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How to find a geometric construction of a TQFT

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Assume that the surface $\sum$ is equipped with the structure of a smooth algebraic curve over $\mathbb{C}$. We denote by $H^0(M_\sum,\mathcal{L}^{\otimes k})$ the space of sections of $\mathcal{L}^{\otimes k}$ on $M_\sum$, where $M_\sum$ is the moduli space of semi-stable rank N bundles with trivial determinant over $\sum$ , and $\mathcal{L}$ is the determinant line bundle on $M_\sum$. It is known that $H^0(M_\sum,\mathcal{L}^{\otimes k})$ is isomorphic to $V(\sum)$ of a $TQFT (V,Z)$ derived from the quantum group $U_q(sl_N)$ at a $(k + N)$-th root of unity. In this sense, $H^0(M_\sum,\mathcal{L}^{\otimes k})$ gives a geometric construction of such a $V(\sum)$.
How can we find a geometric way to associate a vector in $H^0(M_\sum,\mathcal{L}^{\otimes k})$ to a $3$ manifold $M$ with $\delta M = \sum$.
In physics one can obtain such vector by applying infinite dimensional analogue of geometric invariant theory and sympletic quotients of Chern-Simons integral. We would like to make mathematical sense in that argument.

This post imported from StackExchange MathOverflow at 2017-02-02 22:59 (UTC), posted by SE-user Soutrik
asked Jan 29 in Theoretical Physics by Soutrik (25 points) [ no revision ]
retagged Feb 2
You might find the answr in The Geometry and Physics of Knots by M. Atiyah.

This post imported from StackExchange MathOverflow at 2017-02-02 22:59 (UTC), posted by SE-user abx
@abx Yes, I know the reference. But there it is written in "physics" sense. I am looking for some pure mathematical sense (like higher category theory, Hochschild homology etc) to construct TQFT using the space of sections of tensoring k times the determinant line bundle on the moduli space of semistable rank N bundles with trivial determinant over the surface.

This post imported from StackExchange MathOverflow at 2017-02-02 22:59 (UTC), posted by SE-user Soutrik
see mathoverflow.net/questions/86792/… and references there.

This post imported from StackExchange MathOverflow at 2017-02-02 22:59 (UTC), posted by SE-user user25309

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