TITLE: Can Chern class/character be categorified?
QUESTION [19 upvotes]: The Chern character sends the class of a locally free sheaf to the cohomology ring of the underlying variety X. And it is a ring homomorphism from K to H^*. I saw people write its source as the bounded derived category too, which make sense if the underlying variety is smooth (sending a bounded complex to the alternating "sum" of the Chern characters of its cohomology sheaf).
My question is, if I want to think $D^b(X)$ as a certain categorification of $K_0(X)$, is it possible to categorify the chern character map? What will be a good candidate of the target category? (Or is there a heuristic showing this is not likely to be true?)

REPLY [10 votes]: The paper http://arxiv.org/abs/0804.1274 of Toën-Vezzosi is about categorifying the Chern character. Let me try to summarize their strategy. 
First of all they introduce a triangulated $2$-category $Dg(X)$ of derived categorical sheaves on a (derived) scheme $X$. It is based on a the idea that a categorification of the theory of modules on a commutative ring $k$ is given by $k$-linear categories: they argue that dg-categories can be used in order to categorify homological algebra in a similar but better way (better in the sens that the non-dg setting seems to be too rigid to allow push-forwards). 
The second step is to use, for a given (derived) scheme $X$, the pull-back along the projection $LX\to X$. For a categorical sheaf $F$ on $X$ on consider its pull-back $p^*F$, which naturally come equipped with a self-equivalence $u$. The rough idea to see this is to consider the pull-back (a-k-a >transgression) along the evaluation map $S^1\times LX\to X$, and then to observe that categorical sheaves on $S^1\times LX$ are categorical sheaves on $LX$ together with a $\mathbb{Z}$-action. 
Finally, they conjecture the existence of an $S^1$-equivariant trace $Tr^{S^1}(u)\in D^{S^1}(LX)$. Its $K_0$ would be a candidate for the (categorified) Chern character of $F$. 
Why does this categorify the Chern character ?
If we do the same construction starting with a sheaf of $X$, then we get in the end an element in $\pi_0(\mathcal O_{LX}^{S^1})=HC_0^{-}(X)$ (while the non-$S^1$-invariant trace takes values in $\pi_0(\mathcal O_{LX})=HH_0(X)$). 
One can show that this constructs the ususal Chern character. The main difficulty is the (conjectural) existence of the $S^1$-invariant trace. 
Follow-up
A complete treatment of this approach (together with a proof of the conjecture) has been done by the above mentioned authors in a long paper in french.