a Riesz regularisation

R(σ)(z)(ξ) = σ(ξ) |ξ|−z ∀|ξ| ≥ 1,

for which we have −∫R

Rd σ = −∫R

Rd σ by Exercise 43, infer from Theorem 11 the results ofProposition 13 and Theorem 6 for classical symbols.Hint: The result follows from combining α′(0) = −1 with

(σ A)′(0)(ξ)− (σ′(0) A)(ξ) = (log |Aξ| − log |ξ|)σ(Aξ),

which by the covariance of the residue leads to:

|detA| res ((σ A)′(0)− (σ′(0) A)) = res(ξ 7→ (log |ξ| − log |A−1ξ|)σ(ξ)

)= −

∫Sd−1

log |A−1ξ|σ−d(ξ)dξ,

(∂iσ)′(0)(ξ)− (∂iσ′)(0)(ξ) = ξi |ξ|−2 σ(ξ),

and(t∗ησ)′(0)(ξ)− t∗η(σ′(0))(ξ) = (log |ξ + η| − log |ξ|) t∗ησ(ξ),

which by the translation invariance of the residue leads to:

res((t∗ησ)′(0)− t∗η(σ′(0))

)= res

(ξ 7→ (log |ξ + η| − log |ξ|) t∗ησ(ξ)

)=∫Sd−1

log |ξ−η|σ−d(ξ) dξ.

tu

62

4 The canonical sum on non-integer order classicalsymbols

Just as we extended the ordinary integral to the algebra of classical symbols, usinga cut-off or Hadamard finite part procedure on balls of increasing radius, we extendthe ordinary discrete sum to the whole algebra of classical symbols, using a cut-offor Hadamard finite part procedure on polytopes of increasing size. When restrictedto the set of non-integer order symbols, cut-off regularised sums obtained in this wayturn out to be independent of the chosen polytope and Zd-translation invariant (seeTheorem 12) just as cut-off integrals of non-integer order symbols are independentof the rescaling of the radius of the ball they are integrated on (see Proposition 12).On the one hand we show that Zd-translation invariant linear forms on the set ofnon-integer order classical symbols are proportional to the canonical sum obtainedfrom any of these regularised sums (see Theorem 13). On the other hand we extendthe ordinary sum to regularised discrete sums using holomorphic regularisations, forwhich we establish a regularised Euler-Maclaurin formula in Theorem 14. We thenapply the regularised Euler-Maclaurin formula to express certain Hurwitz zeta valuesin Proposition 22 (based on joint work with D. Manchon [MP]) and certain zeta valuesassociated with quadratic forms in Theorem 15.

4.1 The Euler-Maclaurin formula

A large part of this section is dedicated to the construction of the cut-off discretesum on CScc(Rd) which when restricted to non-integer order symbols, provides a Zd-translation invariant extension of the ordinary integral on L1∩CScc(Rd), thus provingthe existence of such an extension.The classical Euler-Maclaurin formula which relates a sum to an integral, involves theBernoulli numbers defined by the following Taylor expansion at t = 0 (see e.g [Ca]):

t

et − 1:=

∞∑n=0

Bntn

n!. (83)

Since tet−1 + t

2 = t2et2 +e−

t2

et2−e−

t2

is an even function, B1 = − 12 and B2k+1 = 0 for any

positive integer k.

Remark 13 In view of a generalisation to higher dimensions, it is useful to observethat t

et−1 = Td(−t) where Td(t) := t1−e−t is the Todd function so that

Td(t) =∞∑n=0

(−1)nBntn

n!=t

2+∞∑k=0

B2kt2k

(2k)!= 1 +

t

2+∞∑k=2

(−1)kBkk!. (84)

Exercise 52 Check the following values of the Bernoulli numbers (see e.g. [Ca]):B0 = 1, B1 = − 1

2 ; B2 = 16 , B4 = − 1

30 ; B6 = 142 ; B8 = − 1

30 ;B10 = 566 .

Bernoulli polynomials are defined similarly by:

∞∑n=0

Bn(x)tn

n!=

t et x

et − 1, (85)

so that in particular, Bn(0) = Bn.

63

Exercise 53 Show that this initial condition combined with the differential equationsobtained from differentiating (85) with respect to x

∂xBn(x) = nBn−1(x), (86)

completely determine the Bernoulli polynomials and that:

Bn(x) =n∑k=0

(n

k

)Bn−k x

k. (87)

Check that B1(x) = − 12 + x.

Furthermore,∑∞n=0 (Bn(1)−Bn(0)) t

n

n! = tet−tet−1 = t so that

B1(1) = B1(0) + 1 and Bn(1) = Bn(0) ∀n ≥ 2.

Since Bn(1) = Bn for any n ≥ 2, setting x = 1 we have

Bn =n∑k=0

(n

k

)Bn−k =

n∑k=0

(n

k

)Bk ∀n ≥ 2. (88)

Let us recall the Euler-Maclaurin formula (see e.g. [Ha]).

Exercise 54 Show that for any function f in C∞(R) and any two integers M < N

N∑n=M

f(n) =f(M) + f(N)

2+∫ N

M

f(x) dx

+K∑k=2

(−1)kBkk!

(f (k−1)(N)− f (k−1)(M)

)+

(−1)K−1

K!

∫ N

M

BK(x) f (K)(x) dx (89)

with Bk(x) = Bk (x− [x]) and where K is any positive integer larger than 1.Hint: Set for convenience:

SK(f) :=K∑k=2

(−1)kBkk!

(f (k−1)(N)− f (k−1)(M)

),

RK(f) :=(−1)K−1

K!

∫ N

M

BK(x) f (K)(x) dx

and observe thatSK(f) +RK(f) = SK+1(f) +RK+1(f).

The proof then follows by induction.

Remark 14 The index K in the Euler-Maclaurin formula can be chosen arbitrarilylarge.

The following exercises serve as a preparation for the higher-dimensional generalisationof the Euler-Maclaurin formula. The first one shows that the remainder term RK(f)vanishes if f is polynomial.

64

Exercise 55 Show that if f is polynomial, then

N∑n=M

f(n) =∫ N

M

f(x) dx+

[(Td (∂h1) Td (∂h2)− Id)

∫ N+h2

M−h1

f(ξ) dξ

]|h1=h2=0

.

In the following exercise we investigate the limit CK(f) = (−1)K−1

K!

∫R BK(x) f (K)(x)

(when it exists) of the remainder term.

Exercise 56 Let σ be a classical symbol on R.

1. Show that for large enough K,

CK(σ) =(−1)K−1

K!

∫RBK(x)σ(K)(x) dx (90)

is an absolutely convergent integral independent of K.Hint: Use integration by parts and the fact that derivation lowers the order ofthe symbol.

2. Setting C(σ) := CK(σ) for large enough K, show that provided the order of σhas real part small enough we have:∑

n∈Zσ(n) =

∫Rσ(ξ) dξ + C(σ).

Hint: Use the Euler-Maclaurin formula (89).

The higher dimensional Euler-Maclaurin formula takes place on convex polytopesinstead of intervals in the one dimensional case.

Definition 16 A compact convex polytope ∆ in Rd is a compact set obtained asthe intersection of finitely many half-spaces ∆ = H1 ∩ · · · ∩Hm where

Hi = x ∈ Rd, 〈ui, x〉+ ai ≥ 0, i = 1, · · · ,m

for some vectors ui ∈ Rd, i = 1, · · · ,m and some positive real numbers ai, i = 1, · · · ,m.Alternatively, a compact convex polytope is the convex hull of a finite set of points inRd.

Example 3 Clearly the hypercube [−R,R]d is a convex polytope in Rd. Here m =2d,ai = R for i ∈ 1, · · · , 2d, ui = ei, ui+d = −ei for i ∈ 1, · · · , d, wheree1, · · · , ed is the orthonormal basis of Rd.

We restrict to integral polytopes which correspond to polytopes whose vertices arepoints of the lattice Zd and further to simple integral polytopes, namely to integralpolytopes that have exactly d edges (and d facets) emanating from each vertex. Incertain cases we specialise to a subclass thereof, regular integral polytopes, integralpolytopes whose edges emanating from each vertex are along lines generated by a Z-vertex of the lattice Zd. A regular integral polytope is therefore defined in terms ofequations 〈ui, x〉+ai ≥ 0, i = 1, · · · ,m where the ui’s are primitive lattice vectors,i.e. such that no ui can be expressed as a multiple of a lattice vector by an integergreater than 1, and the ai’s are positive integers.

65

We consider polytopes ∆ with the origin 0 in the interior of ∆. The expanded polytopeR∆ for positive R reads:

〈x, ui〉+Rai ≥ 0 i = 1, · · · ,m.

The extended Euler-Maclaurin formula compares the discrete sum

P∆N (f) :=

∑~n∈Zd∩N ∆

f(~n)

on integer points of the expanded polytope N ∆ for some positive integer N with theintegral

P∆N (f) :=

∫N ∆

f(ξ) dξ

over the expanded polytope for any smooth function f on Rd. We only quote theresults of [GSW] in higher dimensions without proofs, indicating references whereproofs can be found. For this purpose we introduce further notations borrowed from[GSW] which generalise the Taylor expansion (84) at zero to higher dimensions:

Todd (ξ) :=d∏i=1

ξi1− eξi

=∑|α|≥0

(−1)|α|Bαα!ξα, ∀ξ ∈ Rd

for some constants Bα and where we have set α! := α1! · · ·αd! and |α| = α1 + · · ·+αd.The Khovanskii-Pukhlikov formula [KP] relates the discrete sum of a polynomial func-tion over integer points of a polytope with its integral by:

P∆N (f)− P∆

N (f) =(

(Todd(∂h)− Id) P∆N,h(f)

)|h=0,

where we have setP∆N,h(f) :=

∫∆N,h

f(ξ) dξ,

with the deformed polytope ∆R,h defined by the following set of inequalities:

〈x, ui〉+Rai + hi ≥ 0 ∀i = 1, · · · ,m,

for any h = (h1, · · · , hm) ∈ Rm.The Khovanskii-Pukhlikov formula was generalised to classical symbols in [GSW] (for-mula (15), see also [AW], [KSW1] and [KSW2] for previous results along these lines)in which case the formula is not exact anymore but only holds asymptotically.

Proposition 16 [GSW] Given a symbol σ ∈ CScc(Rd) and a convex polytope ∆, thereis a constant C(σ) independent of ∆ such that

P∆N (σ)− P∆

N (σ) ∼N→∞ (Todd(∂h)− Id) P∆N,h(σ)|h=0 + C(σ).

More precisely, there are polynomials M [K],K ∈ N on Rd such that

P∆N (σ)− P∆

N (σ) =((M [K](∂h)− Id

)P∆N,h(σ)

)|h=0

+R∆K,N (σ) (91)

66

where R∆K,N (σ) tends as N goes to infinity (see formula (18) in [GSW]) to a constant

independent of ∆

CK(σ) =∫

Rd

|α|=dK∑|α|=K

φα,K(ξ) ∂ασ(ξ)

dξ

as N →∞, where φα,K are bounded piecewise smooth periodic functions as describedin [KSW1] for a regular polytope ∆ and in [KSW2] for a simple polytope.As in the one-dimensional case, the constant CK(σ) is also independent of K for Kchosen sufficiently large and we denote it by C(σ).

Remark 15 In the one-dimensional case we have α = K and φα,K = (−1)K−1

k! BK .

Exercise 57 Show that

fpN→∞((M [K](∂h)− Id

)P∆N,h(σ)

)|h=0

= 0 ∀σ ∈ CS /∈Z(Rd).

One can hope for a more precise description of the functions φα,K entering the descrip-tion of C(σ) by specialising to a regular integral polytope since C(σ) is independentof the choice of polytope. In the case of a regular integral polytope, the higher dimen-sional Euler-Maclaurin formula follows from the one dimensional one since the localcone at each vertex can be transformed by an integral unimodular affine transforma-tion of the form

Tv,α : (x1, · · · , xd) 7→ x = v + x1 α1 + · · ·+ xd αd

into a neighborhood of the origin in the standard orthant Rd≥0 :=∏di=1 R≥0. For

a regular integral polytope ∆, the results of [KSW1] express the remainder termR∆K,1(σ) corresponding to a simple integral polytope as a sum of integrals over orthants

of bounded periodic functions (which generalise the functions PK ’s built from theBernoulli polynomials) times various partial derivatives of σ of order no less than Kd(see formula (47) in [KSW2])

RRd≥0K,1 (σ) (92)

=∑

I⊂1,··· ,d

(−1)(K−1)(d−|I|)∫Rd≥0

∏i∈I

L2k (−∂i)∏i/∈I

PK(ξi)∏i/∈I

∂Ki σ(ξ) dξ1 · · · dξd

over the standard closed orthant, where we have set σ = σ Tv,α. Here k =[K2

]and

L(x) =x/2

tanh(x/2)= 1 +

∞∑k=1

B2k

(2k)!x2k =

12

(Td(x) + Td(−x)) ,

where tanh stands for the hyperbolic tangent.

4.2 Cut-off discrete sums on Zd subordinated to convex poly-topes

Combining the cut-off regularised integral built in the previous chapter with theEuler-Maclaurin- Khovanskii-Pukhlikov formula for classical symbols, we build a lin-ear extension regularised discrete sum on CS /∈Zcc (Rd) of the ordinary discrete sum onL1(Rd) ∩ CScc(Rd).

67

Proposition 17 Let σ ∈ CScc(Rd) and let ∆ be a convex polytope in Rd. Themap R 7→

∫R∆

σ(ξ) dξ has the same type of asymptotic expansion as the map R 7→∫B(0,R)

σ(ξ) dξ as R tends to ∞. The constant term

−∫ ∆

Rdσ(ξ) dξ := fpR→∞

∫R∆

σ(ξ) dξ,

called the cut-off integral of σ subordinated to the polytope ∆ reads:

−∫ ∆

Rdσ(ξ) dξ = −

∫Rdσ(ξ) dξ +

∫∆

χ(ξ)σ−d(ξ) dξ −∫B(0,1)

χ(ξ)σ−d(ξ) dξ (93)

for any σ in CScc(Rd) independently of the smooth cut-off function χ which vanishesin a neighborhood of zero and is one on the complement of B(0, 1) ∪∆.Consequently,

−∫ ∆

Rdσ(ξ) dξ = −

∫Rdσ(ξ) dξ ∀σ ∈ CS /∈Z(Rd), (94)

where −∫

Rd is the cut-off integral defined previously as the constant term in the asymp-totic expansion of

∫B(0,R)

σ(ξ) dξ as R→∞.

Proof: Writing σ =∑N−1j=0 σa−j + σ(N) as in (18) and setting AB := A− (A∩B),

for R chosen large enough, we can estimate the difference:∫R∆

σ(ξ) dξ −∫B(0,R)

σ(ξ) dξ

=N−1∑j=0

∫R∆B(0,R)

σa−j(ξ) dξ −N−1∑j=0

∫B(0,R)R∆

σa−j(ξ) dξ

+∫R∆B(0,R)

σ(N)(ξ) dξ −∫B(0,R)R∆

σ(N)(ξ) dξ. (95)

Since σ(N) is a classical symbol of order a−N we have

|σ(N)(ξ)| ≤ C(1 + |ξ|2)Re(a)−N

2

for some constant C where |·| stands for the Euclidean norm. For N chosen sufficientlylarge, the exponent Re(a)−N is negative so that outside the ball B(0, R) the followingupper bound holds, (1 + |ξ|2)

Re(a)−N2 ≤ (1 +R2)

Re(a)−N2 and∣∣∣∣∣

∫R∆B(0,R))

σ(N)(ξ) dξ

∣∣∣∣∣ ≤ C ′ (1 +R2)Re(a)−N

2 Vol (R∆B(0, R)))

≤ C ′Rd(1 +R2)Re(a)−N

2 Vol (∆)

≤ C ′ (1 +R2)Re(a)+d−N

2 Vol (∆) .

Since the expanded polytope R∆ contains a ball B(0, λR) for some positive λ, outsidethe expanded polytope and for N chosen sufficiently large the following upper boundholds, (1 + |ξ|2)

Re(a)−N2 ≤ C∆(1 +R2)

Re(a)−N2 for some constant C∆ depending on ∆,

68

hence there exists a constant C ′∆ such that∣∣∣∣∣∫B(0,R)B(0,R)

σ(N)(ξ) dξ

∣∣∣∣∣≤ C ′∆ (1 + dR)Re(a)−NVol (B(0, R)− (C(0, R) ∩B(0, R)))≤ C ′∆Rd(1 +R)Re(a)−NVol (B(0, 1))≤ C ′∆ (1 +R)Re(a)+d−NVol (B(0, 1)) .

Consequently, we can choose N sufficiently large so that∫C(0,R)B(0,R)

σ(N)(ξ) dξ −∫B(0,R)C(0,R)

σ(N)(ξ) dξ = O((1 +R)Re(a)+d−N ).

This settles the case of integrals involving the remainder term σ(N). As for integralsof homogeneous symbols

∫R∆B(0,R)

σa−j(ξ) dξ, we have∫R∆B(0,R)

σa−j(ξ) dξ =∫

∆B(0,1)

σa−j(Rη)Rd dη

= Ra−j+d∫

∆B(0,1)

σa−j(η) dη,

which shows they are homogeneous of degree a− j + d.Similarly,

∫B(0,R)R∆

σa−j(ξ) dξ = Ra−j+d∫B(0,1)∆

σa−j(η) dη.Combining these results shows that R 7→

∫R∆B(0,R)

σ(ξ) dξ defines a classical symbolof order a+ d with constant term given by

fpR→∞

∫R∆B(0,R)

σ(ξ) dξ =∫

∆B(0,1)

σ−d(η) dη −∫B(0,1)∆

σ−d(η) dη

=∫

∆

χ(η)σ−d(η) dη −∫B(0,1)

χ(η)σ−d(η) dη

independently of the smooth cut-off function χ which vanishes in a neighborhood ofzero and is one on the complement of B(0, 1) ∪ ∆. Thus the map R 7→

∫R∆

σ(ξ) dξdefined by (95) has the same type of asymptotic expansion as R→∞ as

∫B(0,R)

σ(ξ) dξand its finite part differs from fpR→∞

∫B(0,R)

σ(ξ) dξ by fpR→∞∫R∆B(0,R)

σ(ξ) dξ =∫∆σ−d(η) dη −

∫B(0,1)

σ−d(η) dη which vanishes when σ has non-integer order. tu

Exercise 58 Show that for any polynomial P in d variables,

−∫ ∆

RdP = −

∫RdP

is independent of ∆.

The following results are straightforward consequences of Proposition 17.

Corollary 7 On non-integer order symbols, the cut-off integral subordinated to a poly-tope is independent of the choice of polytope and we have:

−∫ ∆

Rdσ = −

∫Rdσ ∀σ ∈ CS /∈Z(Rd).

69

Corollary 8 Let σ be a smooth rational function with rational coefficients and let∆ be a regular integral polytope. The regularised integral −

∫∆

Rd σ subordinated to thepolytope ∆ is a period.

Proof: This follows from (93). Indeed, by Corollary 2, we know that −∫

Rd σ is a period.On the other hand, the homogeneous component σ−d is like σ, a rational function withrational coefficients, so that

∫B(0,1)

σ−d is clearly a period and since ∆ is defined interms of integral coefficients, so is

∫∆σ−d a period. tu

Combining Propositions 16 and 17 shows that for any symbol σ in CScc(Rd), themap

N 7→ P∆N (σ) =

∑N ∆∩Zd

σ

has an asymptotic expansion as N →∞ of type (38).The constant term fpN→∞

∑N ∆∩Zd σ in the expansion gives rise to a linear form

which extends the ordinary discrete summation map∑

Zd on L1(Rd)∩CScc(Rd) as aresult of the Euler-Maclaurin/ Khovanskii-Pukhlikov formula for symbols. It is definedas follows.

Definition 17 Given a convex polytope ∆ in Rd, we call the linear form defined by

−∆∑Zd

: CScc(Rd) → C

σ 7→ fpN→∞∑

N ∆∩Zdσ

the cut-off regularised sum of σ subordinated to the polytope ∆.

Notation: Let us denote by −∑

Zd the discrete cut-off sum subordinated to the hyper-cube = [−1, 1]d, which similarly to −

∫Rd , will be used as a reference regularised sum.

By Proposition 16 the cut-off regularised sum subordinated to ∆ relates to the cut-offregularised integral subordinated to ∆ by:

−∆∑Zd

σ −−∫ ∆

Rdσ = fpN→∞

((M [K](∂h)− Id

)∫P∆N,h

σ

)|h=0

+ C(σ), (96)

for large enough integer K with the notations of Proposition 16.

Exercise 59 Show that for a polynomial function P in d-variables,

1. C(P ) = 0 and −∑

Zd P := −∑∆

Zd P = P (0) independently of ∆.Hint: Observe that the sum

∑N ∆∩Zd P is polynomial in N so that only the

constant terms survive in the finite part procedure as N →∞.

2. Deduce that the cut-off regularised sum subordinated to some polytope ∆ is nottranslation invariant on CScc(Rd).

Exercise 60 When σ lies in L1(Rd) ∩ CScc(Rd) show that∑

Zd∩N∆ σ tends to thefinite sum

∑Zd σ as N tends to infinity and that (96) reduces to the following Euler-

Maclaurin type formula ∑Zd

σ =∫

Rdσ + C(σ). (97)

70

The constant C(σ) is rather mysterious. For a smooth function ψ with compactsupport in a Euclidean ball centered at zero and strictly contained in the polytope∆, we have C(ψ) = ψ(0) −

∫Rd ψ so that it is not identically zero on smoothing

symbols. Thus C(σ) depends on the whole symbol, not just on a finite number ofhomogeneous components of the symbol. Also, because the expression of C(σ) involvesboth discrete sums and integrals, it is neither expected to be continuous in the L1(Rd)nor in the L1(Zd) topology; however we shall see that limz→0 C(σ(z)) = C(σ(0)) fora holomorphic perturbation σ(z) of σ around zero.Whether or not C(σ) is a period for a smooth rational function σ in CScc(Rd) withrational coefficients remains an open question. This issue is closely related to whetheror not the cut-off sum of such a function is a period. It is easy to see from (97) thatthe sum

∑Zd σ of a smooth rational function σ in L1(Rd) ∩ CScc(Rd) with rational

coefficients defines a period whenever the constant C(σ) does. This generalises tocut-off discrete sums.

Corollary 9 For a smooth rational function σ on Rd with rational coefficients, theconstant C(σ) is a period whenever there is a convex polytope ∆ such that the cut-offregularised sum −

∑∆Zd σ subordinated to ∆ is a period.

Proof: Since C(σ) does not depend on the choice of polytope ∆, by (96) it is a periodif and only if there is a polytope ∆ such that

−∆∑Zd

σ −−∫ ∆

Rdσ − fpN→∞

((M [K](∂h)− Id

)∫P∆N,h

σ

)|h=0

defines a period. By Corollary 8, we know that the expression −∫∆

Rd σ is a period. Sinceσ is a rational function with rational coefficients, the term

fpN→∞

((M [K](∂h)− Id

)∫P∆N,h

σ

)|h=0

defines a period. It follows from (96) that C(σ) is a period if and only if −∑∆

Zd σ is. tu

4.3 Zd-translation invariant linear forms on symbols

Just as Rd-translation invariance relates to Stokes’ property (see Proposition 11),Zd-translation invariance relates to a discrete version of Stokes’ property. For anynon-negative integer i in 1, · · · , d and any function f : Zd → C we set:

∂Zdi f(~n) := f(~n+ ~ei)− f(~n),

where ~ei is the vertex in Zd with vanishing coordinates except for the i-th which isone. Let S ⊂ CScc(Rd) be a subset stable under Zd- translations i.e.:

σ ∈ S =⇒ t?~pσ ∈ S ∀~p ∈ Zd.

Definition 18 A linear form λ : S → C is said to be Zd -translation invariantwhenever it is invariant under the action of the translation group:

t∗~pλ = λ ∀~p ∈ Zd,

where we have sett?~pλ(σ) := λ(t?~pσ) ∀σ ∈ CScc(Rd).

71

The following proposition relates Zd- translation invariance and a discrete Stokes’ typeproperty. Let S be a subset of CScc(Rd) stable under Zd- translations.

Proposition 18 A linear form λ on S is Zd-translation invariant if and only ifλ ∂Zd

i = 0 for any i ∈ 1, · · · , d.

Proof: If λ is Zd-translation invariant, then clearly for any i in 1, · · · , d we have

λ ∂Zdi (f) = λ (f(·+ ei))− λ (f) = 0.

Conversely, the discrete Stokes property λ (f(·+ ei)) = λ (f) applied to t∗−eif impliesthat λ

(t∗−eif

)= λ

(t∗−eif(·+ ei)

)= λ(f) for any i ∈ 1, · · · , d. By induction on

pi ∈ N it follows that λ (f(·+ pi ei)) = λ (f) for any pi ∈ Z and more generally that

λ

(f(·+

d∑i=1

pi ei)

)= λ (f) ∀(p1, · · · , pd) ∈ Zd.

Hence t∗~pλ = λ for any ~p ∈ Zd. tuThe following result relates Zd- and Rd-translation invariance. Let S denote an Rd-translation invariant subset of CScc(Rd) which is also stable under partial differenti-ation.

Proposition 19 Let λ be a linear form on S which vanishes on L1-symbols.The following conditions are equivalent:

1. It is Zd- translation invariant.

2. It vanishes on derivatives.

3. It is Rd-translation invariant.

Proof: By the results of Proposition 11, we only need to prove that (1) implies (2).Let ~p ∈ Zd and let σ be a symbol in S. From the proof of Proposition 10 we know thatthere is an integer K suuch that RηK(σ) := t∗~pσ −

∑K−1|α|=0 ∂

βσ ~pβ

β! lies in L1(Rd). Thus,under the assumptions of the proposition condition (a) implies that for any ~p ∈ Zd wehave

0 = λ(t∗~pσ − σ

)=

K−1∑|α|=1

λ(∂βσ)~pβ

β!+ λ(R~pK(σ)) =

K−1∑|α|=1

λ(∂βσ)~pβ

β!.

Choosing ~p = ei, the i-th orthonormal basis vector and β = (0, · · · , 0, 1, 0 · · · , 0) withthe 1 at the i-th place yields λ(∂iσ) = 0 and hence condition (b). tu

Example 4 The noncommutative residue gives rise to a Zd-translation invariant lin-ear form on CScc(Rd) since it vanishes on L1-symbols and satisfies Stokes’ property.

On the other hand, all cut-off regularised sums subordinated to a polytope coincide onnon-integer order classical symbols where they define Zd-translation invariant linearforms.

Theorem 12 The cut-off regularised sum

−∑Zd

σ := −∆∑Zd

σ ∀σ ∈ CS /∈Zcc (Rd)

72

is independent of the choice of polytope ∆ on non-integer order symbols and we have

−∑Zd

σ = −∫

Rdσ + C(σ) ∀σ ∈ CS /∈Z

cc (Rd). (98)

On L1(Rd) ∩ CScc(Rd) this reduces to:∑Zd

σ =∫

Rdσ + C(σ).

The cut-off regularised sum −∑

Zd σ on non-integer order symbols is furthermore trans-lation invariant:

−∑Zd

t∗~nσ = −∑Zd

σ ∀~n ∈ Zd, ∀σ ∈ CS /∈Zcc (Rd)

and so is the map σ 7→ C(σ) a Zd-translation invariant map on CS /∈Zcc (Rd):

C(t∗~nσ) = C(σ) ∀~n ∈ Zd, ∀σ ∈ CS /∈Zcc (Rd). (99)

Proof: By Exercise 57

fpN→∞((M [j](∂h)− Id

)Pt∗−~n∆

N,h (σ))|h=0

= 0 ∀σ ∈ CS /∈Z(Rd).

By Corollary 7, −∫∆

Rd = −∫

Rd is independent of ∆ on non-integer order symbols. On theother hand, from Proposition 16 we know that the constant C(σ) in formula (96) isalso independent of the choice of polytope. We therefore conclude that for any symbolσ in CS /∈Z

cc (Rd)

−∆∑Zd

σ = −∫ ∆

Rdσ(ξ) dξ + C(σ)

= −∫

Rdσ(ξ) dξ + C(σ)

is independent of the choice of ∆. Since t∗−~n∆ is also a polytope and t∗−~n∆ ∩ Zd =t∗−~n(∆ ∩ Zd) for any ~n ∈ Zd, wededuce that

−∑Zd

t∗~nσ = −∆∑Zd

t∗~nσ = fpN→∞∑

N ∆∩Zdt∗~nσ

= fpN→∞∑

t∗−~n(N ∆∩Zd)

σ = fpN→∞∑

t∗−~n(N ∆)∩Zdσ

= fpN→0

∑N t∗−~n∆∩Zd

σ = −t∗−~n∆∑

Zdσ

= −∑Zd

σ.

Since both −∑

Zd and −∫

Rd are invariant under translation by ~n ∈ Zd on non-integer ordersymbols, so is the map σ 7→ C(σ). Consequently

C (t∗~nσ) = C(σ) ∀σ ∈ CS /∈Zcc (R) ∀~n ∈ Zd

which shows (98). tu

73

Exercise 61 Let P be a complex polynomial in d variables.

1. Let P (z)(ξ) := P (ξ) 〈ξ〉z, where we have set as before 〈ξ〉 =√

1 + |ξ|2. Using(96) show that the map z 7→ −

∑Zd P (z) is holomorphic at zero.

Hint: Use Exercise 81 combined with the holomorphicity of z 7→ C(P (z)) atzero.

2. Deduce that limz→0−∑P (z) = 0 and compare with the result of Exercise 59 to

show that the map −∑∆ is not continuous on continuous families z 7→ σ(z).

4.4 A characterisation of the noncommutative residue as aZd-translation invariant linear form

We want to characterise Zd-translation invariant linear forms on certain subsets ofCScc(Rd). We first consider L1-symbols.

Exercise 62 Let L1(Zd) := f : Zd → C,∑~n∈Zd |f(~n)| < ∞. Let σ be a symbol

in CScc(Rd).

1. Show that σ lies in L1(Zd) if and only if it has order with real part smaller than−d.

2. Deduce that σ|Zd ∈ L1(Zd)⇐⇒ σ ∈ L1(Rd).

Just as Rd-translation invariant linear forms on L1-classical symbols on Rd are pro-portional to the ordinary integration map, the following proposition shows that Zd-translation invariant linear forms on L1-classical symbols on Zd are proportional tothe ordinary discrete summation map.

Proposition 20 Any Zd-translation invariant linear form λ on L1(Zd) is propor-tional to the discrete sum

t∗~pλ = λ ∀~p ∈ Zd =⇒ ∃ c ∈ C, λ(f) = c∑~n∈Zd

f(~n) ∀f ∈ L1(Z).

Proof: We write a function f in L1(Zd) as f =∑~n∈Zd f(~n) δ~n with δ~n : Zd → C

vanishing outside the point ~n where it takes the value 1. Translation invariance impliesthat λ(δ~n) = λ(δ~0) which we set to be equal to some constant c. For any non-negativeinteger N , let us set fN :=

∑~n∈[−N,N ]d∩Zd f(~n) δ~n. We then deduce from the linearity

of λ thatλ (fN ) =

∑~n∈[−N,N ]d∩Zd

f(~n)λ (δ~n) = c∑

~n∈[−N,N ]d∩Zdf(~n).

Thus, for any non-negative integer p

|λ (fN+p − fN )|

≤ |c|

∣∣∣∣∣∣∑

~n∈[−(N+p),N+p]d∩Zdf(~n)−

∑~n∈[−N,N ]d∩Zd

f(~n)

∣∣∣∣∣∣≤ |c|

∑~n∈([−(N+p),N+p]d−[−N,N ]d)∩Zd

|f(~n)|

74

tends to zero as N tends to infinity since f lies in L1(Zd). Hence λ is continuous inthe L1(Zd) topology so that

λ(f) = limN→∞

λ(fN ) = c∑

~n∈[−N,N ]d∩Zdf(~n) = c

∑~n∈Zd

f(~n).

tuThe following characterisation of the noncommutative residue is similar to that ofProposition 9.

Proposition 21 The noncommutative residue is (modulo a multiplicative factor) theonly Zd-invariant linear form on CScc(Rd) which vanishes on L1-symbols.

Proof: A Zd-invariant linear form λ on CScc(Rd) which vanishes on L1-symbols fulfilsassumption (a) of Proposition 11. We deduce that λ satisfies Stokes’ property whichimplies, using Proposition 9, that it is proportional to the noncommutative residue.tu

Theorem 13 The canonical discrete summation map −∑

Zd is the unique ( modulo amultiplicative factor) Zd-translation invariant linear map on CS /∈Z

cc (Rd).

Proof: Let λ be a Zd-translation invariant linear map on CS /∈Zcc (Rd). Its restriction

to Zd induces a Zd-translation invariant linear map λ|Zd . By the same proof as inProposition 20 we prove the existence of a constant c such that:

λ|Zd = c∑Zd

.

The difference λ − c −∑

Zd , which by construction vanishes on non-integer order L1-symbols, actually vanishes on CS /∈Z

cc (Rd)∩L1(Rd) by Exercise 62. Thus, by Proposition21, it is proportional to the noncommutative residue. Hence there exists a constant µsuch that

λ = c −∑Zd

+µ res.

Since the noncommutative residue vanishes on non-integer order symbols, we havethat

λ = c −∑Zd

.

tu

4.5 Regularised discrete sums on symbols

Using holomorphic perturbations σ(z) on symbols σ combined with regularised eval-uators at zero, we build linear extensions of the canonical sum −

∑Zd on non-integer

order symbols to the whole algebra CScc(Rd) of symbols. The construction is verysimilar to that carried out previously, of linear extensions of the canonical integral −

∫Rd

on non-integer order symbols to the whole algebra CScc(Rd) of symbols.Let us recall from [GSW], that given such a holomorphic perturbation, the mapz 7→ C(σ(z)) is holomorphic at zero21, a fact which yields the translation invarianceof the linear form σ 7→ C(σ), as the following exercise shows.

21Their proof which holds for holomorphic families of symbols of order a−z can easily be generalisedto our more general setup of holomorphic families with any non-constant affine order a− qz, q 6= 0.

75

Exercise 63 Show that for any vector η in Rd,

C(t∗ησ) = C(σ) ∀σ ∈ CScc(Rd).

Hint: Consider a holomorphic perturbation R(σ) : σ 7→ σ(z) with non-constant affineorder and deduce from (99) the following identity of meromorphic functions:

C(t∗η (σ(z))) = C(σ(z))

to show that

C(t∗η (σ)) = limz→0

C((t∗ησ)

(z))

= limz→0

C(t∗η (σ(z))

)= limz→0

C(σ(z)) = C(σ).

The following result shows that the discrete sum on L1 symbols extends to a mero-morphic function on CScc(Rd) with simple poles.

Theorem 14 Let R(σ) : z 7→ σ(z) be a holomorphic regularisation of σ ∈ CScc(Rd)(so that σ(0) = σ) with non-constant affine order α(z).

1. The mapz 7→ −

∑Zd

σ(z)

is meromorphic and satisfies the following identity of meromorphic functions:

−∑Zd

σ(z) = −∫

Rdσ(z)(ξ) dξ + C(σ(z)). (100)

It has a discrete set of simple poles in α−1 ([−d,∞[∩Z) and complex residue atz = 0 given by :

Res0

(−∑Zd

σ(z)

)= − (2π)d

α′(0)res(σ). (101)

2. Consequently, given any regularised evaluator λ0 at zero, there is a constant µsuch that

λ0

(−∑Zd

σ(z)

)= −R∑Zd

σ(z) + µ res(σ). (102)

where the R-regularised discrete sum of σ is defined by

−R∑Zd

σ := evreg0

(−∑Zd

σ(z)

).

3. The R-regularised discrete sum relates to the corresponding R-regularised inte-gral via the following regularised Euler-Maclaurin formula:

−R∑Zd

σ = −∫ R

Rdσ(ξ) dξ + C(σ), (103)

where as before −∫R

Rd σ(ξ) dξ = ev0

(−∫

Rd σ(z)(ξ) dξ)

and C(σ) = limz→0 C(σ(z)).

76

4. Whenever the order of σ has real part smaller than −d (resp. is non-integer),the map z 7→ −

∑Zd σ(z) is holomorphic at z = 0 and converges to the ordinary

sum∑

Zd σ (resp. cut-off regularised sum −∑

Zd σ) as z → 0 so that in that case

−R∑Zd

σ =∑Zd

σ =∫

Rdσ(ξ) dξ+C(σ),

(resp. −

R∑Zd

σ = −∑Zd

σ = −∫

Rdσ(ξ) dξ + C(σ).

)

Remark 16 The term C(σ), which is independent of R, arises here as a differenceof regularised integrals, thus confirming a result of [GSW].

Proof:

1. By Theorem 12, outside the set α−1 (Z), the symbol σ(z) has non-integer orderand we have the following identity:

−∑Zd

σ(z) = −∫

Rdσ(z)(ξ) dξ + C(σ(z)) ∀z /∈ α−1(Z). (104)

On the one hand, by Theorem 9 we know that the map z 7→ −∫

Rd σ(z)(ξ) dξ ismeromorphic with a discrete set of simple poles in α−1 ([−d,∞[∩Z) and that atzero

Res0 −∫

Rdσ(z)(ξ) dξ = − (2π)d

α′(0)res(σ). (105)

It therefore follows from (104) that the left hand side of (104) defines a meromor-phic map z 7→ −

∑Zd σ(z) with a discrete set of simple poles in α−1 ([−d,∞[∩Z)

and complex residue at z = 0 given by

Res0 −∑Zd

σ(z) = − (2π)d

α′(0)res(σ).

2. Relation 102 then follows from Theorem 8.

3. Taking finite parts at z = 0 in (104) yields (103) since limz→0 C(σ(z)) = C(σ).

4. When the order of σ has real part smaller than −d (resp. is non-integer), themap z 7→ −

∫Rd σ(z)(ξ) dξ is holomorphic at 0 since σ has vanishing residue. Its

limit at z = 0 coincides with the ordinary integral∫

Rd σ(ξ) dξ (resp. the cut-offregularised integral −

∫Rd σ(ξ) dξ.). By (104) the holomorphicity at zero of the

map z 7→ C(σ(z))induces that of the map z 7→ −∑

Zd σ(z) =∑

Zd σ(z), whoselimit at zero reads:

−R∑Zd

σ = limz→0

∑Zd

σ(z) = limz→0

∫Rdσ(z)(ξ) dξ+C(σ(z)) =

∫Rdσ(ξ) dξ+C(σ) =

∑Zd

σ,

(resp. −

R∑Zd

σ = limz→0−∑Zd

σ(z) = −∫

Rdσ(ξ) dξ + C(σ) = −

∑Zd

σ

)where the last sum is an ordinary sum (resp. a cut-off regularised sum).

tu

77

Exercise 64 Check that for any classical symbol σ with non integer order a and forany holomorphic perturbation σ(z) (so that σ(0) = σ) of order −qz + a with q 6= 0such that σ(0) = σ the map z 7→ −

∑Zd σ(z) is holomorphic at z = 0 and we have

limz→0−∑Zd

σ(z) = −∑Zd

σ = ev0

(−∑Zd

σ(z)

).

Hint: Use Exercise 38.

4.6 The (Hurwitz) zeta function and zeta functions associatedwith quadratic forms

We now view the well-known zeta and Hurwitz zeta functions (see e.g [Ca]) as regu-larised sums of a symbol.Given a subset A ⊂ Zd, there is some smooth function χA ∈ C∞(Rd) which is identi-cally 1 on A and zero on Zd −A. For a symbol σ ∈ CScc(Rd), we set:

−∑A

σ = −∑Zd

χA σ,

which is clearly independent of the specific choice of smooth function χA satisfyingthe above requirements. We now fix d = 1 and choose A = N setting

−∞∑1

σ = −∑Zd

χN σ.

Applying Theorem 9 to the holomorphic family σs(z)(ξ) = (ξ + v)−s−z χN(ξ) leads tothe following result.

Proposition 22 For any real number v and any complex number s, the map z 7→−∑∞n=1(n+ v)−s+z is meromorphic with a simple pole at z = 0 for s = 1 given by 1 and

with finite part

ζ(s; v) := evreg0

(−∞∑n=1

(n+ v)−s+z).

When v = 0, ζ(s) := ζ(s; 0) is called the zeta function at argument s. For positiveq, ζ(s, q) := ζ(s; q − 1) is called the Hurwitz zeta function at argument s withparameter q.

Let us recall the known fact that the values at negative integers are rational providedthe parameter v is rational.

Proposition 23 [MP] Let z 7→ γ(z) be a holomorphic function with non-vanishingfirst derivative at zero. The map z 7→ −

∑∞n=1(n + v)a−γ(z) is holomorphic at zero for

any a 6= −1. For any a ∈ N and any non-negative rational v its limit at zero:

evreg0

(−∞∑n=1

(n+ v)a−λγ(z)

)= limz→0−∞∑n=1

(n+ v)a−λγ(z) = −Ba+1(1 + v)a+ 1

= ζ(−a, 1 + v),

(106)is a rational number. When v = 0 this yields ζ(−a) = −Ba+1

a+1 (see e.g. [Ca]). Whena = −1 the residue at 0 reads 1

γ′(0) .

78

Proof: Applied to f(x) = (x+ v)a with a ∈ C, the classical Euler-Maclaurin formula(89) gives:

∑0<n≤N

(n+ v)a =(N + v)a + (1 + v)a

2+∫ N

1

(x+ v)a dx

+2K∑k=2

Bk[a]k−1

k!((N + v)a−k+1 − (1 + v)a−k+1

)+

[a]2K+1

(2K + 1)!

∫ N

1

B2K+1(x) (x+ v)a−2K−1 dx

= (1− δa+1)(N + v)a+1

a+ 1− (1− δa+1)

(1 + v)a+1

a+ 1

+(N + v)a + (1 + v)a

2+ δa+1 (log(N + v)− log(1 + v))

+2K∑k=2

Bk[a]k−1

k!((N + v)a−k+1 − (1 + v)a−k+1

)+

[a]2K+1

(2K + 1)!

∫ N

1

B2K+1(x) (x+ v)a−2K−1 dx. (107)

Let us set

RK(a) :=[a]2K+1

(2K + 1)!

∫ N

1

B2K+1(x) (x+ v)a−2K−1 dx; SK(a)

:=2K∑k=2

Bk[a]k−1

k!(1 + v)a−k+1.

Replacing a in (107) by a− λγ(z) and taking finite parts as N →∞ we have:

−∞∑1

(n+ v)a−λγ(z) = − (1 + v)a−λγ(z)+1

a− λγ(z) + 1(108)

+(1 + v)a−λγ(z)

2− SK

(a− λγ(z)

)+RK

(a− λγ(z)

).

Hence Res0 −∑∞

1 (n+ v)a−λγ(z) = δa+11λ . Taking the finite part at 0 then yields:

evreg0

(−∞∑1

(n+ v)a−λγ(z)

)= −(1− δa+1)

(1 + v)a+1

a+ 1

− δa+1 log(1 + v) +(1 + v)a

2− SK(a) +RK(a),

which for a non-negative integer a gives:

limz→0−∞∑0

(n+ v)a−λγ(z) = −a+1∑k=0

(a+ 1k

)Bk (1 + v)a+1 = −Ba+1(1 + v)

a+ 1.

tu

79

Remark 17 Whereas ζ(−a) is rational for non-positive integers a, the derivativesζ(b)(−a) are not expected to be rational for non-positive integers a.For example, ζ ′(0) = − 1

2 log 2π [Ca] is not rational but it is a period.

tuCut-off regularised sums −

∑Zd are useful in building meromorphic extensions of ordi-

nary sums of holomorphic families of symbols; we recover in this way the existence ofmeromorphic extensions of zeta functions associated with quadratic forms.To a positive definite quadratic form q(x1, · · · , xd) and a smooth cut-off function χwhich vanishes in a small neighborhood of 0 and is identically one outside the unitEuclidean ball, we assign the classical symbol

ξ 7→ σq,s(ξ) := q(ξ)−s χ(ξ) ∈ CScc(Rd). (109)

Theorem 15 Given any complex number s, the holomorphic map

z 7→∑Zd

σq,s+ z2

=∑

Zd−0

q−(s+ z2 )

which is holomorphic on the half plane Re(z) > 2(d − s), extends to a meromorphicmap

z 7→ −∑

Zd−0

σq,s+ z2

= −∑Zd

q−(s+ z2 )

with simple pole at z = 0 if s = d2 given by:

Res0 −∑

Zd−0

q−(s+ z2 ) = δ2s=d

∫|ω|=1

q(ω)−d/2 dSω

and the constant term at z = 0 is given by

Zq(s) := evreg0

−∑Zd−0

q−(s+−d/2)

= evreg0 −∫

Rdσq,s+ z

2+ C (σq,s) . (110)

Proof: The meromorphicity follows from Theorem 14 applied to σq,s and the holo-morphic regularisation (compare with (75))

R(σ)(z)(ξ) = σ(ξ) q(ξ)−z2 χ(ξ) + σ(ξ) (1− χ(ξ)).

80

By (104) the pole at z = 0 is given by the pole of −∫

Rd R(σ)(z)(ξ) dξ. We write

Res0 −∫

RdR(σ)(z)(ξ) dξ

= Res0

∫0≤|ξ|≤1

χ(ξ) q(ξ)−s dξ + Resz=0

(fpR→∞

∫1≤|ξ|≤R

q(ξ)−s−z2 dξ

)

= Res0

(fpR→∞

∫|ω|=1

∫ R

1

q(rω)−s−z2 rd−1 dSω

)

= Res0

((fpR→∞

∫ R

1

r−2s−z+d−1 dr

) (∫|ω|=1

q(ω)−s dSω

))

= Res0

(fpR→∞

R−2s−z+d − 1−2s− z + d

) (∫|ω|=1

q(ω)−s dSω

)

= Resz=01

2s+ z − d

(∫|ω|=1

q(ω)−s dSω

)

= δ2s−d

(∫|ω|=1

q(ω)−s dSω

). (111)

Hence, there is a pole at z = 0 only if s = d/2 in which case the residue coincideswith

∫|ω|=1

q(ω)−s dSω. The finite part Zq(z) at z = 0 is given by the regularisedEuler-Maclaurin formula (103). tu

Example 5 For d = 2 and q(x, y) = ax2 + bxy + cy2 with 4ac− b2 > 0, Zq(s) yieldsa meromorphic extension of Epstein’s ζ-function

∑(m,n)∈Z−0(am

2 + bmn+ cn2)−s

(see e.g.[CS]).When a = c = 1, b = 0, Zq(s) provides a meromorphic extension of the zeta functionof Z[i] given by (see e.g. [Ca])

Z4(s) :=∑

z∈Z[i]−0

|z|−2s =∑

(m,n)∈Z−0

m2 + n2.

When a = b = c = 1, Zq(s) provides a meromorphic extension of the zeta function ofZ[j] given by (see e.g. [Ca])

Z3(s) :=∑

z∈Z[j]−0

|z|−2s =∑

(m,n)∈Z−0

m2 +mn+ n2.

Proposition 24 Whenever Re(s) ≤ 0

1. Zq(s) = C(ξ 7→ q(ξ)−s).

2. Specifically, Zq(−k) = 0. for any non-negative integer k.

3. Moreover, Zq is holomorphic at s = −k for any non-negative integer k andZ ′q(−k) = ∂s|s=−kC(q−s) where the derivative at k = 0 stands for the derivativeof the map C(q−s) restricted to the half plane Re(s) ≤ 0 22 .

22In contrast to the value at s = −k which vanishes, one does not expect the derivative to vanishin general.

81

Proof:

1. When Re(s) ≤ 0, the map ξ 7→ q(ξ)−s can be extended by continuity to ξ = 0by23

σq,s(ξ) := q(ξ)−s ∀ξ 6= 0, σq,s(0) = 0.

In that case, there is no need to introduce a cut-off function χ at 0 and we have:

Zq(s) = −∑~n∈Zq

q(~n)−s = −∫

Rdq(ξ)−s dξ + C(ξ 7→ q(ξ)−s)

along the lines of the proof of the previous proposition. Using polar coordinatesξ = r ω with r > 0 and ω in the unit sphere, the result then follows from thefact that the cut-off regularised integral vanishes if Re(z) ≤ 0 since we have

−∫

Rdq(ξ)−2s dξ = fpR→∞

∫|ξ|≤R

q(ξ)−2s dξ

= fpR→∞

∫|ω|=1

∫ R

0

q(rω)−2s rd−1 dξ

=

(fpR→∞

∫ R

0

r−2s+d−1 dr

) (∫|ω|≤1

q(ω)−2s dSω

)

=(

fpR→∞R−2s+d

−2s+ ddr

) (∫|ξ|≤1

q(ω)−2s dSω

)if − 2s+ d 6= 0

= 0 if − 2s+ d 6= 0.

where µS is the volume measure on the unit sphere induced by the canonicalmeasure on Rd.

2. When s = −k, we also have C(ξ 7→ qk(ξ)) = 0 since C vanishes on polynomialsso that Zq(−k) = 0.

3. By Theorem 15 there is no pole at s = −k (the presence of the cut-off functionχ does not affect poles) since the only pole corresponds to s = d/2. The mapZq is therefore holomorphic at s = −k with derivative given by the derivative ofthe map C(ξ 7→ q(ξ)−s) at s = −k.

tu

23Note that this extension is not smooth at 0 so that it does not define a symbol. It neverthelesshas the same asymptotic behaviour as |ξ| → ∞ as ξ 7→ χ(ξ) q(ξ)−s which is enough for our needs.

82

5 The canonical trace and noncommutative residueon operators

The noncommutative residue on classical symbols and the canonical integral on non-integer order classical symbols give rise respectively to the noncommutative residue onclassical pseudodifferential operators and the canonical trace on those which have non-integer order. We give a preliminary characterisation of the noncommutative residuein Theorem 16.Passing from pseudodifferential symbols to pseudodifferential operators first requiresextending the notion of pseudodifferential symbol on Rd to symbols with varyingcoefficients on an open subset of Rd. When patched up using a partition of unity,these lead to pseudodifferential operators on a closed manifold (see e.g. [Gi], [Sh],[Ta], [Tr2]).Pseudodifferential operators generalise differential operators with smooth coefficients,namely linear operators P acting on smooth functions on an open subset U of Rddefined by

Pu(x) =∑|α|≤a

cα(x)Dαxu(x), u ∈ C∞(U), cα ∈ C∞(U),

where the non-negative integer a stands for the order of P and where we have setDαx := (−i)|α|∂α1

x1· · · ∂αdxd and α = (α1, · · · , αd) ∈ Zd≥0 is a multiindex.

To the differential operator P one associates the polynomial

σ(P )(x, ξ) =∑|α|≤a

cα(x) ξα

called the symbol of P , which is related to P via a Fourier transform

Pu(x) =∫ei〈x,ξ〉 σ(P )(x, ξ) u(ξ) dξ ∀x ∈ U.

Similarly, we define a pseudodifferential operator via its symbol; we therefore firstneed to describe the class of symbols under consideration.In the following V denotes a finite dimensional complex linear space, M a closedd-dimensional manifold and E a finite rank Hermitian vector bundle over M .

5.1 From pseudodifferential symbols to pseudodifferential op-erators

The following definition generalises Definition 17 insofar as the coefficients of thesymbol now depend on a variable x. Such a generalisation is useful in view of gaugetheory in physics or geometry in mathematics.

Definition 19 A smooth function σ in C∞(T ∗U) is a scalar symbol on U wheneverthe following condition is satisfied. There is some real constant a such that for anymultiindices γ, δ ∈ Zd+, for any compact subset K ⊂ U there exists a constant Cγ,δ,K ∈R+ such that for any x ∈ K, ξ ∈ Rd

|∂γx∂δξσ(x, ξ)| ≤ Cγ,δ,K 〈ξ〉a−|δ| (112)

where we have set 〈ξ〉 :=√

1 + |ξ|2.

83

Exercise 65 Show that polynomials

σ(x, ξ) =∑|α|≤a

cα(x)ξα

with cα ∈ C∞(U) are symbols.

Let Sa(U) ⊂ C∞(T ∗U) denote the set of scalar valued symbols on U which fulfilcondition (112). Let us further set Sa(U, V ) := Sa(U) ⊗K End(V ) for any K-linearspace V and

S(U, V ) :=⋃a∈RSa(U, V ).

With these notations we have Sa(U) = Sa(U,K); S(U) = S(U,K).

Remark 18 For two real numbers a1, a2, we have

a1 ≤ a2 =⇒ Sa1(U, V ) ⊂ Sa2(U, V ).

We call a symbol in the intersection

S−∞(U, V ) :=⋂a∈RSa(U, V )

a smoothing symbol.

Exercise 66 Show that symbols σ(x, ξ) with compact support in ξ are smoothing sym-bols.

Equality modulo smoothing symbols defined by

σ ∼ σ′ ⇐⇒ σ − σ′ ∈ S−∞(U, V )

defines an equivalence relation in S(U, V ).For symbols σ ∈ S(U, V ), σk ∈ Sak(U, V ), k ∈ N0 where ak is a sequence of symbolswhose orders have decreasing real parts as k tends to infinity, we also set(

σ ∼∑k∈N0

σk

)

⇐⇒

∀α ∈ R, ∃K(α) ∈ N, s.t. K ≥ K(α)⇒ σ −∑k≤K

σk ∈ Sα(U, V )

.

We define classical and log-polyhomogeneous symbols on U as in Definition 1.

Definition 20 Let a be a complex number.

1. A symbol σ ∈ S(U, V ) is classical (or polyhomogeneous) of order a if for anyx ∈ U the map ξ 7→ σ(x, ξ) is an End(V )-valued classical (or polyhomogeneous)symbol of order a.

2. A symbol σ ∈ S(U, V ) is log-polyhomogeneous of log-type k and order a forsome complex number a and some non-negative integer k if for any x ∈ U themap ξ 7→ σ(x, ξ) is an End(V )-valued log-polyhomogeneous symbol of log-type kand order a.

84

Let CSa,k(U, V ) ⊂ S(U, V ) denote the class of log-polyhomogeneous symbols of ordera and log-type k and let us set CSa,∗(U, V ) :=

⋃k∈N0

CSa,k(U, V ). Then CSa(U, V ) :=CSa,0(U, V ) corresponds to the set of classical symbols of order a.

Definition 21 The top degree homogeneous component σa in C∞(U,End(V )) of aclassical symbol σ in CSa(U, V ) is called the leading symbol of the symbol.

The star-product of two symbols σ ∈ CSa(U, V ) and τ ∈ CSb(U, V ) is defined by:

σ ? τ ∼∑γ

(−i)|γ|

γ!∂γξ σ(x, ξ) ∂γxτ(x, ξ), (113)

or equivalently

σ ? τ −N−1∑|γ|=0

(−i)|α|

γ!∂γξ σ(x, ξ) ∂γxτ(x, ξ) ∈ CSa+b−N (U, V ) ∀N ∈ N,

where stands for the composition in End(V ). It is a classical symbol of order a+ b.

Exercise 67 Check that the leading symbols σa and τb of σ and τ mutliply under thisproduct:

(σ ? τ)a+b = σa · τb.

We consider the algebras

CS(U, V ) := 〈∪a∈CCSa(U, V )〉; ((resp. CS∗,∗(U, V ) := 〈∪a∈CCS

a,∗(U, V ))〉

generated by the corresponding unions of classical (resp. log-polyhomogeneous) sym-bols on U .The set CSa,∗(U, V ) inherits a Frechet topology similar to that described in (22)given by the following semi-norms labelled by multiindices α, β and integers j ≥ 0, N ,l ∈ 0, · · · , k and i ∈ N corresponding to a countable family Ki, i ∈ N of compactsets covering U (see [Ho]):

supx∈Ki,ξ∈Rd(1 + |ξ|)−Re(a)+|β| ‖∂αx ∂βξ σ(x, ξ)‖;

supx∈Ki,ξ∈Rd |ξ|−Re(a)+N+|β|∥∥∥∂αx ∂βξ (σ − N−1∑

j=0

χ(ξ)σa−j)

(x, ξ)∥∥∥;

supx∈Ki,|ξ|=1‖∂αx ∂βξ σa−j,l(x, ξ)‖, (114)

where χ any smooth function which vanishes around zero and is identically one outsidethe unit ball and σa−j =

∑kl=0 σa−j,l.

Remark 19 For σ and τ in CS(U), the equivalence (113) translates to

σ ? τ = limN→∞

N−1∑|γ|=0

(−i)|α|

γ!∂γξ σ(x, ξ) ∂γxτ(x, ξ)

in the Frechet topology of symbols of constant order (here a+ b).

85

Let F(u)(ξ) = u(ξ) =∫

Rd e−i〈y,ξ〉u(y) dy denote the Fourier transform of an L1 func-

tion u on Rd. To a symbol σ ∈ S(U, V ) ⊂ C∞(U × Rd) ⊗ End(V ) corresponds alinear operator acting on the algebra C∞cpt(U, V ) of smooth V -valued functions onwith compact support in U :

Op(σ) : C∞cpt(U, V ) → C∞(U, V )

u 7→ (Op(σ)(u)) (x) := F−1 (σ(x, ·) u)

called a pseudodifferential operator on U with coefficients in End(V ). We call itclassical (resp. log-polyhomogeneous) of order a ∈ C (resp. and of log-type k ∈ N0) ifthe symbol σ has order a and lies in the corresponding class of symbols.

Exercise 68 Show that

Op

∑|α|≤a

cα ξα

=∑|α|≤a

cαDαx

for coefficients cα in C∞(U,End(V )) and where we have set Dαx = (−i)α∂αx .

For any symbol σ,

(Op(σ)u) (x) =∫

Rdei〈x,ξ〉σ(x, ξ)u(ξ)dξ

=∫

Rd

∫Rdei〈x−y,ξ〉σ(x, ξ)u(y)dy dξ (115)

=∫

RdK(x, y)u(y) dy,

whereK(x, y) :=

∫Rdei〈x−y,ξ〉σ(x, ξ)dξ

is the kernel of the operator Op(σ). In general, this is a distribution kernel, whosesingularities can be shown to be located on the diagonal. In particular, for two smoothfunctions χ, χ with disjoint compact supports, the operator χOp(σ)χ has smoothkernel.

Definition 22 A linear operator

A : C∞cpt(U, V ) → C∞(U, V )

u 7→ Au(x) :=∫

RdK(x, y)u(y) dy

defined by a smooth kernel K is called a smoothing operator.

5.2 Basic properties of pseudodifferential operators

We state some basic properties of pseudodifferential operators acting on functions withsupport in an open subset U of Rd. These properties easily extend to pseudodifferentialoperators on vector valued functions with support in a compact set U . Let V denotea linear space.

86

Definition 23 An End(V )-valued pseudodifferential operator A on U is a linearoperator A : C∞cpt(U, V )→ C∞(U, V ) of the form

A = Op(σ(A)) +R(A)

for some smoothing operator R(A) and some symbol σ(A) ∈ S(U, V ), called the symbolof A which is determined modulo a smoothing symbol.

Remark 20 Unlike a differential operator A which is local in the sense that if uvanishes on U then Au also vanishes on U , pseudodifferential operators are not localsince they are defined by Fourier transforms which smear out the support. However,a pseudodifferential operator A is pseudo-local in the following sense [Gi]. Given anyopen subset W ⊂ U and χ, χ ∈ C∞cpt(W ), then

χu ∈ C∞cpt(W,V ) =⇒ χ Au ∈ C∞cpt(W,V ) ∀u ∈ C∞(U, V ).

The operator Op(σ) is not expected to send a function u in C∞cpt(U, V ) to anotherfunction in C∞cpt(U, V ) if K(x, y)u(y) arising in (115) is not compactly supported inU × U , hence the need for properly supported operators (see e.g. [Sh], [Ta], [Tr2]).These can be defined in different ways; here is one characterisation [Tr2] (Proposition3.1).

Definition 24 A pseudodifferential operator A with distribution kernel K is prop-erly supported in U , if for any χ ∈ C∞cpt(U), both χ(x)K(x, y) and K(x, y)χ(y)have compact support in U × U .

Clearly, a properly supported operator stablises the set C∞cpt(U).

Example 6 If A is a pseudodifferential operator on U ; then for any χ ∈ C∞cpt(U) andχ ∈ C∞cpt(U) the operator χA χ is properly supported for its kernel χ(x)K(x, y) χ(y)where K is the kernel of A, has compact support in U × U .

Exercise 69 Differential operators are properly supported operators.

A properly supported operator A reads A = Op(σ(A)) where σ(A) is the smoothfunction on T ∗U defined by

σ(A)(x, ξ) = e−i〈x,ξ〉A(ei〈x,ξ〉

).

We call σ(A)(x, ξ) the symbol of the properly supported operator A. Since anypseudodifferential operator differs from a properly supported operator by a smoothingoperator (see e.g. Proposition 3.4 in [Tr2] or Proposition 3.3 in [Sh]), the symbol σ(A])of a properly supported operator A] = Op(σ(A])) that differs from A by a smoothingoperator can be chosen as a symbol of A. Consequently, Op(σ(A) in (23) can bereplaced by Op(σ(A]) so that we can henceforth legitimately assume that Op(σ(A))in (23) is properly supported.

Remark 21 This can actually serve as a definition of a symbol (see e.g. [Sh]): thesymbol of a pseudodifferential operator is then defined (modulo smoothing symbols) asthe symbol of any properly supported operator which differs from A by a smoothingoperator.

We now single out a subclass of pseudodifferential operators.

87

Definition 25 A pseudodifferential operator is classical (or polyhomogeneous)(resp. log-polyhomogeneous of type k) if its symbol σ(A) is classical (resp. log-polyhomogeneous of type k).

Thus, in contrast to a differential operator which has well-defined symbol, the symbolof a pseudodifferential operator only makes sense modulo smoothing symbols.The leading symbol of a classical pseudodifferential operator A of order a on U isdefined by

σL(A)(x, ξ) = limt→∞

t−aσA(x, tξ) ∀(x, ξ) ∈ T ∗U.

Definition 26 A pseudodifferential operator is called elliptic if σL(x, ξ) ∈ Aut(V )for any x ∈ U, ξ ∈ T ∗xU − 0.

Exercise 70 Let A =∑|α|≤a aα(x)Dα

x and B =∑|β|≤b bβ(x)Dβ

x be two differen-tial operators whose symbols read σ(A)(x, ξ) =

∑|α|≤a aα(x)Dα

x and σ(B)(x, ξ) =∑|β|≤b bβ(x)Dβ

x . Their product AB has symbol given by:

σ(AB)(x, ξ) =∑

|γ|≤|α|+|β|

(−i)γ

γ!∂γξ σ(A)(x, ξ) ∂γxσ(B)(x, ξ) = σ(A)(x, ξ) ? σ(B)(x, ξ).

Since a properly supported pseudodifferential operator sends functions with compactsupport to functions with compact support, one can compose two properly supportedoperators. The above formula generalises to properly supported pseudodifferentialoperators.

Proposition 25 ([Sh] Th. 3.4) The product of two classical pseudodifferential oper-ators A of order a, B of order b both properly supported in U with symbols σ(A) andσ(B), is a properly pseudodifferential operator of order a + b with symbol σ(AB) =σ(A) ? σ(B), where

σ(A) ? σ(B) ∼∞∑|γ|=0

(−i)γ

γ!∂γξ σ(A)(x, ξ) ∂γxσ(B)(x, ξ). (116)

In particular, the leading symbol is multiplicative

σL(AB) = σL(A)σL(B).

In other words,

σ(AB) = limN→∞

N−1∑|γ|=0

(−i)γ

γ!∂γξ σ(A)(x, ξ) ∂γxσ(B)(x, ξ)

in the Frechet topology of symbols of constant order (here a+ b).

Exercise 71 Given a properly supported operator A = Op(σ(A)) on an open subsetU , show that for a smooth function φ with compact support in U , the bracket [φ,A] issmoothing and that the symbol of the commutator [xj , A] = [φxj , A]− xj [φ,A] reads

σ ([xj , A]) ∼ i∂ξjσ(A).

88

The symbol of a pseudodifferential operator does not transform covariantly under achange of parametrisation.For two open subsets U,U ′ of Rd, a pseudodifferential operator A on U and a diffeo-morphism κ : U → U ′, we set

κ∗A = κ∗ A κ∗ (117)

where κ∗u := u κ and κ∗u = u κ−1 for any u ∈ C∞(U). Note that dκ(x) : TxU →Tκ(x)U

′ and (dκ(x))t : T ∗κ(x)U′ → T ∗xU .

Let us recall the following well-known result (see e.g. Th. 5.1 in [Ta] or Lemma 1.3.3in [Gi]) which relates κ∗Op(σ) to Op(κ∗σ) where

κ∗σ(x′, ξ′) := σ(κ−1(x′), dκ(κ−1(x′))t ξ′). (118)

Proposition 26 Let A be a properly supported pseudodifferential operator on U withsymbol σ. Then κ∗A is a pseudodifferential operator on U with symbol

κ∗σ(x′, ξ′) := κ∗σ(x′, ξ′) +∑|α|>0

1α!φα(κ−1(x), η) ∂α2 σ

(κ−1(x′), (dκ(κ−1(x′))t η

).

(119)Here φα(x′, ξ′) is a polynomial in ξ of degree ≤ |α|2 with φ0(x′, ξ′) = 1.

Remark 22 In particular, κ∗σ(x′, ξ′)− κ∗σ(x′, ξ′) is of the form∑|β|≤ |α|2

aα,β(x) ξβ ∂αξ σ(·, ξ),

for some smooth functions aα,β and where we have set x′ = κ(x), ξ = dκ(κ−1(x′))t ξ′,an observation which turns out to play a crucial role in the following.

Even though the symbol of an operator is modified by a coordinate change, some ofits features are preserved.

Corollary 10 Let A be a properly supported oeprator on U with symbol σ.

1. κ∗A and A have same order given by the order of

ord (κ∗σ) = ord (κ∗σ) = ord (σ) ,

where ord(τ) is the order of the symbol τ .

2. If A is smoothing, so is κ∗A.

3. If σ is a classical (resp. log-polyhomogeneous of type k), then so is κ∗σ, so thatthe type of A is invariant under a coordinate change.

4. The leading symbol of A transforms covariantly

σL(κ∗A) = κ∗σL(A).

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5.3 Pseudodifferential operators on manifolds

Recall that a coordinate chart (U, φ) on M is an open subset of U of M and a diffeo-morphism φ : U → φ(U) ⊂ Rd. An atlas is a collection Ui, φi, χi, i ∈ I where theopen subsets Ui, i ∈ I form an open cover of M , for each i ∈ I, (Ui, φi) is a coordinatechart and χi, i ∈ I is a partition of unity subordinated to the covering. Following[Gi], we set the following definition.

Definition 27 We call localisation subordinated to a chart (U, φ) around apoint x in M of a linear operator A : C∞(M) 7→ C∞(M) any map χA χ, where χand χ are smooth functions with compact support in U which are identically one in aneighborhood of x.

Exercise 72 Check that for two compactly supported smooth functions χ and χ withdisjoint supports the operator χA χ is smoothing.

Definition 28 A linear operator A : C∞(M) 7→ C∞(M) on M is called a pseudo-differential operator if given any local chart (U, φ) and any localisation AU := χA χsubordinated to this chart, the induced localised operator

φ∗AU : f 7→ φ∗ AU φ∗(f), (120)

where φ∗f := f φ; φ∗f := f φ−1, is a pseudodifferential operator on φ(U).The symbol σφ(A)(x, ·) of A in a given local chart (U, φ) around x ∈ U is defined bythe symbol of φ∗AU .

With these definitions at hand, we write a pseudodifferential operator

A =∑

Supp(χi)∩Supp(χj)6=∅

χiAχj +∑

Supp(χi)∩Supp(χj)=∅

χiAχj

=∑

Supp(χi)∩Supp(χj)6=∅

φ∗iOp(σij) +R(A), (121)

where Op(σij) are properly supported operators on φi(Ui) ∩ φj(Uj) and R(A) =∑Supp(χi)∩Supp(χj)=∅ χiAχj is a smoothing operator.

Features of pseudodifferential operators of the type Op(σ) which are preserved underdiffeomorphisms can be extended to pseudodifferential operators on manifolds. Letus make this statement more precise. If (U, φ) and (U ′, φ′) are two local coordinatecharts, then setting κ := φ′ φ−1, on the intersection U ∩ U ′ we have

φ′?AU∩U ′ = κ∗ φ∗AU∩U ′ κ? = κ? (φ?AU∩U ′) . (122)

From Proposition 26 and its Corollary 10 it follows that Aφ′(U) and Aφ′(U) are ofthe same type (classical, log-polyhomogeneous), have the same order and differ bya pseudodifferential operator of strictly smaller order. It therefore makes sense tointroduce the following definitions.

Definition 29 Let A be a pseudodifferential operator on M .

1. A is classical (resp. log-polyhomogeneous of type k) if, given any lo-cal chart (U, φ) and any localisation AU subordinated to this chart, the inducedlocalised operator Aφ(U) is classical (resp. log-polyhomogeneous of type k).

90

2. A has order a if, given any local chart (U, φ) and any localisation AU subordi-nated to this chart, the induced localised operator φ∗AU has order a.

3. We call a linear operator smoothing if given any local chart (U, φ), any locali-sation AU subordinated to this chart is smoothing.

4. The leading symbol σL(A)(x) at a point x ∈ U is given by σL (φ∗AU ) (φ(x)) forany local chart (U, φ) and any localisation AU := χ Aχ subordinated to this chartwhere χ and χ are identically one in a neighborhood of x.

5. A pseudodifferential operator A with invertible leading symbol σL(A)(x, ξ) forany x ∈M and any ξ 6= 0 is called an elliptic pseudodifferential operator.

On the grounds of these definitions we introduce the set C`a(M) (resp. C`a,k(M))of classical (resp. log-polyhomogeneous) pseudodifferential operators on M of order a(and log-type k).

Exercise 73 1. Deduce from Proposition 25 that the product AB of two classical(resp. log-polyhomogeneous) pseudodifferential operators A and B on M of ordera and b (resp. and log types k and l) is a pseudodifferential operator of ordera+ b (resp. and log-type k + l).

2. Show that the product is elliptic if A and B are elliptic as a result of the multi-plicativity of leading symbols.

We can therefore define the algebras

C`(M) = 〈⋃a∈C

C`a(M)〉, resp. C`∗,∗(M) = 〈⋃a∈C

C`a,∗(M)〉 =⋃k∈N0

〈⋃a∈C

C`a,k(M)〉

generated by all classical (resp. log-polyhomogeneous) pseudodifferential operators onM for the product of operators. Here 〈S〉 stands for the algebra generated by the setS. We also consider the algebra C`−∞(M) :=

⋂a∈C C`

a(M) of smoothing operatorswhich is a two-sided ideal in C`(M) and C`∗,∗(M).

Exercise 74 With these notations show that

L1(M) ∩ C`∗,∗(M) = ∪Re(a)<−dC`a,∗(M),

where L1(M) denotes the first Schatten class of L2(M) (see e.g. [ReSi]).

These definitions extend to linear operators acting on smooth sections C∞(M,E) ofa smooth vector bundle E over M . If V is the model space for the vector bundle E,the above definitions and properties generalise, replacing coordinate charts (U, φ) onM by local trivialisations (U, φ,Φ):

E|U → φ(U)× V(x, v) 7→ (φ(x),Φ(v)) .

This leads to algebras defined in a similar manner to the above algebras:

C`(M,E) = 〈⋃a∈C

C`a(M,E)〉, C`−∞(M,E) :=⋂a∈C

C`a(M,E),

C`∗,∗(M,E) = 〈⋃a∈C

C`a,∗(M,E)〉 =⋃k∈N0

〈⋃a∈C

C`a,k(M,E)〉.

91

As before, since the product of two operators with orders a and b has order a + b,the algebra C`−∞(M,E) of smoothing operators is a two-sided ideal in C`(M,E) andC`∗,∗(M,E).

We now equip these infinite dimensional sets of operators with the Frechet topol-ogy of constant-order symbols. For any complex number a and any non-negativeinteger k, the linear space C`a,k(M,E) of classical pseudodifferential operators oforder a and log-type k can be equipped with a Frechet topology which is simi-lar to the one described in (22) in the case k = 0. For this, one equips the setCSa,k(U, V ) = CSa,k(U)⊗End(V ) of log-type k symbols of order a on an open subsetU of Rd with values in a Euclidean space V (with norm ‖ · ‖) with a Frechet struc-ture. The following semi-norms labelled by multiindices α, β and integers j ≥ 0, N ,l ∈ 0, · · · , k and i ∈ N corresponding to a countable family Ki, i ∈ N of compactsets covering U , give rise to a Frechet topology on CSa,k(U, V ) (see [Ho]):

supx∈Ki,ξ∈Rd(1 + |ξ|)−Re(a)+|β| ‖∂αx ∂βξ σ(x, ξ)‖;

supx∈Ki,ξ∈Rd |ξ|−Re(a)+N+|β|∥∥∥∂αx ∂βξ (σ − N−1∑

j=0

χ(ξ)σa−j)

(x, ξ)∥∥∥;

supx∈Ki,|ξ|=1‖∂αx ∂βξ σa−j,l(x, ξ)‖,

where χ is any smooth function which vanishes around zero and is identically oneoutside the unit ball, and σa−j =

∑kl=0 σa−j,l.

In this topology, the set C∞(U)⊗(CSa,kcc (Rd)⊗ End(V )

)is dense in CSa,k(U, V )

This Frechet structure on CSa,k(U, V ) induces one on C`a,k(M,E). Indeed, as in (121)given an atlas (Ui, φi)i∈I onM and local trivialisations Φi : E|Ui ' φi(Ui)×V, i ∈ I (forsome finite set I) compatible with the charts, using a partition of unity subordinatedto the chosen atlas, we write an operator A in C`a,k(M,E) as follows:

A =∑

Supp(χi)∩Supp(χj)6=∅

Φi?Op(σij) +R(A),

where R(A) =∑

Supp(χi)∩Supp(χj)=∅ χiAχj is smoothing and Op(σij) are properlysupported pseudodifferential operators on φi(Ui ∩ Uj).The countable family of semi-norms built from

1. a countable family of semi-norms given by the supremum norm of the kernel ofR(A) and its derivative on a countable family of compact subsets,

2. the countable family of semi-norms on Op(σi) induced by the ones on the symbolsσi described above,

provide a Frechet topology on C`a,k(M,E).

5.4 From closed linear forms on symbols to traces on operators

Let U be an open subset of Rd and D(U) a subset of CS∗,∗(U) stable under partialderivatives ∂xi , ∂ξi , i = 1, · · · , d and under multiplication by functions in C∞(U). IfD(U) is a linear subset 24 of CS∗,∗(U), this makes it a C∞(U)-submodule of CS∗,∗(U).The set

S := σ ∈ CScc(Rd),∃f ∈ C∞(U), f ⊗ σ ∈ D(U)24We shall also consider subsets such as CS /∈Z,∗(U) := ∪a/∈ZCS

a,∗(U) which are not linear spaces.

92

is stable under partial differentiation ∂ξi since D(U) is.

Example 7 Corresponding to the algebras

D1(U) = CS∗,∗(U), D2(U) = CS(U), D3(U) = CS /∈Z(U),

we have the following sets of symbols with constant coefficients

S1 = CS∗,∗cc (Rd), S2 = CScc(Rd), S3 = CS /∈Zcc (Rd).

Corresponding to the sets

D4(U) = CS /∈Z,∗(U); D5(U) = CS∗,k(U); D6(U) = CSa(U),

we have the following set of symbols with constant coefficients

S4 = CS /∈Z,∗cc (Rd), S5 = CS∗,kcc (Rd); S6 = CSacc(Rd).

Provided C∞(U)⊗S is dense in D(U), i.e., for any complex number a the set C∞(U)⊗Sa is dense in Da(U) := D(U)∩C`a(U) for the Frechet topology on constant symbols,a continuous linear form λ : S → C induces a continuous linear map 25 λ defined by

λ(f ⊗ τ) = f λ(τ) ∀f ∈ C∞(U), ∀τ ∈ CScc(Rd).

With these notations, we have: λ(f ⊗ τ)(x) = f(x)λ(τ) = λ(f(x) τ) ∀x ∈ U so thatfor any σ in CScpt(U) we have

λ(σ)(x) = λ(σ(x, ·)).

Integrating it along U yields a linear form:

λU : Dcpt(U) → C

σ 7→∫U

λ(σ(x, ·)) dx,

where Dcpt(U) is the subset of symbols in D(U) with compact support in U , whichis a C∞cpt(U)-module. Given a finite dimensional complex linear space V , it furtherinduces a linear form

λU tr : Dcpt(U)⊗ End(V ) → C

σ 7→∫U

λ(tr(σ(x, ·))) dx.

Example 8 1. If λ is the closed linear form resk on CS∗,kcc (Rd), then λU is thek-th higher order noncommutative residue on CS∗,kcpt(U):

resk(σ) =∫S∗U

σ−d,k(x, ξ) dSξ dx

and λU tr yields a linear form:

resk : CS∗,kcpt(U)⊗ End(V ) → C

σ ⊗ u 7→ resk(σ) =∫S∗U

tr(σ−d,k(x, ξ)) dSξ dx,(123)

where CS∗,kcpt(U) is the set of symbols in CS∗,k(U) with compact support in U .When k = 0 we simply write res instead of res0.

25As before, if S is not linear, we assume that λ, resp. λU only preserve linear combinations lyingin S, resp.D(U).

93

2. If λ is the closed linear form −∫

Rd on CS /∈Z,∗cc (Rd), then λU is the cut-off regularised

integral on CS∗,kcpt(U):

−∫T∗U

σ = −∫T∗U

σ(x, ξ) dξ dx

and λU tr yields a linear form:

CS∗,kcpt(U)⊗ End(V ) → C

σ ⊗ u 7→ −∫T∗U

tr(σ) = −∫T∗U

tr(σ(x, ξ)) dξ dx. (124)

Proposition 27 With the above notation we assume that S is stable under partialdifferentiations and invariant under the action σ 7→ σ T of the linear group GLd(R).Then for any diffeomorphism κ : U → U ′,

σ ∈ D(U) =⇒ (κ?σ ∈ D(U ′) and κ?σ ∈ D(U ′))

with the notation of Proposition 26.Let λ : S → C be a linear form. Provided λ is covariant, i.e.:

|det(T )|λ(σ T ) = λ(σ) ∀T ∈ GLd(R) ∀σ ∈ S, (125)

andωλ (Op(σ)) := λ(σ(x, ·)) dx1 ∧ · · · ∧ dxd

transforms covariantly under a diffeomorphism κ : U → U ′, i.e. for any σ in D(U):

κ∗ωλ (κ?Op(σ)) = ωλ (Op(σ)) ∀σ ∈ D(U).

Equivalently, ωλ(σ) := ωλ(Op(σ)) satisfies the following property:

ωλ(κ?σ) = ωλ(σ).

Proof: Recall from Proposition 26 that the symbol κ?σ of κ?Op(σ) differs from κ?σby a polynomial expression in ξ,

φα(x, ξ) =∑

0<|β|≤ |α|2

aα,β(x) ξβ ∂αξ σ(·, ξ).

Since D(U) is stable under partial differentiation and multiplication by smooth com-plex valued functions with compact support, φα lies in D(U). Since S is invariantunder linear transformations, κ?σ also lies in D(U). It follows that κ?σ lies in D(U).We now need to check that λ(φα(x, ·)) = 0. Since the sum in the expression for φαruns over |β| ≤ |α|2 < |α| and λ is closed, it follows from (31) that

λ(φα(x, ·)) = 0 ∀x ∈ U.

Combining this with dx′ = |det (dκ(x))| dx and the covariance property (125) of λapplied to T = (dκ(x))t, at the point x′ = κ(x) we have:

ωλ (κ?(Op(σ))) (x′) = λ (κ?σ(x′, ·)) dx′1 ∧ · · · ∧ dx′d= λ

(σ(κ−1(x′),

(dκ(κ−1(x′))

)t ·)) dx′1 ∧ · · · ∧ dx′d= λ

(σ(κ−1(x′), ·

))dx1 ∧ · · · ∧ dxd

= λ (σ(x, ·) ) dx1 ∧ · · · ∧ dxd= ωλ (Op(σ)) (x).

tu

94

Remark 23 For future reference, let us observe that the two essential ingredients inthe proof are the closedness and covariance of the linear form λ.

In view of this covariance property, ωλ (σ) (x) defines a global density on the closedmanifold M which can be integrated over M to build a linear form on a set D(M) ofoperators in C`∗,∗(M) determined by a purely symbolic condition in terms of the setS.Given a local coordinate chart (U, φ) on M and a symbol σ on φ(U), the operatorOp(σ) is not a priori properly supported in U and φ∗Op(σ) therefore does not apriori extend to a pseudodifferential operator on M . However, there is a properlysupported operator Op](σ) which differs from Op(σ) by a smoothing operator, forwhich φ∗Op](σ) does extend to a pseudodifferential operator on M .

Definition 30 We say, for a subset D(M) ⊂ C`∗,∗(M) which contains smoothingoperators and is stable under multiplication by functions in C∞(M), that it is sym-bolically defined by S provided the set

S := τ ∈ CS∗,∗cc (Rd), φ∗Op] (f ⊗ τ) ∈ D(M) ∀f ∈ C∞cpt(φ(U)) (126)

of symbols on Rd with constant coefficients is independent of the local chart (U, φ) onM , where we have set φ∗ =

(φ−1

)?

(with the notation of (118)).More generally, given a finite rank vector bundle E over M modelled on a linear spaceV , we say of a subset D(M,E) of CS∗,∗(M,E) which is stable under multiplication byfunctions in C∞(M), that it is defined symbolically by the set S⊗End(V ) provided

S ⊗ End(V ) = σ = τ ⊗ u ∈ CS∗,∗cc (Rd)⊗ End(V ),Φ∗Op] (f ⊗ σ) ∈ D(M,E) ∀f ∈ C∞cpt(U) (127)

of symbols on Rd with matrix coefficients is independent of the local trivialisation(U, φ,Φ : EU → φ(U) × V ) of E over the local open set U , where we have set Φ∗ =(Φ−1

)?

(with the notation of (118)).

Remark 24 Since D(M) contains smoothing operators, S contains smoothing sym-bols.

Example 9 The sets D(M) = C`∗,k(M), resp. D(M) = C`/∈Z,k(M) are defined interms of the symbol sets S = CS∗,kcc (Rd), resp. S = CS /∈Z,k

cc (Rd).

Integrating the covariant volume form ωλ(Op(σ)) introduced in Proposition 27 leads tolinear forms on certain classes of operators built from linear forms on the correspondingclasses of symbols.More precisely and with the notation of Definition 30, if S is stable under partialdifferentiation and invariant under the action of the linear group GLd(R), any closedand covariant linear form λ on S yields an associated linear form

Λ : D(M,E) → C

A 7→∫M

λ(tr(σ(A)(x, ·))) dx. (128)

Remark 25 If λ is continuous in the Frechet topology of constant order symbols itinduces a uniformly continuous map σ 7→ λ(σ(x, ·)) in x on compact sets so that Λ iscontinuous in the Frechet topology of constant order operators.

95

Example 10 The higher noncommutative residue λ = resk on S = CS∗,kcc (Rd) givesrise to the higher noncommutative residue resk on C`∗,k(M):

resk(A) =∫M

resk(σ(A)(x, ·)) dx =∫S∗M

tr (σ−d,k(A)(x, ξ)) dSξ dx, (129)

where S∗M ⊂ T ∗M stands for the cotangent unit sphere of M .

When k = 0, this yields the well-known noncommutative or Wodzicki residue firstintroduced by Wodzicki [Wo1, Wo2]26

Definition 31 The noncommutative residue of an operator A ∈ C`(M,E) isdefined by:

res(A) =∫M

res(σ(A)(x, ·)) dx =∫S∗M

tr (σ−d(A)(x, ξ)) dSξ dx, (130)

One can read off from this definition various straightforward but important propertiesof the noncommutative residue on classical pseudodifferential operators.

Definition 32 Let D(M,E) be a set of pseudodifferential operators such that

C`−∞(M,E) ⊂ D(M,E) ⊂ C`∗,∗(M,E).

We call singular, a linear form on D(M,E) which vanishes on smoothing operatorsand we say it is regular when it coincides on smoothing operators with the ordinarytrace

Tr(A) :=∫M

tr (K(x, x)) dx, (131)

where K stands for the kernel of A.It is called continuous when it is continuous in the Frechet topology of operators ofconstant order, i.e. it is continuous on Da(M,E) := D(M,E) ∩ C`a,∗(M,E) for anycomplex number a.

• res is local since it only depends on a finite number of positively homogeneouscomponents, namely the −d-th degree one, of the symbol of the operator.

• res is singular since it vanishes on operators whose order has real part smallerthan −d and hence on smoothing operators,

• res vanishes on operators of non-integer order since the −d-th positively homo-geneous component of their symbol vanishes,

Exercise 75 Show that the residue res is continuous in the Frechet topology of op-erators of constant order.Hint: It follows from the continuity of the residue res(σ) =

∫Sd−1 σ−d on symbols.

Example 11 The cut-off integral λ = −∫

Rd on CS∗,∗cc (Rd), whose restriction to S =CS /∈Z,k

cc (Rd) ⊂ Ker(resk) is closed and covariant, gives rise to a linear form which, incontrast with the residue, extends the L2-trace.

26see also [Ka] for a pedagogical review.

96

This leads to the canonical trace popularised by Kontsevich and Vishik in [KV] inthe context of the multiplicative anomaly for determinants which we shall investigatelater in these lectures.

Definition 33 The canonical trace TR of an operator A ∈ C`/∈Z,∗(M,E) is definedby

TR(A) =∫M

−∫T∗xM

tr (σ(A)(x, ·)) dx = −∫T∗M

tr (σ(A)(x, ξ)) dξ dx. (132)

In other words, the local density:

ωKV (A)(x) :=

(−∫T∗xM

trx (σA(x, ξ)) dξ

)dx (133)

patches modulo a global density on M whenever the operator A in C`∗,∗(M,E) hasnon-integer order or has order < −d so that the canonical trace reads:

TR(A) :=∫M

ωKV (A)(x). (134)

Properties of the canonical trace contrast with those of the residue.

• TR is a priori non-local since it depends on an infinite number of positivelyhomogeneous components of the symbol of the operator.

• TR is regular since it coincides with the ordinary trace

Tr(A) =∫M

∫T∗xM

tr (σ(A)(x, ·))) dx (135)

on operators whose order has real part smaller than −d and hence on smoothingoperators,

• We shall see later on that TR canonically extends Tr to non-integer order oper-ators.

Exercise 76 Show that the linear form TR is continuous on C`/∈Z(M,E) in theFrechet topology of operators of constant order.Hint: It follows from the continuity of the cut-off regularised integral on non-integerorder symbols with fixed order, a property which can be seen from (39).

Definition 34 • A trace on a Lie algebra L is a linear form Λ on L which vanisheson brackets of operators in L:

Λ ([a, b]) = 0 ∀a, b ∈ L.

• A graded trace on a graded Lie algebra

GrL = ⊕∞k=0GrkL

is a family parametrised by k ∈ Z+ of linear forms Λk on GrkL, which vanisheson graded brackets:

Λk+l ([a, b]) = 0 ∀a ∈ Lk, b ∈ Ll.

97

Proposition 28 A linear form Λ on D(M,E) built from a continuous, closed andcovariant linear form λ as in (128) vanishes on brackets:

[A,B] ∈ D(M,E) =⇒ Λ([A,B]) = 0 ∀A,B ∈ D(M,E).

If D(M,E) is an algebra the linear form Λ is a trace on D(M,E).

Proof: By (116) the symbol of the bracket [A,B] reads

σ(A), σ(B)? ∼∑α

(−i)|α|

α!(∂αξ σ(A)∂αx σ(B)− ∂αξ σ(B)∂αx σ(A)

), (136)

which is of order a+ b where a is the order of A, b the order of B.Combining the closedness of λ which yields “integration by parts formulae” in ξ andStokes’ theorem on M which yields “integration by parts formulae” in x with thecontinuity of the linear form λ on symbols of constant order, we have:

Λ([A,B])

=∫M

λ (tr (Op(σ(A), σ(B)?)) dx

= limN→∞

∫M

λ(

tr(

Op(σ(A), σ(B)(N)?

))dx

= limN→∞

∑|α|≤N−1

(−i)|α|

α!

∫M

λ(tr(∂αξ σ(A)∂αx σ(B)− ∂αξ σ(B)∂αx σ(A)

))dx

= limN→∞

∑|α|≤N−1

(−i)|α|

α!

∫M

λ(tr(∂αξ σ(A)∂αx σ(B)− ∂αx σ(B)∂αξ σ(A)

))dx

= limN→∞

∑|α|≤N−1

(−i)|α|

α!

∫M

λ(tr(∂αξ σ(A)∂αx σ(B)− ∂αξ σ(A) ∂αx σ(B)

))dx

= 0,

where, in the identity before last, we used the cyclicity of the ordinary trace tr onmatrices. tu

Example 12 The higher order residues resk, k ∈ Z+ define a graded trace on

GrC`∗,∗(M,E) =∞∑k=0

GrkC`∗,∗(M,E),

where Grk C`∗,∗(M,E) = C`∗,k(M,E)/C`∗,k−1(M,E). since [L1]

A ∈ C`∗,k(M,E), B ∈ C`∗,l(M,E) =⇒ resk+l ([A,B]) = 0,

resp. a trace on C`(M,E) since for k = l = 0 this implies

A ∈ C`(M,E), B ∈ C`(M,E) =⇒ res ([A,B]) = 0,

as a consequence of Proposition 28 applied to λ = resk on S = CS∗,kcc (Rd), resp.λ = res on S = CScc(Rd).

98

tu

Example 13 The canonical trace TR on C`/∈Z,∗(M,E) deserves this name since itvanishes on brackets of operators in C`/∈Z,∗(M,E)

[A,B] ∈ C`/∈Z,∗(M,E) =⇒ TR ([A,B]) = 0 ∀A,B ∈ C`/∈Z,∗(M,E),

resp. in C`/∈Z(M,E)

[A,B] ∈ C`/∈Z(M,E) =⇒ TR [A,B]) = 0 ∀A,B ∈ C`/∈Z(M,E),

as a consequence of Proposition 28 applied to λ = −∫

Rd on S = CS /∈Z,∗cc (Rd), resp.

S = CS /∈Zcc (Rd).

5.5 First characterisation of the noncommutative residue

We characterise the noncommutative residue as the unique ( modulo a multiplicativeconstant) linear form on the algebra CScc(Rd) which vanishes both on smoothingsymbols and on partial derivatives (Stokes’ property). As will become clear laterin these notes, the first assumption, called “singularity” of the linear form, is notnecessary to characterise the residue. At this stage, it is nevertheless convenient towork with.

Definition 35 If E is a finite rank vector bundle over M modelled on a linear spaceV , we call admissible a subset D(M,E) of C`(M,E) defined in terms of a symbol setS ⊗End(V ) with S ⊂ CScc(Rd) as in (126), whenever the symbol set S is admissible.

Exercise 77 Show that the algebra C`−∞(M,E) of smoothing operators and the setC`/∈Z(M,E) of non-integer order classical pseudodifferential operators on M are ad-missible subsets of C`(M,E).

Exercise 78 Show that the stability of S under partial derivations implies that ofD(M,E) under brackets [xi, ·] in local coordinate charts.Hint: Use Exercise 71.

Recall from Exercise 75 that the noncommutative residue is a continuous singulartrace on the algebra C`(M,E) and from Exercise 76 that the canonical trace is acontinuous linear form on C`/∈Z(M,E) which vanishes on non-integer order brackets.The following theorem characterises continuous singular traces on admissible sets.

Theorem 16 Let D(M,E) be a closed27 admissible subset of C`(M,E) which con-tains smoothing operators.Continuous singular traces on D(M,E) are proportional to the noncommutative residue,meaning by this that for any linear form Λ on D(M,E) which vanishes on C`−∞(M,E)we have

∀A,B ∈ C`(M,E), Λ ([A,B]) = 0 if [A,B] ∈ D(M,E)=⇒ ∃C ∈ C, Λ(A) = C · res(A) ∀A ∈ D(M,E).

27i.e. for every real number a, the intersection Da(M,E) := D(M,E) ∩C`a(M,E) is closed underthe Frechet topology of operators of constant order.

99

Proof: Let us assume that the subset D(M,E) is defined in terms of a symbol setS ⊗ End(V ) with S ⊂ CScc(Rd) as in (126). For any real number a we set Sa :=S ∩ CSacc(Rd) which is closed in CSacc(Rd) for the Frechet topology of symbols ofconstant order as a result of the closedness of Da(M,E) in C`a(M,E) for the Frechettopology of operators with constant order.Let (U, φ) be a local trivialising chart and Φ : E|U → φ(U) × V a trivialisation overU .

1. With the notation of Definition 30, for any real number a, we introduce thebilinear form

λΦ : C∞cpt(φ(U))⊗ (Sa ⊗ End(V )) → C

(f, τ) 7→ Λ(

Φ∗Op](f ⊗ τ)).

Since Λ is singular, this expression is well-defined independently of the choice ofthe properly supported operator Op](f ⊗ τ) which differs from Op(f ⊗ τ) by asmoothing operator. We can therefore drop the symbol ] and write Λ (Φ∗Op(f ⊗ τ)).The linear form Λ is continuous as a result of the continuity of Λ on Da(M,E)and therefore lies in the dual

(C∞cpt(φ(U))⊗ (Sa ⊗ End(V ))

)∗.2. Let E, resp. F , be two Frechet spaces whose topology is given by a countable

family of semi-norms ‖ · ‖En , n ∈ N, resp.‖ · ‖Fn , n ∈ N. Their topological dualsE∗, resp. F ∗, equipped with the countable family of semi-norms ‖L‖E∗n :=supx∈E,|u|n≤1‖L(u)‖, resp. |L|F∗n with E replaced by F , are also Frechet spaces.They obey the inclusion E∗ ⊗ F ∗ ⊂ (E ⊗ F )∗ since the tensor product of twocontinuous linear forms yields a continuous bilinear form on the tensor product.Equality E∗⊗F ∗ = (E⊗F )∗ holds when replacing the ordinary tensor product⊗ by the completed tensor product ⊗ which equips the tensor product V ⊗W oftwo Frechet spaces with the finest locally convex topology such that ⊗ : V ×W →V ⊗W is continuous. Applying this to the Frechet spaces E = C∞cpt(φ(U)) andF = Sa ⊗ End(V ) for any real number a we deduce that

λΦ ∈(C∞cpt(φ(U))

)∗ ⊗ (Sa ⊗ End(V ))∗ . (137)

It is therefore sufficient to determine λΦ as a linear form in each of the variables,with the other one fixed.

3. λΦ satisfies Stokes’ property in both variables y ∈ φ(U) and ξ ∈ Rd. Indeed, forany function f in C∞cpt(φ(U)) and any symbol τ in S ⊗ End(V ) we have

λΦ(∂xi(f ⊗ τ)) = Λ(Φ∗Op(∂xj (f ⊗ τ))

)= −iΛ ([ξj , φ∗Op(f ⊗ τ)]) = 0

and provided ∂jτ lies in S we have:

λΦ(∂ξj (f ⊗ τ)) = Λ(Φ∗Op(∂ξj (f ⊗ τ))

)= −iΛ ([xj ,Φ∗Op(f ⊗ τ)]) = 0

since Λ vanishes on brackets.Moreover, since Λ is tracial, λΦ vanishes on tensor products f ⊗ τ ⊗ [u, v] forany f ∈ C∞cpt(φ(U)), τ ∈ CScc(Rd) and any endomorphisms u and v of V .

4. In view of Stokes’ property in the variable ξ, for a fixed function f in C∞cpt(U),by Exercise 33 we know that τ 7→ λΦ(f ⊗ τ) is proportional to τ 7→ f · res tr(τ).Similarly, in view of Stokes’ property in the variable x, for fixed τ in S⊗End(V ),it is known that the map f 7→ λΦ(f ⊗ τ) is proportional to f 7→

(∫φ(U)

f)· τ .

100

5. Combining these facts yields the existence of a constant CΦ such that for any fin C∞cpt(φ(U)) and any τ in S ⊗ End(V ) we have

Λ (Φ∗Op(f ⊗ τ)) = CΦ

(∫φ(U)

f

)res (tr(τ)) .

Since C∞cpt(φ(U))⊗CScc(Rd) is dense in CScpt(φ(U)) for the topology of symbolsof constant order, we deduce that for any symbol σ in CScpt(φ(U)) ⊗ End(V )such that Φ∗Op(σ) lies in D(M,E) we have:

Λ (Φ∗Op(σ)) = CΦ res (tr(σ)) .

6. Recall from (121) that an operator A in C`(M,E) can be written as a finite sum

A =∑

Supp(χi)∩Supp(χj) 6=∅

χiAχj +∑

Supp(χi)∩Supp(χj)=∅

χiAχj

=∑

Supp(χi)∩Supp(χj)6=∅

Φi?Op(σij) +R(A)

where R(A) =∑

Supp(χi)∩Supp(χj)=∅ χiAχj lies in C`−∞(M,E) and the Op(σij)are properly supported pseudodifferential operators in C`(M,E) on φi(Ui∩Uj).It follows from the previous steps in the proof that there are constants Cij suchthat

Λ(Φ?iOp(σij)) = Cij · res (tr (σij))

so that by linearity of Λ and since it vanishes on smoothing operators we have:

Λ(A) =∑

Supp(χi)∩Supp(χj) 6=∅

Cij · res (tr(σij)) + Λ(R(A))

=∑

Supp(χi)∩Supp(χj) 6=∅

Cij · res(Aij).

where we have set Aij := Φ?iOp(σij).

7. The constants Cij are independent of the indices i and j. Indeed, let χ′k, k ∈ Kbe a partition of unity subordinated to a refined atlas U ′k, k ∈ K so that eachopen subset of Ui of the original atlas is a finite union of open subsets U ′ik of thenew atlas and each function χi is a finite sum of χik . Thus,

Aij =∑k,l

χ′kAijχ′l

=∑

supp(χ′k)∩supp(χ′l)6=∅

A′ijkl +R′(Aij)

where we have set A′ijkl = χ′kAijχ′l and R′(Aij) =

∑supp(χ′k)∩supp(χ′l)=∅

χ′kAijχ′l.

On the one hand, using the linearity of Λ combined with the existence of con-stants C ′ijkl such that Λ

(A′ijkl

)= C ′ijklres(A′ijkl) we have

Λ(Aij) =∑

supp(χ′k)∩supp(χ′l)6=∅

Λ(A′ijkl

)=

∑supp(χ′k)∩supp(χ′l)6=∅

C ′ijklres(A′ijkl

).

101

On the other hand, by linearity of res

Λ(Aij) = Cijres (Aij)

= Cij∑

supp(χ′k)∩supp(χ′l)6=∅

res(A′ijkl

)=

∑supp(χ′k)∩supp(χ′l)6=∅

Cij res(A′ijkl

)

since the residue vanishes on smoothing operators. Comparing the two epxres-sions of Λ(Aij) for some Aij = A′ijkl compactly supported in U ′k ∩U ′l shows thatC ′ijkl = Cij is independent of the chosen refinement and the Cij independent ofthe choice of atlas and subordinated partition. Thus,

Λ(A) = C∑

Supp(χi)∩Supp(χj)6=∅

res(Aij) = C res(A)

for some constant C. tu

Remark 26 Uniqueness of traces on sets D(M,E) actually amounts to uniquenessof traces on D(M) via the following classical argument (see e.g. [RLL] p.41). By aresult of [Sw] , the vector bundle E over M is a direct summand in a trivial vectorbundle i.e., there is a vector bundle F over M and a large enough integer N such thatE⊕F ∼M×CN . Thus, for each point x in M , Ex⊕Fx ' CN . Let e ∈ glN (C∞(M))be the projection map defined pointwise by e(x) : CN → Ex. Then E = Ee, where

Ee := (x, v) ∈M × CN , v ∈ e(x)(CN ). (138)

With these identifications the distribution kernel K(x, y) of an operator A in C`(M,E)lies in glN (D′(M×M)), where D′(M×M) stands for the set of distributions on M×M .On the one hand, the embedding f(x) 7→ f(x) Ix where Ix is the identity matrix in thefibre Ex = e(x)

(CN)

gives rise to the map:

ι : C`(M) → C`(M,E)(f 7→

∫M×M

K(x, y)f(y)dy)7→

(u 7→

∫M×M

K(x, y)Ix ⊗ Iyu(y)dy).

Note that ι(AB) = ι(A) ι(B) since they both have kernel(∫MKA(x, z)KB(z, y) dz

)Ix⊗

Iy. Let Λ be a linear form on D(M,E) which vanishes on commutators; henceΛ ([A,B]) = 0 for any A,B in D(M,E) such that [A,B] also lies in D(M,E). Inparticular,

Λ ι ([A,B]) = Λ ([ι(A), ι(B)]) = 0

for any A,B in D(M), provided the bracket [A,B] also lies in D(M). Thus, Λ ιvanishes on commutators in D(M); since ι transforms a smooth kernel into a smoothkernel, Λι vanishes on smoothing operators. Assuming we already know that the onlylinear forms on D(M), which vanish both on commutators and smoothing operatorsare proportional to the noncomutative residue, it follows that

∃c ∈ C, Λ ι = c res.

102

On the other hand, the matrix trace tr induces a map

tr : C`(M,E) → C`(M)(u 7→

∫M×M

K(x, y)u(y)dy)7→

(f 7→

∫M×M

tr (K(x, y)) f(y)dy).

Since the noncommutative residue res vanishes on commutators,the linear form restralso vanishes on commutators by the trace property of the matrix trace. It follows that

Λ ι tr = c res tr. (139)

SinceKer (ι tr) = Ker (tr) = [glN (C`(M)), glN (C`(M))] ,

for any operator A ∈ C`(M,E) the difference A−N−1 ιtr(A), which lies in Ker (ι tr)also lies in [glN (C`(M)) , glN (C`(M))] . Consequently, Λ vanishes on the differenceA−N−1 ι tr(A). By (139), it follows that for any operator A in C`(M,E) we have

Λ(A) = N−1 Λ (ι tr(A)) = c′ res tr(A) ∀A ∈ D(M,E),

where we have set c′ = cN .

103

6 The noncommutative residue on operators as acomplex residue and zeta regularisation

The L2-trace on trace-class pseudodifferential operators extends to linear forms calledweighted traces and defined on the whole algebra of classical (or log-polyhomogeneous)operators by means of a zeta regularisation (see Theorem 19). This is an instance ofmore general holomorphic regularisation procedures which amount to embedding anoperator A in a holomorphic family A(z) and picking the constant term of a mero-morphic extension TR(A(z)) built by means of the canonical trace TR (see Theorems17 and 18). Weighted traces obtained in this way using zeta regularisation presentdiscrepancies (see Proposition 34) which can be expressed in terms of the noncommu-tative residue. Although weighted traces are expected to be non-local in general, theweighted trace of a differential operator is local insofar as it can be expressed in termsof a noncommutative residue. In particular, the weighted trace (investigated in The-orem 20) and supertrace (investigated in Theorem 21) of the identity map are bothlocal quantities. We end the chapter with a characterisation (see Theorem 22) of thecanonical trace and noncommutative residue which confirms known results relative tothe uniqueness of these concepts28 but our approach to which is new to our knowledgeand uses weighted traces.

6.1 Complex powers

We review the construction and properties of complex powers of elliptic operators.

An operator A ∈ C`(M,E) has principal angle θ if for every (x, ξ) ∈ T ∗M−0,the leading symbol σL(A)(x, ξ) has no eigenvalue on the ray Lθ = reiθ, r ≥ 0; inthat case A is elliptic.

Definition 36 We call an operator A ∈ C`(M,E) admissible with spectral cut θ ifA has principal angle θ and the spectrum of A does not meet Lθ = reiθ, r ≥ 0. Inparticular such an operator is invertible and elliptic. Since the spectrum of A does notmeet Lθ, θ is called an Agmon angle of A.

Remark 27 In applications, an invertible operator A is usually obtained from anessentially self-adjoint elliptic operator B ∈ C`(M,E) by setting A = B+πB using theorthogonal projection πB onto the kernel ker(B) of B corresponding to the orthogonalsplitting L2(M,E) = ker(B) ⊕ R(B) where R(B) is the (closed) range of B. HereL2(M,E) denotes the closure of C∞(M,E) with respect to a Hermitian structure onE combined with a Riemannian structure on M .

Let A ∈ C`(M,E) be admissible with spectral cut θ and positive order a. Outside asmall disk around zero (see e.g. Theorem 9.3 in [Sh]), for any real number s there isa constant Cs such that

‖(A− λI)−1‖s ≤ Cs |λ|−1

where ‖·‖s is the operator norm on the Sobolev closure Hs(M,E) of C∞(M,E). Thus,for negative Re(z), the complex power Azθ of A is defined by the Cauchy integral [Se]

Azθ =i

2π

∫Γr,θ

λzθ (A− λ)−1 dλ (140)

28concerning the uniqueness of the noncommutative residue see [Wo1, Wo2], [BrGe], [L2] and[FGLS] in the boundary case and concerning the uniqueness of the canonical trace [MSS], [L2].

104

where λzθ = |λ|zeiz(argλ) with θ ≤ argλ < θ + 2π.Here

Γr,θ = Γ1r,θ ∪ Γ2

r,θ ∪ Γ3r,θ (141)

whereΓ1r,θ = ρ eiθ,∞ > ρ ≥ r

Γ2r,θ = ρ ei(θ−2π),∞ > ρ ≥ r

Γ3r,θ = r eit, θ − 2π ≤ t < θ,

is a contour along the ray Lθ around the non-zero spectrum of A. The constant r isany small positive real number such that Γr,θ ∩ Sp(A) = ∅, where Sp(A) stands forthe spectrum of A.For Re(z) smaller than k we set Azθ := Az−kθ Ak. Since A0

θ = A0 = I, this extends theabove definition to the whole complex plane.The operator Azθ is a classical pseudodifferential operator of order az with homoge-neous components of the symbol of Azθ given by

σaz−j(Azθ)(x, ξ) =i

2π

∫Γθ

λzθ b−a−j(x, ξ, λ) dλ

where the components b−a−j are the positive homogeneous components of the resol-vent (A− λI)−1 in (ξ, λ

1a )).

Exercise 79 Check the expected multiplicative property, i.e. Az1θ Az2θ = Az1+z2

θ , andAkθ = Ak, for k ∈ Z.

Exercise 80 Show that the leading symbol of Azθ is given by σL(Azθ) =(σL(A)

)zθ

andconclude that Azθ is elliptic.

Complex powers of operators depend on the choice of spectral cut in the followingmanner, a result we state without proof.

Proposition 29 [Wo1, Wo2, Po1] Let θ and φ be two spectral cuts for an admissibleoperator A in C`(M,E) such that 0 ≤ θ < φ < 2π. The complex powers for these twospectral cuts are related by

Azθ −Azφ =(1− e2iπz

)Πθ,φ(A)Azθ, (142)

where we have set Πθ,φ(A) = A(

12iπ

∫Γθ,φ

λ−1(A− λ)−1 dλ)

where Γθ,φ is a contourinside the cone

Λθ,φ := ρ eit,∞ > ρ ≥ r, θ < t < φ. (143)

Formula (142) generalises to spectral cuts θ and φ such that 0 ≤ θ < φ + 2kπ <(2k + 1)π for some non-negative integer k by

Azθ −Azφ = e2ik πz I +(1− e2iπz

)Πθ,φ(A)Azθ. (144)

If the cone Λθ,φ defined by (143) delimited by the angles θ and φ does not intersect thespectrum of the leading symbol of A, it only contains a finite number of eigenvaluesof A and Πθ,φ(A) is a finite rank projection and hence a smoothing operator. Ingeneral (see Propositions 3.1 and 3.2 in [Po1]), Πθ,φ(A), which is a pseudodifferential

105

projection, is a zero order operator with leading symbol given by πθ,φ(σL(A)) definedsimilarly to Πθ,φ replacing A by the leading symbol σLA of A so that:

σLΠθ,φ(A) = πθ,φ(σL(A)) := σL(A)

(1

2iπ

∫Γθ,φ

λ−1(σL(A)− λ)−1 dλ

).

for any (x, ξ) ∈ T ∗M and any B ∈ Cl(M,E).

6.2 A fundamental formula

Given an operator A in C`∗,∗(M,E) and an admissible elliptic operator Q in C`(M,E)of positive order and spectral cut θ, the family z 7→ AQ−zθ yields a holomorphic familyof log-polyhomogeneous pseudodifferential operators in the following sense.

Definition 37 Let (A(z))z∈Ω be a family of pseudodifferential operators in C`∗,∗(M,E)with distribution kernels (x, y) 7→ KA(z)(x, y). The family is holomorphic if

1. the order α(z) of A(z) is holomorphic in z,

2. in any local trivialisation of E, we can write A(z) in the form A(z) = Op(σ(z))+R(z), for some holomorphic family of End(V )-valued symbols (σ(z))z∈Ω whereV is the model space of the fibres of E, and some holomorphic family (R(z))z∈Ω

of smoothing operators i.e. given by a holomorphic family of smooth Schwartzkernels,

3. the (smooth) restrictions of the distribution kernels KA(z) to the complementof the diagonal ∆ ⊂ M × M , form a holomorphic family with respect to thetopology given by the uniform convergence in all derivatives on compact subsetsof M ×M −∆.

A holomorphic family of log-polyhomogeneous operators of holomorphic order α(z)parametrised by Ω has order with real part smaller than −d on the set

Ω ∩ α−1 (Z∩]−∞,−d[) .

On that set, the canonical trace TR (A(z)) defines a holomorphic map in the complexvariable z.

Theorem 17 [KV][L1] Let, for any non-positive integer k, z 7→ A(z) ∈ C`α(z),k(M,E)be a holomorphic family of log-polyhomogeneous symbols of order α(z) on a domainΩ ⊂ C with α′(0) 6= 0.The map z 7→ TR (A(z)) is meromorphic with poles at points zj in the set

Ω ∩ α−1 ([−d,+∞[∩Z)

of order no larger than k + 1.For any A ∈ C`∗,k(M,E) the following expression defined in terms of the complexresidue of order at most k + 1 at 0:

resk(A) := (−1)k+1 (α′(0))k+1 Resk+10 TR (A(z)) , (145)

only depends on the holomorphic family A(z) such that A(0) = A via its order throughα′(0).

106

In particular, when k = 0, i.e. for A ∈ C`(M,E) the pole is of order 1 and its residueis given by

Res0TR (A(z)) = − 1α′(0)

res(A). (146)

Proof: Applying Theorem 10 relative to the residue of cut-off regularised integralsof holomorphic families of symbols to the holomorphic family of symbols tr (σ(A(z))),yields the meromorphy of the map z 7→ −

∫Rd tr (σ(A)) (z)(x, ξ) dξ with poles of order at

most k + 1. Moreover, at a pole z0 we have

Resk+1z0 −

∫Rd

tr (σ (A(z))) (x, ξ) dξ =(−1)k+1

(α′(z0))k+1resx,k (σ (A(z0)(x, ·))) . (147)

Outside the poles, the operators have non-integer order so that the expression

−∫

Rdtr (σ (A(z))) (x, ξ) dξ

integrates over M to TR(A(z)). Since z 7→ −∫

Rd tr (σ (A(z))) (x, ξ) dξ defines a mero-morphic map with poles of order at most k + 1 so does z 7→ TR(A(z)). Integrating(147) over M leads to (145). tu

6.3 Zeta regularised traces

The definition of holomorphic regularisation scheme for symbols (see Definition 13)generalises to operators.

Definition 38 A holomorphic regularisation scheme on C`(M,E), resp. onC`∗,∗(M,E) is a linear map

R : A 7→ (z 7→ A(z))

which sends an operator A in C`(M,E), resp. in C`∗,∗(M,E), to a holomorphicfamily of operators A(z) in C`(M,E), resp. in C`∗,∗(M,E), parametrised by z ∈ Cof non-constant affine order α(z) = α′(0) z + α(0) such that A(0) = A.

By Theorem 17, given a holomorphic regularisation R, the map TR (A(z)) is mero-morphic with poles of order at most k + 1 for A in C`∗,k(M,E).

Example 14 ζ- regularisation

RQ : A 7→ A(z) := AQ−zθ

with Q an admissible operator in C`(M,E) of positive order q and spectral cut θ yieldstypical (and very useful) examples of holomorphic regularisations. The meromorphicmaps given by

ζmer(A,Q)(z) := TR(RQ(A)(z)

)(148)

are called generalised ζ-functions.

We can therefore pick the finite part in the Laurent expansion TR (A(z)) by means ofa regularised evaluator evreg

0 and set the following definition.

107

Definition 39 A holomorphic regularisation scheme Rθ : A 7→ A(z) on a subset Aof C`∗,∗(M,E), induces a linear map:

TrR : A → CA 7→ TrR(A) := evreg

0 (TR (A(z))) ,

which we call the R-regularised trace.

Exercise 81 Show that TrR coincides with the ordinary L2-trace Tr on operatorswhose order has real part smaller than −d

TrR(A) = limz→0

TR (A(z)) = Tr(A).

Show that TrR coincides with the canonical trace TR on operators of non-integer order

TrR(A) = limz→0

TR (A(z)) = Tr(A).

Hint: Use the results of Exercise 44.

Regularised traces associated with a ζ-regularisation scheme correspond to weightedtraces which one comes across in the physics and mathematics literature in variousguises29.

Definition 40 Let Q be an admissible operator in C`∗,∗(M,E) of positive order andspectral cut θ. The Q-weighted trace of an operator A in C`∗,∗(M,E) is defined by:

TrQθ (A) := evreg0

(TR

(RQθ (A)(z)

))= evreg

0 (ζmerθ (A,Q)(z)) , (149)

with the notation of (148).

These correspond to generalised zeta values:

ζθ(A,Q)(0) := TrQθ (A), ζθ(Q)(0) := TrQθ (I).

Remark 28 Whenever the spectral cut is irrelevant we shall drop it from the notation,writing ζθ(A,Q)(0) and ζθ(Q)(0).

When A has order with real part smaller than −d, then TrQ(A) = Tr(A) so that theweighted trace coincides with the L2-trace. Weighted traces therefore provide linearextensions of the L2-trace.

Exercise 82 With these notations show that TrQt

(A) = TrQ(A) for any positive num-ber t.

6.4 Logarithms of admissible operators

Given a Banach (unital) algebra A, and a in A, let θ ∈ R be such that the spectrumof a, i.e. the set of complex scalars λ such that a− λ1 is not invertible in A, does notmeet the ray Rθ = reiθ, r ≥ 0. The map (see e.g. [Hil])

z 7→ azθ :=i

2π

∫Γθ

λzθ (a− λI)−z dλ

29see e.g. [N], [Q] where they are concealed behind logarithmic variations of determinants, see e.g.[BF], [Fr] where they are used to define regularised first Chern classes.

108

defines a holomorphic function on the complex plane with values in A where we haveset λzθ = |λ|s eizArg(λ) for θ ≤ Argλ < θ+ 2π. Here Γθ is any bounded contour aroundthe spectrum of a which does not intersect the ray Rθ.Hence, the logarithm of a

logθ a :=(d

dzazθ

)|z=0

=i

2π

(d

dz

∫Γ

λzθ (a− λI)−z)z=0

=i

2π

∫Γ

logθ λ (a− λI)−z dλ

lies in A, where we have set logθ λ = log |λ|+ iArg(λ) for θ ≤ Argλ < θ + 2π.This applies to the Banach algebra A = B(H) of bounded linear operators on a Hilbertspace H equipped with the operator norm ‖A‖ = sup‖x‖=1‖Ax‖ where ‖ · ‖ standsfor the norm on H. Thus, given an admissible operator A in C`(M,E) of order zeroand spectral cut θ, its complex powers give rise to a holomorphic map z 7→ Azθ on thecomplex plane with values in B(Hs(M,E)) for any real number s, where Hs(M,E)stands for the Hs-closure of the space C∞(M,E) of smooth sections of E (se e.g.[Gi]). The logarithm of A is the bounded operator on Hs(M,E) defined in terms ofthe derivative at z = 0 of this complex power:

logθ A := (∂zAzθ)|z=0

=i

2π

(∂z

∫Γr,θ

λzθ (A− λI)−z dλ

)|z=0

=i

2π

∫Γr,θ

logθ λ (A− λI)−z dλ

with the notations of (140).The notion of logarithm extends to an admissible operator A of positive order a andspectral cut θ in the following way. For any positive ε the map z 7→ Az−εθ of ordera(z − ε) defines a holomorphic function on the half plane Re(z) < ε with values inB (Hs(M,E)) for any real number s. Thus we can set

logθ A = Aεθ (∂z (Azθ − ε))|z=0= Aεθ

(∂z

(i

2π

∫Γr,θ

λz−εθ (A− λ)−1 dλ

))|z=0

. (150)

for any positive ε the operator logθ AA−ε = A−ε logθ A lies in B (Hs(M,E)) for anyreal number s. It follows that logθ A defines a bounded linear operator from Hs(M,E)to Hs−ε(M,E) for any positive ε.

Exercise 83 Show that this definition of logθ A is independent of the choice of ε > 0.

Just as complex powers do, the logarithm depends on the choice of spectral cut [O1].However, in order to simplify notations we shall often drop the explicit mention of thespectral cut.The following proposition shows how the symbol of the logarithm of an operator differsfrom a classical one.

Proposition 30 Let A ∈ C`(M,E) be an admissible operator with spectral cut θ. Ina local trivialisation, the symbol of logθ A reads:

σ(logθ A)(x, ξ) = a log |ξ|I + σ0(logθ A)(x, ξ) (151)

109

where a denotes the order of A and σ0(logA) a symbol of order zero.Moreover, the leading symbol of σA0 is given by

σL0 (logθ A)(x, ξ) = logθ

(σL(A)

(x,

ξ

|ξ|

))∀(x, ξ) ∈ T ∗M − 0. (152)

Proof: Given a local trivialisation over some local chart, the symbol of Azθ has theformal expansion σ(Azθ) ∼

∑j≥0 b

(z)az−j where a is the order of A and b

(z)az−j is a pos-

itively homogeneous function of degree az − j. Since logθ A = A(∂zA

z−1θ

)|z=0

(herewe take ε = 1), by (116) the formal expansion of the symbol of logθ A is

σ(logθ A) ∼ σ(A) ? σ((∂zA

z−1θ

)|z=0

).

Suppose that ξ 6= 0; using the positive homogeneity of the components, we haveb(z−1)az−a−j(x, ξ) = |ξ|az−a−jb(z−1)

az−a−j

(x, ξ|ξ|

)and hence

∂zb(z−1)az−a−j(x, ξ) = a log |ξ| |ξ|az−a−jb(z−1)

az−a−j

(x,

ξ

|ξ|

)+ |ξ|az−a−j∂zb(z−1)

az−a−j

(x,

ξ

|ξ|

).

It follows that(∂zb

(z−1)az−a−j(x, ξ)

)|z=0

= a log |ξ| b(−1)−a−j(x, ξ) + |ξ|−a−j

(∂zb

(z−1)az−a−j

(x,

ξ

|ξ|

))|z=0

.

Hence(∂zA

z−1θ

)|z=0

has symbol(∂zb

(z−1)(x, ξ))|z=0

of the form a log |ξ|σ(A−1)(ξ) +τ(A)(x, ξ) with τ(A) a classical symbol of order −a whose homogeneous componentof degree −a− j reads:

τ−a−j(A)(x, ξ) = |ξ|−a−j(∂zb

(z−1)az−a−j

(x,

ξ

|ξ|

))|z=0

.

Thus, logθ A = A(∂zA

z−1θ

)|z=0

has a symbol of the form a log |ξ| + σ0(logθ A)(x, ξ)where

σ0(logθ A)(x, ξ) ∼∞∑k=0

∑i+j+|α|=k

(−i)|α|

α!∂αξ σa−i(A) ∂αx τ−a−j(A)(x, ξ)

is a classical symbol of order zero. Its leading symbol reads

σL0 (logθ A)(x, ξ) = σL(A)(x, ξ) |ξ|−a(∂zb

(z−1)az−a (x,

ξ

|ξ|))|z=0

= σL(A)(x, ξ)

(∂z

(σL(A)

(x,

ξ

|ξ|

))z−1

θ

)|z=0

= logθ σL(A)

(x,

ξ

|ξ|

)for any (x, ξ) in T ∗M − 0. tuUsing logarithmic operators, one can build various useful classical ones, as these ex-ercises are meant to show.

110

Exercise 84 Show that for two admissible operators Q1 and Q2 of the same ordersthe difference logQ1− logQ2 is classical. Show that if Q1 and Q2 have positive ordersq1 and q2, then the weighted difference logQ1

q1− logQ2

q2is classical.

Hint: Use the fact that the symbol of the difference only involves σ0(logQ1) andσ0(logQ2).

Exercise 85 Show that the operator bracket [A, logQ] is classical for any admissibleoperator Q whenever A is classical.Hint: Use the fact that the star bracket [σ(B), σ0(logQ)]? of the symbol σ(B) of Band the classical part σ0(logQ) of the symbol of logQ is classical.

Given three admissible operators A,B and AB in C`(M,E) we set

L(A,B) := log(AB)− logA− logB (153)

Remark 29 Strictly speaking, one should specify the spectral cuts θ of A, φ of B andψ of AB in the expression L(A,B) setting instead

Lθ,φ,ψ(A,B) := logψ(AB)− logθ A− logφB.

Up to a modification of the operators A and B, one can actually choose fixed spectralcuts θ and φ by the following argument of Okikiolu [O1]:

Lθ,φ,ψ(A,B) = Lπ,π,ψ−(θ+φ)(ei(π−θ)A, ei(π−φ)B).

Keeping this observation in mind, in order to simplify notation we assume that A andB have spectral cuts π and drop the explicit mention of the spectral cuts.

Exercise 86 Show that L(A,B) is classical.Hint: Express the symbol of the operator L(A,B) in terms of the classical partsσ0(logA), σ0(logB) and σ0(logAB) of the symbols of logA, logB, and log(AB).

In [O1] Okikiolu proves a Campbell-Hausdorff formula for pseudodifferential operatorswith scalar leading symbol which provides an expression of L(A,B) as an infinite sumof iterated brackets with decreasing order.In [Sc2], Scott proves that L(A,B) has vanishing noncommutative residue (see Exercise94) which amounts to the multiplicativity of the residue determinant that we shallintroduce in (170).

6.5 Weighted traces of differentiable families

This rather technical paragraph comprises various useful properties of traces of differ-entiable families of operators.Let I be a non-empty open interval in R. We call differentiable a family At ∈C`(M,E), t ∈ I that satisfies the same requirements as a holomorphic family seeDefinition 37, with holomorphic replaced by differentiable.

Proposition 31 Let I be a non empty open interval in R and At, t ∈ I be a differen-tiable family of C`(M,E) of constant order a.

1. The noncommutative residue commutes with differentiation

d

dtres(At) = res(At), (154)

where we have set At = ddtAt.

111

2. If the order a is non-integer, the canonical trace commutes with differentiation

d

dtTR(At) = TR(At). (155)

3. For any weight Q with order q and spectral cut α,

d

dtTrQα (At) = TrQα (At). (156)

Proof: Using (18) we write the symbol σ(At) of At as follows:

σ(At)(x, ξ) =N−1∑j=0

χ(ξ)σa−j(At)(x, ξ) + σ(N)(At)(x, ξ).

1. By assumption, the map t 7→ tr (σ−d(At)(x, ·)) is differentiable, leading to a dif-ferentiable map t 7→

∫S∗xM

tr (σ−d(At)(x, ·)) after integration over the compact

set S∗xM with derivative t 7→∫S∗xM

tr(σ−d(At)(x, ·)

), where σj(At) = σj(At)

stands for the derivative of σj(At) at t. Thus, the map t 7→ res(At) is differen-tiable with derivative given by (154).

2. By (133) and (134) to prove formula (155) we need to check the differentiabilityof the map t 7→ −

∫T∗xM

tr (σ(At)(x, ·)) and to show that

d

dt−∫T∗xM

tr (σ(At)(x, ·)) = −∫T∗xM

tr(σ(At)(x, ·)

).

The cut-off integral involves the whole symbol which we denote by σt := σ(At)in order to simplify notation. Since the family σt has constant order the in-teger N can be chosen independently of t in the asymptotic expansion. Thecorresponding cut-off integral can be computed explicitly (see (39)):

−∫T∗xM

tr(σ(At)(x, ξ)) dξ =∫T∗xM

tr(σ(N)(At)(x, ξ)

)dξ

+N−1∑j=0

∫|ξ|≤1

χ(ξ) tr (σa−j(At)(x, ξ)) dξ

−N−1∑

j=0,a−j+n 6=0

1a− j + n

∫|ξ|=1

tr (σa−j(At)(x, ω)) dSω.

The map t 7→∫T∗xM

tr(σ(N)(At)(x, ξ)

)dξ is differentiable at any point t0 since

by assumption the maps t 7→ tr(σ(N)(At)(x, ξ)

)are differentiable with modulus

bounded from above ∣∣∣tr((σt)(N) (x, ξ))∣∣∣ ≤ C|ξ|Re(a)−N

by an L1 function provided N is chosen large enough and with the constant Cchosen independently of t in a compact neighborhood of t0. Its derivative isgiven by t 7→

∫T∗xM

tr(σ(N)(At)(x, ξ)

)dξ. The remaining integrals∫

|ξ|≤1

χ(ξ) tr (σa−j(At)(x, ξ)) dξ and∫|ξ|=1

tr (σa−j(At)(x, ω)) dSω

112

are also differentiable as integrals over compact sets of integrands involving dif-ferentiable maps t 7→ tr

((σt)a−j

). Their derivatives are given by∫

|ξ|≤1

χ(ξ) tr(σa−j(At)(x, ξ)

)dξ and

∫|ξ|=1

tr((

σ(At))a−j

(x, ω))dSω.

Thus, t 7→ −∫T∗xM

trx(σ(At)(x, ξ)) dξ is differentiable with derivative given by−∫T∗xM

tr(σAt(x, ξ)) dξ.

3. By the defect formula (168) we have

TrQα (At)

=∫M

dx

(−∫T∗xM

tr (σ(At)(x, ξ)) dξ −1q

∫S∗xM

tr (σ−d(At logαQ)(x, ξ)) dξ

),

which reduces the proof of the differentiability of t 7→ TrQα (At) to that of thetwo maps t 7→ −

∫T∗xM

tr (σ(At)(x, ·)) and t 7→∫S∗xM

tr (σ−d(At logαQ)(x, ξ)).Differentiability of the first map was shown in the second item of the proof. Letus first investigate the second map. By (113) we have

σ−d(At logαQ) =∑

|α|+a−j−k=−d

(−i)|α|

α!∂αξ σa−j(At) ∂

αx σ−k(logαQ).

By assumption, the maps t 7→ σa−j(At) are differentiable so that the map t 7→∫S∗xM

tr (σ−d(At logαQ)) (x, ξ) dSξ is differentiable with derivative

t 7→∫S∗xM

tr(σ−d(At logαQ)

)(x, ξ) dSξ.

Integrating over the compact manifold M then yields that the map t 7→ TrQα (At)is differentiable with derivative given by

TrQα (At)

=∫M

dx

(−∫T∗xM

tr(σAt(x, ξ)

)dξ − 1

q

∫S∗xM

tr(σ(At logαQ)(x, ·)

)dSξ

).

tuWe also need a technical result derived in [OP] for which we use the following easypreparatory exercises.

Exercise 87 Check that for any A and B in C`(M,E) and any complex number λin the resolvent sets of A and B (i.e. such that (λ−A)−1 and (λ−B)−1 is a boundedoperator), we have

(λ−A)−1 − (λ−B)−1 = −(λ−B)−1 (A−B) (λ−A)−1,

where we have set adB(A) := [B,A].

113

Exercise 88 Show that for any A and B in C`(M,E) and any complex number λ inthe resolvent set of A (i.e. such that (λ−A)−1 is a bounded operator) we have

[(λ−A)−1, B

]=

K∑k=1

adkA(B)(λ−A)−(k+1) +[(λ−A)−1, adK−1

A (B)]

(λ−A)−K .

(157)

Proposition 32 Let At, t ∈ I be a differentiable family of admissible operators inC`(M,E) with constant spectral cut θ.Their logarithms form a differential family of pseudodifferential operators, which isclassical of order zero if At has constant order. For any positive integer K we have

d

dtlogθ At =

K∑k=0

(−1)k

k + 1adkAt(At)A

−(k+1)t +RK(At, At) (158)

where we have set At := ddtAt and

RK(At, At) := − i

2π

∫Γθ

logθ λ[(λ−At)−1, adKAt(At)

](λ−At0)−K−1 dλ, (159)

since At commutes with (λ−At)−k.Here Γθ is a contour around the spectrum as in (141).

Remark 30 If At commutes with At then ddt logαAt = AtA

−1t .

If At has scalar leading symbol and order at, then for fixed t, the operators

adkAt(At)A−(k+1)t

have decreasing order ord(At)− at − k as k grows so that (158) yields an asymptoticexpansion in operators of decreasing order.

Proof: Since the spectral cut is constant we drop it in the notation, setting logAt =logθ At.By (4.3) in [O1] we first observe that for any t in a compact neighborhood Kt0 of t0,one can bound the modulus of the order α(t) from above by some integer k, in whichcase

∃C > 0, ∀t ∈ Kt0 , ‖(At − λ)−1‖s,s−k ≤ |λ−1|,where ‖ · ‖s,s′ stands for the operator norm of bounded operators form the Sobolevclosure Hs(M,E) to the Sobolev closure Hs′(M,E) of C∞(M,E). Moreover, thefamily (At − λ)−1 is differentiable at t0 with derivative given by:

d

dt

∣∣∣∣t=t0

(At − λ)−1 = −(At0 − λ)−1 At0 (At0 − λ)−1

as a consequence of (87) applied to A = At and B = At0 . where we have set ∆t :=At−At0t−t0 .

For operators At of zero order this leads to

d

dt

∣∣∣∣t=t0

logAt =i

2π

∫Γθ

logθ λd

dt

∣∣∣∣t=t0

(At − λ)−1 dλ

= − i

2π

∫Γθ

logθ λ (At0 − λ)−1 At0 (At0 − λ)−1 dλ,

114

where we have set At0 = ddt

∣∣∣∣t=t0

At.

In order to generalise this to higher order operators, we need to consider the family(see (150)):

logAtA−1t =

i

2π

∫Γθ

logθ λλ−1 (At − λ)−1 dλ

for which we can also write:

d

dt

∣∣∣∣t=t0

(logAtA−1

t

)=

i

2π

∫Γθ

logθ λλ−1 d

dt

∣∣∣∣t=t0

(At − λ)−1 dλ

= − i

2π

∫Γθ

logθ λλ−1 (At0 − λ)−1 At0 (At0 − λ)−1 dλ.

This leads to the expected formula

d

dt

∣∣∣∣t=t0

logAt =d

dt

∣∣∣∣t=t0

(logAtA−1

t

)At +

(logAtA−1

t

)At

= − i

2π

∫Γθ

logθ λλ−1 (At0 − λ)−1 At0 (At0 − λ)−1 At0 dλ

+i

2π

∫Γθ

logθ λλ−1 (λ−At0)−1 dλ

= − i

2π

∫Γθ

logθ λλ−1 (At0 − λ)−1 At0 (At − λ)−1 (At0 + (At0 − λ)) dλ

= − i

2π

∫Γθ

logθ λ (At0 − λ)−1 At0 (At0 − λ)−1 dλ.

Using (157) applied to A = At0 and B = At0 yields

d

dt

∣∣∣∣t=t0

logAt = − i

2π

∫Γθ

logθ λ [(At0 − λ)−1 , At0 ] (At0 − λ)−1 dλ

− i

2π

∫γθ

logθ λ At0 (At0 − λ)−2dλ

=K∑k=0

adkAt0 (At0)i

2π

∫Γθ

logθ λ (λ−At0)−(k+2)

+i

2π

∫Γθ

logθ λ[(λ−At0)−1, adK−1

At0(At0)

](λ−At0)−K−1 dλ.

Iterated integrations by parts then yield

d

dt

∣∣∣∣t=t0

log(At) =K∑k=0

(−1)k

k + 1adkAt0 (At0)A−(k+1)

t0 +RK(At0 , At0),

where we have set

RK(At0 , At0) = − i

2π

∫Γθ

logθ λ[(λ−At0)−1, adKAt0 (At0)

](λ−At0)−K−1 dλ

which yields (158). tu

115

Exercise 89 Show that if A and B commute, then the operator L(A,B) vanishes.Since L(A,B) is classical by Exercise 86, we can apply Proposition 31 to the familyA− t = L(At, Bµ) for any complex power µ. When combined with Proposition 32, thisleads to the following identities.Hint: Use Proposition 32 applied to the family At = AtB.

Proposition 33 Let A and B be two admissible operators of positive orders a andb in C`(M,E) such that their product AB is also admissible. We have the followingidentities for weighted traces:

d

dt

∣∣∣∣t=0

TrB(L(At, Bµ)) = 0,d

dt

∣∣∣∣t=0

TrA(L(At, Bµ)) = 0 ∀µ ∈ C,

where we have dropped the explicit mention of the spectral cut.

Proof: Let us prove the result for the B-weighted trace; a similar proof yields theresult for the A-weighted trace. Proposition 31 implies that weighted traces and theresidue commute with differentiation on constant order operators so that

d

dt

∣∣∣∣t=0

TrQ(L(At, Bµ)

)= TrQ

(d

dt

∣∣∣∣t=0

L(At, Bµ))

Butd

dt

∣∣∣∣t=0

L(At, Bµ) =d

dt

∣∣∣∣t=0

log(AtBµ)− d

dt

∣∣∣∣t=0

logAt.

We therefore apply Proposition 32 to At := AtBµ so that A0 = Bµ, including thecase µ = 0 for which At = At and A0 = I. Since A0 = logABµ and A0A

−10 = logA,

implementing the weighted trace TrB yields

d

dt

∣∣∣∣t=0

TrB(log(AtBµ)

)= TrB(logA) +

K∑k=1

(−1)k

k + 1TrB(adkBµ(logABµ)B−µ(k+1)) + trB(RK(Bµ, logABµ))

for arbitrary large K, with remainder term

RK(Bµ, logABµ)

= − d

dz

(i

2π

∫Γα

λz[(λ−Bµ)−1, adKBµ(logABµ)

](λ−Bµ)−(K+1) dλ

)|z=0

= −adKBµ

(d

dz

(i

2π

∫Γα

λz[(λ−Bµ)−1, logABµ

](λ−Bµ)−(K+1) dλ

)|z=0

),

since B commutes with Bµ.For any positive integer k, by (161) we have

TrB(adkBµ(ABµ)B−µ(k+1))) = TrB(

adBµ(adk−1Bµ (ABµ))B−µ(k+1))

)= TrB

(adBµ

(adk−1Bµ (ABµ)B−µ(k+1)

))= −1

bres(

adk−1Bµ (ABµ)B−µ(k+1) [Bµ, logB]

)= 0,

116

since B commutes with logB. A similar computation shows that

trB(RK(Bµ, logABµ)) = 0.

Thusd

dt |t=0

TrB(log(AtBµ)

)= TrB (logA) .

It follows that ddt |t=0

TrB (log(AtBµ)) = TrB (logA) independently of µ so that

d

dt |t=0TrB

(L(At, Bµ)

)= 0.

tu

6.6 Discrepancies

Regularised traces present discrepancies, amongst which are their dependence on thechoice of regularisation and their non-cyclicity which conflicts with the work “tracial-ity/cyclicity” used in the nevertheless convenient terminology “regularised traces”.To simplify the presentation we restrict to classical pseudodifferential operators, butthe formulae could be extended to log-polyhomogeneous operators. We also focus onζ-regularisation schemes. The following results can be found in [MN] and [CDMP].

Proposition 34 1. Given an operator A in C`(M,E) and two admissible opera-tors Q1 and Q2 in C`(M,E) of positive orders q1 and q2 and spectral cuts θ1

and θ2,

TrQ1θ1

(A)− TrQ2θ2

(A) = res(A

(logθ1 Q1

q1−

logθ2 Q2

q2

))(160)

which is a local expression.

2. Given two operators A and B in C`(M,E) and an admissible operator Q inC`(M,E) of positive order q and spectral cut θ,

TrQθ ([A,B]) =1q

res (A [B, logθ Q]) . (161)

Proof: For simplicity we drop the explicit mention of the spectral cuts in the nota-tions.

1. Recall from Exercise 84 that the difference logQ2q2− logQ1

q1is classical. On the

other hand, for any admissible operator Q in C`(M,E) of positive order q wehave:

TrQ(A) = evreg0

(TR

(AQ−z

))= evreg

0

(TR

(AQ−

zq

))= TRQ

1q (A).

The result then follows from (146) applied to A(z) =A

„Q− zq1

1 −Q− zq2

2

«z which is a

holomorphic family of operators with holomorphic order α(z) = a−z and which

117

coincides with the classical operator A(

logQ1q1− logQ2

q2

)at z = 0. Indeed, by

the result of Exercise 82 we can write

TrQ1(A)− TrQ2(A) = TrQ1q11 (A)− TrQ

1q22 (A)

= evreg0

(TR

(A(Q− zq1

1 −Q− zq2

2

)))= Res0

(TR

(z−1

(A(Q− zq1

1 −Q− zq2

2

))))= res

(A

(logθ1 Q1

q1−

logθ2 Q2

q2

)).

2. We simplify notation, dropping the explicit mention of spectral cut. Recall fromExercise 85 that the bracket [B, logQ] is classical. The statement then followsfrom (146) applied to the holomorphic family A(z) = [B,Q−z ]A

z of operatorsof holomorphic order a + b − qz which coincides at z = 0 with the operator−[B, logQ]A. The cyclicity of the canonical trace on non-integer order operatorswhich yields TR ([C(z), D(z)]) = 0 as an identity of meromorphic maps forholomorphic families C(z) and D(z) with order c− q z, d− q z and q > 0 leadsto the following identities:

TrQ ([A,B]) = evreg0

(TR

([A,B]Q−z

))= evreg

0

(TR

(ABQ−z

))− evreg

0

(TR

(BAQ−z

))= evreg

0

(TR

(Q−z/2ABQ−z/2

))− evreg

0

(TR

(Q−z/2 BAQ−z/2

))= evreg

0

(TR

(BQ−z A−Q−z BA

))= evreg

0

(TR

([B,Q−z]A

))= Res0

(TR

([B,Q−z]A

z

))=

1q

res ([B, logQ]A)

=1q

res (A [B, logQ]) .

tuThese discrepancies are local for, in the same way as the residue, they only depend ona finite number of positively homogeneous components of the symbols of the operatorsinvolved.

Exercise 90 Show thatTrQ+Rθ (A) = TrQθ (A)

for any smoothing operator R and any admissible operator Q of positive order.Hint: Use the fact that logθ(Q+R) = logθ Q+R′ for some other smoothing operatorR′ and that the residue vanishes on smoothing operators.

6.7 Regularised traces; locality versus non locality

In general, the expression of the finite part evreg0 (TR(A(z))) involves both local and

global terms. The following theorem is a direct consequence of Theorem 10.

118

Theorem 18 [PS]

1. Let us for some non-negative integer k, consider a holomorphic family A(z)in C`α(z),k(M,E) of log-polyhomogeneous operators parametrised by z ∈ Ω, adomain of C of non-constant affine order, α(z) = −qz + α(0)with q 6= 0. Thenwe have

evregz0 (TR (A(z))) (162)

=∫M

(−∫T∗xM

tr (σ(A(z0))) (x, ξ) dξ +k∑l=0

1(l + 1) ql+1

resx,l(A(l+1)(z0)

))dx,

where we have set

resx,l(A) :=∫S∗xM

tr(σ−d,l(A))(x, ξ) dSξ.

2. In particular, for a holomorphic family z 7→ A(z) ∈ C`α(z)(M,E) of classicaloperators, we have

evregz0 (TR (A(z))) (163)

=∫M

(−∫T∗xM

tr (σ(A(z0))) (x, ξ) dξ +1q

resx(A′(z0))

)dx,

where we have set

resx(A) :=∫S∗xM

tr(σ−d(A))(x, ξ) dSξ.

Any other regularised evaluator λz0 at z0 yields:

λz0 (TR (A(z)))

=∫M

(−∫T∗xM

tr (σ(A(z0))) (x, ξ) dξ +1q

resx (A′(z0))) dx

)+ µ res(A(z0)),

for some complex constant µ.

Remark 31 It is not a priori clear that the r.h.s. of (162) and (163) are well defined;it follows here from the fact that the l.h.s. is well defined.

Proof: The result follows from the corresponding formulae (65) derived on the levelof symbols applied to σ(z) := tr(σ(A(z))):

evregz0

(−∫

Rdtr(σ(A(z)))(x, ξ) dξ

)= −

∫Rd

tr (σ(A(z0))) (x, ξ) dξ

+k∑l=0

1(l + 1) ql+1

resx,l(A(l+1)(z0)

).

Indeed, since the l.h.s. integrates over M to a well-defined quantity, we deduce thefollowing identity

evregz0 (TR (A(z))) = evreg

z0

(∫M

(−∫T∗xM

tr (σ(A(z))) (x, ξ) dξ

)dx

)and the rest of the theorem easily follows. tu

119

Exercise 91 With the notations of the theorem show that if A(z0) has order whosereal part is smaller than −d or has non-integer order or is a differential operator thenthe map z 7→ TR(A(z)) is holomorphic at z0 and that in the first two cases we have:

limz→z0

TR(A(z)) = TR(A(z0)).

In particular, show that given an admissible operator A in C`(M,E) with spectral cutθ and of positive order, the map z 7→ TR(A−zθ ) is holomorphic at z = 0.

Corollary 11 With the notations of Theorem 18 given a holomorphic regularisationscheme R : A 7→ A(z) on C`∗,∗(M,E) that sends an operator A to a holomorphicfamily A(z) with non-constant affine order α(z) = −qz + α(0) we have

TrR (A) =∫M

(−∫T∗xM

tr (σ(A)(x, ξ)) dξ +k∑l=0

1(l + 1) ql+1

resx,l(A(l+1)(0)

))dx.

(164)

In particular, for any holomorphic regularisation R : A 7→ A(z) on classical symbolsthat sends an operator A to a holomorphic family A(z) with non-constant affine orderα(z) = −qz + α(0), we have

TrR (A) =∫M

−∫T∗xM

(tr (σ(A)) (x, ξ) dξ +

1q

resx(A′(0)))dx. (165)

Thus TrR(A) is an integral over M of a sum of a local term

k∑l=0

1(l + 1) ql+1

resx,l(A(l+1)(0)

)involving only a finite number of components of the symbol of A and a global term−∫

Rd tr (σ(A)) (x, ξ) dξ. As we shall see for zeta regularisation, in special cases TrR(A)can reduce either to a local or to a global term.

These results apply to the ζ-regularisation scheme RQ for some admissible opera-tor Q in C`(M,E) with spectral cut θ and of positive order q. Formula (167) in thefollowing theorem shows how the weighted trace of A splits into local terms involvingthe (higher) noncommutative residues resx,l and a global term involving the canonicaltrace density TRx(A)

Theorem 19 For any A ∈ C`∗,k(M,E) and any admissible operator Q ∈ C`(M,E)with spectral cut θ and of positive order q, ζθ(A,Q)(z) = TR(AQ−zθ ) has poles oforder no larger than k + 1 in the discrete set −a+d−k

q , k ∈ N0 and we have

Resk+10 ζθ(A,Q)(z) = qk+1 resk(A). (166)

The finite part

TrQθ (A) := evreg0 (ζθ(A,Q)) = evreg

0

(TR(AQ−zθ )

)reads:

TrQθ (A) =∫M

(TRx(A) +

k∑l=0

(−1)l+1

(l + 1) ql+1resx,l

(A logl+1

θ Q))

dx (167)

120

where we have set TRx(A) := −∫T∗xM

tr (σA(x, ξ)) dξ.If A is classical this amounts to

TrQθ (A) =∫M

(TRx(A)− 1

qresx (A logθ Q)

)dx. (168)

Exercise 92 Show that if A has non-integer order or if the real part of its order issmaller than −d then

TrQθ (A) =∫M

TRx(A) dx = TR(A)

independently of the choice of the weight Q.

Exercise 93 Show that if A is a differential operator, then TRx(A) = 0 and resx (A logθ Q)integrates over M to res(A logθ Q) :=

∫M

resx (A logθ Q) dx, which is therefore welldeifned. Deduce from (168) the following local formula for the weighted trace:

ζθ(A,Q)(0) = TrQθ (A) = −1q

res (A logθ Q) . (169)

Hint: Use the fact that the cut-off integral vanishes on polynomials (see Exercise 29)to prove that TRx(A) = 0 if A is a differential operator.

Let us now investigate

ζθ(Q)(0) := TrQθ (I) = −1q

res (logθ Q)

for some admissible operator Q in C`(M,E) with spectral cut θ and positive order q.We first observe that

eres(logθ Q) = e−q ζθ(Q)(0) (170)

is the residue determinant introduced by Wodzicki for zero order operators andgeneralised by Scott in [Sc2] (see also [Sc3] and [Sc4]) for positive order operators.The following lemma computes its logarithmic variation.

Lemma 4 Given a differentiable family At, t ∈ I of admissible operators in C`(M,E)with constant spectral cut α, we have

d

dtres (logαAt) = res

(A−1t At

). (171)

Proof: We drop the explicit mention of the spectral cut and first observe thatddt res (logAt) = res

(ddt logAt

). Indeed, the map t 7→ tr (σ−d(logAt)(x, ·)) is differ-

entiable leading to a differentiable map t 7→∫S∗xM

tr (σ−d(logAt)(x, ξ)) dξ after inte-gration over the compact set S∗xM , whose derivative reads:∫

S∗xM

tr(σ−d

(d

dtlogAt

)(x, ξ)

)dξ = res

(d

dtlogAt

).

121

Thus, the map t 7→ res(logAt) is differentiable with derivative given by t 7→ res(ddt logAt

).

By (158) and with the notation of Proposition 32 we have

d

dtres (logAt) = res

(d

dtlogAt

)= res

(AtA

−1t

)+

K∑k=1

(−1)k

k + 1res(

adkAt(At)A−(k+1)t

)+ res

(RK(At, At)

)= res

(AtA

−1t

),

using the cyclicity of the noncommutative residue. tu

Exercise 94 Using the fact that as a result of Proposition 31, differentiation com-mutes with the residue on classical operators combined with (171) and the fact thatL(A,B) =

∫ 1

0ddtL(At, B), show that

res(L(A,B)) = 0, (172)

thus confirming a result by Scott [Sc2].

Theorem 20 Let E →M be a vector bundle over a closed manifold M . There existsan admissible operator Q in C`(M,E) with spectral cut θ for which ζθ(Q)(0) does notvanish.

Remark 32 The number ζθ(Q)(0) can therefore be seen as the “Q-dimension” of theinfinite dimensional space of smooth sections C∞(M,E). When M is a point, sothat E is a linear space which coincides with the fibre V , ζθ(Q)(0) coincides with thedimension of V .

Proof: We first note the existence of an operator T of order −d with non-vanishingresidue, e.g. T = (∆ + 1)−d/2 where ∆ = −tr(∇∗∇) is a generalised Laplacian (seee.g. [BGV]) with ∇ a connection on the bundle E and its dual ∇∗ obtained via aRiemannian metric on the manifold M and with the connection and the metric builtusing a partition of unity on M .With the existence of such an operator T in hand, we can build another admissibleoperator Q(1 + T ) since Q has positive order q. Were

ζθ(Q)(0) = −1q

res(logθ Q)

to vanish for any admissible elliptic operator Q, then the same property would holdfor Q(1 + T ) leading to

0 = res(logθ(Q(1 + T ))

=∫ 1

0

d

dt(res(logθ(Q(1 + t T ))) dt

=∫ 1

0

d

dtres(QT (1 + t T )−1Q−1

)dt

=∫ 1

0

d

dtres(T (1 + t T )−1

)dt

= res (logπ(1 + T ))

122

using (171) combined with the cyclicity of the noncommutative residue.On the other hand, since T has order −d,

res (logπ(1 + T )) =∞∑k=0

(−1)k

k + 1res(T k+1

)= res (T )

as the other terms which are residues of operators with order smaller than −d inthe log expansion do not come into play. Hence, assuming that ζθ(Q)(0) = 0 andζθ(Q(1 + T ))(0) = 0 would lead to res (T ) = 0 thus contradicting the assumption onT. tuWhen the underlying manifold M is a point ∗ in which case the vector bundle Eover M reduces to the model space V , the space C∞(M,E) of smooth sections of Ereduces to the linear space V . In that case weighted traces of the identity TrQθ (I) boildown to the ordinary trace of the identity and hence the dimension of V . They cantherefore be interpreted as Q-weighted dimensions of the infinite dimensional spaceC∞(M,E).

Exercise 95 Let Q1, Q2 in C`(M,E) be admissible operators of positive orders q1,q2 and common spectral cut which we omit to write explicitly.

1. Show that the operatorlogl+1Q1

ql+11

− logl+1Q2

ql+12

lies in C`∗,l(M,E).

2. With these notations, show that for any admissible operator A in C`(M,E) andany positive integer k

TrQ1

(logk A

)− TrQ2

(logk A

)(173)

=∫M

(k∑l=0

(−1)l+1

(l + 1)resx,l

(logk A

(logl+1Q1

ql+11

− logl+1Q2

ql+12

)))dx.

3. We specialise to the case k = 1. Let a be the order of A, which we assume to bepositive. Let Q in C`(M,E) be an admissible operator of positive order q andthe same spectral cut as A. Show that

TrA (logA)− TrQ (logA) =∫M

[−resx,0

(logA

(logAa− logQ

q

))+

12

resx,1

(logA

(log2A

a2− log2Q

q2

))]dx

and discuss the locality of this expression.

We end this subsection with a continuity property.

Proposition 35 Weighted traces are continuous in the Frechet topology of operatorsof constant order.

123

Proof: We show this continuity property in the classical case; the proof easily extendsto the log-polyhomogeneous case. By the defect formula (168) we have

TrQα (A) =∫M

dx

(−∫T∗xM

tr(σ(A)(x, ·))− 1q

∫S∗xM

tr (σ(A logαQ)(x, ·))−d

).

From the continuity of the cut-off regularised integral −∫T∗xM

in the Frechet topology ofsymbols of constant order we deduce the continuity of the mapA 7→ −

∫T∗xM

tr (σ(A)(x, ·)).Let us now investigate the continuity of the remaining map

A 7→∫S∗xM

tr (σ−d(A logαQ)(x, ·)) .

By (113) we have

σ−d (A logαQ) =∑

|α|+a−j−k=−d

(−i)|α|

α!∂αξ σa−j(A) ∂αx σ−k(logαQ).

The continuity of the map A →∫S∗xM

tr (σ−d(A logαQ)) therefore follows from thatof the maps A 7→ σa−j(A)(x, ·) and A 7→ ∂αξ σ−k(logαQ)(x, ·) which are clearly con-tinuous in the topology of operators of constant order. tu

6.8 The index as a superresidue

When the weight Q has positive leading symbol σL(Q), we can choose π as a spectralcut30 and the Q-weighted trace TrQ relates to the heat-kernel regularised trace we areabout to define. We quote the following result from [GS] in the classical case, [L1] inthe log-polyhomogeneous case.

Proposition 36 Let A ∈ C`a,k(M,E) and let ∆ ∈ C`(M,E) be an elliptic operatorof positive order q and non-negative leading symbol. The map ε 7→ tr

(A e−ε∆

)which

is defined for any ε > 0, is exponentially decreasing at infinity:

∃A > 0,∃λ > 0, ∃C > 0,∣∣tr (A e−ε∆

)∣∣ ≤ Ce−ελ ∀ε > A,

and has the following asymptotic behaviour as ε→ 0:

Tr(A e−ε∆

)∼ε→0

∞∑j=0

εj−n−a

q Pj(log ε) +∞∑j=0

γjεj , (174)

where Pj is a polynomial of degree ≤ k if j−n−aq /∈ N0 and ≤ k + 1 if j−n−a

q ∈ N0.

Heat-kernel regularisation is an alternative regularisation procedure.

Definition 41 Let A ∈ C`a,k(M,E) and let ∆ ∈ C`(M,E) be an elliptic operatorwith positive order q and non-negative leading symbol. We call the constant term inthe asymptotic expansion of ε 7→ tr

(A e−ε∆

)as ε→ 0

TrHK,∆(A) := evreg0

(Tr(A e−ε∆

))the heat-kernel regularised trace of A.

30We then drop the subscript π for simplicity.

124

Exercise 96 Let A ∈ C`(M,E) and let ∆ ∈ C`(M,E) be an elliptic (essentially)self-adjoint operator with positive order and non-negative leading symbol. Let π∆ bethe orthogonal projection onto the kernel of ∆. Show that

Tr∆+π∆(A) = TrHK,∆(A) + γ res(A).

where γ is the Euler constant.In particular, if ∆ is a differential operator we have:

Tr∆+π∆(A) = TrHK,∆(A).

Hint: First show that the Mellin transform

z 7→ M(f)(z) :=1

Γ(z)

∫ ∞0

εz−1f(ε) dε

of a function f ∈ C∞(]0,+∞[) which is exponentially decreasing at infinity defines ameromorphic map on the complex plane with poles of order ≤ 1 at 0 and that

evreg0 M(f)(z) = evreg

0 f(ε) + γ Res0M(f)(z). (175)

Weighted traces can be extended to weighted supertraces.Let E = E+⊕E− be a Z2-graded vector bundle (see e.g [BGV]) over a closed manifoldM and let Q+ ∈ C`(M,E+), Q− ∈ C`(M,E−) be two admissible operators with thesame spectral cut θ. Setting Q := Q+ ⊕Q− we define the weighted supertrace of aneven operator A = A+ ⊕A−, with A+ in C`(M,E+), A− in C`(M,E−)

sTrQθ (A) := TrQ+θ (A+)− TrQ−θ (A−),

which clearly extends the ordinary supertrace Tr(A) = Tr(A+)−Tr(A−) on trace-classoperators. IfD+ : C`(M,E+)→ C`(M,E−) is an elliptic operator in C`

(M,E∗+ ⊕ E−

)then its (formal) adjoint D− := D∗+ : C`(M,E−)→ C`(M,E+) is an elliptic operatorin C`

(M, (E−)∗ ⊗ E+

)and ∆ = ∆+ ⊕∆− with ∆+ := D−D−, ∆− := D+D− are

non-negative (formally) self-adjoint elliptic operators.The following theorem which combines formulae due to McKean and Singer [MS] andSeeley [Se], expresses the index of D+:

ind(D+) := dim (Ker(D+))− dim (Ker(D−))

in terms of the superweighted trace of the identity. Interpreting the residue as aweighted supertrace of the identity and using the local formula (169) for weightedtraces of differential operators provides a local formula for the index in terms of thenoncommutative residue (see [Sc2]).

Theorem 21 The superresidue

sres (log(∆ + π∆)) :=∫M

(resx

(log ∆+ + π∆+

)− resx

(log ∆− + π∆−

))dx,

where as usual resx(B) =∫S∗xM

tr (σ−d(B)(x, ξ)) dξ is well defined and we have

ind(D+) = sTr∆+π∆(I) = sTr(e−ε∆

)= − 1

ord(∆)sres (log(∆ + π∆)) ∀ε > 0, (176)

where π∆ , π∆+ , π∆− denote the orthogonal projections onto the kernels of ∆, ∆+

and ∆− and ord(∆) is the order of ∆.

125

Remark 33 Since the operators ∆, ∆+ and ∆− are non negative, the standard choiceπ of spectral cut is not explicitely mentioned.

Proof: Let us first check that

sres (log(∆ + π∆)) :=∫M

(resx

(log ∆+ + π∆+

)− resx

(log ∆− + π∆−

))dx

is well defined.Since the identity is a differential operator, the expression sTRx(I) := TRx(I+) −TRx(I−) vanishes, with I+ the identity bundle map on E+ and I− the identity bundlemap on E−. Formula (169) applied to ∆+ + π∆+ and ∆− + π∆− of order q = ord(∆)then shows that

sresx (log(∆ + π∆)) := resx(log(∆+ + π∆+)

)− resx

(log(∆− + π∆−)

)integrates to a global density

sres (log(∆ + π∆)) =∫M

sresx (log(∆ + π∆)) dx

=∫M

resx(log(∆+ + π∆+

)dx−

∫M

resx(log(∆− + π∆−)

)dx

= res(log(∆+ + π∆+)

)− res

(log(∆− + π∆−)

),

and that we have

sTr∆+π∆(I) := Tr∆++π∆+ (I+)− Tr∆−+π∆− (I−) = − 1ord∆

sres(log(∆ + π∆)).

We further observe that

Spec(∆+)− 0 = Spec(∆−)− 0. (177)

Indeed,

∆+u+ = λ+u+ ⇒ ∆−(D+u+) = λ+D+u+ ∀u+ ∈ C∞(M,E+)

so that an eigenvalue λ+ of ∆+ with eigenvector u+ is an eigenvalue of ∆− witheigenvector D+u+ provided the latter does not vanish. The converse holds similarly.Let us denote by λ+

n , n ∈ N the discrete set of eigenvalues of ∆+ and by λ−n , n ∈ Nthe discrete set of eigenvalues of ∆−. For any complex number z:

sTr((∆ + π∆)−z

)=

∑n∈N

(λ+n + δλ+

n

)−z−∑n∈N

(λ−n + δλ−n

)−z=

∑λ+n 6=0

(λ+n

)−z − ∑λ−n 6=0

(λ−n)−z + dimKer∆+ − dimKer∆−

= ind(D+) using (177).

Taking the finite part at z = 0 therefore yields:

ind(D+) = fpz=0sTr((∆ + π∆)−z

)= sTr∆+π∆(I)

= − 1ord(∆)

sres (log(∆ + π∆)) .

126

Here is an alternative heat-kernel approach via the McKean-Singer formula. For anypositive ε

sTr(e−ε∆

)=

∑n∈N

e−ε λ+n −

∑n∈N

e−ε λ−n

=∑λ+n 6=0

e−ε λ+n −

∑λ−n 6=0

e−ε λ−n + ind(D+)

= ind(D+) using (177).

Using Proposition 96 (which easily extends to supertraces) and taking the limit asε → 0 leads to sTr∆(I) = sTrHK,∆(I) = ind(D+). This completes the proof of thetheorem. tu

This formula gives an a priori local formula for the index; the price to pay for thislocality is the occurrence of a logarithm whose residue is difficult to compute in prac-tice. Some specific cases were worked out in yet unpublished joint work with JoukoMickelsson. The Riemann-Roch theorem was derived from this formula by S. Scottand D. Zagier [Sc3] and [Sc4].

6.9 Uniqueness of the canonical trace and noncommutativeresidue on operators

We end this chapter with a characterisation of the canonical trace and noncommutativeresidue. Let us first quote without proof a result of Guillemin31 [Gu1].

Lemma 5 Any trace on the algebra C`−∞(M,E) of smoothing operators is propor-tional to the L2-trace Tr introduced in (131).

This leads to a characterisation of the canonical trace on certain classes of classicalpseudodifferential operators and of the residue on the whole algebra of classical pseu-dodifferential operators. Our approach, similar in spirit to the one adopted in [MSS],stresses the role of Stokes’ property of the canonical integral on non-integer ordersymbols.

Definition 42 If E is a finite rank vector bundle over M modelled on a linear spaceV , we call admissible a subset D(M,E) of C`(M,E) defined in terms of a symbolset S ⊗ End(V ) ⊂ CScc(Rd) ⊗ End(V ) as in (127), whenever the symbol set S isadmissible.

Exercise 97 Show that the algebra Cl−∞(M,E) of smoothing operators and the setC`/∈Z(M,E) of non-integer order classical pseudodifferential operators on M are ad-missible subsets of C`(M,E).

Exercise 98 Show that the stability of S under partial differentation implies that ofD(M,E) under brackets [xi, ·] in local coordinate charts.Hint: Use the fact that in a local coordinate chart (U, φ) teh following identity holds;Op(∂ξiσ) = −i [xi,Op(∂ξiσ)] for any σ in CScpt(φ(U)).

31Theorem A.2 in [Gu1] applies to the Hilbert space Hs(M,E) of Sobolev Hs-sections of the vectorbundle E over M , see Remark 1 following the proof of Theorem A.2.

127

The following theorem characterises the canonical trace and the noncommutativeresidue, which by Exercises 75 and 76 define continuous linear forms. Whereas theresult is well known (see [Wo1], [Gu2] for the residue and [MSS], [L2] for the canonicaltrace), the rather straightforward proof used here is new to our knowledge. It never-theless uses a continuity assumption needed to link up this characteristation with thecharacterisation derived in Theorem 16 of the residue as a continuous singular trace.

Theorem 22 Let D(M,E) be an admissible subset of C`(M,E) which contains smooth-ing operators.

1. If D(M,E) is contained in Ker(res) and the canonical trace TR defines a con-tinuous linear form32 on D(M,E) which vanishes on brackets of operators inD(M,E), then any continuous linear form Λ on D(M,E) which vanishes onbrackets:

Λ ([A,B]) = 0 ∀A,B ∈ C`(M,E) s.t. [A,B] ∈ D(M,E) (178)

is proportional to the canonical trace:

∃C ∈ C, Λ(A) = C · TR(A) ∀A ∈ D(M,E).

2. If D(M,E) = C`(M,E) then any continuous trace is proportional to the non-commutative residue

∃C ∈ C, Λ(A) = C · res(A) ∀A ∈ C`(M,E).

Proof: Let Λ be a continuous linear form on D(M,E); by assumption it restricts toa trace on C`−∞(M,E) so that by Lemma 5 there is a constant C such that:

Λ|C`−∞(M,E)= C Tr,

where as before, Tr denotes the L2-trace on C`<−d(M,E). Let us assume there is acontinuous linear extension Tr of Tr to D(M,E) in which case

Λ := Λ− C Tr

is a singular linear form since it vanishes on C`<−d(M,E). Let us now investigate thetwo cases spelled out in the theorem.

1. Since the canonical trace defines a continuous linear form on D(M,E), we canchoose Tr := TR. Since TR vanishes on brackets, so does the linear form Λ.Being moreover continuous, by Theorem 16 it is proportional to the noncom-mutative residue. Since D(M,E) lies in Ker(res), it follows that Λ vanishesidentically so that there is a constant C such that Λ = C TR.

2. We now consider the case D(M,E) = C`(M,E).Using a partition of unity, we can equip M with a Riemannian metric and Ewith a connection ∇. Let ∆ = −tr(∇∗∇) be the associated Laplacian, in whichcase the operator Q := ∆+π∆, where π∆ denotes the orthogonal projection onto

32By linear we mean here that Λ(αA + βB) = αΛ(A) + βΛ(B) whenever A,B and αA + βB ∈D(M,E)

128

the kernel of ∆, is an invertible elliptic operator of order 2. Since the weightedtrace TrQ extends the L2-trace, we can choose Tr := TrQ and set:

Λ := Λ− C TrQ.

Let (U, φ) be a local trivialising chart of M and Φ : E|U → U × Ck a localtrivialisation over U . For a smooth function f with compact support in φ(U),we define a linear form on CScc(Rd) by

λΦ,f (τ) := Λ (Φ∗Op(f ⊗ τ)) .

Since Λ is singular, so is λΦ,f singular.We follow the line of proof of Theorem 4 and introduce the symbols τ(ξ) =χ(ξ) |ξ|−d where χ is any smooth function which vanishes in a small neigh-borhood of zero and is one outside the unit ball and τj(ξ) = ξj |ξ|−d so that∑dj=1 ∂jτj = 0. Since λφ,f vanishes on smoothing symbols and since Λ vanishes

on brackets, with the notation of Theorem 16 we have

0 = λf,Φ

χ d∑j=1

∂jτj

= λf,Φ

d∑j=1

(χ∂jτj)

= Λ

Φ∗Op

d∑j=1

∂j(χ τj)

= −i

d∑j=1

Λ ([xj ,Φ∗Op(χ τj)])

= i C

d∑j=1

TrQ ([xj ,Φ∗Op(χ τj)])

Hence, by (161) it follows that

0 = −i C2

d∑j=1

res (Φ∗Op(χ τj) [xj , logθ Q])

=C

2

d∑j=1

(res(ξj |ξ|−d ∂ξj log |ξ|2

)+ res

(ξj |ξ|−d ∂ξjσ0(logθ Q)

))=

C

2

2d∑j=1

(res(ξ2j |ξ|−d−2

)+ res

(ξj |ξ|−d ∂ξj logθ σL(Q)(ξ/|ξ|)

))= C res

(|ξ|−d

)since σL(Q)(ξ) = |ξ|2

= CVol(Sd−1)

(2π)d.

129

Thus C = 0 and Λ = Λ so that the linear form Λ is both singular and vanisheson brackets. Being moreover continuous, by Theorem 16 it is proportional tothe noncommutative residue, which ends the proof of the theorem.

tu

Remark 34 Results 33 of Wodzicki34[Wo1] and Guillemin [Gu2] actually show thatall (not necessarily a priori continuous) traces on C`(M,E) are proportional to thenoncommutative residue.

33see also [BG] and see [L1, L2] for extensions of this characterisation to other classes of pseudo-differential operators.

34This result was proved in Wodzicki’s thesis written in Russian but the main steps of the proofcan be found in [L1]. An alternative proof can be found in [Po2] which also encompasses the onedimensional case and which uses arguments similar in spirit to those underlying the proof in theboundary case carried out in [FGLS] and those underlying the proof of the uniqueness of the canonicaltrace on non-integer order operators carried out in [MSS].

130

7 Multiplicative and conformal anomaly of regu-larised determinants

We describe two types of regularised determinants, weighted determinants and theζ-determinant, which are related by a local formula. Weighted and ζ-determinantsare not multiplicative but, as is well-known for zeta determinants since the work ofOkikiolu [O2] on the one hand and since the work of Kontsevich and Vishik [KV] on theother hand, the corresponding multiplicative anomaly which measures the obstructionto the multiplicativity is local in a sense we make precise in Theorem 24 for weighteddeterminants and in Theorem 25 for zeta determinants. This first part of the chapter isbased on joint work with M.-F. Ouedraogo [OP]. Neither is the zeta determinant of aconformally covariant operator, conformally invariant: in Theorem 26 we compute thewell-known conformal anomaly of conformally covariant Laplacians after establishingin Theorem 23 a just as well-known formula for the logarithmic variation of the zetadeterminant. The second part of the chapter is based on joint work with StevenRosenberg [PayR].

7.1 Weighted and zeta determinants

An admissible operator A ∈ C`(M,E) with spectral cut θ and positive order has welldefined Q-weighted determinant [Duc] (see also [FrG]) where Q in C`(M,E) is aweight of order q with the same spectral cut θ:

DetQθ (A) := eTrQθ (logθ A).

Since the weighted traces restrict to the ordinary trace on trace-class pseudodifferentialoperators namely those whose order has real part smaller than −d, this determinantextends the ordinary determinant Detθ(A) = eTr(logθ A) on pseudodifferential operatorsin the determinant class, namely those that differ from the identity by an operatorwhose order has real part smaller than −d. In general, it involves residue terms, ascan be seen from the following exercise.

Exercise 99 Using (167) show that

log DetQθ (A) (179)

=∫M

(TRx(logθ A)− 1

qresx,0 (logθ A logθ Q) +

12 q2

resx,1(logθ A log2

θ Q))

dx.

The weighted determinant, as well as being dependent on the choice of spectral cut θ,also depends on the choice of spectral cut α.

Exercise 100 Let 0 ≤ θ < φ < 2π be two spectral cuts for the admissible operator A.If there is a cone Λθ,φ (see 143) which does not intersect the spectrum of the leadingsymbol of A show that

DetQθ (A) = DetQφ (A).

Hint: Use the fact that the cone Λφ,θ defined as in Proposition 29, contains only afinite number of points in the spectrum of A.

Given an admissible operator A in C`(M,E) with spectral cut θ and positive order,by Exercise 91 the map z 7→ TR(A−zθ ) is holomorphic at z = 0 so that the operatorA has well defined ζ-determinant:

Detζ,θ(A) := e−ζ′A,θ(0) = eTrAθ (logθ A).

131

In mathematics, zeta determinants were first introduced by Ray and Singer in [RaSi]to describe analytic torison, where in physics they first arose in pioneering work byHawkins [Haw].

Exercise 101 1. Show that

logDetζ,θ(A)

DetQθ (A)

=∫M

[−resx,0

(logA

(logAa− logQ

q

))+

12

resx,1

(logA

(log2A

a2− log2Q

q2

))]dx (180)

where a is the order of A and deduce that zeta and weighted determinants differby a local expression.Hint: Use (174).

2. Deduce from there the more compact formula (see [Duc], Proposition III.1.7)

logDetζ,θ(A)

DetQθ (A)= −a

2res

((logAa− logQ

q

)2). (181)

Hint: Use the fact that

logA(

logAa− logQ

q

)=a

2

(logAa− logQ

q

)2

− a

2

(log2A

a2− log2Q

q2

).

The ζ-determinant generally depends on the choice of spectral cut. However, it isinvariant under mild changes of spectral cut in the following sense.

Exercise 102 Let 0 ≤ θ < φ < 2π be two spectral cuts for the admissible operator A.Show that if there is a cone Λθ,φ (see 143) which does not intersect the spectrum ofthe leading symbol of A then

Detζ,θ(A) = Detζ,φ(A).

Hint: Use the fact that the cone Λφ,θ defined as in Proposition 29, contains only afinite number of points in the spectrum of A.

We end this subsection with a classical and moreover very useful formula for thelogarithmic variation of the zeta determinant of a differentiable family of admissibleoperators.

Theorem 23 The zeta determinant Detζ,θ(At) of a differentiable family At, t ∈ I ofinvertible admissible operators of constant order and with fixed spectral cut θ parametrisedby a non-empty open interval I in R is differentiable and

d

dtlog Detζ,θ(At) = TrAtθ (A−1

t At), (182)

where we have set At := ddtAt.

132

Proof: By (181) we have[d

dt,TrAtθ

](logAt) := lim

t→0

TrAtθ (logAt)− TrA0θ (logAt)

t

= − 12α

limt→0

res

((logθ At − logθ A0)2

t

)= 0,

where α stands for the (constant) order of At. The result then follows from (158),which yields

d

dtlog Detζ,θ(At) = TrAtθ

(d

dtlogAt

)+[d

dt,TrAtθ

](logAt)

= TrAtθ

(d

dtlogAt

)= TrAtθ (AtA−1

t ) +K∑k=1

(−1)k

k + 1TrAtθ (adkAt(At)A

−(k+1)t ) + TrAtθ (RK(At, At))

= TrAtθ (AtA−1t ),

since by (161) we have

TrAtθ (AdAt(B)) =1α

res (B [At, logθ At]) = 0 ∀B ∈ C`(M,E).

tu

7.2 Multiplicative anomaly of weighted determinants

Unlike ordinary matrix determinants, weighted determinants are not multiplicative.The multiplicative anomaly for Q-weighted determinants of two admissible operatorsA, B with spectral cuts θ, φ such that AB has spectral cut ψ is defined by:

MQθ,φ,ψ(A,B) :=

DetQψ (AB)

DetQθ (A) DetQφ (B),

which we writeMQ(A,B) to simplify notation, thus dropping the explicit mention ofthe spectral cuts in what follows.

Theorem 24 For two admissible operators A,B ∈ C`(M,E) of positive orders a andb such that their product AB is also admissible, we have

res(L(A,B)) = 0. (183)

Moreover, there is an operator

W (τ)(A,B) :=d

dt |t=0L(At, AτB) (184)

in C`0(M,E) depending continuously on τ such that

TrQ(L(A,B)) =∫ 1

0

res(W (τ)(A,B)

(log(AτB)aτ + b

− logQq

))dτ (185)

where Q is any weight of order q.

133

Proof: By Proposition 33, we know that

d

dt |t=0res(L(At, B)) =

d

dt |t=0TrQ(L(At, B)) = 0.

We want to compute ddt |t=τ res(L(At, B)) = d

dt |t=0res(L(At+τ , B)) and

d

dt |t=τTrQ(L(At, B)) =

d

dt |t=0TrQ(L(At+τ , B)).

For this we observe that

L(AB,D)− L(A,BD) = − log(AB)− log(D) + logA+ log(BD)= L(B,D)− L(A,B).

Replacing A by At, B by Aτ and D by B, we get

L(At+τ , B)− L(At, AτB) = L(Aτ , B)− L(At, Aτ ) = L(Aτ , B).

Implementing the noncommutative residue, in view of Proposition 33 we have:

d

dt |t=τres(L(At, B)) =

d

dt |t=0res(L(At+τ , B))

=d

dt |t=0res(L(At, AτB))

= 0.

Hence

res(L(A,B)) =∫ 1

0

d

dt |t=τres(L(At, B)) dτ + res(L(I,B)) = 0, (186)

since L(I,B) = 0.If instead we implement the weighted trace trQ, we have:

d

dt |t=τTrQ(L(At, B)) =

d

dt |t=0TrQ(L(At+τ , B))

=d

dt |t=0TrQ(L(At, AτB)).

Since A and B have positive order so has Aτ B, so that applying Proposition 33 withweighted traces trA

τ B yields:

d

dt |t=τTrQ(L(AtB) =

d

dt |t=0TrQ(L(At, AτB))

=d

dt |t=0TrA

τB(L(At, AτB))

+d

dt |t=0

(TrQ(L(At, AτB))− trA

τB(L(At, AτB)))

=d

dt |t=0

(TrQ(L(At, AτB))− trA

τB(L(At, AτB))).

134

Applying (160) to Q1 = Q and Q2 = AτB yields

d

dt |t=0

(TrQ(L(At, AτB))− TrA

τB(L(At, AτB)))

=d

dt |t=0res(L(At, AτB)

(log(AτB)aτ + b

− logQq

))= res

(W (τ)(A,B)

(log(AτB)aτ + b

− logQq

)),

where q is the order of Q and where we have set W (τ)(A,B) := ddt |t=0

L(At, AτB).Since L(I,B) = 0, we finally find that

TrQ(L(A,B)) = TrQ(L(A1, B))− TrQ(L(A0, B))

=∫ 1

0

res(W (τ)(A,B)

(log(AτB)aτ + b

− logQq

))dτ. (187)

tu

Proposition 37 Let A and B be two admissible operators with spectral cuts θ and φin [0, 2π[ such that there is a cone delimited by the rays Lθ and Lφ which does notintersect the spectra of the leading symbols of A, B and AB. Then the product AB isadmissible with a spectral cut ψ inside that cone and for any weight Q with spectralcut, dropping the explicit mention of the spectral cuts we have:

logMQ(A,B) =∫ 1

0

res(W (τ)(A,B)

(log(AτB)aτ + b

− logαQq

))dτ. (188)

Proof: Since the leading symbol of the product AB has spectrum which does notintersect the cone delimited by Lθ and Lφ, the operator AB only has a finite numberof eigenvalues inside that cone. We can therefore choose a ray ψ which avoids boththe spectrum of the leading symbol of AB and the eigenvalues of AB, in which casethe weighted determinants DetQθ (A), DetQφ (B) and DetQψ (AB) do not depend on thechoices of spectral cuts satisfying the requirements of the proposition.Since

logMQ(A,B) = log DetQ(AB)− log DetQ(A)− log DetQ(B) = TrQ(L(A,B)),

the logarithm of the multiplicative anomaly for weighted determinants is a local quan-tity (185) derived in Theorem 24 tu .

Exercise 103 Show that weighted determinants are multiplicative on commuting op-erators.Hint: Use the fact that by Exercise 89, we have log(AB) = logA+ logB when A andB commute.

7.3 Multiplicative anomaly of the zeta determinant

The ζ-determinant is not multiplicative 35. Indeed, let A and B be two admissible op-erators of positive order and spectral cuts θ and φ and such that AB is also admissible

35It was shown in [LP] that all multiplicative determinants on elliptic operators can be built fromtwo basic types of determinants; they do not include the ζ-determinant.

135

with spectral cut ψ. The multiplicative anomaly

Mθ,φ,ψζ (A,B) :=

Detζ,ψ(AB)Detζ,θ(A) Detζ,φ(B)

,

was proved to be local, independently by Okikiolu [O2] for operators with scalar lead-ing symbol and by Kontsevich and Vishik [KV] for operators “close to the identity”.For simplicity, we drop the explicit mention of θ, φ, ψ and write Mζ(A,B).

Recall from (181) that

logDetζ,θ(A)

DetQθ (A)= −a

2res

((logAa− logQ

q

)2).

Exercise 104 Using (188), check that

logMζ(A,B)

= logMQ(A,B)− a+ b

2res

((logABa+ b

− logQq

)2)

+a

2res

((logAa− logQ

q

)2)

+b

2res

((logBb− logQ

q

)2)

(189)

is also a local expression.

Setting Q = A yields the following explicit local formula for the multiplicative anomalyindependently of Okikiolu’s assumption that the leading symbols be scalar.

Theorem 25 [OP] Let A and B be two admissible operators in C`(M,E) of positiveorders a, b and with spectral cuts θ and φ in [0, 2π[ such that there is a cone delimitedby the rays Lθ and Lφ which does not intersect the spectra of the leading symbols of A,B and AB. Then the product AB is admissible with a spectral cut ψ inside that coneand the multiplicative anomaly Mθ,φ,ψ

ζ (A,B) is local as a noncommutative residue,independently of the choices of θ, φ, and ψ satisfying the above requirements.Explicitly, and dropping the explicit mention of the spectral cuts, there is a classicaloperator W (τ)(A,B) given by (184) of order zero depending continuously on τ suchthat:

logMζ(A,B) =∫ 1

0

res(W (τ)(A,B)

(log(AτB)aτ + b

− logAa

))dτ

− a+ b

2res

((logABa+ b

− logAa

)2)

(190)

+b

2res

((logBb− logA

a

)2),

Exercise 105 When A and B commute the multiplicative anomaly reduces to:

logMζ(A,B) = −res(

12(a+ b)

log2(AB)− 12a

log2A− 12b

log2B

)=

ab

2(a+ b)res

[(logAa− logB

b

)2]. (191)

136

Hint: Use the fact that by Exercise 89, we have log(AB) = logA+ logB when A andB commute.

Remark 35 For commuting operators, (191) gives back the results of Wodzicki aswell as formula (III.3) in [D]:

logMζ(A,B) =res(log2(AbB−a)

)2ab (a+ b)

.

Exercise 106 Derive similar local formulae setting Q = B in (189) instead of Q = A.

7.4 Conformally covariant operators

We view the Laplace-Beltrami operator ∆g associated with a Riemannian metric g asan example of a more general class of conformally covariant operators.Given a vector bundle E over a closed n-dimensional manifold M , le us consider maps

Met(M) → Cl(M,E)g 7→ Ag,

where Met(M) denotes the space of Riemannian metrics on M .

Definition 43 The operator Ag ∈ Cl(M,E) associated to a Riemannian metric gis conformally covariant of bidegree (a, b) if the pointwise scaling of the metricg = e2fg, for f ∈ C∞(M) yields

Ag = e−bfAgeaf = e(a−b)f A′g, for A′g := e−af Ag e

af , (192)

for constants a, b ∈ R.

We survey known conformally covariant differential and pseudodifferential opera-tors; more details are in Chang [Ch].Operators of order 1. (Hitchin [Hit]) For M spin, the Dirac operator Dg := γi · ∇giis a conformally covariant operator of bidegree

(n−1

2 , n+12

).

Operators of order 2. If dim(M) = 2, the Laplace-Beltrami operator ∆g is confor-mally covariant of bidegree (0, 2). It is well known that in dimension two

Rg = e−2f (Rg + 2∆gf) , (193)

where Rg is the scalar curvature, and by the Gauss-Bonnet theorem (see e.g. [Ros])∫M

Rg d volg = 2πχ(M), (194)

with the Euler characteristic χ(M) (much more than) a conformal invariant.On a Riemannian manifold of dimension n, the Yamabe operator, also called the

conformal Laplacian,Lg := ∆g + cnRg,

is a conformally covariant operator of bidegree(n−2

2 , n+22

), where cn := n−2

4(n−1) .Operators of order 4. (Paneitz [Pan, BO]) In dimension n, the Paneitz operators

Png := Png + (n− 4)Qng

137

are conformally covariant scalar operators of bidegree(n−4

2 , n+42

). Here Png := ∆2

g +d∗ ((n− 2) Jg g − 4Ag·) d with

Jg :=Rg

2(n− 1), Ag =

Ricg − Rgn g

n− 2+Jgng,

Ag the homomorphism on T ∗M given by φ = (φi) 7→ (Ag)ji φj , and

Qng :=nJ2

g − 4|Ag|2 + 2∆g Jg

4

is Branson’s Q-curvature [B1], a local scalar invariant that is a polynomial in the co-efficients of the metric tensor and its inverse, the scalar curvature and the Christoffelsymbols. Note that Ag = 1

nJg g precisely when g is Einstein.

The Q-curvature generalizes the scalar curvature Rg in the following sense. On a4-manifold, we have

Q4g = e−4f

(Q4g +

12P 4g f

)(cf. (193)), and

∫MQ4gdvolg is a conformal invariant (cf. (194)), as is

∫MQngdvolg in

even dimensions [B2].Operators of order 2k. (Graham, Jenne, Mason and Sparling [GJMS]) Fix k ∈ Z+

and assume either n is odd or k ≤ n. There are conformally covariant (self-adjoint)scalar differential operators Png,k of bidegree

(n−2k

2 , n+2k2

)such that the leading part

of Png,k is ∆kg and such that Png,k = ∆k

g on Rn with the Euclidean metric.Png,k generalises Png , since Png = Png,2, and satisfies

Png,k = Png +n− 2k

2Qng

where Png = d∗Sng d for a natural differential operator Sng on 1-forms.Note that Png,k has bidegree (a, b) with b − a = 2k independent of the dimension

and in particular has bidegree (0, 2k) in dimension 2k.Pseudodifferential Operators. (Branson and Gover [BrGo], Peterson [Pe]) Peter-son has constructed ΨDOs , Png,k, k ∈ C, of order 2Re(k) and bidegree((n − 2k)/2, (n + 2k)/2) on manifolds of dimension n ≥ 3 with the property thatPng,k − e−bfPng,keaf is a smoothing operator. Thus any conformal covariant built fromthe total symbol of Png,k is itself a conformal covariant of Png,k. The family Png,kcontains the previously discovered conformally covariant pseudodifferential operatorsassociated to conformal boundary value problems [BrGo].

7.5 Conformal anomalies

This subsection based on [PayR] and [AJPS] derives the conformal anomaly of thezeta determinant of a conformally covariant operator.Let M be a closed Riemannian manifold andMet(M) denote the space of Riemannianmetrics on M . The spaceMet(M) is trivially a Frechet manifold as the open cone ofpositive definite symmetric (covariant) two-tensors inside the Frechet space

C∞(T ∗M ⊗s T ∗M) := h ∈ C∞(T ∗M ⊗ T ∗M) : hab = hba

138

of all smooth symmetric two-tensors. The Weyl group W(M) := ef : f ∈ C∞(M)acts smoothly on Met(M) by Weyl transformations

W (g, f) = g := e2fg,

and given a reference metric g in Met(M), a functional F :Met(M)→ C induces amap

Fg = F W (g, ·) : C∞(M) → C,f 7→ F(e2fg).

Definition 44 A functional F on Met(M) is conformally invariant for a referencemetric g if Fg is constant on a conformal class, i.e.

F(e2fg) = F(g) ∀f ∈ C∞(M).

A functional F on Met(M) is conformally invariant if it is conformally invariantfor all reference metrics. A functional F : Met(M) ×M → C is called a pointwiseconformal covariant of weight w if

F(e2fg, x) = w · f(x)F(g, x) ∀f ∈ C∞(M), ∀x ∈M.

A functional F :Met(M)→ C which is Frechet differentiable has a differential

dF(g) : TgMet(M) = C∞(T ∗M ⊗s T ∗M)→ C,

dF(g).h :=d

dt

∣∣∣t=0

F(g + th)−F(g)t

.

For such a funcitonal F , the differentiability of the Weyl map implies that the compo-sition Fg : C∞(M) → C is differentiable at 0 with differential dFg(0) : T0C

∞(M) =C∞(M)→ C.

Definition 45 The conformal anomaly for the reference metric g of a differentiablefunctional F on Met(M) is dFg(0). In physics notation, the conformal anomaly inthe direction f ∈ C∞(M) is

δfFg := dFg(0).f = dF(g).2f g

= limt=0

F(g + 2tfg)−F(g)t

=d

dt

∣∣∣t=0F(e2tfg).

Remark 36 F is conformally invariant if and only if dFg(0).f = 0 for all g ∈Met(M), f ∈ C∞(M).

For a fixed Riemannian metric g = (gab), we equip C∞(M) with the L2 metric

(f, f)g =∫M

f(x)f(x)d volg(x).

We define an L2 metric on Met(M) by

〈h, k〉g :=∫M

gac(x)gbd(x)hab(x) kcd(x) d volg(x) =∫M

hcd(x) kcd(x) d volg(x) (195)

139

with (gab) = (gab)−1 and hab(x) := gac(x)gbd(x)hcd(x). The L2 metric induces aweak L2-topology onMet(M), and L2(T ∗M⊗sT ∗M), the L2-closure of C∞(T ∗M⊗sT ∗M) with respect to 〈 , 〉g, is independent of the choice of g modulo Hilbert spaceisomorphism. The choice of a reference metric yields the inner product (195) on thetangent space TgMet(M) = C∞(T ∗M ⊗s T ∗M), giving the weak L2 Riemannianmetric on Met(M), and forming the completion of each tangent space.

The various inner products are related as follows:

Exercise 107 For g in Met(M), h in C∞(T ∗M ⊗s T ∗M) and f in C∞(M), showthat

〈h, f g〉g = (trg(h), f)g

where we have set: trg(h) := hbb = gabhab.

tu

Definition 46 If the differential dF(g) : C∞(T ∗M ⊗s T ∗M) → C extends to a con-tinuous functional dF(g) : L2(T ∗M ⊗s T ∗M) → C, then by Riesz’s lemma there is aunique two-tensor Tg(F) with

dF(g) · h = 〈h, Tg(F)〉g, ∀h ∈ L2(T ∗M ⊗s T ∗M).

Tg(F) is precisely the L2 gradient of F at g.

Proposition 38 Let F be a functional on Met(M) which is differentiable at themetric g and whose differential dF(g) extends to a continuous functional dF(g) :L2(T ∗M ⊗s T ∗M) → C. Then the differential dFg(0) also extends to a continuousfunctional dFg(0) : L2(M) → C. Identifying the conformal anomaly at g with afunction in L2(M), we have

dFg(0) = 2 trg (Tg(F)) .

In particular, the functional F is conformally invariant iff trg (Tg(F)) = 0 for allmetrics g.

Proof: The differential d(Fg)0 extends to a continuous functional because

dFg(0) · f = dF(g)(2fg) =⇒ dFg(0) · f = dF(g)(2fg)

By Exercise 107,

dFg(0) · f = dF(g) · (2fg) = 〈Tg(F), 2f g〉g = 2 (trg(Tg(F)), f)g ,

as desired. tu

Definition 47 Under the assumptions of the Proposition, the function

x 7→ δxFg := 2trg (Tg(F)) (x)

is called the local anomaly of the functional F at the reference metric g.

140

Exercise 108 This is taken from bosonic string theory.Let (M, g) be a closed Riemann surface and X : M → Rd a smooth map. Let

∆g := − 1√detg

∂i√

detg gij∂j

(where as before det g stands for the determinant of the metric matrix (gij) and (gij)its inverse) denote the Laplace-Beltrami operator on M . The classical Polyakovaction [Po] (see also [AJPS] and references therein for a review) for bosonic stringreads

A(X, g) := 〈∆gX,X〉g. (196)

1. Show that it yields a conformal invariant action depending on X.

2. For any h ∈ C∞(M,T ∗M ⊗s T ∗M), show that

dA(X, ·)(g) · h = 〈h, Tg(x)〉g

where Tg is the two-covariant tensor Tij := ∂iXµ ∂jX

µ − 12 trg(∂iXµ∂jX

µ) gcalled the energy-momentum tensor.

3. Check that trg(Tg) = 0 as a consequence of the conformal invariance of A(X, ·).

7.6 Conformal anomaly of the zeta determinant

If there were a well defined determinant “det” on differential operators with the usualproperties (192), a conformally covariant operator Ag of bidegree (a, b) would yield afunction

“det”(Ae2fg) = “det”(e−b fAgea f )

= “det”(e−b f ) “det”(Ag) “det”(ea f )

= “det”(e(a−b) f ) “det”(Ag),

where ec f is treated as a multiplication operator for c ∈ R. This function would suffera conformal anomaly:

δf log “det”(Ag) = δf log “det”(e(a−b) f ) = (a− b) “tr”(f),

where “tr” is a hypothetical trace associated to “det”.

The ζ-determinant Detζ on operators is used by both physicists and mathematiciansas a substitute for the usual determinant on matrices. The following well-known resultshows that the above heuristic derivation holds replacing the hypothetical trace “tr”by a weighted trace TrAg .The above discussion hints to the fact that regularisation procedures involved in the ζ-determinant are not responsible for the conformal anomaly which arises as soon as oneuses the conformally covariant operator Ag associated to the originally conformallyinvariant classical action A(g). The conformal anomaly computed in the subsequenttheorem therefore has nothing to do with the multiplicative anomaly investigatedpreviously.

141

Theorem 26 [BO], [ParR], [Ros] Let Ag ∈ C`(M,E) be a conformally covariantadmisible elliptic operator of positive order α independent of g. Then the conformalanomaly of the ζ-determinant of Ag is a local expression given by:

δf log Detζ(Ag) = (a− b) TrAg (f) = − 1α

res (f logAg) ∀f ∈ C∞(M).

Proof: By (195), setting At := Ag+2tfg) and A0 = ddt |t=0

At we have

δf log Detζ(Ag) =(d

dtlog Detζ(At)

)t=0

= TrA0

(A0A

−10

)= TrAg

(δfAg A

−1g

)= (a− b) TrAg (f)

since δfAg := (a − b)f Ag. Since the multiplication operator by f is a differentialoperator, it follows from (169) that trAg (f) = − 1

α res (logAg) . tu

Remark 37 In [PayR], we use the Kontsevich-Vishik canonical trace to produce aseries of conformal spectral invariants (or covariants or anomalies) associated to con-formally covariant pseudodifferential operators. Although only one covariant is new,the use of canonical traces provides a systematic treatment of these covariants.

Example 15 The conformal anomaly of the ζ-determinant of the conformally co-variant operator ∆g on a Riemann surface (M, g) is given by the Polyakov formula[Po]:

δf (Detζ(∆g)) =1

24π(〈f,∆gf〉g + 2〈Rg, f〉g) ,

where as before Rg is the scalar curvature. It contributes (see e.g. [AJPS]) to the con-formal anomaly of the partition function for bosonic strings derived from the Polyakovaction (196).

142

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