# Calculate $Hom$ in derived category

Suppose $X$ is a smooth variety over $\mathbb{C}$. Let $K^{b}(X)$ be the homotopy category of bounded complex of coherent sheaves, and $D^{b}(X))$ be the derived category of bounded complex of coherent sheaves. One can define: $$Hom^{\cdot}(A^{\cdot}, B^{\cdot}): K^{b}(X) \to K(Ab)$$ as $$Hom^{i}(A^{\cdot},B^{\cdot}):= \oplus Hom(A^k,B^{k+i}),$$ $$d(f):=d_B \circ f - (-1)^{i} f \circ d_A.$$ Then we can pass to the derived categories, and define the right derived functor: $$RHom^{\cdot}(A^{\cdot},): D^{b}(X ) \to D(Ab).$$

My questions is: How to compute $RHom(F,k(x))$ in the derived category? Here $F$ is a sheaf on $X$(viewed as a complex in $D(X)$ concentrated in degree $0$), and $k(x)$ is a skyscraper sheaf on $x \in X$. In particular how to compute $RHom(k(x),k(x))$?

I think by definition, one has to do an injective resolution to $k(x)$, and perform the computation in the homotopy category.

• It is better to replace $\oplus$ with $\prod$ in the definition of $Hom^i$. – Sasha Mar 13 '13 at 20:51

For computations you it is better to replace the first argument by a locally free resolution. This allows to compute the local $R{\mathcal H}om$. Then you can use the local-to-global spectral sequence to compute the global $RHom$. In your particular example, as everything happens in a neighborhood of $x$ and since $x$ is a locally complete intersection in $x$, you can find a vector bundle $E$ and it section in a neighborhood of $x$ such that its zero locus is $x$. Then the Koszul complex $$0 \to \Lambda^n E^* \to \dots \to \Lambda^2 E^* \to E^* \to O_X \to 0$$ is a locally free resolution of the skyscraper. This shows that $${\mathcal E}xt^i(k(x),k(x)) = \Lambda^i E_x$$ and hence $$Ext^i(k(x),k(x)) = \Lambda^i E_x$$ as well. Moreover, one has $E_x \cong T_xX$, so the final answer is $$RHom(k(x),k(x)) = \oplus_{i=0}^n \Lambda^iT_xX[-i].$$
EDIT. To compute local Ext's you just apply ${\mathcal H}om$ to the Koszul resolution. You will get a complex $$0 \to k(x) \to E\otimes k(x) \to \Lambda^2E \otimes k(x) \to \dots \to \Lambda^nE \otimes k(x) \to 0.$$ Since the map in the Koszul complex are given by wedging with the section of $E$ which vanish at $x$, it follows that in the above complex all maps are zero, so its cohomology sheaves are just its terms.
• Thank you very much! I can follow your argument except a few cases.(1) When using local-global spectral sequence, do you need some degeneracy condition to argue $E_{2}^{p,q}=E_{\inf}^{p,q}$? (2)why $\mathcal{Ext}(k(x),k(x))=\wedge ^{i}E_x$? Thank you again, and I really appreciate your time! – Li Yutong Mar 14 '13 at 0:07