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If I understand correctly, Dijkgraaf-Witten TQFT in dimension 2 is the following. Fix any finite group $G$, we define a field over a closed 2-manifold to be a principle $G$ bundle (it's automatically flat since $G$ has trivial Lie algebra). Fixing an action functional, we get a number assigned to the closed 2-fold. In this question, I will not alter the action functional.

To go to codimension 1, we assign the circle the space of functions over the space of fields, i.e.

$$Z_G(S^1) = \mbox{Function}(\{G\mbox{-bundles over circle}\}),\mathbb{C})$$

So any 2-fold with boundary being a circle is just a function in $Z_G(S^1)$ in the natural way. So far, it gives a $(2,1)$ TQFT, and it is natural to me.

And then we attempt to go to codimension 2 (i.e. dimension 0). By assigning the category of representations of $G$ to the positively oriented point $pt_+$, one can prove that we get a $(2,1,0)$ TQFT. I know how to verify this. However, this construction does not seem natural to me. Thus I'd like to ask:

Question

  1. Is this extension from $(2,1)$ to $(2,1,0)$ unique? Namely, is this the only way that extends the already defined $(2,1)$ theory?

  2. If so, how would one argue? And also if this is the case, how can one go to higher codimension, if we have started from dimension 3 or higher?

  3. Can one canonically get a fully extended $n$-TQFT from this procedure?

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    $\begingroup$ 1. The extension should definitely be unique because the whole theory can be computed from the value of the point. $\endgroup$
    – Daniel Barter
    Commented Oct 14, 2019 at 23:29
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    $\begingroup$ 3. Dikkgraaf-Witten theory is fully extended in all dimensions, so the answer is yes $\endgroup$
    – Daniel Barter
    Commented Oct 14, 2019 at 23:30
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    $\begingroup$ @DanielBarter The fact that the whole theory can be computed from the value of the point does not seem go guarantees uniqueness -- perhaps we can assign another category to $pt_+$ that still gives the same assignments to dimension $1$ and $2$ spaces. $\endgroup$
    – Student
    Commented Oct 15, 2019 at 2:53
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    $\begingroup$ Yes, this is true. For example, Rep(G) and Vec(G) define the same TQFT. If two categories are 2-Morita equivalent, they have the same associated TQFTs $\endgroup$
    – Daniel Barter
    Commented Oct 15, 2019 at 18:25
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    $\begingroup$ Sorry, Vec(G) is the category of G-graded vector spaces. And yes, my claim is that if C and D are Morita equivalent fusion categories, then the corresponding Turaev-Viro TQFTs are equivalent. Unfortunately, I don't have a good reference. It is probably hidden somewhere in one of Kevin Walker's papers though. It is much easier to justify if you think about TQFT from a physicists perspective $\endgroup$
    – Daniel Barter
    Commented Oct 15, 2019 at 21:33

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