Lecture 14 - Stanford Universitysporadic.stanford.edu/quantum/lecture14.pdf · 2019-05-29 ·...

Modular Tensor Categories (III) The SL(2, Z) action

Lecture 14

Daniel Bump

May 29, 2019

∑j

dj θ

ji k

= p+ × θ−1 θ−1

i k


Review: Modular tensor categories

We recall the definition of a modular tensor category. This is asemisimple ribbon category with a finite number of simpleobjects Vi (i ∈ I). We assume that the ground ring K = End(I) isa field, where I is the unit object. This implies that I is itself asimple object. Moreover, we assume that End(Vi) = End(K) forall simple objects Vi. Finally, we assume that the S-matrix (̃sij) isinvertible, where s̃ij is the scalar (i.e. endomorphism of I)defined by the Hopf link:

Vi V∗iVj V∗j


Review: Symmetry of the S-matrix

We now observe that s̃ij = s̃ji. To see this use the crossingidentities to move the link around:

Vi V∗iVj V∗j =Vi V∗i

Vj V∗j= Vi V∗iVj V∗j


The Fusion Ring

The monoidal structure gives the Grothendieck group of thecategory a multiplication that makes it into a ring. This ring is afree abelian group on the isomorphism classes of simplemodules. Thus if i ∈ I let [i] = [Vi] be the class of arepresentative simple module Vi. We can decompose Vi ⊗ Vj

into simple modules Vk with structure constants Nkij. Thus

[i][j] =∑

k

Nkij[k], Vi ⊗ Vj =

⊕k

NkijVk.

The Nkij are nonnegative integers. Take the quantum dimension,

which is multiplicative by Lecture 4:

didj =∑

k

Nkijdk, di = dim(Vi).


Review: Alternative trace principle

Let f : V → V be a morphism in a ribbon category. We cancompute the trace in two different ways:

fV

V∗ = fV

V∗

To see this, replace evV by evV(1⊗ θ−1V )cV,V∗ in the first figure,

and in the second, replace coevV by cV,V∗(θV ⊗ 1) coevV .

fV

V∗

θ−1V

θ−1V

VV∗

f


Review: Consequences of the alternative trace principle

If f : Vi → Vi is any morphism, since Vi is simple, End(Vi) = K.Thus f is just a scalar. In particular, θVi is a scalar which we willdenote θi.

We will denote the quantum dimension of Vi as di. Applying thealternative trace principle to 1V : V → V gives di = di∗, where i∗is the index such that Vi∗ = V∗i .

Remembering that θV∗i

= θ∗Vi, the alternative trace principle

implies that θi = θi∗.

θi

ViV∗i = θV∗

i

V∗iVi


Review: Alternative definition of s̃ij

As in Lecture 13s̃ij = tr

(c−1

Vi∗,Vjc−1

Vj,V∗i

).

Indeed use the principle of the alternative trace to flip one circle:

Vi V∗iVj V∗j

V∗i Vj


Reminder: the ribbon axiom

c−1U,W ◦ c−1

W,U = (θ−1U ⊗ θ

−1W ) ◦ θU⊗W = θU⊗W ◦ (θ−1

U ⊗ θ−1W )

U W

U W

=

θU⊗W

θ−1U θ−1

W

U W

U W


Review: Another formula for the S-matrix

Proposition

s̃ij = θ−1i θ−1

j tr(θVi∗⊗Vj) = θ−1i θ−1

j

∑k

Nki∗,jθkdk.

Vi V∗iVj V∗jV∗i Vj

θ−1V∗

iθ−1

Vj

θV∗i ⊗Vj

V∗i Vj

For the second formula use V∗i ⊗ Vj =∑

k Nki∗,jVk and θi∗ = θi.


Review: The group SL(2,Z)

The group SL(2,Z) is the mapping class group of the torus, i.e.the group of group of homeomorphisms modulo isotropy. Thelarger group SL(2,R) acts on the upper half planeH = {z = x + iy ∈ C|y > 0} via(

a bc d

): z 7→ az + b

cz + d.

The subgroup Γ = SL(2,Z) acts discontinuously.

A modular form is a function f that satisfies

f (z) = (cz + d)−kf(

az + bcz + d

)for(

a bc d

)in SL(2,Z) or a subgroup of finite index.


Review: Two important elements

Note that −I ∈ SL(2,Z) acts trivially on H, so the action is notfaithful. Let

S =

(−1

1

), T =

(1 1

1

)These generate SL(2,Z). Since S2 = −I, S has order 4 as anelement of the group but order 2 in its action on H.

T : z→ z + 1 is the translation by 1. If a holomorphic function fis invariant under SL(2,Z) it is invariant under T and so it has aFourier expansion:

f (z) =∑

anqn, q = e2πiz.

Although T has infinite order, ST has order 3 in its action on Hor 6 as an element of SL(2,Z).

S2 = −I, (ST)3 = −I.


Review: the fundamental domain

Here is the well-known fundamental domain for SL(2,Z):

e2πi/3 i

We have marked the fixed points i and e2πi/3 of S and ST.

S =

(−1

1

), T =

(1 1

1

)

S : z→ −1z, T : z→ z + 1.


The mapping class group of a torus

The group SL(2,Z) has another significance: it is the mappingclass group of a torus. A torus M may be obtained by gluingopposite edges of a parallelogram:

A homeomorphisms may be obtained by obtained by cuttingthe torus apart along the dotted line twisting and regluing.Another complementary twist will serve to generate themapping class group. These Dehn twists correspond to(

1 10 1

),

(1 01 1

).


Relations in a MTC (Following Bakalov and Kirillov Chapter 3)

Letp+ = θid2

i , p− = θ−1i d2

i

and define matrices t = (δijθi), c = (δi,j∗). We will prove:

Proposition

ct = tc, c̃s = s̃c, c2 = 1,

(̃st)3 = p+s̃2, (̃st−1)3 = p−s̃2c,

s̃2 = p+p−c.

The relation ct = tc follows from θi = θi∗, a consequence ofθ∗V = θV∗ and the alternative trace principle. The identity c2 = 1is obvious and c̃s = s̃c is not hard.


Significance

From Lecture 13, SL(2,Z) has generators S and T

S =

(−1

1

), T =

(1 1

1

)with relations

S2 = (ST)3 = −I, (−I)S = S(−I), (−I)T = T(−I).

Compare this with s = c̃s, where c = (δi,j∗) is the “chargeconjugation matrix” and t = (δi,jθi) is the matrix of twists. Thematrix c is central and

s2 = s̃2 = constant× c, (st)3 = constant× c,

and c is central. So we have a projective representationSL(2,Z)→ GL|I|(K) such that

π(S) = s, π(T) = t, π(−I) = c.


Significance (continued)

Turaev showed that there is an action of the mapping classgroup of an orientable surface on the fusion category. Themapping class group of a torus is SL(2,Z).In WZW conformal field theories, MTC categories areconstructed in which the fields are modular forms. Hencethere is an action of SL(2,Z). These categories areassociated with integrable representations of affine Liealgebras, whose characters are modular forms by Kac andPeterson (1984).

In this class we will be concerned with MTC coming fromrepresentations of quantum groups. The two constructions arerelated in papers of Finkelberg, Andersen and Paradowski, andSawin.


Significance (continued)

Before we launch into the construction of the SL(2,Z) action,we explain further this quote from Bakalov and Kirillov.

The appearance of the modular group in tensor categories mayseem mysterious; however there is a simple geometricalexplanation, based on the fact that to each modular tensorcategory one can associate a 2 + 1 dimensional TQFT. Thisshows that in fact we have an action of the mapping classgroup of any oriented 2-dimensional surface on the appropriateobjects in MTC. This is the key idea in [Turaev’s book].

To explain this further, both Turaev’s book and the book ofBakalov and Kirillov are largely concerned with 3 equivalentthings: Modular Tensor Categories, 3D TQFT and 2D ModularFunctors.


Modular Functors

The notion of a modular functor is due to Moore and Seibergand Graeme Segal. Consider the category of 2-dimensionalorientable manifolds with morphisms homotopy classes ofhomeomorphisms. This is a monoidal category, thecomposition law being disjoint union. To oversimplify, a 2Dmodular functor is a monoidal functor τ from this category tothe category of finite-dimensional vector spaces. (See Bakalovand Kirillov Chapter 5.) If M is an object in the category thenthe mapping class group which is the group of automorphismsof M in this category acts on τ(M).


Following Bakalov and Kirillov

Recall thatp+ = θid2

i , p− = θ−1i d2

i .

We will show that

∑j

dj

θVj

Vj

V∗j

Vi

= p+

Vi

θ−1Vi

∑j

dj

θ−1Vj

Vj

V∗j

Vi

= p−

Vi

θVi


Proof

We prove the first one only. Multiply both sides by θi. We need

∑j

djθi

θVj

Vj

V∗j

Vi

= p+

Vi

Since Vi is simple, Hom(Vi,Vi) is one-dimensional so the twosides are proportional. Trace of the left side:

∑j

dj

θ

θj

i


Reminder: the ribbon axiom

c−1U,W ◦ c−1

W,U = (θ−1U ⊗ θ

−1W ) ◦ θU⊗W = θU⊗W ◦ (θ−1

U ⊗ θ−1W )

U W

U W

=

θU⊗W

θ−1U θ−1

W

U W

U W


Proof, continued

We are computing the trace:

tr

∑

j

dj

θVj

ji

=∑

j

djθi

θ

θ

j

i


Proof, continued

Using the principle of the alternative trace to flip the red circle,then the ribbon axiom:

∑j

dj

Vj Vi V∗i V∗j

=∑

j

dj

θVj⊗Vi

j i i∗ j∗


Proof, continued

Now we substitute

Vj ⊗ Vi =∑

k

NkjiVk

to obtain ∑j,k

djNkji tr(θkVk) =

∑j,k

djNkjiθkdk.

Now Nkji is the multiplicity of Vk in Vj⊗Vi, or the multiplicity of the

unit object in V∗k ⊗ Vj ⊗ Vi. Thus Nkji = Nj∗

k∗,i. Now remembering∑k

Nkijdk = didj, dj = dj∗, dk = dk∗,

we get ∑j,k

dj∗Nj∗k∗,iθkdk = di

∑k

θkd2k = dip+.


Proof, concluded

To summarize, we showed that

∑j

tr

djθi

θVj

Vj

V∗j

Vi

= dip+ = p+ tr(1Vi).

Since End(Vi) = K, this implies

∑j

dj

θVj

Vj

V∗j

Vi

= p+

and the other identity is similar.


Our story so far

With p+ = θid2i , p− = θ−1

i d2i and V = Vi irreducible,

∑j

dj θVj

Vj

V∗j

V

= p+V

θ−1V,

∑j

dj θ−1Vj

Vj

V∗j

V

= p−V

θV

By linearity, this identity is true even if V is not irreducible.


This implies ...

We will show that this implies

∑j

dj θ

i k

= p+θ−1 θ−1

i k

Indeed by the last slide, LHS equals

Vi ⊗ Vk

θ−1

and the statement follows using the ribbon axiom.


Towards (st)3

Remember we need:

ct = tc, c̃s = s̃c, c2 = 1,

(̃st)3 = p+s̃2, (̃st−1)3 = p−s̃2c,

s̃2 = p+p−c.

and the first formula is easy. We now prove (̃st)3 = p+s̃2,following Bakalov and Kirillov. We prove this by evaluating

E =∑

j

dj θ

i

j

k

two ways.


First way

By the calculation we just did

E =∑

j

dj θ

ji k

= p+ × θ−1 θ−1

i k

To evaluate this scalar take the trace and divide by dim(Vi):

E = p+d−1i θ−1

i θ−1k × i k = d−1

i θ−1i θ−1

k s̃ik


Second way

Remember how we proved the symmetry s̃ij = s̃ij:

i j = ij

= i j

Similarly:

E =∑

j

dj θ

ji k

=∑

j

djθj

jk

i


Second evaluation, continued

Now

k

j

= d−1j s̃kj × j = d−1

j s̃kj

as we see by taking the trace. Using this twice:

E =∑

j

djθj j

k i

=∑

j

djθjd−1j s̃kj

ji

=∑

j θj̃skj̃sjid−1i


conclusion

Comparing our two evaluations for E:

p+d−1i θ−1

i θ−1k s̃ik =

∑j

θj̃skj̃sjid−1i .

Multiplying by di and writing in terms of the matrices s̃ = (̃sij),t = diag(ti):

p+t−1s̃t−1 = s̃t̃s.

Therefore(̃st)3 = p+s̃.

A similar calculation gives

(̃st−1)3 = p−s̃c,

where c = (δi∗,j) is the charge conjugation matrix.


Concluding the proof of the proposition

So far we have:

ct = tc, c̃s = s̃c, c2 = 1,

(̃st)3 = p+s̃2, (̃st−1)3 = p−s̃2c,

We want to deduce:s̃2 = p+p−c. (1)

First let us show that s̃2 commutes with t. It is enough to showthat t commutes with p+s̃2. We have

t(p+s̃2)t−1 = t(̃st)3 t−1 = (t̃s)3 = s̃−1(̃st)3 s̃ = s̃−1(p+s̃2)̃s = p+s̃2.

A step in the last calculation gives (t̃s)3 = p+s̃2. So

p+p−s̃4c = (t̃s)3(̃st−1)3 = t̃st̃st̃s̃st−1s̃t−1s̃t−1.

Since s̃2 commutes with t, rearrange and get s̃6, whence (1).

Lecture 14 - Stanford Universitysporadic.stanford.edu/quantum/lecture14.pdf · 2019-05-29 ·...

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Transcript of Lecture 14 - Stanford Universitysporadic.stanford.edu/quantum/lecture14.pdf · 2019-05-29 ·...