Simon Catterall- Exact lattice supersymmetry

8/3/2019 Simon Catterall- Exact lattice supersymmetry

1/29

Exact lattice supersymmetry

Simon Catterall

Syracuse University

Exact lattice supersymmetry p.


2/29

Lattice SUSY

Old problem.

Difficult. SUSY extends Poincar broken bydiscretization.

Folklore: Impossible to put SUSY on lattice exactly.

Leads to (very) difficult fine tuning lots of relevant

SUSY breaking counterterms...

Way out!



3/29

Motivations ?

Rigorous definition of SUSY QFT - like lattice QCD.

Dynamical SUSY breaking. Predicting soft terms inMSSM ...

Gauge-gravity duality ? Eg. large N strongly coupledN= 4 SYM and type II string theory in 5d AdS.



4/29

New ideas

Topological twisting

Orbifolding/deconstruction (D. B. Kaplan, M. Unsl, A.Cohen, ...)

Focus on former. Emphasizes geometry. Continuumlimit clear.

Warning: Tricks work only for no. SUSYs Q multiple 2D

... In D = 4 unique theory: N= 4 SYM



5/29

Example: Twisting in 2D

Simplest theory contains 2 fermions iGlobal symmetry: SOLorenz(2) SOR(2)Twist: decompose under diagonal subgroup

Consider fermions as matrix

i

Natural to expand:

=

2I+ + 1212

scalar, vector and tensor (twisted) components!



6/29

Twisted supersymmetry

Twisted theory has scalar SUSY Q.

{Q, Q} = p implies:

Q2

= 0{Q,Q} = p

Plausible: S = Q(, )

Basic idea of lattice theory: discretize twisted formulation,

exact (scalar) SUSY only requires Q2 = 0



7/29

Example: Q = 4 SYM in 2D

In twisted form (adjoint fields AH generators)

S =1

g2QTr F + [D,D] 1

2d

Q A =

Q = 0

Q A = 0

Q = FQ = d

Q d = 0

Note: complexified gauge field A = A + iB



8/29

Action

Q-variation, integrate d:

S =1

g2Tr FF + 1

2[D,D]

2 D[] DRewrite as

S = 1g2

TrF2 + 2BDDB [B, B]2 + LF

where

LF =

122

D2 iB2 D1 + iB1D1 iB1 D2 iB2

1

2



9/29

Moral

Twisting changes spins fields:

Scalars become vectors. Naturally embedded incomplexified connection

Fermions integer spins. Form components ofKhler-Dirac field.

Twisted entire Lorentz symmetry with R-symmetry

maximal twist. Necessary for lattice.

Flat space - twisting just change of variables.



10/29

Lattice ?

A(x) U(n). Complexified Wilson links.

Natural fermion assignment on sites, links, 12diagonal links of cubic lattice.

Fields pick up non-standard U(N) gaugetransformations:

(x) G(x)(x)G(x)

(x) G(x)(x)G(x + )

(x) G(x + + )(x)G(x)

U(x) G(x)U(x)G(x + )

U(x) G(x + )U(x)G(x)

Choice of orientations ensure G.IExact lattice supersymmetry p.


11/29

Lattice supersymmetry

As in continuum:

Q U =

Q = 0Q U = 0

Q = FL

Q = d

Q d = 0

Note: Q2 = 0 still.



12/29

Derivatives

D(+) f(x) = U(x)f(x + ) f(x)U(x + )

D()

f(x) = f(x)U(x) U(x )f(x )

For U(x) = 1 + A(x) + . . . reduce to adjoint covariantderivatives

F = D(+) U(x) = U(x)U(x + ) U(x)U(x + )

Remarkably satisfy exact Bianchi identity:

D(

+)F = 0



13/29

Recap

Discretize twisted version of continuum SYM

Need subgroup of R-symmetry to match SO(D).

Ensures all fermions represented by integer spin forms.Natural map to lattice.

In flat space: twisted formulation completely equivalent

to usual theoryAbsence of fermion doubling twisted fermions fill outKhler-Dirac field (like staggered quarks)

Lattice theory G.I, possesses exact Q and a point groupsymmetry which is subgroup of twisted rotationalsymmetry.



14/29

Bonuses

Topological subsector:

< O(x1) . . . O(xN) > independent of coupling g2, and

points x1 . . . xN if QO = 0. Eg

< O >

g2=< Q(O) >= 0

Novel gauge invariance properties of lattice theorystrongly constrains possible counter terms reducessubstantially fine tuning needed to get full SUSY in

continuum limit.



15/29

Q = 16 SYM in 4D

Twist: diagonal subgroup of SOLorenz(4) SOR(4)

Again after twisting regard fermions as 4 4 matrix.

To represent 10 bosons of N= 4 theory with complexconnections is most natural in five dimensions.

Fermion counting requires multiplet (, a, ab) where

a, b = 1 . . . 5Action contains same Q-exact term as for Q = 4 plusnew Q-closed piece.



16/29

Details

Dimensional reduction to 4D A5 plus imag parts ofA, = 1 . . . 4 yield 6 scalars of N= 4

Fermions: ab

,

a

S = Q 18

abcdedeDcab

Twisted action reduces to Marcus topological twist of

N= 4 (GL-twist). Equivalent to usual theory in flatspace.

Identical to Q = 16 orbifold action (Kaplan, Unsl)



17/29

Transition to lattice

Introduce cubic lattice with unit vectorsia =

ia, a = 1 . . . 4. Additional vector

5 = (1,1,1,1).

Notice:

a a = 0. Needed for G.I.

Assign fields to links in cubic lattice (plus diagonals). Egab(x) lives on link from (x + a + b) x.

Derivatives similar to Q = 4. eg

D(+)a f(x) = Ua(x)f(x + a) f(x)Ua(x)


Si l i


18/29

Simulations

Integrate out fermions. Resulting Pf [MF(A)] simulatedusing RHMC alg. (lattice QCD)

Use pbc SUSY exact. Z = W Witten index -

Q-invariance exhibits topological invariance W.

Preliminary results from single core code. Parallel codenow finished..

Test SUSY, I.R divergences, check sign problems.D = 2 with Q = 4 and D = 4 with Q = 16.


S t i W d id tit


19/29

Supersymmetric Ward identity

Q-exactness ensures that lnZpbc = 0

where = 1g2LD

V4D

Ensures: < SB >=12 V(N

2

1)(nbosons 1)Example: D = 0 SU(2)

SB exact

1.0 4.40(2) 4.5

10.0 4.47(2) 4.5

100.0 4.49(1) 4.5

SB exact

1.0 13.67(4) 13.5

10.0 13.52(2) 13.5

100.0 13.48(2) 13.5

Q = 4 Q = 16


V t bilit fl t di ti


20/29

Vacuum stability - flat directions

Is integration over moduli space [B, B] = 0 divergent ?

0 5 10 15 20 25 30lambda

0

0.5

1

1.5

2

2.5

3

P(lambda)

Q=4 D=0 SU(2)

-2 -1 0 1 2 3 4 5lambda

0

2

4

6

8

10

P(lambda)

Q=16 D=0 SU(2)

Q = 4 Q = 16

D = 0. SU(2). Periodic bcs. Eigenvalues of UU 1Scalars localized close to origin. Power law tails.p(Q = 4) 3, p(Q = 16) 15 (Staudacher et al.)


Pfaffian phase


21/29

Pfaffian phase

Simulation uses |P f(U)|. Measure phase (U).

< O >=< Oe >phase quenched

< e

>phase quenched

SU(2) D = 2: 42.

Q SqB SB SeB cos 4 70.61(4) 65(5) 72.0 -0.016(6)

16 214.7(4) 214.6(3) 216.0 0.999994(3)

< ei(U) >phase quenched pbc= W = 0 for Q = 4 ?

SUSY breaking (Tong et. al) ?


Fermion eigenvalue distribution


22/29

Fermion eigenvalue distribution

SU(2) D = 2: 22

Q = 4 Q = 16

Non-zero density for Q = 4 close to origin linked to logdivergence of < 2 > ?Potential Goldstino ?


N 4 SYM i f di i


23/29

N = 4 SYM in four dimensions

Initial results encouraging: 6000 trajs on SU(2) 24 lattice(1000 hrs)SB/S

exactB = 0.98 < cos() >= 0.98(1)

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

lambda

0

2

4

6

8

10

P(lambda)

Q=16 D=4 SU(2)

Larger lattices currently under study using parallel code.


Applications: holography


24/29

Applications: holography

0 2 4 6 8 10 12

T

0

5

10

15

20

(1

/N2

)E/T

SU(5) quenched

SU(3)

SU(5)

SU(8)

black hole

Example of gauge-gravity duality: Thermodynamics of

N, T 0 AdS5 black hole reproduced by N= 4 SYMtheory reduced to D = 1


Renormalization


25/29

Renormalization

Lattice symmetries:

Gauge invariance

Q-symmetry.Point group symmetry - eg. natural lattice for N= 4 isA4.

Exact fermionic shift symmetry.Conclusion: Renormalized action contains same operatorsas bare theory except for SUSY mass term.

Examine flows at 1-loop - in progress (with J. Giedt)


Future


26/29

Future

Nonperturbative exploration N= 4 YM. Tests ofAdSCFT. Supersymmetric Wilson loops.

But what residual fine tuning needed to get full SUSY

as a 0 ?

Dimensional reductions duality between strings withDp-branes and (p + 1)-SYM ?

Add fermions in fundamental .. (Matsuura, Sugino inD = 2 recently).

Break N= 4 to N= 1 a la Strassler ..


Marcus twist


27/29

Marcus twist

Reduces to:

S = Tr FF +1

2D,D

2+

1

2,

2+ (D)

(D)

D[] D [, ]

D , D ,


Vacuum stability - trace mode


28/29

Vacuum stability - trace mode

Correspondance to continuum requiresU = 1 + aA + O(a2).

For U(N) this is not true < 1NTr U(x)U(x) > 0.5

det(U(x)U(x)) 0!

Vacuum instability det(UU) eB0 implies B0

0 50 100 150 200 250 300MC time

-1

0

1

2

3

4

5

6

7

8

9

10

scalareigenvalue

lambda1

lambda2

Q=4 D=0 U(2) m=0.1


Truncation


29/29

Truncation

Cannot cure with mass m2

Tr (UU I)2

m

0.01 0.45(2)0.1 0.57(6)

0.5 0.38(2)

SB(eB0U) e4B

0S(U) any {U}

Exponential effective potential for B0.

Fix ? - truncate to SU(N) S 1N2 O(a)Also removes exact 0 mode in fermion op.


Documents

Simon Catterall- Exact lattice supersymmetry