Notes on Groups SO(3) and SU(2) - New Mexico Institute of ...iavramid/notes/su2.pdf · Notes on...
Transcript of Notes on Groups SO(3) and SU(2) - New Mexico Institute of ...iavramid/notes/su2.pdf · Notes on...
Notes on Groups S O(3) and S U(2)
Ivan G. Avramidi
New Mexico Institute of Mining and Technology
October, 2011
Author: IGA File: su2.tex; Date: November 29, 2011; Time: 10:07
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1 Group S O(3)
1.1 Algebra so(3)
The generators Xi = Xi, i = 1, 2, 3, of the algebra so(3) are 3×3 real antisymmetricmatrices defined by
X1 =
0 0 0
0 0 1
0 −1 0
, X2 =
0 0 −1
0 0 0
1 0 0
, X3 =
0 1 0
−1 0 0
0 0 0
, (1.1)
or(Xi)k
j = −εki j = εik j (1.2)
Raising and lowering a vector index does not change anything; it is done just forconvenience of notation. These matrices have a number of properties
XiX j = E ji − Iδi j , (1.3)
andεi jkXk = Ei j − E ji (1.4)
where(E ji)kl = δk
jδli (1.5)
Therefore, they satisfy the algebra
[Xi, X j] = −εki jXk (1.6)
The Casimir operator is defined by
X2 = XiXi (1.7)
It is equal toX2 = −2I (1.8)
and commutes with all generators
[Xi, X2] = 0 . (1.9)
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A general element of the group S O(3) in canonical coordinates yi has the form
exp (X(y)) = I cos r + P(1 − cos r) + X(y)sin r
r(1.10)
= P + Π cos r + X(y)sin r
r, (1.11)
where
r =√
yiyi, θi =yi
r(1.12)
X(y) = Xiyi, and the matrices Π and P are defined by
Pij = θiθ j (1.13)
Πij = δi
j − θiθ j (1.14)
In particular, for r = π we have
exp (X(y)) = P − Π (1.15)
Notice thattr P = 1, tr Π = 2 (1.16)
and, therefore,
tr exp (X(y)) = 1 + 2 cos r . (1.17)
Also,
detsinh X(y)
X(y)=
(sin r
r
)2
(1.18)
1.2 Representations of S O(3)
Let Xi be the generators of S O(3) in a general irreducible representation satisfyingthe algebra
[Xi, X j] = −εi jkXk (1.19)
They are determined by symmetric traceless tensors of type ( j, j) (with j a positiveinteger)
(Xi)l1...l jk1...k j = − jε(l1
i(k1δl2k2· · · δ
l j)k j)
(1.20)
ThenX2 = − j( j + 1)I (1.21)
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whereIl1...l j
k1...k j = δ(l1(k1· · · δ
l j)k j)
(1.22)
Another basis of generators is (J+, J−, J3), where
J+ = −iX1 + X2 (1.23)J− = −iX1 − X2 (1.24)J3 = −iX3 (1.25)
so thatJ†+ = J− , J†3 = J3. (1.26)
They satisfy the commutation relations
[J+, J−] = 2J3, [J3, J+] = J+, [J3, J−] = −J− . (1.27)
The Casimir operator in this basis is
X2 = −J+J− − J23 + J3 = −J−J+ − J2
3 − J3 (1.28)
From the commutation relations we have
J3J+ = J+(J3 + 1) (1.29)J3J− = J−(J3 − 1) (1.30)J−J+ = −X2 − J2
3 − J3 (1.31)J+J− = −X2 − J2
3 + J3 (1.32)
Since J3 commutes with X2 they can be diagonalized simultaneously. Sincethey are both Hermitian, they have real eigenvalues. Notice also that since Xi areanti-Hermitian, then
X2 ≤ 0 . (1.33)
Let |λ,m〉 be the basis in which both operators are diagonal, that is,
X2|λ,m〉 = −λ|λ,m〉 (1.34)J3|λ,m〉 = m|λ,m〉 (1.35)
There are many ways to show that the eigenvalues m of the operator J3 are integers.Now, since
X21 + X2
2 = X2 − X23 = X2 + J2
3 ≤ 0 (1.36)
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then the eigenvalues of J3 are bounded by a maximal integer j,
|m| ≤ j (1.37)
that is, m takes (2 j + 1) values
− j,− j + 1, · · · ,−1, 0, 1, · · · , j − 1, j (1.38)
¿From the commutation relations above we have
J+|λ,m〉 = const |λ,m + 1〉 (1.39)J−|λ,m〉 = const |λ,m − 1〉 (1.40)
Let |λ, 0〉 be the eigenvector corresponding to the eigenvalue m = 0, that is
J3|λ, 0〉 = 0 . (1.41)
Then for positive m > 0
|λ,m〉 = aλmJm+ |λ, 0〉 (1.42)
|λ,−m〉 = aλ,−mJm− |λ,−m〉 (1.43)
Then
J+|λ, j〉 = 0 (1.44)J−|λ,− j〉 = 0 (1.45)
Therefore,
0 = J−J+|λ, j〉 =(−X2 − J2
3 − J3
)|λ, j〉 (1.46)
0 = J+J−|λ,− j〉 =(−X2 − J2
3 + J3
)|λ,− j〉 (1.47)
Thus, (λ2 − j2 − j
)|λ, j〉 = 0 (1.48)
So, the eigenvalues of the Casimir operator X2 are labeled by a non-negative inte-ger j ≥ 0,
λ j = j( j + 1) . (1.49)
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These eigenvalues have the multiplicity
d j = 2 j + 1 . (1.50)
In particular, for j = 1 we get λ1 = 2 as expected before.From now on we will denote the basis in which the operators X2 and J3 are
diagonal by
|λ j,m〉 =
∣∣∣∣∣ jm
⟩. (1.51)
The matrix elements of the generators J± can be computed as follows. First of all,⟨ jm
∣∣∣∣∣ J3
∣∣∣∣∣ jm′
⟩= δmm′m (1.52)
We havej( j + 1) =
⟨ jm
∣∣∣∣∣ J+
∣∣∣∣∣ jm − 1
⟩ ⟨ jm − 1
∣∣∣∣∣ J−∣∣∣∣∣ jm
⟩+ m2 − m (1.53)
and ⟨ jm
∣∣∣∣∣ J+
∣∣∣∣∣ jm − 1
⟩=
⟨ jm − 1
∣∣∣∣∣ J−∣∣∣∣∣ jm
⟩(1.54)
This gives ⟨ jm
∣∣∣∣∣ J+
∣∣∣∣∣ jm′
⟩= δm′,m−1
√( j + m)( j − m + 1) (1.55)
⟨ jm
∣∣∣∣∣ J−∣∣∣∣∣ jm′
⟩= δm′,m+1
√( j − m)( j + m + 1) (1.56)
The matrix elements of the generators Xi are⟨ jm
∣∣∣∣∣ X1
∣∣∣∣∣ jm′
⟩= δm′,m−1
i2
√( j + m)( j − m + 1)
+δm′,m+1i2
√( j − m)( j + m + 1) (1.57)⟨ j
m
∣∣∣∣∣ X2
∣∣∣∣∣ jm′
⟩= δm′,m−1
12
√( j + m)( j − m + 1)
−δm′,m+112
√( j − m)( j + m + 1) (1.58)⟨ j
m
∣∣∣∣∣ X3
∣∣∣∣∣ jm′
⟩= δmm′im (1.59)
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Thus, irreducible representations of S O(3) are labeled by a non-negative inte-ger j ≥ 0. The matrices of the representation j in canonical coordinates yi will bedenoted by
D jmm′(y) =
⟨ jm
∣∣∣∣∣ exp[X(y)
] ∣∣∣∣∣ jm′
⟩(1.60)
where X(y) = Xiyi.The characters of the irreducible representations are
χ j(y) =
j∑m=− j
D jmm(y) =
j∑m=− j
eimr = 1 + 2j∑
m=1
cos(mr) (1.61)
2 Group S U(2)
2.1 Algebra su(2)
Pauli matrices σi = σi, i = 1, 2, 3, are defined by
σ1 =
0 1
1 0
, σ2 =
0 −i
i 0
, σ3 =
1 0
0 −1
. (2.1)
They have the following properties
σ†i = σi (2.2)σ2
i = I (2.3)detσi = −1 (2.4)
trσi = 0 , (2.5)σ1σ2 = iσ3 (2.6)σ2σ3 = iσ1 (2.7)σ3σ1 = iσ2 (2.8)
σ1σ2σ3 = iI (2.9)(2.10)
which can be written in the form
σiσ j = δi jI + iεi jkσk . (2.11)
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They satisfy the following commutation relations
[σi, σ j] = 2iεi jkσk (2.12)
The traces of products are
trσiσ j = 2δi j (2.13)trσiσ jσk = 2iεi jk (2.14)
Also, there holdsσiσi = 3I . (2.15)
The Pauli matrices satisfy the anti-commutation relations
σiσ j + σ jσi = 2δi jI (2.16)
Therefore, Pauli matrices form a representation of Clifford algebra in 3 dimen-sions and are, in fact, Dirac matrices σi = (σi
AB), where A, B = 1, 2, in 3 dimen-
sions. The spinor metric E = (EAB) is defined by
E =
0 1
−1 0
. (2.17)
Notice thatE = iσ2 (2.18)
and the inverse metric E−1 = (EAB) is
E−1 = −E (2.19)
ThenσT
i = −EσiE−1 (2.20)
This allows to define the matrices Eσi = (σi AB) with covariant indices
Eσ1 = σ3 =
1 0
0 −1
, (2.21)
Eσ2 = iI =
i 0
0 i
, (2.22)
Eσ3 = −σ1 =
0 −1
−1 0
. (2.23)
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Notice that the matrices Eσi are all symmetric
(Eσi)T = Eσi (2.24)
Pauli matrices satisfy the following completeness relation
σiA
BσiC
D = 2δADδ
CB − δ
ABδ
CD (2.25)
which impliesσi
(A(Bσi
C)D) = δ(A
(BδC)
D) (2.26)
This givesσi ABσi CD = 2EADECB − EABECD (2.27)
In a more symmetric form
σi ABσi CD = EADECB + EBDECA (2.28)
A spinor is a two dimensional complex vector ψ = (ψA), A = 1, 2 in the spinorspace C2. We use the metric E to lower the index to get the covariant components
ψA = EABψB (2.29)
which defines the cospinor ψ = (ψA)
ψ = Eψ (2.30)
We also define the conjugated spinor ψ = (ψA) by
ψA = ψA∗ (2.31)
On the complex spinor space there are two invariant bilinear forms
〈ψ, ϕ〉 = ψϕ = ψA∗ϕA (2.32)
andψϕ = ψAϕ
A = EABψBϕA = ψ1ϕ2 − ψ2ϕ1 (2.33)
Note that〈ψ, ψ〉 = ψψ = ψA∗ψA = |ψ1|2 + |ψ2|2 (2.34)
andψψ = ψAψ
A = 0 (2.35)
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The matricesXi =
i2σi (2.36)
are the generators of the Lie algebra su(2) with the commutation relations
[Xi, X j] = −εi jkXk (2.37)
The Casimir operator isX2 = XiXi (2.38)
It commutes with all generators
[Xi, X2] = 0 . (2.39)
There holdsX2 = −
34
I (2.40)
tr XiX j = −12δi j (2.41)
Notice thatX2 = − j( j + 1)I (2.42)
with j = 12 , which determines the fundamental representation of S U(2).
An arbitrary element of the group S U(2) in canonical coordinates yi has theform
exp( i2σ jy j
)= I cos
r2
+ iσ jθj sin
r2, (2.43)
This can be written in the form
exp (X(y)) = I cos(r/2) + X(y)sin(r/2)
r/2. (2.44)
where X(y) = Xiyi. In particular, for r = 2π
exp (X(y)) = −I. (2.45)
Obviously,tr exp (X(y)) = 2 cos(r/2) (2.46)
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2.2 Representations of S U(2)
We define symmetric contravariant spinors ψA1...A2 j of rank 2 j. Each index herecan be lowered by the metric EAB. The number of independent components ofsuch spinor is precisely
2 j∑i=0
1 = (2 j + 1) (2.47)
Let Xi be the generators of S U(2) in a general irreducible representation sat-isfying the algebra
[Xi, X j] = −εi jkXk (2.48)
They are determined by symmetric spinors of type (2 j, 2 j)
(Xk)A1...A2 jB1...B2 j = i jσk
(A1(B1δ
A2B2· · · δ
A2 j)B2 j)
(2.49)
ThenX2 = − j( j + 1)I (2.50)
whereIA1...A2 j
B1...B2 j= δ(A1
(B1· · · δ
A2 j)B2 j)
(2.51)
Another basis of generators is
J+ = −iX1 + X2 (2.52)J− = −iX1 − X2 (2.53)J3 = −iX3 (2.54)
They have all the properties of the generators of S O(3) discussed above exceptthat the operator J3 has integer eigenvalues m. One can show that they can beeither integers or half-integers taking (2 j + 1) values
− j,− j + 1, . . . , j − 1, j , |m| ≤ j (2.55)
with j being a positive integer or half-integer. If j is integer, then all eigenvaluesm of J3 are integers including 0. If j is a half-integer, then all eigenvalues m arehalf-integers and 0 is excluded.
We choose a basis in which the operators X2 and J3 are diagonal. Let ψ =
ψA1...A2 j be a symmetric spinor of rank 2 j. Let e1 = (e1A) and e2 = (e2
A) be thebasis cospinors defined by
e1A = δA
1 , e2A = δA
2 . (2.56)
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Let
e j,mA1...A j+mB1...B j−m = e1
(A1 · · · e1A j+me2
B1 · · · e2B j−m)
= δ(A11 · · · δ
A j+m
1 δB12 · · · δ
B j−m)2 (2.57)
Then this is an eigenspinor of the operator J3 with the eigenvalue m
J3e j,m = me j,m (2.58)
Then the spinors ∣∣∣∣∣ jm
⟩=
((2 j)!
( j + m)!( j − m)!
)1/2
e j,m (2.59)
form an orthonormal basis in the space of symmetric spinors of rank 2 j.The matrix elements of the generators Xi are given by exactly the same formu-
las ⟨ jm
∣∣∣∣∣ X1
∣∣∣∣∣ jm′
⟩= δm′,m−1
i2
√( j + m)( j − m + 1)
+δm′,m+1i2
√( j − m)( j + m + 1) (2.60)⟨ j
m
∣∣∣∣∣ X2
∣∣∣∣∣ jm′
⟩= δm′,m−1
12
√( j + m)( j − m + 1)
−δm′,m+112
√( j − m)( j + m + 1) (2.61)⟨ j
m
∣∣∣∣∣ X3
∣∣∣∣∣ jm′
⟩= δmm′im (2.62)
but now j and m are either both integers or half-integers.Thus, irreducible representations of S U(2) are labeled by a non-negative half-
integer j ≥ 0. The matrices of the representation j in canonical coordinates yi
areD j
mm′(y) =
⟨ jm
∣∣∣∣∣ exp[X(y)
] ∣∣∣∣∣ jm′
⟩(2.63)
where X(y) = Xiyi. The representations of S U(2) with integer j are at the sametime representations of S O(3). However, representations with half-integer j donot give representations of S O(3), since for r = 2π
D jmm′(y) = −δmm′ . (2.64)
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The characters of the irreducible representations are: for integer j
χ j(y) =
j∑m=− j
D jmm(y) =
j∑m=− j
eimr = 1 + 2j∑
m=1
cos(mr) (2.65)
and for half-integer j
χ j(y) =
− 12 + j∑
k=− 12− j
D j12 +k, 1
2 +k(y) =
− 12 + j∑
k=− 12− j
exp[i(12
+ k)
r]
= 2j∑
k=1
cos[(
12
+ k)
r]
(2.66)Now, let Xi and Y j be the generators of two representations and let
Gi = Xi ⊗ IY + IX ⊗ Yi (2.67)
Then Gi generate the product representation X ⊗ Y and
exp[G(y)] = exp[X(y)] exp[Y(y)] (2.68)
Therefore, the character of the product representation factorizes
χG(y) = χX(y)χY(y) (2.69)
2.3 Double Covering Homomorphism S U(2)→ S O(3)
Recall that the matrices Eσi = (σi AB) are symmetric and the matrix E = (EAB) isanti-symmetric. Let ψAB be a symmetric spinor. Then one can form a vector
Ai = σi BAψAB = tr (Eσiψ) . (2.70)
Conversely, given a vector Ai we get a symmetric spinor of rank 2 by
ψAB =12
Aiσi AB (2.71)
orψ =
12
AiσiE−1 (2.72)
More generally, given a symmetric spinor of rank 2 j we can associate to it asymmetric tensor of rank j by
Ai1...i j = σi1 A1B1 · · ·σi j A jB jψA1...A jB1...B j (2.73)
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One can show that this tensor is traceless. Indeed by using the eq. (??) and thefact that the spinor is symmetric we see that the tensor A is traceless. The inversetransformation is defined by
ψA1...A jB1...B j =12 j A
i1...i jσi1 A1B1 · · ·σi j A jB j (2.74)
The double covering homomorphism
Λ : S U(2)→ S O(3) (2.75)
is defined as follows. Let U ∈ S U(2). Then the matrices UσiU−1 satisfy all theproperties of the Pauli matrices and are therefore in the algebra su(2). Therefore,
UσiU−1 = Λ ji(U)σ j . (2.76)
That is,
Λij(U) =
12
trσiUσ jU−1 (2.77)
The matrix Λ(U) depends on the matrix U and is, in fact, in S O(3). For infinites-imal transformations
U = exp(Xkyk) = I +i2σkyk + · · · (2.78)
the matrix Λ isΛi j = δi j + εi jkyk + · · · =
[exp(Xkyk)
]i j
(2.79)
It is easy to see thatΛ(I) = Λ(−I) = I . (2.80)
So, in shortΛ(exp(Xiyi)) = exp(Λ(Xi)yi) (2.81)
where[Λ(σk)]i
j = 2iεik j (2.82)
3 Invariant Vector FieldsLet yi be the canonical coordinates on the group S U(2) ranging over (−π, π). Theposition of the indices on the coordinates will be irrelevant, that is, yi = yi . Weintroduce the radial coordinate r = |y| =
√yiyi , so that the geodesic distance
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from the origin is equal to r, and the angular coordinates (the coordinates on S 2)θi = yi = yi/r .
Let us consider the fundamental representation of the group S U(2) with gen-erators Ta = iσa, where σa are the Pauli matrices satisfying the algebra
[Ta,Tb] = −2εcabTc. (3.83)
Let T (y) = Tiyi. Then
T (y)T (p) = −(y, p) − T (y ∧ p) (3.84)
where (y, p) = yi pi, (y ∧ p)k = εi jkyi p j and
[T (y)]2 = −|y|2I (3.85)
Therefore,exp[rT (y)] = cos r + T (y) sin r (3.86)
where y is a unit vector. Next, we compute
exp[T (F(rq, ρp))] = cos r cos ρ − (q, p) sin r sin ρ+ cos r sin ρT ( p) + cos ρ sin rT (q)− sin r sin ρT (q ∧ p) (3.87)
This defines the group multiplication map F(q, p) in canonical coordinates.This map has a number of important properties. In particular, F(0, p) =
F(p, 0) = p and F(p,−p) = 0. Another obvious but very useful property ofthe group map is that if q = F(ω, p), then ω = F(q,−p) and p = F(−ω, q). Also,there is the associativity property
F(ω, F(p, q)) = F(F(ω, p), q) (3.88)
and the inverse property
F(−p,−ω) = −F(ω, p). (3.89)
This map defines all the properties of the group. We introduce the matrices
Lαb(x) =∂
∂yb Fα(y, x)∣∣∣∣y=0
, (3.90)
Rαb(x) =
∂
∂yb Fα(x, y)∣∣∣∣y=0
, (3.91)
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Let X and Y be the inverses of the matrices L and R, that is,
LX = XL = RY = YR = I . (3.92)
Also letD = XR , D−1 = YL (3.93)
so thatL = RD−1, X = DY (3.94)
Let Ca be the matrices of the adjoint representation defined by
(Ca)bc = −2εb
ac (3.95)
and C = C(x) be the matrix defined by
C = Caxa . (3.96)
Then one can show that
X =exp(C) − I
C, Y =
I − exp(−C)C
, (3.97)
L =C
exp(C) − I, R =
CI − exp(−C)
. (3.98)
It is easy to see also that
xa = Labxb = Ra
bxb = Xabxb = Ya
bxb . (3.99)
andX = YT , L = RT (3.100)
and
XY = YX =
(sinh(C/2)
C/2
)2
, (3.101)
LR = RL =
(C/2
sinh(C/2)
)2
, (3.102)
as well as
D = XR = RX = exp(C) , DT = YL = LY = exp(−C) . (3.103)
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Note thatC2 = −4r2Π , (3.104)
where r = |x| andΠi
j = δij − θ
iθ j (3.105)
with θi = xi/r. Therefore,
C2n = (−1)n(2r)2nΠ , C2n+1 = (−1)n(2r)2nC (3.106)
Therefore, the eigenvalues of the matrix C are (2ir,−2ir, 0), and for any analyticfunction
f (C) = f (0)(I − Π) + Π12
[f (2ir) + f (−2ir)
]+
14ir
C[f (2ir) − f (−2ir
]and, therefore,
tr f (C) = f (0) + f (2ir) + f (−2ir) (3.107)det f (C) = f (0) f (2ir) f (−2ir) (3.108)
This enables one to compute
Y = I − Π +sin r
rcos rΠ −
12
sin2 rr2 C (3.109)
X = I − Π +sin r
rcos rΠ +
12
sin2 rr2 C (3.110)
R = I − Π + r cot rΠ +12
C (3.111)
L = I − Π + r cot rΠ −12
C (3.112)
D = I − Π + Π cos(2r) +sin(2r)
2rC (3.113)
Then one can show that∂
∂xµFα(x, y) = Lαγ(F(x, y))Xγ
µ(x) , (3.114)
∂
∂xµFα(y, x) = Rα
γ(F(y, x))Yγµ(x) . (3.115)
These matrices satisfy important identities
∂µXaν − ∂νXa
µ = −2εabcXb
µXcν , (3.116)
∂µYaν − ∂νYa
µ = 2εabcYb
µYcν . (3.117)
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17
and
Lµa∂µLνb − Lµb∂µLνa = 2εcabLγc , (3.118)
Rµa∂µRν
b − Rµb∂µRν
a = −2εcabRν
c . (3.119)
Let us define the one-forms
σa− = Xa
µ(x)dxµ , σa+ = Ya
µ(x)dxµ .
They satisfy important identities
dσa− = −εa
bcσb− ∧ σ
c− , (3.120)
dσa+ = εa
bcσb+ ∧ σ
c+ . (3.121)
Next, we define the vector fields.
K−a = Lµa(x)∂µ , K+a = Rµ
a(x)∂µ .
These are the right-invariant and the left-invariant vector fields on the group. Theyare related by
K−a = DabK+
b (3.122)
They form two representations of the group S U(2) and satisfy the algebra
[K+a ,K
+b ] = −2εc
abK+c (3.123)
[K−a ,K−b ] = 2εc
abK−c (3.124)[K+
a ,K−b ] = 0 (3.125)
That is, the left-invariant vector fields commute with the right-invariant ones.Now, let Ta be the generators of some representation of the group S U(2) sat-
isfying the same algebra. Then, of course,
exp[T (ω)] exp[T (p)] = exp[T (F(ω, p))] (3.126)
First, one can derive a useful commutation formula
exp[T (ω)]Tb exp[−T (ω)] = DabTa , (3.127)
where the matrix D = (Dab) is defined by D = exp C .
More generally,
exp[T (ω)] exp[T (p)] exp[−T (ω)] = exp[T (D(ω)p)] . (3.128)
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18
Next, one can show that
exp[−T (ω)]∂
∂ωi exp[T (ω)] = YaiTa , (3.129)
This means that
∂
∂ωi exp[T (ω)] = Yai exp[T (ω)]Ta = Yi
aTa exp[T (ω)] , (3.130)
The Killing vectors have important properties
K+a exp[T (ω)] = (exp[T (ω)])Ta (3.131)
K−a exp[T (ω)] = Ta(exp[T (ω)]) (3.132)
This immediately leads to further important equations
exp[K+(p)] exp[T (ω)] = exp[T (ω)] exp[T (p)] (3.133)
andexp[K−(q)] exp[T (ω)] = exp[T (q)] exp[T (ω)] (3.134)
where K±(p) = paK±a . More generally,
exp[K+(p) + K−(q)] exp[T (ω)] = exp[T (q)] exp[T (ω)] exp[T (p)] (3.135)
and, therefore,
exp[K+(p) + K−(q)] exp[T (ω)]∣∣∣∣ω=0
= exp[T (q)] exp[T (p)] = exp[T (F(q, p))](3.136)
In particular, for the adjoint representation this formulas take the form
DCb = DabCaD (3.137)
K+a D = DCa (3.138)
K−a D = CaD (3.139)
exp[K+(p)]D(x) = D(x)D(p) (3.140)
exp[K−(q)]D(x) = D(q)D(x) (3.141)
exp[K+(p) + K−(q)]D(x) = D(q)D(x)D(p) (3.142)
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19
Then for a scalar function f (ω) the action of the right-invariant and left-invariant vector fields is simply
exp[K+(p)] f (ω) = f (F(ω, p)), exp[K−(q)] f (ω) = f (F(q, ω)). (3.143)
Since the action of these vector fields commutes, we have more generally
exp[K+(p) + K−(q)] f (ω) = f (F(q, F(ω, p))). (3.144)
Therefore,exp[K+(p) + K−(q)] f (ω)
∣∣∣∣ω=0
= f (F(q, p)). (3.145)
The bi-invariant metric is defined by
gαβ = δabYaαYb
β = δabXaαXb
β (3.146)gαβ = δabLαaLβb = δabRα
aRβb , (3.147)
that is,
g =
(sinh[C/2]
C/2
)2
= I − Π +sin2 r
r2 Π . (3.148)
The Riemannian volume elements of the metric g is defined as usual
dvol = g1/2(x)dx1 ∧ dx2 ∧ dx3 (3.149)
where
g1/2 = (det gµν)1/2 =sin2 r
r2 . (3.150)
Thus the bi-invariant volume element of the group is
dvol (x) =sin2 r
r2 dx = sin2 r dr dθ (3.151)
where dx = dx1dx2dx3, and dθ is the volume element on S 2. The invariance ofthe volume element means that for any fixed p
dvol (x) = dvol (F(x, p)) = dvol (F(p, x)) (3.152)
Notice also, that |F(p, q)| is nothing but the geodesic distance between the pointsp and q.
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20
We will always use the orthonormal basis of the right-invariant vector fieldsK+
a and one-forms σa+ and denote the covariant derivative with respect to K+
a sim-ply by ∇a. The Levi-Civita connection ∇ of the bi-invariant metric in the right-invariant basis is defined by
∇aK+b = εc
baK+c (3.153)
so that the coefficients of the affine connection are
ωabc = σa
+(∇cK+b ) = εa
bc . (3.154)
so that for an arbitrary vector
∇aV =[(K+
a Vb) + εbcaVc
]K+
b (3.155)
Then∇aK−d = Mb
daK+b (3.156)
where
Mbda = K+
a Ddb + εb
caDdc
= −2εbcaDd
c + εbcaDd
c = −εbcaDd
c (3.157)
so that∇aK−d = −εb
caDdcK+
b (3.158)
Now, the curvature tensor is
Rabcd = −ε f
cdεa
f b . (3.159)
The Ricci curvature tensor and the scalar curvature are
Rab = 2δab , R = 6 . (3.160)
The scalar Laplacian is
∆0 = δabK−a K−b = δabK+a K+
b . (3.161)
One can show that∆0|ω|
sin |ω|=|ω|
sin |ω|(3.162)
Further,∆0 exp[T (ω)] = (exp[T (ω)])T 2 = T 2 exp[T (ω)] (3.163)
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21
where T 2 = TaT a. This means that
exp[t∆20] exp[T (ω)] = exp(tT 2) exp[T (ω)] (3.164)
The Killing vector fields K±a are divergence free, which means that they areanti-self-adjoint with respect to the invariant measure dvol (ω). that is, Thus theLaplacian ∆0 is self-adjoint with respect to the same measure as well.
Let ∇ be the total connection on a vector bundle V realizing the representationG. Let us define the one-forms
A = σa+Ga = GaYa
µdxµ. (3.165)
and F = dA +A∧A. Then, by using the above properties of the right-invariantone-forms we compute
F =12Fabσ
a+ ∧ σ
b+ =
12FabYa
µYbνdxµ ∧ dxν (3.166)
whereFab = εc
abGc (3.167)
ThusA is the Yang-Mills curvature on the vector bundle V with covariantly con-stant curvature.
The covariant derivative of a section of the bundle V is then (recall that ∇a =
∇K+a )
∇aϕ = (K+a + Ga)ϕ (3.168)
and the covariant derivative of a section of the endomorphism bundle End(V)along the right-invariant basis is then
∇aQ = K+a Q + [Ga,Q] (3.169)
Therefore, we find∇aGb = εc
baGc + [Ga,Gb] = 0 . (3.170)
Then the derivatives along the left-invariant vector fields are
∇K−a ϕ = (K−a + Ba)ϕ (3.171)
∇K−a Q = K−a Q + [Ba,Q] (3.172)
whereBa = Da
bGb (3.173)
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22
The Laplacian takes the form
∆ = ∇a∇a = (K+
a + Ga)(K+a + Ga) = ∆0 + 2GaK+
a + G2 (3.174)
where G2 = GiGi.We want to rewrite the Laplacian in terms of Casimir operators of some rep-
resentations of the group S U(2). The covariant derivatives ∇a do not form a rep-resentation of the algebra S U(2). The operators that do are the covariant Liederivatives. The covariant Lie derivatives along a Killing vector ξ of sections ofthis vector bundle are defined by
Lξ = ∇ξ −12σa
+(∇bξ)εbcaGc (3.175)
By denoting the Lie derivatives along the Killing vectors K±a by K±a this gives forthe right-invariant and the left-invariant bases
K+a = LK+
a = ∇a + Ga = K+a + 2Ga . (3.176)
K−a = LK−a = ∇K−a − Ba = K−a . (3.177)
It is easy to see that these operators form the same algebra S U(2) × S U(2)
[K+a ,K
+b ] = −2εc
abK+c (3.178)
[K−a ,K−b ] = 2εc
abK−c (3.179)
[K+a ,K
−b ] = 0 (3.180)
Now, the Laplacian is given now by the sums of the Casimir operators
∆ = K2 −G2 (3.181)
whereK2 =
12K2
+ +12K2− (3.182)
and K2± = K±aK
±a .
4 Heat Kernel on S 3
Our goal is to evaluate the heat kernel diagonal. Since it is constant we can eval-uate it at any point, say, at the origin. We use the geodesic coordinates yi defined
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23
above. That is why, we will evaluate the heat kernel when one point is fixed at theorigin. We will denote it simply by U(t; y). We obviously have
exp(t∆) = exp(−tG2
)exp
(tK2
). (4.1)
Therefore, the heat kernel is equal to
U(t, y) = exp(−tG2
)Ψ
( t2, y
)(4.2)
where Ψ(t, y) = exp(2tK2)δ(y) is the heat kernel of the operator 2K2.
4.1 Scalar Heat KernelLet
Φ(t, ω) = (4πt)−3/2et∞∑
n=−∞
|ω| + 2πnsin |ω|
exp(−
(|ω| + 2πn)2
4t
)(4.3)
One can show by direct calculation that this function satisfies the equation
∂tΦ = ∆20Φ (4.4)
with the initial conditionΦ(0, ω) = δS 3(ω) (4.5)
Therefore, this is the scalar heat kernel on the group S U(2) and, therefore, on S 3.Now, let us compute the integral
Ψ(t) =
∫S U(2)
dvol (ω)Φ(t, ω) exp[T (ω)] (4.6)
Obviously, Ψ(0) = 1. Next, we have
∂tΨ =
∫S U(2)
dvol (ω) exp[T (ω)]∆0Φ(t, ω) (4.7)
Now, by integrating by parts we get
∂tΨ =
∫S U(2)
DωΦ(t, ω)∆0 exp[T (ω)] (4.8)
which gives∂tΨ = ΨT 2 (4.9)
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24
Thus, we obtain a very important equation
exp(tT 2) =
∫S U(2)
dvol (ω) Φ(t, ω) exp[T (ω)] (4.10)
This is true for any representation of S U(2).We derive two corollaries. First, we get
exp(tT 2) =
∫S U(2)
dvol (ω) Φ(t, ω) exp[T (p)] exp[T (ω)] exp[−T (p)] (4.11)
which means that for any p
Φ(t, ω) = Φ(t, F(−p, F(ω, p)). (4.12)
orΦ(t, F(p, q)) = Φ(t, F(q, p)). (4.13)
Also, we see that
exp[(t + s)T 2] =
∫S 3×S 3
dvol (q)dvol (p)Φ(t, q)Φ(s, p) exp[T (q)] exp[T (p)] (4.14)
By changing the variables z = F(q, p) and p 7→ −p we obtain
exp[(t + s)T 2] =
∫S U(2)
dvol (z)∫
S U(2)dvol (p)Φ(t, F(p, z))Φ(s, p) exp[T (z)]
(4.15)which means that∫
S U(2)dvol (p)Φ(t, F(p, z))Φ(s, p) = Φ(t + s, z) (4.16)
In particular, for z = 0 and s = t this becomes∫S U(2)
dvol (p)Φ(t, p)Φ(t, p) = Φ(t, 0) (4.17)
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