7
$\begingroup$

Let $X$ be a complex manifold (not necessarily Kahler or even compact). For the first Chern class $c_1(E)\in H^2(X,Z)$ of a holomorphic vector bundle $E\to X$ there is an obvious lift to $H^1(X, O^*)$, namely the determinant.

Question: Is there a natural (in some sense) lift of the second Chern class to $H^3(X, O^*)$?

(I have heard that there is a lift to the Deligne cohomology group, which must be related. However, as I know nothing about Deligne cohomology, I have no clue if it is possible to lift from there to $H^3(X, O^*)$.)

$\endgroup$
1
  • 1
    $\begingroup$ $H^3(X, \mathcal{O}_X^{\times})$ is some kind of categorified Brauer group; it should classify objects living in a $3$-category of some sort over $X$, which maybe look something like sheaves of vertex algebras over $X$... $\endgroup$ Commented Nov 29, 2015 at 9:02

1 Answer 1

5
$\begingroup$

There are various instances of the first Chern class:

  • In Betti cohomology $c_1\in H^2(X, \mathbf{Z})$,
  • In Dolbeault cohomology $c_1\in H^1(X, \Omega^1)$,
  • In de Rham cohomology $c_1\in H^2_{dR}(X)$.

These can be combined to the class in Deligne cohomology $c^D_1\in H^2_D(X, \mathbf{Z}(1))\cong H^1(X, \mathcal{O}^\times)$ which is the determinant that you mention. From that class you can recover the previous classes: the morphism $\mathrm{dlog}\colon\mathcal{O}^\times\rightarrow (\Omega^1\rightarrow\Omega^2\rightarrow...)$ sends $c_1^D$ to a class in $F^1 H^2_{dR}(X)$ which is a subgroup of $H^2_{dR}(X)$ and it also projects to $H^1(X, \Omega^1)$. The connecting homomorphism for the exponential sequence gives $H^1(X, \mathcal{O}^\times)\rightarrow H^2(X, \mathbf{Z})$.

This story continues for higher Chern classes. You have a de Rham Chern class $c_n\in F^nH^{2n}_{dR}(X)$ and it has a refinement to the Deligne cohomology class $c_n^D\in H^{2n}_D(X, \mathbf{Z}(n))$. I believe Chern classes in Deligne cohomology are defined by appealing to the splitting principle by writing everything in terms of first Chern classes that you already know and then using multiplication on the Deligne cohomology.

For $n=2$ this gives the second Chern class $c_2^D\in H^3(X, \mathcal{O}^\times\rightarrow \Omega^1)$ which projects to $H^3(X, \mathcal{O}^\times)$.

If $X$ is algebraic, one instead has Chern classes in Chow groups $c_n\in \mathrm{CH}^n(X)$ which gives the same answer for $n=1$, but totally different for $n=2$.

$\endgroup$
9
  • $\begingroup$ Naturally, I mean topological Chern classes. You say that the Chern class $\endgroup$ Commented Nov 29, 2015 at 10:39
  • $\begingroup$ In Deligne cohomology projects to $H^3(X,O^∗)$, but how? Do you mean that there is a map $H^{2p}_D(X,Z(p))\to H^{2p-1}(X,O^*)$? $\endgroup$ Commented Nov 29, 2015 at 10:42
  • $\begingroup$ The Deligne cohomology group $H^{2p}_D(X, \mathbf{Z}(p))$ is isomorphic to $H^{2p-1}(X, \mathcal{O}^\times\rightarrow \Omega^1\rightarrow ...\rightarrow \Omega^{p-1})$ which projects to $H^{2p-1}(X, \mathcal{O}^\times)$. $\endgroup$ Commented Nov 29, 2015 at 10:50
  • $\begingroup$ It is good, but I want a reference for this. $\endgroup$ Commented Nov 29, 2015 at 10:51
  • $\begingroup$ Deligne cohomology is defined in Section 1 of wwwmath.uni-muenster.de/u/pschnei/publ/beilinson-volume/…, Section 8 talks about Chern classes. Using the quasi-isomorphism $(\mathbf{Z}\rightarrow\mathcal{O})\cong \mathcal{O}^\times[-1]$ you get the hypercohomology group I mentioned. $\endgroup$ Commented Nov 29, 2015 at 10:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .