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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.

My question is, is it known when $\mathcal{H}_q(d)$ cannot be given the structure of a bialgebra?

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The braid group $\mathcal{B}_d$ generators $\sigma_i$ act on any $\mathcal{H}_q(d)$ module by an operator with (at most) two eigenvectors. So if $V$ and $W$ are two $\mathcal{H}_q(d)$ irreps and $\mathcal{H}_q(d)$ had a bialgebra structure then the braid group would act on $V\otimes W$. In the case of $q=1$ this $\B_d$ action is diagonal, but in the $q$ generic (or root of unity) case the diagonal action would give us the wrong eigenvalues for $\sigma_i$, assuming a bialgebra structure. This is not a proof that such a bialgebra structure does not exist, but it does seem problematic.. – Eric Rowell Feb 19 '12 at 21:20
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 . 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. – Joe Loubert Feb 20 '12 at 21:59

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