Chiral Boundary States for FermionsDavid Tong

Based on work with Philip Boyle Smith
arXiv:1912.01602
arXiv:2005.11314
arXiv:2006.07369

What’s to Come

1. Review of the Mod 2 Anomaly
2. Chiral Boundary States
3. Boundary RG Flows
4. The Z8 Classification of d=2+1 SPTs
5. An Open Problem: ‘t Hooft Lines

The Mod 2 Anomaly

{ i,

j} =

ij

SU (2)R ⇥ SU (2)I ! U (1)R ⇥ U (1)ISU (2)RThe⇥ SU(2)I 2!
U (1)R ⇥ U (1)IMod
Anomaly
2 2
Z6F
ZMajorana
Dirac==12h+ 8 =3 Tr(1)2h3 = 2
2
3
3

A single, quantum mechanicall2h, is inconsistent:
6F =Majorana
12h +fermion,
8

ZMajoranah ! h=

h!

p

2

h

But there is no problem with two Majorana fermions, or a Dirac fermion:{ i , j } = ij
{ i , j } = ij
and

2
ZDirac = ZMajorana= Tr(1) = 2

described in [4], this same anomaly is lurking when we attempt to place fermion{ i , boundary.
= 1 + 1 dimensions on a manifold with
(A beautifully clear explanatioj } = ij
The Mod 2 Anomaly in Boundary Conditionss from the continuum perspective can be found in the talk [5].) Consider a singlve Majorana fermion , now in d = 1 + 1 dimensions. There are two possib
2
dary conditions that one Z
canimpose,reflecting
theleft-moving fermion L int
=
Z
=
Tr(1)
=
2
Dirac
Majorana
Consider a massive Majorana fermion c in d=1+1 dimensions in the presence
ght-moving
fermionWe Rhave, two choices of boundary conditions:of a boundary.
L = ± Rp

ZMajorana =

2

(1.2

One suchchoice has
a localised
zero formode;
thechoiceother does
not there is a single Mang the Diracequation,onefinds thatoneof signa zero mode localised on the boundary,while for the other there is not. The sig!
L
e that gives rise to the zeromode
is= therefore(Which boundar=
exp (⌥mx)inconsistent.
(0)tion su↵ers a zero mode dependsRon both the sign of the fermion mass, and thtation of the boundary.)
The choice with the zero mode has an anomalous boundary theory.

e anomaly manifests itself in a slightly di↵erent way when we consider a comple
fermion = 1 + i 2 . There is no problem if we impose a boundary conditio
preserves the vector U (1)V symmetry,
Kitaev 2000; Dijkgraaf and Witten 2018
V :

=

(1.3

mion = 1 + i 2 . Therenogives
problem
we zeroimposeboundary
choice isthatrise toif themodea istherefore condit
inconsisrves the vector U (1)Vcondition
symmetry,su↵ers a zero mode depends on both the sign of thorientationof theinboundary.)
The Mod
2 Anomaly
Boundary Conditions
!
V L:
(1
L = R
= anomaly= manifests
exp (⌥mx)
The
itself(0)in a slightly di↵erent way wheRDirac
fermion= 1 + iandThere
is no means
problem
if we impo
ConsideramassiveDiracfermion,theThisfollowing
choices
2 ., withlates to the same sign (1.2) on both
.
that,
if
1
2
of boundary conditions:that preserves the vector U (1) symmetry,the bulk, then there are either two boundaryV zero modes or none. Eith
ystem does not su↵er an anomaly.
L = R
V :
L = R

thereimpose
are either
twotranslates
zero modes
none.
ast, weHere
couldboundary
conditionsof theThis
toor thesame
sign form(1.2) on both

1 and

2.

massive in the bulk, then there are either two boundary zero†
A system:(1c.f. Andreev reflection
L =way, thedoesRnot su↵er an anomaly.

contrast,
wemode.
could
impose boundary conditionsof incidethe foHere there isarise
a single
zero
dary conditions
inInMajorana
wires
attached
to superconductors,
where anbounds as a hole, a process known as Andreev reflection.
If =the† bulk ferm
A:
L

R

15
Such boundary
conditions
in wiresofattached
Note: The
anomaly isarise
independent
the mass.to supercond
electron rebounds as a hole, a process known as Andreev reflec–3–

The Mod 2 Anomaly in Boundary Conditions

Ways to cancel the anomaly:• Put the same boundary condition on each end of an interval• But you can’t put different choices on different ends.
• Add an extra boundary Majorana mode in by hand.
• Anomaly inflow through the Arf SPT phase.

c.f. a beautiful D-brane story: Witten ‘18, Kaidi, Parra-Martinez and Tachikawa, ‘19

Multiple FermionsOur interest in this paper lies in boundary conditions for multiple massless fermions.
Here there are many more possibilities, ones that do not involve simple repetitions of
Chiral Boundary Conditions
the boundary conditions (1.2), (1.3) and (1.4).
2N massless
Majorana
or, equivalently,
Dirac fermions
TheseConsider
novel boundary
conditions
canfermions
be distinguishedby theNsymmetries
that theyd=1+1
dimensions.
preserve.in The
twoboundary conditions (1.3) and (1.4) preserve the U (1)V and U (1)Asymmetry of a single, massless Dirac fermion respectively. However, in general it ispossibleQuestion:
to impose •boundary
conditions
preserve chiralsymmetries,
under which
What symmetries
canthatbe preserved
by boundary
conditions?
the left- and right-moving fermions carry di↵erent charges. Indeed, there is a general• Any
don’tboundary
suffer a ‘t Hooft
anomaly.
Answer:expectationthat onecanthatimposeconditions
preserving any symmetry that
does not su↵er a ’t Hooft anomaly. (See, for example, [23, 24].)For example, if we have N left-moving Weyl fermions in d = 1 + 1 with charges Qi
U(1)symmetryand
thatNacts
on N Dirac fermions
under e.g.
a U a(1)symmetry,right-moving
fermionswith
withcharges
chargesobeying
Q̄i , then one can
impose boundary conditions that preserve the U (1) symmetry provided thatleft-movers

N
X
i=1

Q2i =

N
X
i=1

Q̄2i

right-movers

which is the requirement that this symmetry does not su↵er a ’t Hooft anomaly.

pZMajorana = p2

Chiral Boundary
States
ZMajorana= 2!

In general, these boundary conditions cannot
be implemented directly on!
Lthe fermion fields.
== exp (⌥mx) (0)

=

L

R

= exp (⌥mx) (0)We could introduce new boundary degrees
R of freedom, but at low energies
the relevant physics is captured by boundary state.
L =

R
=
L
R

Open-closed string duality

L
R
L
R
O(2N
)
⇥
O(2N
)
⇥
(
1)
⇥
(
1)
O(2NL)L ⇥ O(2N R)R ⇥ ( 1) ⇥ ( 1)

Boundary conditions captured by states |Ai|Aiand|Bi|Bi
Cardy ‘89

N
first introduced by Cardy [17], with the remnant ofg theboundary deg=R2
encoded in the boundary central charge [18]. The applicationof boun
N =4field theory to SPT phases was initiated in [7, 8]. N = 4kn
N

Chiral Boundary Conditions

2Z : R 2noNNIn this paper, we study a large class ofconformal⇤[R]=
2 Z boundary: R 2 Z states= ZN \ that
R 1Z⇤[R] =

Majorana fermions in d = 1+1 dimensions. In the absence of a boundar
SU (2)rot ⇥1
fermions
enjoy
an SO(2N)L ⇥ SO(2N
chiral symmetry, together w
We consider
boundary
conditions
that preservea U(1))NR symmetry.
SU (2)rot ⇥ U (1)k that the boundary preserves a subgroupchiral fermion number symmetry. We restrict ourselves to boundary11,stat
55 ,
N
N
N
N a subgroupwill Uask
theboundarypreserves(1)that
⇢U
(1)
⇥
U
(1)
⇢
SO(2N
)L ⇥ SO(2N)R maximal torus
an anomaly-freeU(1)subgroupof the11 ,(2.1)
55 ,of99SO(2N )L ⇥L
R

77 , 8

N
N
N
U
(1)
⇢
U
(1)
⇥
U
(1)
)Lright-movers
⇥into
SO(2N
)RQ̄↵i respec(2.1)
ng and right-moving
fermions
arepair
assignedchargesQ
77 , 88
L we
R ⇢↵i andSpecifically,
theSO(2N
fermions
N =Dirac
spinors.
assign↵ = 1,We...,N
andCharges:ieft-moving= 1, . . . , N
labelstheQspeciesof left-moving
complexfermion,while↵ = .1,. . .and, N irespectheWeylfermions,
left-movers
andcharges
right-moving
are assigned
chargesandQ̄
↵==Q1,↵i. .and
, N charges. . i↵. , Nto thi↵ to fermions↵i = 1,Q̄
symmetry
The simple
linearconditions
described
Weyl
i =
1,
. . . , N labels
thewhile
fermion,,1)wherei = 1,group.
. . . , Nfermions,labelsthewhere
species
ofboundary
complexfermion,↵ = and1, . . .↵, N= 1, . .
N
ortheexample,if Q= ±symmetry.
Q̄.
Our interestin thislinear
lies
in theconditions
moreinter16U (1)Thissimple
choiceofpaperU (1)symmetry
is non-anomalous
U (1) symmetrygroup.
The
boundary
described proary
conditions
in which
fermionsdi↵erent16
arise,
for example,
if Qthe= left±Q̄. andOurright-moving
interest inN thispaper carry
lies
in
the
more
interN
X
X
seare
chiralboundary
conditions.
g boundary conditions
in whichthe left- and right-moving
fermionsQ↵i Q i =Q̄↵icarryQ̄ i di↵erent
No ‘t Hooft
anomalies
es.thatThesechiralconditions.
ueanyarechoiceof boundary
U (1)N symmetry
can be preservedby thei=1i=1boundary.
ymmetries that do not su↵er a ’t NHooft anomaly give suitable boundarys not true thatItanyof U (1) symmetry
canboundary
be preservedby preserving
the boundary.
is choicestraightforward
to constructstatessuch a cSee, for example, [23, 24].) This means that the charge matrices necessarily
those symmetries
do not includesu↵er a[19,’t Hooft
anomaly
suitable boundary
Eachthatexamples20]. A
numbergiveof propertiesof the genera
ditiontions. (See, forhave
example,24].) Thismeans
been[23,
exploredin [21,22]. that the charge matrices necessarily
Q↵i Q i = Q̄↵i Q̄ i
(2.2)
the condition

esting boundary
in
whichtheinleftandthe
right-moving
fermions carryero mode depends
onesting
bothconditions
the sign
the fermion
mass,
and
boundary
conditions
which
left-theand right-movingfer
N ofs not true thatany These
choiceareof chiral
U (1) boundary
symmetry
can be preserved by the boundary
charges.
conditions.
charges. These are chiral boundary conditions.
oundary.)2
those symmetries
that
do notsu↵er= a2 ’t Hooft
anomaly
give
suitableboundary
ZDirac = Z= Tr(1)
Majorana
ChiralBoundary
Conditions
N
It is not truethatanychoiceofU(1)symmetry
Ncan be preserved by the bo
It
is
not
true
that
any
choiceofU(1)symmetrycannecessarily
be preser
ifests itself
slightly di↵erent
way
when
we consider
complex,tions.
(See,infora example,[23, 24].)
This
means
that theacharge
matrices
Only thoseOnly
symmetries
that do notthatsu↵erdoa not’t Hooft
anomaly
giveanomaly
suitable gib
those
symmetries
su↵er
a
’t
Hooft
i 2 . There is no problem if we impose a boundary conditionthe
1 + condition
conditions.conditions.
(See, for example,[23,24].) This
means
that
the
chargematrices
ne
(See,
for
example,
[23,
24].)
This
means
that
the
charg
p
ector U (1)V symmetry,
theZthecondition
It turns obey
out thatboundarystate= condition
2↵i QisispecifiedQ= Q̄↵i Q̄byi a rational, orthogonal matrix(2.2
Majoranaobey
theQ̄ Qi =
V : objects(1.3)↵i Q thesei = Q̄↵i
L = R constructed Q
ill need a few further
fromFirst,
Qcharges.
Q̄↵ii↵i Q̄ i we introduc

We will need
few needfurther
objects
from these charges.
First,
we in!Wea will
1 constructeda few
further
objects constructedfrom thesechargehe same sign (1.2) on both R
andmeans that, ifis= (Q̄)i↵ Q↵j
1 ij
2 . This
L
1
R
=
(
Q̄
)i↵ Q↵j Either
1ij= either =expboundary(⌥mx) (0)then there aretwozero modes
or none.
R=(Q̄
)i↵ Q↵j preserved byij the charges
rational, orthogonal
matrixwillbesufficienttoencodeRs not su↵er Thisan anomaly.
rational, orthogonal matrix will be sufficient to encode the charges prese

oundary. The boundary
condition(1.3) inmatrix
whichwill
each
left-moving
fermionThis rational,
orthogonalbe sufficient
to encodethei
the boundary. The boundary condition (1.3) in which each left-moving fer
boundary.
The
condition(1.3) inreflection
which eachleuld impose
boundarythe
conditionsof theted
into a right-movercorresponds
to form
Rboundary= 1. Imposing
Andreev(1.4
e.g. intoL a=right-moverreflected
corresponds
to R = 1. Imposing Andreev reflecti
fermion gives
R for eachreflected
into
a1.right-mover corresponds to R = 1. Imposing An
ch fermion on
corresponds
to
R
=
†
each
fermioncorresponds
to
R
=
1.
A : onL each
= Rfermion
corresponds
to R = 1. (1.4)
for eachfermion givesN
also associateachargelattice⇤[R]✓Z
to ourof boundary
condition
We also associate a charge lattice ⇤[R]
✓ choice
ZN to ourchoiceof boundarycoN
We alsoassociate a chargewherelatticean⇤[R]✓ Z to our choice ofitions
arise
in
wires
attachedtosuperconductors,
incident
is defined byThis is defined byThisasis Andreev
definedbyn If the bulkono fermiona hole, a process knownreflection.
o
NN n
N
N
22ZZ : R 2 ZN⇤[R] =
2⇤[R]
Z =: R ⇤[R]
(2.3=
2 Z : R 2 ZN

NIn
words:
thelattice⇤[R]
consistsofallinteger-valued
vectors2
Z
which reN
rds: the lattice⇤[R]
consistsofallinteger-valued
vectors2
Z
which
remainIn words: the lattice ⇤[R] consists of all integer-valued vectors in2
N
Z when rotated by the rational matrix R. As we’ll see, this lattice plays an im

( 1)L | , ¯ i = ( 1) | , ¯ is not true that any choice of U (1)N symmetry can be preserved by the boundary
those symmetries that do notsu↵er
a ’t Hooft
anomaly give suitable boundary
The
ChargeLatticetions. (See, for example, [23, 24].) This means that the charge matrices necessarily¯2R¯( 1) | , i = ( 1) | , ¯ ithe condition
It turns out that the boundary Q
state= Q̄↵i Q̄byi a rational, orthogonal matrix↵i Qisispecified(2.2

r from these charges. First, we introduc
ill need a few further objects constructed
N
1
g
=
R
Rij = (Q̄ )i↵2Q↵jrational, orthogonal matrix will be sufficient to encode the charges preserved byoundary.
Theboundary
condition
(1.3)
in
which
each
left-moving
fermioni
All our results are expressed in terms of the following charge lattice
ted into a right-mover correspondsNto=R4k= 1. Imposing Andreev reflection (1.4ch fermion corresponds to R = 1.
n
o
N
N
1 N
NN
⇤[R]
=
2 Z ⇤[R]: R ✓ 2Z Z to our= Zchoice\ Rof boundaryZ
also associatea charge
latticecondition
is defined byno⇤[R] =
2 ZN : R 2 ZN
(2.3

rds: the lattice ⇤[R] consists of all integer-valued vectors

2 ZN which remain in

These ground states are annihilated by the positive modes, so Ji,n | , ¯ i = Ji,n | , ¯ i = 0
The boundary|Ri thatabove
we’rethelooking
is thena suitableoverthe Ishibash
for n state
1. Excitations
groundforstate
are thengeneratedsumby thenegative
A Quick
Look
atthe
BoundaryStatestates k modes,, ¯ ii withThe
coefficients
of
this
sum
are fixed by the Cardy
Ji, n and 2
J¯i, ⇤[R].
for
n
1.
n

Lewellen sewing
conditions[37,be38].
The
final result
thesector.
boundary
The condition(2.5) mustsolvedseparately
in each forchargeActing stateon the is givebyground states, we haveX
Rij i j R+( ¯)i =i✓·0(2.7)|✓; Ri = gRee k , ¯ = R ii
(2.8

The charge sectors i that obey 2⇤[R]
this equation for some choice of ¯ i are precisely thosethat live in the charge lattice ⇤[R] defined in (2.3). Only these charge sectors arise in
There arenumberstateof newtheaboundary|Ri. ingredients in this expression. The least important is thphase ei R ( ) . An expression for this phase can be found in Appendix B of [33], but
In charge sector 2 ⇤[R], we can construct the Ishibashi state as the usual coherent
withtheaIshibashi
states follows.
given bywill not sum
play
role
in
what
over excitations [34]. We take ¯ = R , to obey (2.7) and write
1
i✓·
X

!

More interesting is the phasefactor e .1 This
arises because
there is not a uniqu¯¯¯k , ii = exp
Rij Ji, n Jj, n | , in that, for each R, we have a manifold osolution to the sewing conditions. This means
n=1
possible boundary states parameterised by N phases ✓i .
ground states labelled by left and right-moving charges

These phases arise even for the simplest boundary conditions, where the reflection of– 10 –
single left-moving fermion into a right-moving fermion can, in general, be implemente
See, e.g. Recknagel and Schomerus, ‘98;by the boundary condition L = ei✓ R . The N phases✓i that Ryu
appearin the boundar
Cho, Shiozaki,
and Ludwig ’16;
Han,aTiwari,
Hsieh
and Ryu ’17 condition R
state (2.8) are generalisation to multiple fermions withchiralboundarysly derived in a somewhat di↵erent context in [36].
p
ZMajorana = 2
The Central Chargepnormalisation factor is important because it coincides
ZMajoranawith= the
2 A✏eck-Ludwig
The
Boundary
CentralChargeThe most
important
new element in (2.8) is the normalisation factor gR . This ischarge,definedbydetermined by insisting that the overlap between any two boundary! states can be!
L a sensible
interpreted,using
modularinvariance,asthepartitionfunction
theory on
g
=
h0,
0
|
✓;
Ri
The normalization R
is important. Thisis the ofboundary
central== exp (⌥mx) (Lcharge,(There
first introducedby Affleck=caveat
and Ludwig
inexp
1991.
It captures
the boundary
the interval.
is an important
in =
this
statement
regardingthe possible
(⌥mx)
(0)
R
to thefreeenergy.
gRexistence
shouldcontribution
beofzeroas amodes;
count thisof thenumberof boundaryof 4.1.) In
of thought
Majorana
willR
be discussed
furtherdegrees
in Section
m. Thisnumbermust
strictly
decrease in any
boundaryRG[33], weshowedthatthis normalisation
factoris given
byflow.
L = R
L = R
p
gR = Vol(⇤[R])
(2.9)

– 11
Lwas
Here Vol(⇤[R]) is the volume ofthe– primitive unit cellof thelattice⇤.
This
result
O(2N
)
⇥
O(2N
)
⇥
(
1)
⇥(
L
R
O(2N )L ⇥ O(2N )R ⇥ ( 1)L ⇥ ( 1) R
previously
derivedin ofa somewhat
context
in [36].
*
i.e. thevolumethe primitivedi↵erent
unit cell of
the lattice
Smith
Tong ’19
The normalisation factor is important because it coincidesBoylewiththeandA✏eck-Ludwig|Ai |Bicentral charge, defined by|Ai |BiThe trivial boundary conditions, corresponding to R = ±1 (or, indeed, te.g.
diagonal
R withentries
±1.)
= to
This
orrespondingto= ±1(or,hasindeed,
• R
gRgR
= h0,
01.| ✓;any
Ri is the smallest value of the cp= 1. This
the smallestof the necessarily
centralharge.
Any ischiralboundaryvalue
conditions2
2
2 has gR > 1. Each such bougR = degreesc• N=2befermionswithcharges
a
+
b
=
c
+
0
Hence,
gshouldthought
of
as
a
count
of
the
numberofboundaryR
ns
necessarily
has gaR number> 1. Each
such boundary
onditionwill
haveof relevant
operators which induce RG flows. Theof r
freedom. This number must strictly decrease in any boundary
RG flow.
pnt
operators
which
induceRG
flows.
The
restof
his paperisconcerned
withunderstanding
these
operators
andflows.
g=c(*This same formula arose as the tension of a D-brane in R
Bachas, Brunner and Roggenkamp ‘12.)
nding these operators and flows.

one class to another,one
andclass
we present
examplesbelow.
We
will explain
whW
toanother,andwepresent
examplesbelow.
logical classes, characterised by a mod 2 anomaly. One might have thought that RG
problematic.
problematic.
flows would remain withina given class. However, as we now describe, our conjecture

Phase
Each
State
(4.2) To
doesWhich
not haveZthis
property.
It is Does
not difficult
to find
RG Belong?
flows that go from2 SPTFirst,
wereviewtheresultof[33]
thatdeterminesthetopologicalclasFirst,
wereviewtheresultof[33]
thatdetermines
one class to another, and we present examples below. We will explain why this is not t
given
boundary state,givenlabelled
by R,state,sits. labelled
Given aby
CFT
on anGiven
interval,wboundaryR, sits.
a CFT
problematic.

0
di↵erent
boundary
conditions
Rend.
and
end
Consider boundary
imposing a boundary
state
oneR
end
of an
interval,and R
theeitherother.
di↵erent
conditions
R on
and
on
eitherIn0 onon[33],
we deriFirst, we review the result of [33] that determines the 0topological class in whicha0formulaforthenumberofgroundstates
G[R,
R
]
of
th
formulaforthenumberofgroundstates
G[R,
R
]
of
this
system:given boundary state, labelled by R, sits. Given a CFT on an interval, we can imposepp0
The numberof groundstates ofRtheandsystemis givendi↵erent
boundary
conditions
R on
eitherbyend. In0Vol(⇤[R])[33],
we derived
a simple0 ]) ppVol(⇤[RVol(⇤[R])Vol(⇤[R])0
0 R0 ] =G[R,formula for the number
G[R, R ] of this system:det0 (1 0RT R0 ) d
G[R,ofRground] = states
Vol(⇤[R, R ])a2 + Vol(⇤[R,
b2 = c2 + 0 R0 ])pVol(⇤[R]) Vol(⇤[R0 ]) p 0
0
0 R0 )
T
0
G[R,
R
]
=
det
(1
R
(4.4)
Here
the
intersection
lattice⇤[R,
R] is definedbe thHere the intersection lattice Vol(⇤[R,⇤[R, R R]p0is]) defined to be those
integertovector
0
N
0
0
N
0
R
=
R
2
Z
.
The
notationdet
the productg=c
R denotes the productdenotes
R = R 2 Z . The notation0 det
over non-vanishing

Here the intersection lattice ⇤[R, R ] is defined to be those integer vectors for which
0 shown0be
RIt can=R
2 ZN .that,
The notation
det
denotesthe degeneracy
product over non-vanishing
eigenvalues.
The
groundstate
hasaninterestingprope
The ground state degeneracy has an interesting property. If the two
bou
0(
R
and
R
lie
in
the
same
class
(i.e.
eitherboth
vector,o0Thegroundstate
degeneracy
hasaninteresting
property.
If
the
two
boundarystatesR and R lie in the same class
(i.e.
both
Z eitherif sameclassvector, or both axial) then0
G[R,
R (i.e.
] 2 either
p bothR and R0 lie in the same
class
vector, asor bothaxial) then the numberofgroundstatesis
integer
expectedof ground states is integer as expected
2Z if di↵erent classof ground states is integer as expected
0
0
G[R,
R
G[R, R ] 2 Z ] 2 Z

G[R, R0 ] 2 ZBoyle Smith and Tong ’19

Flows Between Boundary States

Boundary RG Flows
The d=0+1 boundary behaves, in many ways, like any other QFT. Boundary
operators are classified as:• Irrelevant• Marginal• Relevant

moves among different boundary conditions.
induces an RG flow to a new boundary condition.

The g-theorem: the boundary central charge decreases under RG.

Affleck and Ludwig ’91, Friedan and Konechny ‘03, Casini, Landea and Torroba ‘16where we have introduced the matrix M↵ = Q↵i Q i =he di↵erence from our previous result (3.2) lies in bothX
the explicit µ ⇢dependent
factor2 /2
relating the charges to the energy,
is consistent
with theq
T Q⇢µ↵ Q↵i ⇢iiµ, and in the sum which now runs
overthe=shifted dualZAB(q; µ)
e lattice N
Relevant Operators⌘(⌧ )?
3.2 Boundary Operators
0 ⇢2⇤[R; ✓]
✓ ✓+⇡
?
?
⇤[R; ✓] := ⇤[R] +2⇡ means
Theourstate-operator
The di↵erence fromprevious resultmap(3.2)
lies inthat
boththethe partition
explicit µ functi
depen
iµ↵useQ↵i ⇢theWe canstate-operator
map
todeterminetheboundary
operators.
i
about
the
spectrumofboundary
operators.
To
extract
th
e
,
and
in
the
sum
which
now
runs
over
the
shifteddual
lattice?
he highest
weightstatesarelabelled
byvectors
⇢
2
⇤[R;
✓]
.
From
(3.4),
wecanThey are labelled by:
F
and
drop
the
contribution
of
⇡
from
the
(
1)
factor. The0ad o↵ their charges
✓ ✓+⇡
? ?
?
labelled by⇤[R;⇢ 2 ⇤[R]
Like the✓] :=. ⇤[R]
+ states, the operators hav
2⇡
L0 , again
Q↵ given= Q↵iby⇢i (3.5) and (3.6).
(3.5)
The highest weight states are labelled by vectors ⇢ 2 ⇤[R; ✓]? . From (3This determines their dimension
and charges
Boundary
operators also come in one of two classes: t
nd energy read o↵ their charges
This fermion parity will play a key role in Section 4 where
1 2operators.
1
1 =
by
such
We
pause here to discuss how(3.6)
to cl
L0 = ⇢ = andQ↵ M↵Q↵Q Q↵i ⇢i
2 it is2possible to assign a fermion parity to the la
explain,energy the matrix M↵ = Q↵i Q i = Q̄↵i Q̄ i . This latterNequality,N
here we haveand
introducedFirst, recall that by definition, under a U (1)L ⇥ U (1)Rlating the charges to the energy, is consistent withthe1 Sugawara construction.
1
2
If you turn on a relevant operator, where
L0do
= you
⇢ flow?
= Q↵ M↵ 1 Qiµ↵ Q↵i iµ↵ Q̄↵i
(e
,e
)
2
2
.2 Boundary Operators
NQ i = Q̄↵i Q̄ i . This lattwhere we have introduced
thematrixM↵ =Q↵ibelongingtothepreservedU(1)
the boundar
he state-operator map means that the partition function alsosubgroup,
contains informationiµ↵ Q↵ is consistentiµ↵ Q↵i ⇢i with the Sugawara constru
relating the charges
tophasethe energy,
up
a
e
=
e
Importantly,wethesetbulk
bout the spectrum of boundary operators. To extract this. information,✓ =ferm✓0
nd drop the contribution of ⇡ from the ( 1)F factor. The boundary operators are thenwhere we have introduced the matrix M↵ = Q↵i Q i =We now turn to the main results of this paper. We will follow the RG flow between
relating the charges to the energy, is consistent with the S
di↵erent boundary states.

Boundary
RG
Flows
3.2 Boundary Operators

We start with a given UV boundary state, preserving the U (1)N symmetry characterised by the charge matrix
RU V . As we haveseen,
relevant
operators
areThe state-operator
map
means
thatboundary
the partition
functiolabelled by a vector ⇢ 2about⇤[RU Vthe]? andcarry
chargespectrum of boundary operators. To extract th

We turn on a relevant boundary operator, labelled byF
and drop the Q
contribution
of
⇡
from
the
(
1)
factor. The↵ = Q↵i ⇢ilabelled by ⇢ 2 ⇤[R]? . Like the states, the operators hav
We turn on a single, relevant,bosonic
boundaryof definite charge to initiate
L0 , again
given
by (3.5)operatorand (3.6).

ThisRG
breakssymmetry:anflow.theAlong
the flow, the symmetry is broken to

Boundary operators also come in one of two classes: thN1
U (1)parity! Uwill(1)Nplay
This fermion
a key role in Section 4 whereby such operators. We pause here to discuss how to cla
In what follows, we make the following, important assumption: At the end of the flow
explain, it is possible to assign a fermion parity to the lat
Assumption:Nan emergent U (1) symmetry is again restored. This means that, in the infra-red, theN a di↵erenN
physics is again described First,
by a boundarystate
the form under(2.8), now
withrecall thatby ofdefinition,a U (1)⇥
U(1)
L
R
charge matrix RIR .
iµ↵N
Qsymmetry↵i
At the end of the RG flow, we restore a (typically different)(eU(1), eiµ↵ Q̄↵i )There is, in fact, a unique choice for RIR for each relevant operator labelled by ⇢This follows because of the U (1)N 1 symmetry that exists
N along the RG flow. Thi
belonging to the preserved U (1) subgroup, the boundary
symmetry must be preserved by the iµIR↵ Qboundary
a condition which translateiµ↵ Qstate,↵↵i ⇢iup a phase e=e. Importantly, the bulk ferm
into the simple requirement thatwherewe haveintroduced
theamatrix
M↵which= Q↵i
Q i=
symmetry must be preservedby theIR boundary
state,
conditiontranslate
into the simple requirementrelatingthat the charges to the energy, is consistent with the S

Boundary
RG
Flows
3.2 Boundary Operators
RIR⇢?

= RU V⇢?

(4.1

The state-operator map means that the partition functio
or inWeotherwords, that the two
matricesmust
agree
on the orthogonal
complement⇢about
the spectrum
ofbyboundary
operators.
To extractofth
turn on a relevant boundary
operator,labelled
But for orthogonal matrices,
thisconditionis highly constraining.
particular,
F
and
drop
the contributionof ⇡ from the (In 1)
factor. thereThe
are only two options for Rlabelled
is R
itself,? but this is quickly ruled out by the
IR . One by⇢ U2V ⇤[R]
. Like the states, the operators hav
fact that gIR = gU V , in contraventionof thewhich states that the centra
L0 , again givenby g-theorem(3.5) and (3.6).
chargemust strictly
under
perturbations.
only
leavesthe second
Assumingthat wedecrease
land among
our relevant
general classof boundary This
states,there
is a unique
Boundary
also comeof twocandidate:
option,which is that the matrices
di↵er operators
by a reflection
along intheonevector⇢: classes: thThis fermion parity✓ will play a key◆ role in Section 4 where2byIRsuch
We pausecla(R)ik = operators.
(RU V )ij jk⇢j ⇢khere to discuss how to(4.22⇢explain, it is possible to assign a fermion parity to the lat
The second factor is the matrix implementing the reflection along ⇢.
NFirst, recall that by definition, under a U (1)N
⇥
U
(1)
L
R
One might naively think that the IR central charge is
One might think that the infra-red central charge is, following (2.9),p(eiµ↵ Q↵i , eiµ↵ Q̄↵i )gnaive = Vol(⇤[RIR ])(4.3)

N
belongingtothepreserved
U(1)
subgroup,
theanswer.
boundaryAnd, for some of the RG flows, where no subtleties arise, this indeed
the correctiµ↵ Q↵iµ↵ Q↵i ⇢i this is too quick.
…but
up
a
phase
e
=
e
. Importantly,
bulk
ferm
However, it is not true in general. There are two rather interestinge↵ectsthethat
may– 22 –
occur, both of which leave us with an infra-red central charge larger than (4.3). First,

a2 + b 2 = c 2 + 0

Boundary
RG Flows
p
gR =

c

(
There are a number of subtletiesthat we must address:
0G[R, R ] 2Zp2Zif same class

if di↵erent class

1. The RG flow can move us from one SPT phase to the other! This is only consistent if
the IR boundary statep
is accompanied by an extra Majorana mode. This increases
the central charge by 2 .

4.2

Non-Primitive Boundary States

2. We could start with an extra Majorana mode and use it to dress a fermionic operator
We now turnto theto initiate
an second
RG flow. subtlety in the RG flows. We have seen that turning on asingle, relevant operator in the UV breaks U (1)N ! U (1)N 1 . However, this is not thefull story.
There
also
remnant
discretesymmetry, so that
3. TheRG isflow
mayapreservea discrete
symmetry
U (1)N ! U (1)N 1 ⇥ Zn
Typically
the n
IR is
boundarystateonedoesintroduced
not. We mustin
thensum overthe measures
images. Thisagain
Here, theintegerthe same(4.6),
which
thefailure ofincreases the central charge.
⇢ to be a primitive vector.

This discrete symmetry Zn is preserved along the RG flow. However, one finds thatthe naı̈ve IR boundary state is not invariant under the full Zn symmetry. To rectify
this, the infra-red boundary state must be a linear sum of states of the form (2.8) suchand the ratio of IR to UVpcentral charges is2
s RG Flows
BoundarypgIR
1Vol(⇤IR )= p ⇥n= ⇢2 /2
gU V
Vol(⇤U V )2Once these subtleties are taken into account, we find the following simple result:
The central
charge relation
We perturb by an operator O . Then the infra-red central charge ismportantly, we find the same ratio of central charges for each of the three types of RG
ows described above. Moreover, we recognise L0 = ⇢2 /2 as the dimension of the UVperator O that initiates the RG flow. We learn thatpgIR = gU V dim O

This is the formula (1.5) advertised in the introduction. Since the UV operator isecessarily relevant, we have ⇢2 < 2. This ensures that gIR < gU V , and the g-theorems obeyed.
Note: this immediately satisfies the g-theorem.

More General RG Flowsn our discussion above, we have restricted attention to RG flows initiated by operatorswith a definite charge under U (1)N . This ensures that the original symmetry is brokeno U (1)N 1 , which allowed us to identify the infra-red state (4.2).

The Z8 Classification of d=2+1 SPT Phasesrs as

i ! e

i ! e

i and on the right-movers asi,a2 + b2 =(c2 + 0Z
if same class
0G[R, R ] 2 p
The
View
fromptheEdgee key physics will depend only
on R;
for example,
theof class
2Z collection
if di↵erent
R =erators and their dimensions depend only gon
R. cNonetheless, we

p so require
ges of these operators are inherited from Q and Q̄ andWe could also view our d=1+1 fermions as the edge2of a d=2+1 SPT phase.
yond a knowledge of R.
(

same class
There is a Z8 classification
which
fermions in d=1+1 can
0 meansZthat 8ifMajorana
G[R, R ] 2 p
LR
the Maldacena-Ludwig
statefermion parity
(
1)
⇥
(
1)
be gapped preserving, chiral.
2Z if di↵erent classinvolves N = 4 Dirac fermions. The boundary conditions are,Fidkowski
Kitaev ‘09, Ryu and Zhang ‘12, Qi ‘12
p and symmetry
y described by requiring an SU (4) ⇥ U (1) global
un2vers transform in the 4+1 representation, while the O
right-moversThis,in turn,means thatit shouldon
bethe
possibleto construct
a boundary condition
ere is nolinearboundary
condition
fermions
that
reflects
for 8 Majorana fermions preserving ( 1)L ⇥ ( 1)R .
ct first noted in the context of monopole physics [39, 40]. Instead,on is implemented
theboundarystate
The simplest bysuchboundarystate
was with
constructed by Maldacena and Ludwig in+++preserves an SO(8) symmetry:
1 1995, and, in
0 addition, 1
C
C
C
A

,

B
B+
Q̄↵i = B
@ +

+

C
C
C
A

)

Rij =

O

ij

1
2

(2.11)(and recognized as such by Cho, Shiozaki, Ryu and Ludwig ‘16)

was previously introduced by Maldacena and Ludwig [32]. Its the symmetry of the Cartan subalgebra U (1)4 ⇢ SU (4) ⇥ U (1).

p
2
O
R
Z
if
same
class
⇥ ( 1)
2
G[R, R(0 ] 21) p
L2Z

if di↵erent class

L
R
(
1)
⇥
(
1)
The Z8 Classification

The Z2 ⇥ Z2 p
symmetry( 1)L ⇥ ( 1)R2
O ( 2 1)L | , ¯ i = ( 1) | , ¯ iFrom now on, we specialise to the Z2 ⇥ Z2 symmetry.
condition for thisOtoL be preserved
is
R
3.2

Which of our boundary states preserve ( 1) ⇥ ( 1) ?
Can show:2O2( 1) | , ¯ i = ( 1) |R , ¯ i8¯ 2 ⇤[R]¯ 2¯( 1) | , i = ( 1) | , i = 0 mod
2L¯( 1) | , i = ( 1) | , ¯ iLHere, we will use theOresults of the previous
section to2L¯ i = ( 1) | , ¯ i
2
¯
(
1)
|
,
R
¯
¯
rstatements:
either (of 1)
the| following, i = ( 1)two| ,stronger
i
N
gR =
¯2
R
¯
¯
( 1) | , i = ( 1) | , ivi =2i.e. preservation if the lattice
9v 2 ⇤[R] is even
s.t.
¯2
R
¯
( 1) | , i = ( 1) | , ¯ i(Rv)i =and if the lattice has dimension N = 4k
Claim: This can only happen
Claim: The stable boundary state with this property
9whas
2 g⇤[R]
R =

r

N
s.t.
2

wi =
(Rw)i =

These are respectively the statements that ( 1)F and (An Open Question‘t Hooft Lines in Chiral Gauge Theories• ‘t Hooft lines in d=3+1 are defined by boundary conditions for fields• Decompose fermions in angular momentum modes to reduce to d=1+1• The lowest angular momentum modes of chiral fermions require chiralboundary conditions

Callan ’83, Polchinski ‘84
Affleck and Sagi ‘93; Maldacena and Ludwig ‘95

2

22‘t Hooft Lines in ChiralN
Theories=
NGauge= 4k
4k⇤[R] ==⇤[R]nn

N = 4kooNNN
1 1 NNN : nR 2 ZNN
2
Z
=
Z
\
R
ZZ
2Z : R 2Z=Z \R o⇤[R] =

2 ZN : R 2 ZNe.g. U(1) gauge theory with Weyl fermions with charges 1, 5, -7, -8, 9= ZN \ R 1 Z

SU
⇥
UU(1)
SU(2)
(2)rot⇥(1)
rotchiral boundary conditions preserving
Left movers:
Right movers:SU (2)rot ⇥ U (1) with

11 , 55 , 99
11 , 55 , 99
77 , 88
77 , 8816
16

16
Boundary state unknown!

Thank you for your attention