Trace of a functor (or dimension of a category) in extended 2d TQFTs - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:54:45Z http://mathoverflow.net/feeds/question/109749 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109749/trace-of-a-functor-or-dimension-of-a-category-in-extended-2d-tqfts Trace of a functor (or dimension of a category) in extended 2d TQFTs Ryan Thorngren 2012-10-15T19:14:19Z 2012-10-22T06:14:57Z <p>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.</p> <p>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.</p> <p>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?</p> <p>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.</p> <p>How about for higher dimensional TQFTs? Does someone know a good reference?</p> <p>Thanks.</p> http://mathoverflow.net/questions/109749/trace-of-a-functor-or-dimension-of-a-category-in-extended-2d-tqfts/109759#109759 Answer by Theo Johnson-Freyd for Trace of a functor (or dimension of a category) in extended 2d TQFTs Theo Johnson-Freyd 2012-10-15T20:59:35Z 2012-10-22T06:14:57Z <p>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.</p> <p>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 $A$.</p> <p>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 <em>cochain</em> complex of $A$.</p> <p>There is another important notion of "trace", which is given by an <a href="http://ncatlab.org/nlab/show/end" rel="nofollow">end</a> (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 <em>chains</em> of $A$.</p> <p>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 <em>2-framing</em> 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 <em>every</em> 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.</p> <p>When thought of in this geometric picture, the "Deligne conjecture" that Hochschild <em>chains</em> has a homotopy-$S^1$-action becomes natural, and Hochschild <em>cochains</em> have their $E_2$-algebra structure coming from embedding two disks into a larger disk.</p> <p>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 <em>chosen</em> 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 "Calabiâ€“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 $\operatorname{Spec}(A)$.</p>