Timeline for What is known about the centraliser of the Hecke algebra in the affine Hecke algebra?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 11, 2011 at 19:18 | comment | added | David Hill | I understand now. When $\lambda=0$, each conjugacy class in $W$ gives one class sum, so you don't get a full $|W|$ elements... | |
| Jun 11, 2011 at 7:10 | comment | added | Peter McNamara | Bruce-I have no idea about the commutativity or otherwise of these elements. David-you can't take lambda=0 in the argument. Hopefully my editing makes it less baffling (email me if you're still confused). | |
| Jun 11, 2011 at 7:05 | history | edited | Peter McNamara | CC BY-SA 3.0 |
rewritten answer
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| Jun 11, 2011 at 1:57 | comment | added | David Hill | 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 $z\sum_{w\in W}e^{w\lambda}$. | |
| Jun 10, 2011 at 7:39 | comment | added | Peter McNamara | 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. | |
| Jun 10, 2011 at 6:43 | comment | added | Bruce Westbury | Can you say anything about if or when two of these elements commute? | |
| Jun 10, 2011 at 6:41 | comment | added | Bruce Westbury | There must exist more elements. The issue is to find an explicit construction. This is what you have done; thanks. | |
| Jun 9, 2011 at 20:43 | comment | added | David Hill | 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. | |
| Jun 9, 2011 at 20:19 | history | answered | Peter McNamara | CC BY-SA 3.0 |