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In the classical version of the Cobordism Hypothesis, such as, e.g., in Jacob Lurie's On the Classification of Topological Field Theories, one considers the $\infty$-category of symmetric monoidal functors from the $(\infty,n)$-symmetric monoidal category of (framed) $n$-cobordism to some $(\infty,n)$-symmetric monoidal target category $\mathcal{C}$. One of the first things one proves is that this category of functors is actually an $(\infty,0)$-category. For instance, if $n=1$ one can take $\mathcal{C}=Vect_k$ and one sees that the datum of a 1-dimensional tqft is the choice of a finite dimensional (i.e. fully dualizable) vector space $V$, so that the datum of a natural transformation between two such tqfts is a morphism of finite dimensional vector spaces $f:V\to W$. These are the data attached to the (oriented) point. Next one looks at what happens for the data attached to 1-dimensional manifolds and sees that $f$ is constrained to be an isomorphism (the quickest test here is to notice that $f$ has to induce an isomorphism between the dimension of $V$ and the dimension of $W$, in the category whose objects are elements of $k$ and whose morphism are only the identities). As this simple example shows, what is crucial here is that 1-dimensional manifolds has come into play. This means that if we had restricted our attention to 0-dimensional manifolds instead, i.e., we would have considered a symmetric monoidal functor from the symmetric monoidal $(\infty,0)$-category of 0-dimensional cobordims to the symmetric monoidal $(\infty,1)$-category of vector spaces over $k$, we would have had complete freedom in the choice of $f$.

This suggests that in general $n$-dimensional cobordism "eats" $n$ non-invertible levels in the target, so that the $\infty$-category of of symmetric monoidal functors from the $(\infty,n)$-symmetric monoidal category of (framed) $n$-cobordism to some $(\infty,n+k)$-symmetric monoidal target category $\mathcal{C}$ is actually an $(\infty,k)$-category.

Apart from the interest in this result in itself, I'm interested into it for the following possible application to the characters of finite groups: a representation $(V,\rho)$ of a finite group $G$ can be seen as a $k$-$k[G]$-bimodule, and so as a morphism from $k$ to $k[G]$ in the $(\infty,2)$-category of algebras, bimodules, bimodule morphisms. If this naturally induces a morphism of 1-dimensional tqfts from the tqft $Z_k$ defined by assigning to the point the algebra $k$ and the tqft $Z_{k[G]}$ defined by assigning to the point the algebra $k[G]$, then we would also have a natural morphism $\varphi_\rho:Z_k(S^1)\to Z_{k[G]}(S^1)$, i.e., from $k$ to the class functions of $G$. This should be nothing but the trace of $\rho$.

Is this correct? Is this point of view already expanded in some reference?

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The idea of combining an $n$-manifold with an $(n{+}k)$-category to get a $k$-category is discussed in Example 6.2.4 of . I think there must be something similar in one of Lurie's papers, and perhaps elsewhere as well, but I don't know specifically where. –  Kevin Walker Mar 24 '12 at 18:30
Willerton and Caldararu discuss ideas similar to your last paragraph in . –  Kevin Walker Mar 24 '12 at 18:34
You could instead think of the k-mod-k[G] bimodule as giving you a boundary between the theory attached to k[G] and the theory attached to k (here it's important that your chosen 1-morphism itself satisfies appropriate dualizability conditions). I think in Jacob's paper this is discussed a little in Section 4.3. You can then think of the map you're looking for as being given by a cylinder with one theory at the top, the other theory at the bottom, and a boundary circle in the middle. –  Noah Snyder Mar 24 '12 at 20:06
Hi Kevin, hi Noah, thanks a lot for the references. Noah, you're suggesting to think of the k-mod-k[G] bimodule as a codimension-one defect separating two tqfts, right? That's a nice point of view, thanks for the suggestion. In Jacob's paper this is mentioned in Example 4.3.23, but anyway the literature on tqfts with defects is quite ample, I'll have a look there. –  domenico fiorenza Mar 24 '12 at 20:49

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