Representations of the a ne BMW algebra...Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 4 / 1 Young’s lattice of partitions (is a crystal graph) Monica Vazirani (UC Davis

Representations of the affine BMW algebra

Monica Vazirani

UC Davis

ICERM March 7, 2013

Joint with Kevin Walker

Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 1 / 1

. . . via a mysterious topological construction (from TQFTs)

What is the algebra and combinatorics behind it?


Recall a partition λ with |λ| = n indexes an irreducible representation ofSn.

Example

λ =indexes an irrep of S5.

Hecke algebra

or λ indexes an irrep M(∅, λ, 5) of the finite Hecke algebra Hfin5 of type A.

Hfinn is a q-deformation of C[Sn] with generators

Ti in place of (i , i+1) = si ∈ Sn.


basis of the module M(∅, λ, n)←→ SYT(λ)

1 2 3

4 5

1 2 4

3 5

1 2 5

3 4

1 3 4

2 5

1 3 5

2 4

A standard Young tableau of shape λ (T ∈ SYT(λ)) corresponds to apath of length n = |λ| from ∅ to λ in Young’s lattice of partitions.


Young’s lattice of partitions (is a crystal graph)

∅

......

...Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 5 / 1

a path ∅ to λ

∅

......

...

∅


a path ∅ to λ ←→ T ∈ STY (λ)

∅

......

...

∅

1

1 2

1 23

1 23 4

1 23 4

5


A standard Young tableau of shape λ (T ∈ SYT(λ)) corresponds to apath of length n = |λ| from ∅ to λ in Young’s lattice of partitions.

Why does this index a basis? Induction/Restriction (among otherreasons)


Skew shapes

Example

µ = (2) ⊆ λ = (3, 2)× ×

λ/µ =

Example (SYT)

SYT(λ/µ)1

2 3

2

1 3

3

1 2

T ∈ SYT(λ/µ) corresponds to a path of length n = |λ/µ| from µ to λ inYoung’s lattice of partitions.


a path µ to λ

∅

......

...

1

1 2

1 23


What is the representation theory behind this?

SYT(λ/µ) index a basis of an irrep M(µ, λ, n) of Haffn .

Haffn is the (extended) affine Hecke algebra of type A.

Haffn is a q-deformation of C[Sn n Zn] with generators

Ti in place of (i , i+1) = si ∈ Sn,Xi in place of (0, 0, . . . , 1, . . . 0) ∈ Zn,

As vector spaces, Haffn ' Hfin

n ⊗ C[X±11 , . . . ,X±1n ].


M(µ, λ, n)

Question

Are these all the irreps of Haffn ?

NO

Why

Because boxes are in Z2 (not C2) we only get representations inRepaff

q , the full subcategory on which {Xi} take eigenvalues in

{qk | k ∈ Z}. (like integral weights)

The configuration of boxes for skew shapes yield the X -semisimple(aka calibrated) representations. (see A. Ram)


Definition

M is X -ssl if its restriction to C[X±11 , . . . ,X±1n ] is semisimple,i.e., if

M =⊕β∈Zn

M[β]

weight spaces

Let β ∈ Zn. The β-weight space of M is

M[β] = {v ∈ M | Xiv = qβi v , 1 ≤ i ≤ n}.

Fact

M is X -ssl =⇒ dimM[β] = 1 or 0.

Determines a weight basis; weights encode SYT


X -ssl M ∈ Repaffq

M[β] 3 vT , T ∈ SYT(λ/µ)

βi describes which diagonal i is on

The combinatorics of spptM = {β ∈ Zn | M[β] 6= 0} is that of SYT(λ/µ),i.e. of n-step paths µ to λ.

Action of Haffn generators on basis {vT | T ∈ SYT(λ/µ)}

XivT = qdiagonal ivT

TivT ∈ span{vT , vsiT }

If siT /∈ SYT(λ/µ), set vsiT = 0. (connection of SYT and weights; thealgebra dictates the combinatorics)


Facts

a (directed) n-step path on Young’s lattice ←→ some T ∈ SYT(λ/µ)←→ some weight β

There is a unique X -ssl M ∈ Repaffq with β ∈ spptM (M[β] 6= 0).

i.e., spptM ∩ spptN = ∅.These are all the allowable weights across all X -ssl modules.

Given X -ssl irrep M ∈ Repaffq , (µ, λ) is unique up to diagonal shift.

× ×× × × ×× × ××


finite vs affine

Hfinn -modules have µ = ∅, n = |λ|.

Hfinn is a quotient of Haff

n via

Haffn � Hfin

n

Ti 7→ Ti

X1 7→ 1 = q0

For SYT, means 1 can only be on 0th diagonal


What is this construction?

We started with two irreps: M(∅, µ,m) of Hfinm , M(∅, λ, `) of Hfin

`

and produced an X -ssl irrep M(µ, λ, n) of Haffn , n = `−m (or 0 if µ 6⊆ λ).

fin× fin→ aff


up-down

What if we now allow n-step paths to go up and down?

Then we capture the combinatorics of weights of X -ssl irreps of the affineBMW algebra Baff

n . (See Leduc-Ram, Orellana-Ram)

Bfinn is a deformation of the Brauer algebra, with generators Ti , Ei ,

1 ≤ i < n.En−1 creates a link between Bfin

n and Bfinn−2.

The irreps of Bfinn correspond to λ with n − |λ| ≡ 0 mod 2.

Baffn has generators Ti ,Ei ,Xi .

Haffn is a quotient of Baff

n (Baffn � Haff

n ) via

Ti 7→ Ti

Xi 7→ Xi

Ei 7→ 0


n-step paths µ to λ ←→ basis of the Baffn -module M(µ, λ, n)

Now µ, λ are arbitrary (we drop the requirement µ ⊆ λ)n is fixed, up to parity, independent of |µ|, |λ|.

Example

M((1), (1), 2) has basis indexed by

→ ∅ → , → → , → →

Example

M((1), (1), 3) = 0

Example

M((1, 1), (2), 2) has basis indexed by→ → , → →


What is this construction?

We started with two irreps: M(∅, µ,m) of Bfinm , M(∅, λ, `) of Bfin

`

and produced an X -ssl irrep M(µ, λ, n) of Baffn

(or 0 if |µ|+ |λ|+ n 6≡ 0 mod 2)

In fact, we produced a whole FAMILY (vary n) of irreps—this is reallyONE irrep of the affine BMW category Baff .

We have a functor

Rep(Bfin)× Rep(Bfin)→ Rep(Baff).

It comes from some bi-module. Where does that bi-module come from?


Topology, TQFTsThis construction actually comes from topology (3-manifolds, (3 + 1)-dimTQFTs (actually (3 + ε)-dim TQFT), . . . )

TQFT = topological quantum field theory

a TQFT is a machine for turning topology into algebra.

The BMW TQFT

assigns to a 3-manifold M a vector space V (M) (the BMW skeinmodule). We actually get a family of vector spaces V (M; c)depending on boundary conditions c on M.

assigns a linear category B(Y ) to a surface Y .

If Y is contained in the boundary M, then the various vector spacesV (M; c) afford a representation of B(Y ).

If a pair of surfaces Y1 ∪ Y2 is contained in the boundary of M, thenthe various V (M; c) constitute a (B(Y1),B(Y2))-bimodule.

Gluing 3-manifolds along a surface Y corresponds to taking tensorproduct over B(Y )


Bfin

We often depict T1 ∈ Bfin3 (or Hfin

3 ) by

or .

But here we’ll draw (say T2 and E2 ∈ Bfin5 ) as


Y = D2 disk

Category B(Y ) = Bfin

Pick n framed (oriented) points in Y .This collection c is an object of B(Y ).

End(c) = Bfinn (resp. Hfin

n )


Y = D2 disk

Category B(Y ) = Bfin

We also have morphisms Mor(c,d).


Y = S1 × [0, 1] annulus

Category B(Y ) = Baff

Pick n framed (oriented) points in Y .This collection c is an object of B(Y ).

End(c) = Baffn (resp. Haff

n )

Draw (say T2 and E2 and X2 ∈ Baff5 ) as


Baff corresponds to Y = S1 × [0, 1] annulus

Or we turn the picture sideways and draw (say T2 and E2 and X2 ∈ Baff5 )

as


“hat-box” (or 1-handle) construction

bi-module V (M) for M = D2 × [0, 1]

The boundary of the solid cylinder is TWO disks and an annulus.When we draw the thickened boundary, this is where our morphisms fromBfin × Bfin and Baff can act


“hat-box” (or 1-handle) construction

bi-module

The (Rep(Bfin)× Rep(Bfin),Rep(Baff))-bimodule structure:

Recap: we produce M(µ, λ, n) by imposing the finite irreps µ, λ (oridempotents) on the top/bottom of the hatbox, and then let Baff act onthe remaining cylindrical boundary.We can apply this “machine” to other 3-manifolds . . .


Other constructions, directions

other 3-manifolds M

other ways of slicing up the boundary (to yield a bi-module)

gluing and tensor product

non X -ssl representations?? (topology and category have rigidity)


Documents

Representations of the a ne BMW algebra...Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 4 / 1 Young’s lattice of partitions (is a crystal graph) Monica Vazirani (UC Davis