A lift of the second Chern class 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 $H^3(X, O^*)$.)
 A: 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 $n=2$.
