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The Kauffman bracket skein module $K_t(F\times I)$ (where $t$ is an indeterminant and $F$ is a closed surface) is an associative algebra (the operation being "stacking" links in the $I$ direction). It is a quantum deformation of the coordinate ring of the character variety $\operatorname{Hom}(\pi_1(F),\operatorname{SL}(2))//\operatorname{SL}(2)$. This skein algebra naturally acts on the vector space $\mathcal H_F$, the vector space associated to $F$ by the WRT TQFT.

Is there a direct construction of this vector space (e.g. in terms of generators and relations) which makes the action of $K_t(F\times I)$ manifest?

I know only two constructions of the vector space $\mathcal H_F$, and both are unsatisfactory for the purposes of this question. First, we could pick a complex structure on $F$, and let $\mathcal H_F$ be the global holomorphic sections of a particular line bundle $\mathcal L$ over the $\operatorname{SU}(2)$ character variety of $\pi_1(F)$. In this construction, I don't know of any explicit way to realize the action of $K_t(F\times I)$. Second, we could say the vector space is generated by all $3$-manfolds with boundary $F$, and quotient by the subspace which has zero WFT invariant when paired with all three-manifolds with boundary $-F$. This is unsatisfactory because the relations aren't "local", and they require understanding the WRT invariants for all three-manifolds.

I am looking for a definition of the vector space in the spirit of the definition of $K_t(F\times I)$, i.e. some generators modulo some local "picture" relations.

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up vote 4 down vote accepted

The answer is $K_t(H)$, where $H$ is a handlebody with boundary $F$. If $t$ is a root of 1 and we are taking the usual semisimple quotient, then $K_t(F\times I)$ is isomorphic to a matrix algebra and $K_t(H)$ isomorphic to the standard representation. Also, in this case we can let $H$ be any 3-manifold with boundary $F$ -- it doesn't even need to be connected.

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Cool! An "exponentiated" half-lives half-dies theorem. –  John Pardon Jan 19 '12 at 17:11
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