TITLE: trace of the atiyah class equals chern class
QUESTION [7 upvotes]: In several textbooks ("The Geometry of Moduli Spaces of Sheaves" by Huybrechts and Lehn, "Calcul differentiel et classes caracteristiques..." by Angeniol and Lejeune-Jalabert) it is mentioned that the trace of the p-th atiyah class equals the p-th chern class or the p-th component of the chern character. I could not find a reference where this statement is proven. Thanks for any help.

REPLY [6 votes]: I might be wrong but it seems to me that the $p$-th Atiyah class does not have any reason to agree with the usual $p$-th Chern class unless the manifold under consideration is Kahler. 
Namely, if $X$ is not Kahler then for a holomorphic vector bundle $E\to X$, $c_p(E)\in H^{2p}(X)$ and $at_p(E)\in H^{p}(X,\Omega^p_X)$ live in different spaces. 
The point is that $c_p(E)$ can be defined as the class of $tr(R^p)$, where $R$ is the curvature of an hermitian connection on $E$, while $at_p(E)$ can be defined as the class of $tr(R_{1,1}^p)$, where $R_{1,1}$ is the $(1,1)$-part of the curvature of a $(1,0)$-connection on $E$. 
The point is that if $X$ is Kahler then there exists an Hermitian $(1,0)$-connection with curvature being of type $(1,1)$. The relation between the Atiyah classes and the Chern classes can be made through the Hodge-to-de Rham spectral sequence. 

So, I think that the Chern classes you are talking about are not the usual (i.e. topological) ones, but the Chern classes in Hodge cohomology. Then they coincide with the Atiyah classes almost by definition (by the way, there is a very nice paper of Grothendieck on Chern classes in Hodge cohomology).