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I am trying to learn about some categorical aspects of topological quantum field theories. For concreteness, I am considering Chern Simons theory with gauge group $G$ in three dimensions. As I understand, I can describe the theory as a functor $F$ on a cobordism category where there are $0$, $1$ and $2$ dimensional objects (manifolds) and morphisms between objects. Let me denote some $\mathbb C$-linear $n$-categories as $C_n$, such that the functor $F$ leads to following assignments: $$F(S^3) = C_{-1} =: Z_{S^3}\,, \qquad F(S^2) = C_0 =: \mathcal H_{S^2}\,, \qquad F(S^1) = C_1\,, \qquad F(\{\mathrm{pt}\}) = C_2\,.$$ where $S^n$ is the $n$-dimensional sphere, $\mathcal H_{S^2}$ is a vector space (with an inner product), and $Z_{S^3}$ is a complex number, this question does not concern the 2-category $F(\{\mathrm{pt}\})$. The 1-category $F(S^1)$ is a category of some representations of the affine Lie group $\widehat G$.

Question: How do I compute the number $Z_{S^3}$ given only the 1-category $F(S^1)$?

As a clarification of what I am asking let me explain how I would proceed for the simpler problem of computing $Z_{S^3}$ given the 0-category $F(S^2)$. Given $\mathcal H_{S^2}$, I can construct $S^3$ as $S^3 = \bar D^3 \sqcup_{S^2} \bar D^3$ where I am gluing two closed three dimensional discs along their $S^2$ boundaries and my understanding is that (I would like to know if this is incorrect) $F$ assigns to $\bar D^3$ an element of $F(\partial \bar D^3) = F(S^2) = \mathcal H_{S^2}$, i.e. $F(\bar D^3)$ is a vector and the result of the gluing is that $F(S^3) = \langle F(\bar D^3), F(\bar D^3) \rangle$ where $\langle -, - \rangle$ is an inner product (which can be written as a path integral over $S^3$).

I would like to know what is the analogous procedure if I start from one dimension bellow. If the answer is standard or trivial I will be happy with a reference, and if it helps anyone willing to write an answer, I am more familiar with the physics side of the subject than the math side. Thanks in advance for any help.

Edit: (some attempts) I can try to construct $F(S^2)$ from the knowledge of $F(S^1)$ by gluing two $\bar D^2$'s along their boundary $S^1$. $F(\bar D^2)$ should be an object of the 1-category $F(S^1)$, so $F(\bar D^2)$ is a $\widehat G$-module, therefore the image of two disjoint $\bar D^2$'s (with one "incoming" and one "outgoing" boundary) under $F$ should be $$F\left(\bar D^2 \sqcup \left(\bar D^2\right)^*\right) = F(\bar D^2) \otimes F(\bar D^2)^*\,.$$ Finally from the physical point of view the following seems reasonable: $$F(S^2) = F\left(\bar D^2 \sqcup_{S^1} \left(\bar D^2\right)^*\right) = \left(F(\bar D^2) \otimes F(\bar D^2)^*\right)^{\widehat G}\,. \tag{1}$$ Then I can proceed along the line described earlier for the simpler situation of having the data of $F(S^2)$.

This kind of answers my original question. For the sake of being able to do computation I have two followup questions:

Followup question 1: How exactly do I decide which element of $F(S^2)$ to assign to $F(\bar D^3)$ (without using a path integral definition of a wave functional)? Is it part of the axioms or is there a derivation?

Followup question 2: In (1), if I choose a basis $\{e_i\}$ for $F(\bar D^2)$ (and a dual basis $\{e_i^*\}$ for $F(\bar D^2)^*$), then $F(S^2)$ seems to be the one dimensional space $\mathrm{Span}\left(\sum_i e_i \otimes e_i^*\right)$, if $F(\bar D^2)$ is infinite dimensional then is there a problem with normalization for the vectors in $F(S^2)$?

Perhaps I am making some silly mistakes, sorry for the lengthy post!

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    $\begingroup$ I suspect that to compute $Z(S^3)$, you'll need additional structure on the 1-category attached to $S^1$. In your decategorified example with $\mathcal H_{S^2}$, you need more than a Hilbert space structure to know what $F(\overline D^3)$ is: it's determined by the unit or counit of the Frobenius algebra structure on $\mathcal H_{S^2}$. Similarly, the $1$-category $Z(S^1)$ attached to a 3D TQFT $Z$ should come with additional data. $\endgroup$ – Arun Debray Jan 9 '17 at 16:05
  • $\begingroup$ @ArunDebray You're right, $\mathcal H_{S^2}$ comes with two maps $\eta: \mathbb C \to \mathcal H_{S^2}$ and $\varepsilon: \mathcal H_{S^2} \to \mathbb C$, and $Z(S^3)$ is formally given by $\varepsilon \circ \eta (1)$ (using a morphism $\emptyset \to S^2 \to \emptyset$). I will update my post once I understand the $S^1$ 1-category better. Thank you for the comment. $\endgroup$ – Nafiz Ishtiaque Jan 9 '17 at 23:08

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