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In an extended 2d TQFT $Z$, a point (with orientation + or -) is assigned a category $Z(+)$ or $Z(-)$. This category should be as close to a vector space as possible: $\mathbb{C}$-linear, monoidal, etc. $Z( + \cup +)$ should be something like $Z( + ) \otimes Z( + )$, the empty set of points should get the unit category for this tensor operation, Vect$_\mathbb{C}$, and $Z(+)$ and $Z(-)$ should be dual.

If we consider a circle as broken up into two opposite U shapes, these properties tell us that $Z(S^1)$ (a monoidal $\mathbb{C}$-linear functor $Z(empty)\rightarrow Z(empty)$, ie. $V\otimes -$ for some vector space $V$) is something like the dimension of $Z(+)$ or the trace of the identity functor.

Can we make sense of this enough to compute it for some simple categories? Eg. the category of $\mathbb{C}$-representations of a finite group?

I'm sure that this wouldn't be hard to answer if I knew more about what the tensor product should be when I write $Z(+)\otimes Z(+)$. All I know about this operation is that the unit should be the category of $\mathbb{C}$-vector spaces.

How about for higher dimensional TQFTs? Does someone know a good reference?

Thanks.

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    $\begingroup$ A priori, why do you believe that $Z(pt)$ should be monoidal? In many examples, it simply isn't. $\endgroup$ Commented Oct 15, 2012 at 20:32
  • $\begingroup$ @Theo I believe that it should be the category of boundary conditions for a 2d tqft, and so fusion should give a product. There's no reason to assume this for very general 2d tqfts I suppose. Anyway it's not that pertinent to the question. And thanks for your answer! $\endgroup$ Commented Oct 15, 2012 at 21:13
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    $\begingroup$ The fusion of boundary conditions (if I understand your comment correctly) is what makes Z(pt) into a category to begin with -- it's the associative composition of morphisms between different objects (or endomorphisms of a fixed object), but as Theo says it doesn't make the category monoidal. $\endgroup$ Commented Oct 16, 2012 at 0:58
  • $\begingroup$ I agree in that morphisms should be local boundary-changing operators and fusion of those gives composition. What I meant is that for boundary conditions that arise from some codimension 1 bulk operators, these can be fused as bulk operators onto the boundary. Sometimes all boundary conditions arise this way (eg. Wilson lines for 2d BF theory), but I'm not sure when one can say this for sure. $\endgroup$ Commented Oct 16, 2012 at 1:21
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    $\begingroup$ In very brief: there's a general notion of dualizable object in a symmetric monoidal ($\infty$- or regular) category. Any such has a "dimension" which is an endomorphism of the unit, generalizing the dimension of a vector space, Euler characteristic of a complex, Hochschild homology (or chains) of a category, etc. In general in any TFT crossing with a circle amounts to calculating dimension in this sense. There are numerous interesting examples. I don't know a canonical source but some examples are in my Luminy lecture notes on my webpage ma.utexas.edu/users/benzvi $\endgroup$ Commented Oct 16, 2012 at 1:26

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I won't be able to give any references, so I hope some more experts can help me out, as there is much work on traces of functors. In general, there are two reasonable notions of "trace" of a functor, and they can be different. Throughout, I let $F$ denote an endofunctor of a nice-enough ($\mathbb C$-linear, etc.) category $\mathcal C$, and $\DeclareMathOperator\id{id}\id$ the identity functor.

Then one notion of "trace" is: $$ \operatorname{trace}(F) = \hom(\id,F)$$ where the $\hom$ is taken in the (monoidal) category $\operatorname{End}(\mathcal C)$ of endofunctors of $\mathcal C$, i.e. it is the space of natural transformations. For this definition to make sense, we need only that $\mathcal C$ is small enough for $\operatorname{End}(\mathcal C)$ to be locally small (otherwise, for generic categories, the hom spaces between functors can be proper classes), or at least for $\hom(\id,F)$ to be small. In the $\mathbb C$-linear setting, one expects that $\hom(\id,F) \in \mathrm{Vect}$, and in fact it is a $\hom(\id,\id)$-module. Note that $\hom(\id,\id)$ is always an algebra. In fact, since $\operatorname{End}(\mathcal C)$ is a monoidal category, $\hom(\id,\id)$ is always a commutative algebra. For example, when choose a $\mathbb C$-algebra $A$, and let $\mathcal C$ denote the category of left $A$-modules. Then $\hom(\id,\id)$ is the center of $A$.

There is an important generalization: work not with categories but $(\infty,1)$-categories. Then one can set $\mathcal C$ to be an appropriate "derived" category of chain complexes of $A$-modules, and $\hom(\id,\id)$ is then the Hochschild cochain complex of $A$.

There is another important notion of "trace", which is given by an end (or is it a coend?) of the functor $\hom(-,F-)$. This notion is slightly closer to the idea of "adding up the diagonal entries of a matrix for $F$". In the $A$-module case, this version gives $\operatorname{trace}(\id) = A / [A,A]$, where $[A,A]$ is the subvector space of commutators (and not an ideal or anything), so that the quotient is simply a vector space (with a distinugished element, namely the image of $1\in A$). In the derived setting, one gets the Hochschild chains of $A$.

The two constructions must give canonically-the-same answer if $\mathcal C$ is the image of an oriented but otherwise unframed 2-TQFT. But if you work with framed TQFTs, they can give different answers. Recall that a 2-framing of a 1-manifold $S$ is a framing of $S\times \mathbb R$, and that a framing of an $n$-manifold is a collection of $n$ vector fields which are at every point linearly independent. The first "trace" corresponds to the circle with "outward" framing, i.e. it has a "2-framing" inherited from embedding the circle as a simple closed curve in $\mathbb R^2$. The second trace corresponds to the "product" framing, i.e. the framed circle where one of the two vector fields is parallel to the circle and the other is orthogonal.

When thought of in this geometric picture, the "Deligne conjecture" that Hochschild chains has a homotopy-$S^1$-action becomes natural, and Hochschild cochains have their $E_2$-algebra structure coming from embedding two disks into a larger disk.

Actually, if you have a complete framed 2-dimensional TQFT which assigns $\mathcal C$ to a point, then the two notions of trace must agree for $\id$, at least in dimension. I mean, if you look at the torus (with its unique framing), the value of the torus must be the dimension of each $\operatorname{trace}(\id)$, by cutting the framed torus into an annulus in two different ways. A framed 2-TQFT does not pick out a chosen isomorphism between the two different traces, and I believe that any choice of such an isomorphism is pretty much enough to extend the framed TQFT to an unframed one. Algebras with such a choice are called "Calabi–Yau", at least by some people, because the data of such an isomorphism is roughly the same (when $A$ is commutative) as a trivialization of the canonical line of $\operatorname{Spec}(A)$.

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    $\begingroup$ Incidentally, in your example, namely representations of a finite group, both traces give a vector space that deserves to be called the "class functions". But the fact that both descriptions calculate the same thing should be surprising. Ben-Zvi and Nadler have emphasized in some of their work that most of the classical "Frobenius"-style representation theory of finite groups is well-understood in the TQFT packaging. As an example, it is a good exercise to see what goes wrong (you don't get a TQFT) when you replace $\mathbb C$ by a field of characteristic dividing the order of the group. $\endgroup$ Commented Oct 15, 2012 at 21:11
  • $\begingroup$ I think at least some of this is explained in Willerton's "Two 2-traces" slides: simonwillerton.staff.shef.ac.uk/ftp/TwoTracesBeamerTalk.pdf $\endgroup$ Commented Oct 15, 2012 at 21:12
  • $\begingroup$ The Freed-Hopkins-Lurie-Teleman paper contains some computations of traces for tensor categories arising out of the three-dimensional Chern-Simons theory gauged by a compact abelian Lie group, see Proposition 6.1. $\endgroup$ Commented Oct 15, 2012 at 22:35
  • $\begingroup$ This is a nice answer, thanks again. And thanks Qiaochu and Dmitri for the references. $\endgroup$ Commented Oct 16, 2012 at 18:01
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    $\begingroup$ I guess that makes it pretty standard. But since it's usually "n-framing" and only occasionally specialized to "2-framing", the terminological overlap is not so bad. As I said above, the Witten-Atiyah notion of 2-framing is not very important -- it's just a klunky version of a $p_1$-structure. So on second thought I retract what I wrote above about it being a bad idea to use "2-framing" in this other sense. $\endgroup$ Commented Oct 23, 2012 at 12:01

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