MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What are generators for the kernel of the (k,r)-quotient of the Hecke algebras of type A? Are just the two projections onto the reps. corresponding to Young digarams with 1-row of length r-k+1 and that with 1-column of length k+1 enough?

share|cite|improve this question
Can you define the $(k,r)$-quotient? Also, are you really asking for generators? What's wrong with considering the image of the generators of the Hecke algebra under the canonical map? – Sammy Black May 28 '10 at 6:30
I take it you mean quotient by the ideal generated by these two projections. Then the irreducible representations of the quotient are labelled by partitions which fit in the rectangle. If this is what you want, then the answer is Yes! – Bruce Westbury May 28 '10 at 11:24
The (k,r)-quotient is the quotient by the annihilator of the trace of the Hecke algebra specialized to $q$ an $r$th root of unity and the value of $tr(e_1)=(1-q^{1-k})/((1+q)(1-q^k)$. (As in Wenzl's 1988 Inventiones paper). Bruce is right, I want to know if the annihilator of the trace is generated by the two projections onto the (1-dimensional) representations labeled by these Young diagrams. Certainly it is known for the case $k=2$ (Temperley-Lieb) but I have never seen the general result in the literature. Do you know a reference Bruce? – Eric Rowell May 28 '10 at 13:56
I edited your question for clarity based on your comment. – Noah Snyder May 28 '10 at 14:36
I have not seen this before. I take it that you have shown that the trace annihilates the two projections. Then you have a surjection from the algebra defined by these relations to the $(k,r)$-quotient. My concern is that I think the first algebra has non-zero radical and the second one doesn't. If so, then I expect the kernel is the radical. – Bruce Westbury May 29 '10 at 7:05

Your Answer


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

Browse other questions tagged or ask your own question.