/sci/ - Science & Math

>>8965404
Nope, I think that's actually the part that's doable, since by Heegaar splitting [math]M = H_g \cup_h H_g[/math] it can be shown (see Kohno, Turaev) that the dimensionality of Witten's invariant can be calculated as [math]\operatorname{Dim}(Z_k(M)) = S_{00}^{-g+1}\sum_{\lambda: \{1,\dots,m\}\rightarrow P^+(k)} \operatorname{tr}_\lambda \rho(h)[/math], where the trace is taken over the space of conformal blocks with the highest weights [math]\lambda[/math], purely from the CFT perspective while that of an operator invariant [math]\tau[/math] of a decorated TQFT can be shown to be [math]\operatorname{Dim}(\Psi_t) = (\mathscr{D}^2)^{g-1}\sum_{j\in I} \operatorname{dim}(j)^{2-2g}[/math], where [math]I[/math] is the set of colorings by [math]\{1,\dots,m\}[/math] of the ribbon graph of type [math]t[/math] in the decorated 3-manifold [math]M[/math], purely from the TQFT perspective. I believe I would be able to massage my TQFT so that the the colorings in the latter expression correspond to labelings of 3-valent graphs that satisfy quantum Clebsch-Gordan conditions, which will automatically guarantee that the sums in each expression coincide. And after this dimensionality check there isn't much else that the operator invariant of my TQFT could be other than something that is at least projectively isomorphic to Witten's invariant.
Good question anon.