Intersecting Branes Geometry F.Fucito Introduction ... · Intersecting Branes and Enumerative...

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Intersecting Branes and Enumerative Geometry F.Fucito Introduction Overview D-branes D-branes T-duality Tilted branes Bound states of Branes Bound states Localization The D(-1)-D7 system Branes at angles SW curves Intersecting Branes and Enumerative Geometry F.Fucito, M.Bill` o, M.L.Frau, A.Lerda, J.F.Morales, R.Poghosyan INFN sez. Roma 2 Universit ´ a di Roma, Tor Vergata Trieste ’10 April 23, 2010

Transcript of Intersecting Branes Geometry F.Fucito Introduction ... · Intersecting Branes and Enumerative...

Page 1: Intersecting Branes Geometry F.Fucito Introduction ... · Intersecting Branes and Enumerative Geometry F.Fucito Introduction Overview D-branes D-branes T-duality Tilted branes Bound

Intersecting Branesand Enumerative

Geometry

F.Fucito

IntroductionOverview

D-branesD-branesT-dualityTilted branes

Bound states of BranesBound statesLocalization

The D(-1)-D7 systemBranes at anglesSW curves

Intersecting Branes and EnumerativeGeometry

F.Fucito, M.Billo, M.L.Frau, A.Lerda, J.F.Morales,R.Poghosyan

INFN sez. Roma 2Universita di Roma, Tor Vergata

Trieste ’10 – April 23, 2010

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Overview

I This talk is about non perturbative effects (NP)in SUSY gauge theories with an emphasis onthe mathematics methods

I For us NP=instantonsI Many interesting phenomena are NP:

• confinement (quantum vacuum)• SUSY breaking, massive and Yukawa terms in

the SM (additional terms with fermions)

I NP can be computed a la SW or via instantoncalculus

I Both ways are connected to EnumerativeGeometry: SW invariants or Gromov-Wittenand Donaldson invariants

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D-branes

I The closed string solution (p0 = 1/√

2α′(αµ0 + αµ0 ))

Xµ(τ, σ) = qµ +√

2α′(αµ0 + αµ0 )τ −√

2α′(αµ0 − αµ0 )σ

+i

√α′

2

∑n6=0

(αµnn

e−2in(τ−σ) +αµnn

e−2in(τ+σ)

)= XµL + XµR

I In the non compact caseXµ(σ = 0) = Xµ(σ = π) =⇒ αµ0 = αµ0

I Let X ∼ X + 2πR in one spatial direction

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I Periodicity in σ and X leads toπ√

2α′(αµ0 − αµ0 ) = 2πmR or

α0 =

√α′

2

(nR

+mRα′

)α0 =

√α′

2

(nR−

mRα′

)

I In turn L + L ∼ p2 + M2 and

M2 =2α′

[ ∞∑n=1

(αµ−nαµn + αµ−nαµn )− 2

]+

(nR

)2

+

(mRα′

)2

I The spectrum is invariant underm←→ n; R −→ R ≡ α′/R. This is T-duality

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I Generalizing to αn → αn; αn → −αn; n ∈ Z we seethe spectrum is T dual

I Also the partition function and correlators areT dual

I Finally is is easy to see∂τX → −∂σX = ∂τ X ; ∂σX → −∂τX = ∂σX : T-dualityimplies XR ⇔ −XR ; XL ⇔ XL It is a parity exchangeoperator for XR

I Open strings are not periodic in σ. Is T-dualityacting on them?

I Take an open string with d − p − 1 compactdirections of radius R• Since there are no winding modes for R → 0

we “loose” d − p − 1 directions and theparticle mass →∞

• The theory seems to live in p + 1 dimensions• In the same limit, closed strings have heavy

KK modes, but a continuum of windingmodes. The number of dimensions stays thesame

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I Taking now Xµ = XµL + Xµ

R and substituting theoscillator expansion, we find the open stringsolution with N-N b.c.

I Doing the same with Xµ = XµR − Xµ

L we find theopen string solution with D-D b.c.

I T-duality exchanges D and N b.c.

D-branes D-branes

open string

. . . . . .

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Tilted branes

I Adding gauge d.o.f. means |α >−→ |α, i, j > with|α, i′ >= Ui′ i |α, i > and |α, j′ > U†

j′ j|α, j >

I Let’s include a U(N) gauge fieldA = diag(θ1, . . . , θN)/2πR along a certaincompactified direction. This is pure gauge,

given that g = diag(eiθ1

2πR X , . . . , eiθN2πR X ).

I Given that X(0) ∼ X(π)+ 2πR parallel transportaround the compactified dimension, leads to

U = ei∫ 2πR

0 AdX = diag(eiθ1 , . . . , eiθN )

I Now |α, i, j >= ei(θj−θi )|α, i, j > and p = nR +

(θj−θi )

2πR

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I Going to complex coordinates we find (in theinternal directions)

∂Z i(z) = i√

2α′

∞∑n=1

αn−θiz−n+θi−1 +

∞∑n=0

β†n+θizn+θi−1

[αin−θi

, α†jm−θj

] = (n− θi)δijδn,m, [β i

n+θi, β†j

m+θj] = (n + θi)δ

ijδn,m

bin−θi

, b†jm−θj

= ain+θi

, a†jm+θj = δijδn,m

withαi

n−θi|θ >NS= β i

n+θi|θ >NS= bi

n−θi|θ >NS= ai

n+θi|θ >NS= 0.

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D(−1)−D3 system

I The mass relation

(Lxψ0 + N(Z) + N(ψ) −

12

+12

3∑i=1

θi)|θ >NS= 0

I For NS if θi = 1/2 then b†j1/2−θj=⇒ b†j0 with

bi0, b†j0 = δij a bosonic “spinor”

I Massless states with θ3 = 0, θ1,2 = 1/2 are|θ >NS , b†10 b†20 |θ >NS , b†10 |θ >NS , b†20 |θ >NS︸ ︷︷ ︸.

I Their mass is (this is wα, the instanton radius)

2α′M2 =

−θ1 + θ2 + θ3−θ2 + θ1 + θ3

= 0

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Bound states

I Bound states of branes lead to instantons

I D(-1)-D3 brane systems leads to the usualgauge instantons in 4 dimensions

I The zero mass sector of the D(-1) gives themoduli

I This analysis is equivalent to instanton calculusi.e. a computation of the functional integralfrom the lagrangian of the susy gauge theoryexpanded around the saddle point given byinstanton solutions

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The localization theorem

I Let M be acted upon by a Lie group withalgebra g, ξ∗ = ξα T i

α∂

∂x i is the fundamentalvector field

I Introduce forms α ∈ C[g]⊗ Ω(M) with agradation deg(P ⊗ β) = 2 deg(P) + deg(β)

I Introduce a differential(Dα)(ξ) = d(α(ξ))− iξ∗α(ξ)

I Given an equivariant closed form,α, acompact manifold, M, group, G, and avector field ξ∗ with isolated zeroes∫

Mα(ξ) = (−2π)n/2

∑x0

α0(ξ)(x0)

det12 Lx0

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I With

Lx0(v) = [ξ∗, v ] = −ξα v i

(∂T j

α

∂x i

)x0

∂x j

I The application of this to our case requiressome comments:

1. a susy version must be worked out2. the moduli space of instantons must be

smoothed out3. an auxiliary rotation must be introduced

(torus action)4. an appropriate setting for the application of

the th. must be found

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I We identify our vector field with SUSYQ∗ = (Q∗)i

B∂∂Bj + (Q∗)i

F∂∂F j

Zk =

∫ Dφ

U(k)DBDFe−S =

∫ k∏I=1

dϕI

∏I<J ϕ2

IJ

SdetL

≡∑x0

1

SdetLx0

with (pay attention to the role of ϕI) jump

SdetL =

∂(Q∗)iB

∂F j∂(Q∗)i

B∂Bj

∂(Q∗)iF

∂F j∂(Q∗)i

F∂Bj

=

∏I

1(ϕI − a)(ϕI − a − ε)

∏I<J

(ϕ2IJ − ε2)

(ϕIJ − ε1)(ϕIJ − ε2)

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I Branes at angles are important because theycan reproduce the SM of particle physics

I Their presence modifies the moduli spaces ofinstantons

I We saw before certain bosonic moduli weremissing. Therefore certain cancellations donot happen anymore

I This has been used to generate a mass termfor neutrinos (seesaw mechanism) andYukawa couplings (to generate a hierarchy):the mass of fermions in the SM is always aproblem due to the different interactions ofthe left and right handed Wyel fermions

I Here i will focus on a more mathematicalproblem: are these new effects due to newinstantons? And if yes, how do they look like?

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I To answer the previous questions we havefocalized on the D(-1)D7 brane system. Thereare many advantages to do so

1. The number of N-D mixed b.c. are eight here.Going to the mass formula for the moduli, it iseasy to see this gives the same structure ofthe moduli spaces with branes at angles: theinstantons have radii equal to zero.

2. This system can be described in type I’ stringtheory which is known to be dual to tendimensional heterotic string, compactified ona two torus. For the latter theory, nonperturbative contributions have beencomputed with CFT methods: a check ispossible!!!

3. An eight dimensional gauge instanton isexpected here and a solution is indeedpresent in literature.

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I What is not known:1. an ADHM type contruction is missing. It is

possible to construct holomorphic vectorbundles over complex projective spaces Pn,but the construction is redundant (not true ofn = 2)

2. The explicit solution generalizes the Hopf map

in d = 4, S7 S3

−→ S4, to S15 S7

−→ S8 but otherexplicit solutions of higher winding numberare not known

3. What is the relevant action for the gaugetheory and in the moduli space?

I These questions can be partially coped withfor N = 2 SUSY in a eight dimensionalextension of the work of SW

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I The content of this theory is given by theN = 1 vector multiplet in the adjointrepresentation of SO(8)

Φ(x , θ) = φ(x) +√

2θΛ(x) +12θγµνθFµν(x) + . . .

I In turn from the non abelian extension of theBI Lagrangian LBI =

∫dp+1x

√ηmn + Fmn/2πα′ ∼

a0F2 + a1FD2F + a2F4 + a3F2D2F + . . . we get the actionfor the D7 brane

SD7 =1

128π5α′2 gs

∫d8x Tr

(F2)−

196π3gs

∫d8x Tr

(t8 F4)

+α′

gs

∫d8x L(5)(F , DF) + · · · .

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I In order to keep the quartic term anddescribe the effects of the D-instantons in thefield theory limit, we should take the limitα′ → 0 with gs fixed. This is dangerousbecause the quadratic Yang-Mills termnaively explodes. Now, given that theinstanton is Fµν = − 2ρ2

(x2+ρ2)2 γµν , the quadraticterm becomes

2d/2−1

4π5α′2 gs7

∫ Rd8x

ρ4

(r2 + ρ2)4∼

ρ4

α′2 gslog

(ρ/R

)I Sending ρ → 0 before removing the regulator

the YM vanishes

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I It is now easy to compute t8 = ∓ε8/2 + t± andthe first bosonic term is ∼ F ∧ F ∧ F ∧ F thefourth Chern class with F ∧ F = ∗(F ∧ F)

I Computations can now be carried out andthe final result is in agreement with the stringcomputation for the heterotic stringcompactified on T 2

I This has been generalized also to SO(N) andto the D(−1)−D3−D7 and also in this caseagreement is found with previously knownresults

I This partition function (or equivariant volume)has been instrumental to computations ofDonaldson invariants in four dimensional

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I Can these results be extended to theDonaldson-Thomas invariants in sixdimensions?

I The same results can be also obtained a laSW. Is there any connection with what wediscussed here?

I The last part of this talk will deal with this issue

I The idea is to exponentiate and do thesaddle point of the expression qkZk jump

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qk Zk = qk∫ k∏

I=1

dϕIe∑

I ln(ε1+ε2)

ε1ε2P(φI )P(φI +ε)+

∑I 6=J ln

ϕ2IJ (ϕ2

IJ−ε2)

(ϕIJ−ε1)(ϕIJ−ε2)

=k∏

I=1

dϕIek ln q+k ln

ε1+ε2ε1ε2

−2∑

I ln P(ϕI )−2∑

I 6=Jε1ε2ϕ2

IJ

I Defining ρ(x) = ε1ε2∑

I δ(x − φI) with∫dxρ(x) = kε1ε2, the saddle point is found for

k →∞ and ε1, ε2 → 0. Therefore

qk Zk ∼∫Dρ(x)e

− 2ε1ε2

∫dxρ(x) ln P(x)√

q −2

ε1ε2

∫dxdy ρ(x)ρ(y)

(x−y)2

I The problem has become now to find thefunction ρ(x) which minimizes the expressionat the exponent

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I We can now introduce the profile function

f (x) = −ρ(x) +N∑

i=1

|x − ai |x→∞−−−−→ N|x |

I N is the number of connected pieces of thesupport of f (x). In fact the support of ρ(x) isthe union of N disjoint intervals containingal ∈ [α−l , α+

l ]

I Moreover∫ α+

l

α−lxf ′′(x)dx = al where f ′′(x) is a

charge densityI The functional to be minimized is

H = −14

∫dxdyf ′′(x)f ′′(y)K (x − y)

=14

∫dxdy

[f ′′(x)(x − y)

(ln

x − yΛ− 1

)]δf ′′(x)

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I From the last expression we find fory ∈ [α−l , α+

l ]

F(y) =

∫dxf ′′(x) ln

x − yΛ

= 0

I In turn via a conformal map F(y) can beextended to the complex plane to a complexfunction w = ϕ(z) = U(x , y) + iV (x , y), whereV (x , y) coincides with F(y) on the real axis

z wϕ(z)

α−1 α+1 α−3 α+

3α−2 α+

2

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I The electrostatic analogy: find the potentialin a portion of space in presence ofconductors. As it is well-known charges willdistribute on the conductors in such a way tominimize the energy of the field. On thesurface of the conductors the potential isconstant and the electric field is orthogonalto it

I The mathematical translation of all of this isthe Laplace problem with Neumannboundary conditions or, given a realharmonic function, F(y), find the analyticfunction whose imaginary part is given bythat function

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I Why conformal maps are relevant? They mapanalytic functions into analytic functions. So ifwe map our problem into one for which thesolution is known we are done. Usuallyboundary problems can be solved for certainstandard shapes (half-plane, exterior ofcircle). So we try to map to these.

z w

w = ϕ(z) = cos−1 zk

I Taking w = ln y ′ we get y ′ + 1/y ′ = P(z)/Λβ/2

or y2 = P(z)2 − Λβ/2 with y ′ = (y + P(z))/Λβ/2

and P(z) =∏N

i=1(z − ai), the SW curve

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I This problem has very interesting connectionswith many branches of mathematics

1. Shape and height fluctuations in 2-d randomgrowth models (randomly growing Youngdiagram, first-passage site percolationmodel) This shape goes like ∼ Nχ whereχ = 1/3 and N is the mean of the linear size ofthe shape or the height

2. Indeed the problem of the distribution of thelength of the longest increasing subsequencein a random permutation has the same χ. LetN = 5 and take the permutation (5,1,3,2,4).The longest subsequence are (1,3,4) and(1,2,4). The problem is to find qn,N for N →∞where qn,N = Prob(lN ≤ n) = fn,N/N! wherefn,N = no. permutations with lN ≤ n

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I ..... more

3. Random matrices

4. Asymptotics of reducible representations ofthe symmetric group Sn for n →∞

5. Free probability theory and the calculus ofpartitions

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F.Fucito

IntroductionOverview

D-branesD-branesT-dualityTilted branes

Bound states of BranesBound statesLocalization

The D(-1)-D7 systemBranes at anglesSW curves

I Other interesting cases1. Adjoint masses

F(y) =

∫dxf ′′(x) ln

∣∣∣∣∣ x − y − m2

x − y + m2

∣∣∣∣∣ = const

There is no symmetry for y → y + m but if m isanalitically continued then F(y) = −F(y + im)withy ∈ [α−l + im/2, α+

l + im/2], [α−l − im/2, α+l − im/2]

Page 29: Intersecting Branes Geometry F.Fucito Introduction ... · Intersecting Branes and Enumerative Geometry F.Fucito Introduction Overview D-branes D-branes T-duality Tilted branes Bound

Intersecting Branesand Enumerative

Geometry

F.Fucito

IntroductionOverview

D-branesD-branesT-dualityTilted branes

Bound states of BranesBound statesLocalization

The D(-1)-D7 systemBranes at anglesSW curves

2. Two antisymmetric + four fundamentals

F(y) =

∫dxf ′′(x)

[ln(x − y)−

12

ln(x − y + m1)−12

ln(x − y + m2)

]− ln

(y +

m1

2

)− ln

(y +

m2

2

)+ ln(y + M) = const

If we set m1 = m2 = m = 2M then

F(y) =

∫dxf ′′(x) ln

∣∣∣∣ x − yx − y + m

∣∣∣∣ = const

Then F(−y −m) = −F(y) and a solution ispossible. What about the general case? Itseems that if a symmetry is lacking, theelectrostatic problem is not well defined,since the values of the potential on theconductors is not specified. The presence ofextra point charges also is to be noted.