Some physicists have told me that if you think about an extended ndimensional 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?
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 2category, 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 Ktheory of Z(X) (in great generality there's a map from Ktheory to Hochschild homology), if you want to compare this to another version of decategorification, which is taking Kgroups.

1$\begingroup$ Do you have a good reference for when this is the same as the Kgroup? $\endgroup$ – Ben Webster♦ Nov 7 '09 at 16:42

1$\begingroup$ If your category is the derived category of a smooth projective variety, then Hochschild homology coincides with de Rham cohomology, and the map from Ktheory is a rational isomorphism. (Note Ktheory is an integral invariant, while HH is linear over your ground ring, eg C, so you need to tensor Ktheory 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 Hodgede Rham for this to be the same as Hochschild (cf arXiv:math/0606241 and 0611623). more generally, I have no idea! $\endgroup$ – David BenZvi 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.)