# TQFTs as rigid monoidal functors

In contrast to the usual Atiyah definition of a topological quantum field theory as a monoidal functor between the category of n-dimensional Cobordisms and the category of vector spaces, in some places (e.g. at https://arxiv.org/pdf/0905.3010.pdf or in Wikipedia) I read definitions that also demand it to respect the rigid structure on those categories, i.e. orientation reversal on manifolds and the construction of the dual space. Is this actually an additional requirement, or does it already follow from the usual axioms, at least it seems to be true for 1- and 2-dimensional TQFTs? Thank you in advance for your answers.

## 1 Answer

For two dimensional TQFTs a weaker (but very similar) property holds for free.

If $$F$$ is a 2-dimensional topological quantum field theory then it follows that every linear space in its essential image is finitely dimensional (the whole argument is a little bit long).

Observe that an object $$S$$ of the domain $$2Cob$$ of $$F$$ is a finite disjoint union of circumferences (as $$S$$ is a compact oriented 2-dimensional smooth manifold without boundary). Moreover one can prove that, if two objects $$S,T$$ of $$2Cob$$ are diffeomorphic through an orientation preserving diffeomorphism, then they are isomorphic in $$2Cob$$ and therefore (being $$F$$ a functor) it is the case that $$FS\cong FT$$ as linear spaces. In particular, as a circumference $$C$$ is always diffeomorphic to $$C'$$ ($$C$$ with the opposite orientation) through an orientation preserving diffeomorphism, it is the case that $$FC\cong FC'$$. Hence, if $$S$$ is an object of $$2Cob$$ and $$S'$$ is the manifold $$S$$ with the opposite orientation, then $$FS\cong FS'$$ (just use that $$S$$ is a finite disjoint union of circumferences).

Let's keep on considering an object $$S$$ of $$2Cob$$ and $$S'$$. Before I said that the essential image of $$F$$ is made of finitely dimensial vector spaces. Basically this follows because you can find two arrows $$f \colon S \sqcup S' \to \emptyset$$ and $$g \colon \emptyset \to S'\sqcup S$$ of $$2Cob$$ (that probably you know) such that:

$$(S \xrightarrow{1_S \sqcup g} S\sqcup S'\sqcup S \xrightarrow{f \sqcup 1_S} S)=(S\xrightarrow{1_S}S);$$

as $$F$$ is a monoidal functor, by this equality it follows that:

$$(FS \xrightarrow{1_{FS} \otimes Fg} FS\otimes FS'\otimes FS \xrightarrow{Ff \otimes 1_{FS}} FS)=(FS\xrightarrow{1_{FS}}FS).$$

As I said, this equality can be used to prove that $$FS$$ is finitely dimensional and hence conclude that linear spaces in the image of $$F$$ are finitely dimensional. However now we use it for another argument. In fact, you can check that this last equality implies that the linear map: \begin{aligned} FS' &\to (FS)^* \\ v &\mapsto ((Ff)(v\otimes-)\colon FS \to \mathbb{K}) \end{aligned} is injective and hence (being $$FS$$ and $$FS'$$ finitely dimensional and of the same dimension) an isomorphism (here I used that $$Ff$$ is a linear map $$FS' \otimes FS \to \mathbb{K}$$ and this follows from the monoidality of $$F$$).

Hence what we conclude is that there is a "very natural" isomorphism $$FS'\cong (FS)^*$$. Probably the equality between $$(FS)^*$$ and $$FS'$$ does not follow, but it seems reasonable to assume it, after having seen this fact.