Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question is a sequel to 66602

The Hecke algebra is the quotient of the group algebra of the braid group of type $A_n$ by quadratic relations and the affine Hecke algebra is the quotient of the group algebra of the braid group of type $B_n$ by quadratic relations. This is well known and the relations can be found in the references in the answer and comment to the above question.

This gives an inclusion of the Hecke algebra in the affine Hecke algebra and I am interested in the centraliser algebra. The obvious elements in the centraliser are the central elements in the affine Hecke algebra. This centre is (I don't have the reference) the algebra of symmetric polynomials in the $x_1,\ldots ,x_n$. The other obvious elements are the central elements in the Hecke algebra.

These are all the elements I can construct in the centraliser. Is there any construction which gives more elements in the centraliser? Of course I would really like to have a set of generators.

The motivation is to understand the $6j$ symbols of $GL(n)$ which was also the subject of my question 15800.

share|improve this question
    
By the way, your link was b0rked, so I fixed it. –  Ben Webster Jun 4 '11 at 3:22
add comment

2 Answers 2

This answer is being completely rewritten because it wasn't very comprehensible. We will show that there are more elements in the centraliser than the obvious ones mentioned in the question. At best, we will have an inelegant and impractical way to construct these extra elements.

For most of this answer we'll just talk about the q=1 case, ie group algebras. We need some notation. Let W be the finite Weyl group and W' be the affine Weyl group. This is the semidirect product of W and the root lattice Q. k=ground field (char 0).

The centraliser is the set of elements in the group algebra k[W'] invariant under the conjugation action of W. The general construction is: Given a vector space V with an action of W, then for any v ∈ V, the average $\sum_{w\in W} w.v$ is in the invariant space VW.

So now in the affine Hecke algebra, fix an isomorphism between the finite Hecke algebra and the group algebra k[W] (In type E7 and E8 you'll need a square root of q in k). Then you can find elements in the centraliser using the above construction.

The rest of this answer will show that at q=1, there are more elements in the centraliser than described by Bruce. We need more notation. Z(?)=centre(?). For λ in Q, we write eλ for the corresponding element in the group algebra k[W'] and wλ for the action of w on λ.

We aim to show there are elements in the centraliser not in Z(k[W]) ⊗ Z(k[W']).

Note that Z(k[W']) is spanned by $\sum_{w\in W}e^{w \lambda}$.

Pick λ in a free W-orbit and u ∈ W. We apply our averaging construction to v=ueλ.

So $y=\sum_{w\in W}wxw^{-1}e^{w\lambda}$ lies in the centraliser. If y is also in the tensor product, then we must have $y=z\sum_{w\in W}e^{w \lambda}$ for some $z\in Z(k[W])$.

Let u vary over W and fix λ. We get |W| linearly independent choices of y, yet there is a smaller dimensional space of choices of z (Well if |W|=2, this is false, but a different argument is concoctable). Thus there exist elements in the centraliser that are not constructed in the question.

Initially there was a question about the braid group Bn here whose existence makes David Hill's first comment understandable.

share|improve this answer
    
I think its the same. The braid group of type $B_n$ is generated by elements $b_0,\ldots,b_{n-1}$ subject to the relation $b_ib_j=b_jb_i$ for $|i-j|>1$, $b_ib_{i+1}b_i=b_{i+1}b_ib_{i+1}$ for $i>0$, and $b_0b_1b_0b_1=b_1b_0b_1b_0$. The map $b_0\mapsto X_1$, $b_i\mapsto T_i$ is a surjective homomorphism onto the affine Hecke algebra. –  David Hill Jun 9 '11 at 20:43
    
There must exist more elements. The issue is to find an explicit construction. This is what you have done; thanks. –  Bruce Westbury Jun 10 '11 at 6:41
    
Can you say anything about if or when two of these elements commute? –  Bruce Westbury Jun 10 '11 at 6:43
    
The only construction of more elements I can think of is unpleasant. Fix an isomorphism between the finite Hecke alg and the group alg of the Weyl group (We may need sqrt{q} in our field for this, but maybe not in type A). Then play the sum over the finite Weyl group averaging trick as in my answer. –  Peter McNamara Jun 10 '11 at 7:39
    
Peter- I am completely baffled by your argument. If I take $\lambda=0$ in your definition of $x$, then it seem you get $|W|$ linearly independent elements in $Z(\mathbb{C}[W])$. In fact, unless I am misunderstanding the algebra structure (in which case, I would appreciate some clarification), then your construction gives precisely elements of the form $z\sum_{w\in W}e^{w\lambda}$. –  David Hill Jun 11 '11 at 1:57
show 2 more comments

This is not a complete answer, but does give some information.

In addition to having an embedding of the Iwahori-Hecke algebra into the affine Hecke algebra, $\iota:H_n^{\mathrm{fin}}\hookrightarrow H_n^{\mathrm{aff}}$, there is also a surjective homomorphism $\pi:H_n^{\mathrm{aff}}\twoheadrightarrow H_n^{\mathrm{fin}}$ which is the identity on $H_n^{\mathrm{fin}}\subset H_n^{\mathrm{aff}}$, sends $X_1\mapsto 1=L_1$, and $X_i\mapsto L_i$, for $2\leq i\leq n$, where $$L_i=q^{1-i}T_{i-1}T_{i-2}\cdots T_1T_1\cdots T_{i-2}T_{i-1}$$ is the $i$th Jucys-Murphy element.

Now, the center $Z^{\mathrm{aff}}$ of $H_n^{\mathrm{aff}}$ consists of symmetric Laurent polynomials in the $X_i$, while the center $Z^{\mathrm{fin}}$ of $H_n^{\mathrm{fin}}$ consists of symmetric Laurent polynomials in the Jucys-Murphy elements.

Evidently, the centralizer of $H_n^{\mathrm{fin}}$ in $H_n^{\mathrm{aff}}$ is contained in $\pi^{-1}(Z^{\mathrm{fin}})$.

share|improve this answer
    
That is a nice observation. –  Bruce Westbury Jun 9 '11 at 19:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.