# Trace map $Ext^i(E,E)\rightarrow H^i(X,O_X)$

Let $X$ be a scheme (or complex manifold if you like) and $E$ be a sheaf on $X$. I would like to know the definition of so-called trace map $$Ext^i(E,E)\rightarrow H^i(X,O_X)$$ for $1\le i\le \dim X$.

Edit I am also interested in noncommutative case; is there any traca map which looks like $$Ext_R^i(E,E)\rightarrow ?,$$ where $R$ is a noncommutative algebra and $E$ is a noetherian right $R$-module.

-
add comment

## 1 Answer

Proceeding formally, we have $Ext^i_X(E,E)\cong Ext^i_{X\times X}(\mathcal O_\Delta,\hom(E,E))$, where $\Delta\subset X\times X$ is the diagonal. Now there is a trace map $\hom(E,E)\to\mathcal O_X$ (if $E$ is finitely generated projective over affines, for example), and this map is $\mathcal O_\Delta$ linear, so induces a map $Ext^i_{X\times X}(\mathcal O_\Delta,\hom(E,E))\to Ext^i_{X\times X}(\mathcal O_\Delta,\mathcal O_\Delta)=H^i(X,\mathcal O_X)$.

-
Thank you for the answer. How do you see the first isomorphism? –  M. K. Nov 4 '12 at 5:36
In the noncommutative case (if $R$ is an algebra over a field $k$, say) there is an isomorphism $Ext^i_R(M,M)\cong Ext^i_{R\otimes R^{op}}(R,\hom_k(M,M))$, and you can try to play the same game. –  Mariano Suárez-Alvarez Nov 4 '12 at 5:49
To get the isomorphism, i think you can simply construct it locally over affines. –  Mariano Suárez-Alvarez Nov 4 '12 at 5:50
In noncommutative case, where is the range of the map? –  M. K. Nov 4 '12 at 6:06
Thank you for your kind answer, Mariano :) I also heard that traceless Ext are related to deformation of sheaves with fixed determinant. Does noncommutative case have a similar description? –  M. K. Nov 4 '12 at 6:06
show 1 more comment