Some news from Stoimenow

Back when I was a grad student, the Lawrence-Krammer representations of the braid groups were a relatively new thing. These representations were apparently known to Magnus, although the first person to really study them was Ruth Lawrence, who demonstrated how to compute the Jones polynomial of a knot from a plat closure using the Lawrence representations. Daan Krammer rediscovered the Lawrence representations in his search for a faithful representation of the braid groups, then he and Bigelow proved they were faithful -- by rather beautifully independent methods. Back when I was a grad student I was hopeful that I could use the Lawrence-Krammer representation to construct some new invariants of knots or 3-manifolds. I had observed that the Lawrence-Krammer representation preserved a sesquilinear form, and I was hoping to use that form to "mine" the representation for invariants.

Some specifics. The Lawrence-Krammer representation is the representation from the Braid group on n strands to the general linear group of rank "n choose 2" matrices with entries in the group ring Z[ZxZ]=Z[q,t] -- ie: 2-variable Laurent polynomials with integer coefficients. It comes about from the braid group's action on the Lawrence-Krammer module, which is the 2nd homology group of a ZxZ cover of the configuration space of 2 points in a punctured disc (as many punctures as there are strands in the braid). As a module over the covering transformations, the homology has rank n choose 2, and we label the generators by v_{j,k} where 1 <= j < k <= n. The matrices have the form:

The sigma_i's are the Artin generators of the braid group. The form preserved by the Lawrence-Krammer representation is given by:

One of the striking things about this sesquilinear form (and the representation) is that when the variables "q,t" are specialized unit complex numbers, the representation becomes a complex representation, and the sesquilinear form becomes a non-degenerate Hermitian form. The signature of the form is not constant and it has a "shattered glass" appearance:

The above image is a sketch of the signature vs q and t in the case of the 6-stranded braid group. So when q is close to 1 and t close to i, the representation is negative-definite and one can think of the braid group as a subgroup of a unitary group. This is reminiscent of how the free group on 2 generators is a dense subgroup of SO_3 -- as in the proof of the Banach-Tarski paradox.

Anyhow, why was I studying this again? Matt Zinno had shown that the Lawrence-Krammer representation is irreducible. I was hoping that if one restricted the representation to suitable natural subgroups -- like say the Hilden or "Wicket" subgroup as some people like to say, then the representation would reduce and one might be able to derive knot and link invariants from this, by means of the Birman-Hilden correspondence between links and their plat closures. Alas, the Lawrence-Krammer representation turns out to be irreducible over the Hilden/Wicket subgroup as well.

This result has been rediscovered recently by Stoimenow. Stoimenow has gone much further to show that the image of these representations of the braid group is dense in the unitary groups. So among other things, the above sesquilinear form preserved by the Lawrence-Krammer representation is essentially the "only" one that can be preserved.

Now I'm interested in a different analogy. The representations of the mapping class groups of a surface on the homology of the surface preserve the intersection product (the dual of the cup product via Poincare duality). Moreover, the image of these representations into the automorphisms of the homology is precisely the intersection-product preserving automorphisms. So perhaps something like this theorem is true for the braid group. Ie consider the Lawrence-Krammer representation to be a homomorphism:

B_n --> GL_{Z[q,t]} (n(n-1)/2)

Let L_n be the subgroup of GL_{Z[q,t]} (n(n-1)/2) which preserves the above intersection form. So B_n --> L_n is injective. How far' is this map from being surjective?

L_n is strictly larger than B_n since L_n contains the full centre of the general linear group, yet the image of B_n only contains a slice' of it, as Bigelow observed in his dissertation. Is there much more to this story?