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Let $\mathfrak{g}$ be a finite-dimensional simple Lie algebra over $\mathbb{C}$, and let $U_q(\mathfrak{g})$ be some incarnation of the quantized universal enveloping algebra of $\mathfrak{g}$, where $q \in \mathbb{C}^\times$ is not a root of unity.

Lusztig showed that the braid group $B_{\mathfrak{g}}$ (an infinite cover of the Weyl group of $\mathfrak{g}$ obtained by dropping the relations that the generators are involutive) acts by algebra automorphisms on $U_q(\mathfrak{g})$. (In fact there are several different ways that $B_{\mathfrak{g}}$ can act, but they are closely related and it doesn't matter which one we pick for the purposes of this question.)

Has anybody seen anything addressing the question of whether this action is faithful? I would much appreciate an explanation or a reference that resolves the question either way. Thank you!

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  • $\begingroup$ I do not know the answer, but this paper (not its main result though) of Millson and Toledano-Laredo (www2.math.umd.edu/~millson//paper/mtlfinalversion.ps) might shed some light on this question. $\endgroup$
    – Misha
    Jan 31, 2013 at 21:24
  • $\begingroup$ @Teo B, I am taking my $q$ to be not a root of unity, so $q=\pm 1$ is not allowed. To address your question: $U_q(\mathfrak{g})$ is not quite unique. For any lattice $L$ sitting between the root lattice and the weight lattice of $\mathfrak{g}$, there is a quantized enveloping algebra $U_q^L(\mathfrak{g})$ which contains $K$'s corresponding to all elements of $L$. (The standard choice is when $L$ is just the root lattice.) By "some incarnation" I meant only that it doesn't matter which lattice we take. So I probably should have just left out the words "some incarnation of". $\endgroup$
    – MTS
    Jan 31, 2013 at 22:25
  • $\begingroup$ I should add that the reason that the choice of lattice doesn't matter for me is that the braid group action on the $K_\lambda$'s factors through the usual action of the Weyl group. So, whichever lattice $L$ I choose, the Cartan part of the quantized enveloping algebra is not enough to detect faithfulness. $\endgroup$
    – MTS
    Jan 31, 2013 at 22:46
  • $\begingroup$ @MTS: The question is natural, but the immediate response I got after mentioning it to a well-informed person was "It's not known". Has there been any study of the rank 2 cases? Aside from that, I wonder what interesting consequences there would be, one way or the other? $\endgroup$
    – Jim Humphreys
    Feb 4, 2013 at 23:45
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    $\begingroup$ @Jim Humphreys, my motivation for asking the question actually didn't require the full strength of faithfulness. I came across a suspect claim in a paper, and if the braid group action was faithful that would have given me an easy way to refute the claim. But I have since managed to disprove the claim anyway - I just needed to show that some particular element of the braid group didn't act trivially. But the general question seemed interesting and I thought I'd ask it anyway. I am unaware of any study of the rank 2 cases. Thanks for your interest! $\endgroup$
    – MTS
    Feb 5, 2013 at 0:50

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