6
$\begingroup$

Jones-Wenzl projectors are an explicit combinatorial description of the central projectors of the Temperley-Lieb algebra. They also describe very explicitly the failure of certain representations to exist for $q$ a root of unity and generate the negligible ideal (the ideal of elements $a$ such that $\operatorname{tr}(ab) = 0$ for every $b$.)

Is there a similar combinatorial description of such projectors for the Brauer algebra? Here the Brauer algebra is drawn with the same diagrams as Temperley-Lieb, but the strands are allowed to cross. It is a graphical description of the decomposition of tensor powers ofan irrep of a classical orthogonal or symplectic algebra.

I think the answer is "no" at least for all of the properties I mentioned above, but I'd love to be surprised. I also don't think I need all the properties above: just knowing some elements of the negligible ideal (for whichever value of the loop parameter and whatever number of strands) would be useful.

I think this question is related to mine, except that I looked up the book mentioned and didn't find it helpful. The paper "On central idempotents in the Brauer algebra" shows how to compute some of the idempotents, but non-recursively and without a formula for the trace, so it may be that the question I'm asking doesn't even have a partial answer.

$\endgroup$
2
$\begingroup$

Via email, Noah Snyder pointed me to the papers Tuba and Wenzl - On braided tensor categories of type BCD and Lehrer and Zhang - The second fundamental theorem of invariant theory for the orthogonal group. I wound up going in a different direction than this question was suggesting, so I did not read the papers in detail and try to translate their results to more diagrammatic language, but it seems like that should be possible.

In particular, Section 7.9 of the first paper (by Tuba and Wenzl) has the type of recursive formula I was looking for.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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