Daniel Bump May 28, 2019 coevsporadic.stanford.edu/quantum/lecture9.pdfcoevV or coevV depending on...

Duality in Ribbon Categories The Kauffman bracket as a quantum trace Gaussian binomial coefficients

Lecture 9

Daniel Bump

May 28, 2019

V ∗ V

coevV

=

V ∗ V

θ−1V

coevV


Dual objects

Let V be an object in a rigid monoidal category. We recall theassumptions we made of the dual, which we call the left dualV ∗. It comes with morphisms evV : V ∗ ⊗ V → I andcoevV : I → V ⊗ V ∗ subject to

(1V⊗evV )◦(coevV ⊗1V ) = 1V , (evV ⊗1V∗)◦(1V∗⊗coevV ) = 1V∗ .

V

evV

coevV

V

V ∗=

V

V

V ∗

evV

coevV

V ∗

V=

V ∗

V ∗


The right dual

Dually, we can ask for a right dual ∗V with morphismsevV : V ⊗ ∗V → I and coevV : I → ∗V ⊗ V subject to

(evV ⊗1V ) ◦ (1V ⊗ coevV ) = 1V ,

(1∗V ⊗ evV ) ◦ (coevV ⊗1∗V ) = 1V∗ .

V

evV

coevV

V

∗V=

V

V

∗V

evV

coevV

∗V

V=

∗V

∗V


Left and right duals

The notions of left and right dual are not really different. V ∗ is aright dual of V if and only if V is a left dual of V ∗. If every objectin the category has both a right and a left dual, then bydefinition

∗(V ∗) = V , (∗V )∗ = V ,

since the only difference between the defining properties

V ∗

evV

coevV

V ∗

V =

V ∗

V ∗

V

evV

coevV

V

∗V =

V

V

is the labelling of V , V ∗ and ∗V .


Duals in ribbon categories

PropositionIn a ribbon category, every left dual is also a right dual.

We describe the morphisms evV : V ⊗ V ∗ → I andcoevV : I → V ∗ ⊗ V .

evV = evV ◦(1V∗ ⊗ θ−1V ) ◦ cV ,V∗

V V ∗

evV

=

V V ∗

θ−1V

evV


Duals in ribbon categories (continued)

coevV = cV ,V∗ ◦ (1V∗ ⊗ θ−1V ) ◦ coevV

V ∗ V

coevV

=

V ∗ V

θ−1V

coevV

We will leave it to the reader to check that these definitions ofevV and coevV make the left dual V ∗ satisfy the definition of theright dual.


Knot invariants

Let us pick a module V in a ribbon category with unit object K .We now interpret any framed knot or link as a morphismK → K . The framing is important because the framed knotinvariant we describe depends on the framing. Turning it into aactual knot invariant requires modifying the definition to takeinto account the effect of Type I Reidemeister moves.

Let us project the knot onto a plane, resulting in a knot diagram,with crossings. Then there is a natural framing, namely we maytake the normal vectors to be perpendicular to the plane.


Labeling the edges

We now describe a way of assigning a vector space to an edge.(Turaev and Reshetikin worked in greater generality, but this isenough for the Jones polynomial.)

On every segment that points down, we label every edge by V .If the segment points up we label it by V ∗. Then we interpretevery local maximum of the curve as a coevaluation, eithercoevV or coevV depending on whether the arrow points left orright.


Labelling the trefoil knot

In other words we label segments of the curve where theorientation tangent vector points down by V , and segmentswhere it points up by the dual V ∗:

V V ∗

V V ∗

Reading the diagram from top to bottom gives a morphismK → K , a scalar which should be a framed knot invariant.


The trefoil knot unravelled

V V ∗

V V ∗

K V ⊗ V ∗ V ⊗ V ⊗ V ∗ ⊗ V ∗

K V ⊗ V ∗ ⊗ V ⊗ V ∗ V ⊗ V ⊗ V ∗ ⊗ V ∗

coevV 1V⊗coevV ⊗1V∗

cV ,V⊗cV∗,V∗

coevV ⊗ coevV 1V⊗cV ,V∗⊗1V∗

We use the ribbon element implicitly since we use coevV .


The standard module

Now let us consider the case where V is the standard moduleof Uq(sl2). A simplification here is that V = V ∗ is self-dual. Wewill not completely study this example today. For example wehave not computed the twist. But we have computed theR-matrix, which we denote T today:

T = τR =

q

q − q−1 11

q

.

As we noted, T satisfies the quadratic equation:

T 2 = (q − q−1)T + 1,

which appeared in the definition of the Hecke algebra.


Normalization

Recall that we obtained T in Lecture 6 not by deducing it fromthe universal R-matrix, but by requiring it to be an H-modulehomomorphism V ⊗ V → V ⊗ V , where H = Uq(sl2), and thenimposing the Yang-Baxter equation. This procedure onlydetermines T up to a constant.

Thus let c = cV ,V be the correct ribbon category normalizationof the braiding V ⊗ V → V ⊗ V . In fact c = q−1/2T . We need tojustify this normalization properly, but at the moment we simplynote that it gives cV ,V∗cV∗,V determinant 1. (We are refrainingfrom identifying V = V ∗ for the purpose of this last statement.)


The skein relation

Now

T 2 = (q − q−1)T + 1, T − T−1 = q − q−1

can be written:

q1/2c − q−1/2c−1 = q − q−1.

Let us see what the implications of this are for the quantumtrace. We consider three knots L+, L−, L0 that are the sameexcept for the circled region:

L+ L− L0


The Kauffman bracket as a quantum trace

We have explained how to attach to a knot or link L a morphismK → K , which is after all just a scalar. We will denote thisscalar tr(V ).

Returning to the three knots L+, L− and L0, we are using c toimplement the morphism in L+, c−1 to implement the morphismin L− and 1V⊗V in L0. Taking traces, the quadratic relation givesus

q1/2 trL+ −q1/2 trL+ = (q − q−1) trL0 .

We recognize this as the skein relation for the Kauffmanbracket (from Lecture 8):

a〈L+〉 − a−1〈L−〉 = (a2 − a−2)〈L0〉.

The parameter a has been replaced by q1/2.


Noncommutative geometry

One can generalize some of algebraic geometry to includenoncommutative variables. A common example is the stronganalogy between the symmetric algebra of a vector spaceS(V ), which equals the polynomial ring on the dual space V ∗,and the exterior algebra

∧V , in which variables anticommute:

x ∧ y = −y ∧ x .

A generalization involves variables (say x and y ) that commutethis way:

yx = qxy .

If q = 1, the algebra generated by x and y is a polynomial ring,the affine algebra (coordinate ring) of the complex plane. moregenerally, the algebra generated by x and y such that yx = qxymay be regarded as the quantized function algebra associatedwith the quantum affine plane.


Noncommutative geometry and q-everything

In (commutative) algebraic geometry, the Nullstellensatz tellsus that there is a bijection between the points of an algebraicvariety (over an algebraically closed field) and the maximalideals of its affine algebra. This fails in noncommutativegeometry, and the notion of a point loses meaning to someextent. However the ring of functions survives.

Gaussian binomial coefficients emerge quickly in anycommputation involving x and y that satisfy yx = qxy . Theseare polynomials in q that generalize the usual binomialcoefficients

(mn

). They are the tip of the iceberg in the world of

q-series which includes the work of Ramanujan and manyothers on partitions, so-called basic hypergeometric functions,modular forms, characters of affine Lie algebras, etc.


Gaussian binomial coefficients

The equation yx = qxy leads to identities for Gaussian binomialcoefficients, that are important both for quantized functionalgebras and for quantized enveloping algebras.Gaussian binomial coefficients permeate the subject ofquantum groups. Let q be a parameter. In one commonly usednormalization (Gauss’)(

mn

)(q)

=(qm − 1) · · · (qm−n+1 − 1)

(qn − 1) · · · (q − 1)=

m(q)!

n(q)!(m − n)q!

where we define

m(q) =qm − 1q − 1

, m(q)! =m∏

k=1

k(q).

We define(m

n

)(q) = 0 unless 0 6 n 6 m.


Gaussian binomial coefficients (continued)

This normalization is useful for combinatorial purposes; forexample if q is a prime power it counts the number of points ina Grassmannian over the finite field Fq. That is:

Proposition

The number of vector subspaces of Fmq of dimension n is

(mn

)(q).

Indeed, GL(m,Fq) acts transitively with stabilizer

P =

{(g1 ∗

g2

) ∣∣∣∣g1 ∈ GL(n),g2 ∈ GL(m − n)

}the index is

(mn

)(q). An approach to quantum groups using flag

varieties over finite fields (like this Grassmannian) was given byBeilinson, Lusztig and MacPherson.


Gaussian binomial coefficients (continued)

We will also encounter the normalization[mn

]q

=[m]q!

[n]q![m − n]q!, [m](q) =

qm − q−m

q − q−1 ,

[m]q! =m∏

k=1

[k ]q.

When q → 1 we have m(q) −→ m and [m]q −→ 1 since

m(q) =(qm − 1)

q − 1= qm−1 + . . .+ 1 (m terms).

So(m

n

)q →

(mn

)and the Gaussian binomial coefficients really

are a generalization of the ordinary binomial coefficients.


Some important formulas

The familiar formula for the usual binomial coefficients:(m + 1n + 1

)=

(mn

)+

(m

n + 1

)has a generalization to Gaussian binomial coefficients:(

m + 1n + 1

)(q)

= qm−n(

mn

)(q)

+

(m

n + 1

)(q). (1)

Equivalently[m + 1n + 1

]q

= qn+1[

mn + 1

]q

+ qn−m[

mn

]q


Proof of (1)

Let us prove (1). First let us check that

qm−nn(q) + (m − n)(q) = (m)(q). (2)

indeed

qm−n (qn − 1)

q − 1+

qm−n − 1q − 1

=qm − 1q − 1

.

Now

qm−n(

mn

)(q)

+

(m

n + 1

)(q)

=

qm−n m(q)!

n(q)!(m − n)(q)!+

m(q)!

(n + 1)(q)!(m − n − 1)(q)!.

Put this over a common denominator, use (2) and (1) follows.


A variant

There is a variant of the formula we just proved:(m + 1n + 1

)(q)

= qm−n(

mn

)(q)

+

(m

n + 1

)(q)

The variant is(m + 1n + 1

)(q)

= qn+1(

mn + 1

)(q)

+

(mn

)(q)

Indeed note that(m

n

)(q) =

( mm−n

)(q)

so

m + 1n + 1

(q)

=

m + 1m − n

(q)

= qn+1

mm − n − 1

(q)

+

mm − n

(q)

,

proving the second formula.


The Gaussian binomial theorem

There is a Gaussian binomial theorem. Let x and y benoncommuting variables such that

yx = qxy .

Then

(x + y)m =∑m

n=0

(mn

)(q)

xnym−n.


Proof of the Gaussian Binomial Theorem

By induction

(x + y)m+1 = (x + y)m∑

n=0

(mn

)(q)

xnym−n.

m∑n=0

(mn

)(q)

xn+1ym−n +m∑

n=0

qn(

mn

)(q)

xnym−n+1.

The coefficient of xn+1ym−n is(mn

)(q)

+ qn+1(

mn + 1

)(q)

=

(m + 1n + 1

)(q)

Substituting this and changing n to n − 1 completes the proof.


The comultiplication of Uq(sl2)

As an example, consider Uq(sl2). Recall that

∆E = 1⊗ E + E ⊗ K

where KEK−1 = q2E . So if x = E ⊗ K and y = 1⊗ E then

yx = q−2xy .

Therefore (since ∆ is a homomorphism)

∆Em = (∆E)m =m∑

k=0

(mn

)(q−2)

En ⊗ K nEm−n.

A similar formula applies for ∆F m.


A kind of Frobenius

If q is a root of unity, this has an interesting implication,analogous to properties of the Frobenius map. Remember thatif p is a prime then

(pk

)= 0 in Fp for 0 < k < p, because the

binomial coefficient is p!k!(p−k)! and the numerator is divisible by

p, while the denominator is not. Using the binomial theorem,

(x + y)p = xp + yp.


A kind of Frobenius (continued)

Very similarly, let N be an odd positive integer. If q is a primitiveN-th root of unity, then (N)q = 0 but (k)q 6= 0 if k < N.Moreover the same is true for (p)(q−2) since q2 is also aprimitive N-th root of unity. This means that

(Nk

)(q−2)

= 0 if0 < k < N. Therefore

∆EN = EN ⊗ 1 + EN ⊗ K N ,

and similarly∆F N = F N ⊗ 1 + K−N ⊗ F N .

From this it may be deduced that if I is the ideal in H = Uq(sl2)generated by EN , F N and K N − 1 then ∆(I) ⊆ H ⊗ I ⊕ I ⊕ H.This implies that ∆ induces a comultipliccation on H/I, whichthen becomes a finite-dimensional Hopf algebra,


Exercises

Exercise 1. Prove that coevV and evV as defined in the firstsection by

evV = evV ◦(1V∗ ⊗ θ−1V ) ◦ cV ,V∗ ,

coevV = cV ,V∗ ◦ (1V∗ ⊗ θ−1V ) ◦ coevV

make V ∗ into a right dual.

Exercise 2. (a) Prove that if H is a Hopf algebra and I is anideal such that ∆(I) ⊆ H ⊗ I + I ⊗ H, ε(I) = 0 and S(I) ⊂ I then∆ induces a comultiplication on the quotient ring H/I anddeduce that H/I is a Hopf algebra.(b) Check these conditions for I as above (last page).


Exercises (continued)

Exercise 3. Suppose q is a nonzero complex number that is nota root of unity. Let x and y be noncommuting variables suchthat yx = qxy . Define

eq(x) =∞∑

n=0

xn

n(q)!.

Prove thateq(x + y) = eq(x)eq(y).

Daniel Bump May 28, 2019 coevsporadic.stanford.edu/quantum/lecture9.pdfcoevV or coevV depending on...

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