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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?


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A priori, why do you believe that $Z(pt)$ should be monoidal? In many examples, it simply isn't. –  Theo Johnson-Freyd Oct 15 '12 at 20:32
@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! –  Ryan Thorngren Oct 15 '12 at 21:13
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. –  David Ben-Zvi Oct 16 '12 at 0:58
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. –  Ryan Thorngren Oct 16 '12 at 1:21
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 –  David Ben-Zvi Oct 16 '12 at 1:26
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up vote 9 down vote accepted

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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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. –  Theo Johnson-Freyd Oct 15 '12 at 21:11
I think at least some of this is explained in Willerton's "Two 2-traces" slides: simonwillerton.staff.shef.ac.uk/ftp/TwoTracesBeamerTalk.pdf –  Qiaochu Yuan Oct 15 '12 at 21:12
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. –  Dmitri Pavlov Oct 15 '12 at 22:35
This is a nice answer, thanks again. And thanks Qiaochu and Dmitri for the references. –  Ryan Thorngren Oct 16 '12 at 18:01
Theo, is your use of "2-framing" standard (i.e. does anyone other than you use the term in this sense)? If the answer is yes, that presents a problem, since Witten and Atiyah use "2-framing" to mean something very different in the context of the Chern-Simons TQFT. This is unfortunate, since your definition is natural and useful, while the Witten-Atiyah definition is perhaps best forgotten. Still, they grabbed "2-framing" first, so perhaps we should respect that. –  Kevin Walker Oct 22 '12 at 20:35
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