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Some physicists have told me that if you think about an extended n-dimensional TQFT $F$, then the decategorification is given by $F'(X)=F(X\times S^1)$, which I believe they call "compactification on a circle." Is there any way to make this statement precise?

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up vote 7 down vote accepted

In a general extended TQFT Z, the assignment $Z(X x S^1)$ is the "dimension" of Z(X), in the following sense. Write the circle as an incoming arc followed by an outgoing arc. The incoming arc is a morphism (coevaluation) from the unit (Z(empty set)) to Z(X) tensor its dual $Z(X^{op})=Z(X)^*$, followed by a morphism (evaluation) back to the unit. In particular we learn Z(X) HAS a dual (is dualizable), and these are the two canonical maps that come in the definition of being a dual. The composition is an endomorphism of the unit Z(empty), which is very generally called the dimension of Z(X), or the Hochschild homology of Z(X).

If Z(X) is a vector space, End Z(empty) = numbers and this is the usual dimension. If Z(X) is a category (or an algebra, or a 2-category, or....), this is what is usually known as its Hochschild homology. In particular Hochschild homology is where characters (or traces) of objects in Z(X) live, if it's a category. In simple situations this will be the same as the K-theory of Z(X) (in great generality there's a map from K-theory to Hochschild homology), if you want to compare this to another version of decategorification, which is taking K-groups.

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Do you have a good reference for when this is the same as the K-group? – Ben Webster Nov 7 '09 at 16:42
If your category is the derived category of a smooth projective variety, then Hochschild homology coincides with de Rham cohomology, and the map from K-theory is a rational isomorphism. (Note K-theory is an integral invariant, while HH is linear over your ground ring, eg C, so you need to tensor K-theory with the ground ring, and in char 0 you get an isomorphism). In general the map to HH always factors through cyclic homology, so you need a version of degeneration of Hodge-de Rham for this to be the same as Hochschild (cf arXiv:math/0606241 and 0611623). more generally, I have no idea! – David Ben-Zvi Nov 8 '09 at 16:44

If you think of a TQFT as a functor from cobordisms to vector spaces, then $F(X \times S^1)$ will give you the dimension of the state space of $X$ (or the superdimension or whatever), because it is the trace of the identity. (In the cri du jour of $\infty$-categories, the cobordism functor is not everything, but is something.)

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