Equivalence of Chiral Fermion Formulations

Saturday 22 April 2023Saturday 22 April 2023Workshop on Computational Hadron PhysicsWorkshop on Computational Hadron Physics

Hadron Physics I3HP Topical WorkshopHadron Physics I3HP Topical Workshop

Equivalence of Chiral Equivalence of Chiral Fermion FormulationsFermion Formulations

A D KennedyA D KennedySchool of Physics, The University of EdinburghSchool of Physics, The University of Edinburgh

Robert Edwards, Robert Edwards, Bálint JoóBálint Joó, , Kostas OrginosKostas Orginos ((JLabJLab) ) Urs Wenger Urs Wenger (ETHZ)(ETHZ)

Saturday 22 April 2023 A D Kennedy 2

Contents

On-shell chiral symmetryNeuberger’s OperatorInto Five Dimensions

Kernel

Schur ComplementConstraintApproximation

tanhЗолотарев

RepresentationContinued Fraction Partial FractionCayley Transform

Chiral Symmetry BreakingNumerical StudiesConclusions

Saturday 22 April 2023 A D Kennedy 3

Chiral Fermions

Conventions

We work in Euclidean spaceWe work in Euclidean space

γγ matrices are Hermitianmatrices are Hermitian

We writeWe write

We assume all Dirac We assume all Dirac

operators are operators are γγ55 HermitianHermitian

†5 5D D

D D

Saturday 22 April 2023 A D Kennedy 4

It is possible to have chiral symmetry on the lattice without doublers if we only insist that the symmetry holds on shell

On-shell chiral symmetry: I

Such a transformation should be of the form

(Lüscher)

is an independent field from

has the same Spin(4) transformation properties as

does not have the same chiral transformation

properties

as in Euclidean space (even in the continuum)

y†y

y

551 1;i aD i aDe e

y

†y

†y†y

Saturday 22 April 2023 A D Kennedy 5

On-shell chiral symmetry: II

For it to be a symmetry the Dirac

operator must be invariant 5 51 1i aD i aDD e De D

5 51 1 0aD D D aD For an infinitesimal transformation this implies that

5 5 52D D aD D

Which is the Ginsparg-Wilson relation

Saturday 22 April 2023 A D Kennedy 6

5 5 5†sgnˆ W

w

W W

D MD M

D M D M

Both of these conditions are satisfied if(f?) we define

(Neuberger)

Neuberger’s Operator: I

We can find a solution of the Ginsparg-Wilson relation as follows

† †15 5 5 5 5 52 1 ;ˆ ˆ ˆaD aD aD

Let the lattice Dirac operator to be of the form

This satisfies the GW relation iff 25 1ˆ

It must also have the correct continuum limit

Where we have defined where W WD Z O a 2WZM aZ

2 25 5 52 1 1ˆ WD

D Z aZ O a O aM

Saturday 22 April 2023 A D Kennedy 7

Into Five Dimensions

H Neuberger hep-lat/9806025A Boriçi hep-lat/9909057, hep-lat/9912040, hep-lat/0402035A Boriçi, A D Kennedy, B Pendleton, U Wenger hep-lat/0110070R Edwards & U Heller hep-lat/0005002趙 挺 偉 (T-W Chiu) hep-lat/0209153, hep-lat/0211032, hep-lat/0303008R C Brower, H Neff, K Orginos hep-lat/0409118 Hernandez, Jansen, Lüscher hep-lat/9808010

Saturday 22 April 2023 A D Kennedy 8

Is DN local?It is not ultralocal (Hernandez, Jansen, Lüscher)

It is local iff DW has a gap

DW has a gap if the gauge fields are smooth enough

q.v., Ben Svetitsky’s talk at this workshop (mobility edge, etc.)

It seems reasonable that good approximations to DN

will be local if DN is local and vice versa

Otherwise DWF with n5 → ∞ may not be local

Neuberger’s Operator: II

1, 1 1 sgn

52D H H

N 10 μ

Saturday 22 April 2023 A D Kennedy 9

Four dimensional space of algorithms

Neuberger’s Operator: III

Representation (CF, PF, CT=DWF)

Constraint (5D, 4D)

,

( )sgn( ) ( )

( )n

n mm

P HH H

Q HApproximation

5 WH D MKernel

Saturday 22 April 2023 A D Kennedy 10

Kernel

Shamir kernel

55

5

;2

WT T T

W

a D MH D aD

a D M

Möbius kernel

5 5

55 5

;2

WM M M

W

b c D MH D aD

b c D M

5W WH D M

Wilson (Boriçi) kernel

Saturday 22 April 2023 A D Kennedy 11

1 0 0 111 11 0 0 1

A A B

CA D CA B

1 0 0 1

1 0 0 1

1 0

1 11 0

A B

CA D CA B

Schur Complement

It may be block diagonalised by an LDU factorisation (Gaussian elimination)

1det det

A BAD ACA B

C DIn particular

A B

C D

Consider the block matrixEquivalently a matrix over a skew field = division ring

The bottom right block is the Schur complement

Saturday 22 April 2023 A D Kennedy 12

1

2

2

1

n

n

D

00

00

1

2

52

1

n

n

D

Constraint: I

1

2

2

1

n

n

DU

So, what can we do with the Neuberger operator represented as a Schur complement?Consider the five-dimensional system of linear

equations00

00

1L 1LL

The bottom four-dimensional component is, Nn n DD

Saturday 22 April 2023 A D Kennedy 13

Constraint: II

Alternatively, introduce a five-dimensional pseudofermion field 1 12 n

Then the pseudofermion functional integral is

† 15†

5 ,1

det det det detn

Dj j

j

d d e D LDU D D

So we also introduce n-1 Pauli-Villars fields

†,

11 1

†,

1 1

detj j jj

n nD

j j j jj j

d d e D

and we are left with just det Dn,n = det DN

Saturday 22 April 2023 A D Kennedy 14

Approximation: tanh

Pandey, Kenney, & Laub; Higham; NeubergerFor even n (analogous formulæ for odd n)

11 1

1,11

1tanh tanh

1

nxx

n n nxx

x n x

2

2 2

2

2 2

1

tan

1

12tan

1

n

n

kx

nk

kx

nk

xn

2

2 2 21 12 2cos sin1

2 1n

x k kk n n

xn

ωj

Saturday 22 April 2023 A D Kennedy 15

Approximation: Золотарев

2

2/ 2

21

212

sn ;1

sn / ; sn 2 / ;1sn ; sn ;

1sn 2 / ;

n

m

z k

z M iKm n k

z k M z k

iK m n k

sn(z/M,λ)

sn(z,k)

ωj

Saturday 22 April 2023 A D Kennedy 16

Approximation: Errors

The fermion sgn problem

Approximation over 10-2 < |x| < 1Rational functions of degree (7,8)

ε(x) – sgn(x)

log10 x

0.01

0.005

-0.01

-0.005

-2 1.5 -1 -0.5 0.50

Золотарев tanh(8 tanh-1x)

Saturday 22 April 2023 A D Kennedy 17

Representation: Continued Fraction I

0

1

2

3

1 0 0

1 1 0

0 1 1

0 0 1

A

A

A

A

Consider a five-dimensional matrix of the form

10

01 10 11

1 121 2

1 32

1 0 0 0 1 0 00 0 01 0 0 0 1 00 0 0

0 1 0 0 0 0 0 0 10 0 0

0 0 1 0 0 0 1

SSS SS

S S SS

S

Compute its LDU decomposition

0 01

1; n n

n

S A S AS

where

3 3 3 32

2 21

10

1 1 11 1

1

S A A AS A A

S AA

then the Schur complement of the matrix is the

continued fraction

Saturday 22 April 2023 A D Kennedy 18

Representation: Continued Fraction II

We may use this representation to linearise our rational approximations to the sgn function

1

1 21

1

1 0

1

22

, ( )

n

n

n

n

n

n

cc

H HH

HH

cc

c cc

c

0

01

0 0

0 02

01

21

21 1

22 1 2

21 2 1

0

1

10

c c cn n n

c c cn n n

c c c

c c c c

Hn

Hn

H

H

Hc

as the Schur complement of the five-dimensional matrix

Saturday 22 April 2023 A D Kennedy 19

Representation: Partial Fraction I

2

22

1

1

1

2

2

0 0

0 0 0

0 0

0 0 0

0 0

1 1

1 1

1

1

11

x

p

p x

x

p

q

p

x

q

R

Consider a five-dimensional matrix of

the form (Neuberger & Narayanan)

Saturday 22 April 2023 A D Kennedy 20

Compute its LDU decomposition

So its Schur complement is

12 2

1

22 2

2

p x

x q

p xR

x q

2

2 2 2

2 2

2

1

1 1 1

2

1 2

2

1

1 0 0 0 0

1 0 0 0

0 0 1 0 0

0 0 1 0

1

p

x

p q x q

x x q x q

p

xp q x q

x x q x q

12 2

1 1

2

1

1

22 2

2 2

2

2

2

2 2

2

2 2

1

(

( )

)

0 0 0 0

0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0

x

p

p x q

xq

p x

x q

x

p

p x q

R

xqp x

x q

1

2

1

1 1

2 2

1 1

2

2 2

2 2

2 2

1 0 0

0 1 0 0

0 0 1

0 0 0 1

0 0 0 0 1

p

p

x

x

p

xq x q

x q x qp

xq x q

x q x q

1 11

11 01

1

1 12

21 02

1 0

0 0

0 0

0 0

0 0

0

x

pp x

q

x

R

x

pp

q

Representation: Partial Fraction II

Saturday 22 April 2023 A D Kennedy 21

Representation: Partial Fraction III

This allows us to represent the partial

fraction expansion of our rational function as

the Schur complement of a five-dimensional

linear system

1, 2 2

1

( )n

jn n

j j

pH H

H q

Saturday 22 April 2023 A D Kennedy 22

1

2 1

3 2 1

4 3 2 1

1 0 0 0 1 0 0

0 1 0 0 0 1 0

0 0 1 0 0 0 1

0 0 0 0 0 0 1

A

A A

A A A

C A A A A

1

2 1

3 2 1

4 3 2 1

1 0 0

0 1 0

0 0 1

0 0 0

A

A A

A A A

C A A A A

2

3

4

1 0 0 0

1 0 0

0 1 0

0 0 1

A

A

A

Representation: Cayley Transform I

1

2

3

4

1 0 0

1 0 0

0 1 0

0 0

A

A

A

A C

Consider a five-dimensional matrix of the

formCompute its LDU decomposition

CT 1 2 1n nS C A A A A So its Schur complement is

Neither L nor U depend on C

1CT 2

1 112 51

11 1

1P P T

TS

T

If where , and , then

1 nT T T 1s sA T

1C P P

Saturday 22 April 2023 A D Kennedy 23

Representation: Cayley Transform II

The Neuberger operator is

CTN

CT

,1

P P SD H

S

1 1

( ) ;1 1

T x xx T x

T x x

T(x) is the Euclidean Cayley transform of, ( ) sgn( )n m x x

1

x x T xT x

0 0 0 1T jj

j

xT x

x

For an odd function we have

In Minkowski space a Cayley transform maps between Hermitian (Hamiltonian) and unitary (transfer) matrices

Saturday 22 April 2023 A D Kennedy 24

(1) (1) (1)

(2) (2) (2)

(3) (3) (3)

(4) (4) (4)

00

00

D D P D PD P D D P

D P D D PD P D P D

The Neuberger operator with a general Möbius kernel is related to the Schur complement of

D5 (μ) (1) (1) (1) (1)

(2) (2) (2) (2)

(3) (3) (3) (3)

(4) (4) (4) (4) (4)

0 0 0 00 0 0 0

0 0 0 00 0 0

D D D DD D D D

P PD D D D

D D D D D

Representation: Cayley Transform III

with and( ) ( )

5

( ) ( )5

( ) 1

( ) 1

s ss W

s ss W

D b D M

D c D M

152 1P

( ) ( ) ( ) ( )5 55 5 5 5 5 5;s s s s

s

b cb c b c b c

P-μP-

P+ μP+

Saturday 22 April 2023 A D Kennedy 25

(1) (1) (1) (1)

(2) (2) (2) (2)

5 (3) (3) (3) (3)

(4) (4) (4) (4)

0 0 0 0

0 0 0 0( )

0 0 0 0

0 0 0 0

D D D D

D D D DD P P

D D D D

D D D D

P

1 0 0 0 0 0 0 1

0 1 0 0 1 0 1 0

0 0 1 0 0 1 0 0

0 0 0 1 0 0 1 0

P P

P

Cyclically shift the columns of the right-handed part where

5 5D D P

Representation: Cayley Transform IV

P+P- μP+μP-

Saturday 22 April 2023 A D Kennedy 26

( ) ( ) ( )s s sQ D P D P

Representation: Cayley Transform V

1 0 0 0

0 1 0 0

0 0 1 0

10 0 0 P P

C

(1)

(2)

(3)

(4)

0 0 0

0 0 0

0 0 0

0 0 0

Q

Q

Q

Q

Q1( ) ( )s s s M

ss M

HT Q Q

H

With some simple rescaling

11

12

13

14

1 0 0

1 0 01

0 1 05

10 0

( )

T

T

T

T P P

D

Q PCx

The domain wall operator reduces to the form introduced before

Saturday 22 April 2023 A D Kennedy 27

Representation: Cayley Transform VI

It therefore appears to have exact off-shell chiral symmetryBut this violates the Nielsen-Ninomiya theorem

q.v., Pelissetto for non-local versionRenormalisation induces unwanted ghost doublers, so we cannot use DDW for dynamical (“internal”) propagatorsWe must use DN in the quantum action instead

We can us DDW for valence (“external”) propagators, and thus use off-shell (continuum) chiral symmetry to manipulate matrix elements

5 5( ) (1)D D We solve the equation

Note that satisfies1 1DW ND D a

1 1 1 15 5 5 5 5, , 2 , 2 0DW N N N ND D a D D D a

Saturday 22 April 2023 A D Kennedy 28

Chiral Symmetry Breaking

Ginsparg-Wilson defect5 5 5 52 LD D aD D Using the approximate Neuberger operator 1

52 1aD H

L measures chiral symmetry breaking 212 1La H

The quantity is essentially the usual

domain wall residual mass (Brower et al.)

†

res †

tr

trLG G

mG G

mres is just one moment of L

G is the quark propagator

Saturday 22 April 2023 A D Kennedy 29

Numerical Studies

Used 15 configurations from the RBRC dynamical DWF dataset

316 32 12 2

0.8 1.8 0.02

V n L ns fM

Matched π mass for Wilson and Möbius kernelsAll operators are even-odd preconditionedDid not project eigenvectors of HW

Saturday 22 April 2023 A D Kennedy 33

mres per Configuration

mres is not sensitive to this small eigenvalue

But mres is sensitive to this one

ε

Saturday 22 April 2023 A D Kennedy 34

Cost versus mres

Saturday 22 April 2023 A D Kennedy 35

Conclusions

Relatively goodZolotarev Continued FractionRescaled Shamir DWF via Möbius (tanh)

Relatively poor (so far…)Standard Shamir DWFZolotarev DWF ( 趙 挺 偉 )

Can its condition number be improved?

Still to doProjection of small eigenvalues

HMC5 dimensional versus 4 dimensional dynamicsHasenbusch acceleration

5 dimensional multishift?Possible advantage of 4 dimensional nested Krylov solvers

Tunnelling between different topological sectorsAlgorithmic or physical problem (at μ=0)Reflection/refraction Assassination of Peter of Assassination of Peter of Lusignan Lusignan

(1369)(1369)(for use of wrong chiral formalism?)(for use of wrong chiral formalism?)