User joe loubert - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:12:44Z http://mathoverflow.net/feeds/user/21037 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109071/algorithm-for-reducing-words-in-a-coxeter-group Algorithm for reducing words in a Coxeter group Joe Loubert 2012-10-07T15:58:11Z 2012-10-08T01:29:54Z <p>Let $W$ be a Coxeter group with set of simple reflections $S$. Suppose that I have chosen a preferred reduced decomposition for every element of $W$. Given an arbitrary word in the alphabet $S$, is there an algorithm for reducing this word to my chosen decomposition using the Coxeter relations? That is to say, an algorithm which at each step replaces a subword of the form $s_i s_k s_i \dots$ by $s_k s_i s_k \dots$, or which removes an occurrence of $s_k^2$.</p> <p>I would be happy with an answer in the following situation: $W=\Sigma_n$ is the symmetric group on $n$ letters and the preferred decomposition is given by taking the lexicographically smallest (or largest) reduced decomposition.</p> http://mathoverflow.net/questions/88762/when-is-a-hecke-algebra-not-a-bialgebra When is a Hecke algebra not a bialgebra? Joe Loubert 2012-02-17T20:29:47Z 2012-02-17T20:42:42Z <p>Let $\mathcal{H}_q(d)$ denote the Iwahori-Hecke algebra of type $A$ over a field of characteristic zero. When $q = 1$, this is just the group algebra of the symmetric group on $d$ letters. In this case, it is a bialgebra (Hopf, even). In the more general situation, however, there is no obvious choice of coproduct. However, when $q$ is not a root of unity, the Hecke algebra degenerates into the group algebra of the symmetric group, and so it is in fact a bialgebra in an obscure way.</p> <p>My question is, is it known when $\mathcal{H}_q(d)$ cannot be given the structure of a bialgebra?</p> http://mathoverflow.net/questions/109071/algorithm-for-reducing-words-in-a-coxeter-group/109079#109079 Comment by Joe Loubert Joe Loubert 2012-10-07T23:17:42Z 2012-10-07T23:17:42Z What do you mean by 'greedy'? For example, how does this algorithm deal with $s_1 s_3 s_2 s_1 s_3$? When you reference Garsia, I assume that you are referring to the proof of Theorem 1.1.2. This describes a different algorithm which is probably sufficient for my purposes. http://mathoverflow.net/questions/88762/when-is-a-hecke-algebra-not-a-bialgebra Comment by Joe Loubert Joe Loubert 2012-02-20T21:59:08Z 2012-02-20T21:59:08Z When $q$ is not a root of unity, the Hecke algebra $\mathcal{H}_q(d)$ is isomorphic to the group the algebra of $S_d$. This can be seen using Brundan and Kleshchev's isomorphism theorem in <a href="http://arxiv.org/abs/0808.2032" rel="nofollow">arxiv.org/abs/0808.2032</a> . Unfortunately, the Coxeter presentation does not behave nicely under this isomorphism, so the bialgebra structure is hidden. Additionally, if we use the KLR presentation of the paper above, it is not clear that any Hecke algebra has a bialgebra structure.