representations of the a ne bmw algebra...monica vazirani (uc davis ) reps of bmwaff march 7, 2013 4...
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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?
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 2 / 1
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.
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 3 / 1
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.
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 ) Reps of BMWaff March 7, 2013 5 / 1
a path ∅ to λ
∅
......
...
∅
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 6 / 1
a path ∅ to λ ←→ T ∈ STY (λ)
∅
......
...
∅
1
1 2
1 23
1 23 4
1 23 4
5
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 7 / 1
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)
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 8 / 1
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.
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 9 / 1
a path µ to λ
∅
......
...
1
1 2
1 23
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 10 / 1
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 ].
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 11 / 1
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)
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 12 / 1
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
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 13 / 1
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)
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 14 / 1
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.
× ×× × × ×× × ××
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 15 / 1
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
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 16 / 1
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
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 17 / 1
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
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 18 / 1
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→ → , → →
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 19 / 1
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?
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 20 / 1
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 )
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 21 / 1
Bfin
We often depict T1 ∈ Bfin3 (or Hfin
3 ) by
or .
But here we’ll draw (say T2 and E2 ∈ Bfin5 ) as
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 22 / 1
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 )
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 23 / 1
Y = D2 disk
Category B(Y ) = Bfin
We also have morphisms Mor(c,d).
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 24 / 1
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
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 25 / 1
Baff corresponds to Y = S1 × [0, 1] annulus
Or we turn the picture sideways and draw (say T2 and E2 and X2 ∈ Baff5 )
as
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 26 / 1
“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
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 27 / 1
“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 . . .
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 28 / 1
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)
Monica Vazirani (UC Davis ) Reps of BMWaff March 7, 2013 29 / 1