This question is related to one of my previous question Do we have the following isomorphism for $\mathcal{Ext}$?

Let $X$ be a smooth variety (over $\mathbb{C}$) and $\Delta: X \rightarrow X \times X$ be the diagonal embedding and $p_1: X\times X\rightarrow X, ~p_2: X\times X\rightarrow X$ be the projections to the first and second components. Let $E$ be a finite dimensional vector bundle on $X$. We define $$ E_{\Delta}:=\Delta_*E $$ to be a sheaf on $X\times X$. In particular we have $\mathcal{O}_{\Delta}:=\Delta_*\mathcal{O}$. We have the sheaf Ext functor $\mathcal{Ext}$ on $X\times X$ and we can define ${Rp_1}_*\mathcal{Ext}_{X\times X}^{\bullet}(E_{\Delta},E_{\Delta})$. Sasha told me in my previous question that we have $$ {Rp_1}_*\mathcal{Ext}_{X\times X}^{\bullet}(E_{\Delta},E_{\Delta})\cong {Rp_1}_*\mathcal{Ext}_{X\times X}^{\bullet}(\mathcal{O}_{\Delta},\mathcal{O}_{\Delta})\otimes \mathcal{End}(E). $$

Now we can take the derive global section on the above (complex of) sheaves: $$ R\Gamma(X, {Rp_1}_*\mathcal{Ext}_{X\times X}^{\bullet}(E_{\Delta},E_{\Delta}))\cong R\Gamma(X, {Rp_1}_*\mathcal{Ext}_{X\times X}^{\bullet}(\mathcal{O}_{\Delta},\mathcal{O}_{\Delta})\otimes \mathcal{End}(E)). $$

Notice that when the vector bundle $E=\mathbb{C}$ the trivial bundle, we just get the Hochschild cohomology $HH^{\bullet}(X)$. So we can consider the hypercohomology $R\Gamma(X, {Rp_1}_*\mathcal{Ext}_{X\times X}^{\bullet}(E_{\Delta},E_{\Delta}))$ as a "Hochschild cohomology with coefficients in $E$" and maybe we can denote it by $HH^{\bullet}(X,E)$ (I don't know whether it has already been studied). Nevertheless I think that when $E$ is nontrivial, we do not simply get $HH^{\bullet}(X,E) \cong HH^{\bullet}(X)\otimes \text{End}(E)$.

$\textbf{My question}$ is: could we compute this $HH^{\bullet}(X,E)$? For examply, could we construct a good spectral sequence which convergence to it?

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    $\begingroup$ «Hochschild cohomology with coefficients» is better match to $Ext_{X\times X}(O_X,\mathcal F)$ with $\mathcal F$ some sheave, rather. $\endgroup$ – Mariano Suárez-Álvarez Sep 18 '13 at 11:56
  • $\begingroup$ can't you apply the Grothendieck spectral sequence for $\otimes$ and $R\Gamma$ since that seems to be what you are asking about applied to some situation? $\endgroup$ – user36931 Sep 18 '13 at 15:05
  • $\begingroup$ @ Mariano Oh, yes. Your definition is compatible with the corresponding definition on algebras. $\endgroup$ – Zhaoting Wei Sep 19 '13 at 18:00

By adjunction $Rp_{1*}\mathcal{RHom}^*(\Delta_*E,\Delta_*E)=\mathcal{RHom}^*(\Delta^*\Delta_*E,E)$. By the splitting, you obtain $\mathcal{RHom}^*(\Delta^*\Delta_*E,E)=\mathcal{RHom}^*(E\otimes \Delta^*\Delta_*\mathcal O,E)=\mathcal{RHom}^*(\Delta^*\Delta_*\mathcal O,\mathcal{End}E)$. By HKR, you have $\Delta^*\Delta_*\mathcal O=Sym(\Omega[1])$, yielding $Rp_{1*}\mathcal{RHom}^*(\Delta_*E,\Delta_*E)=Sym(T[-1])\otimes \mathcal{End}(E)$. So $$R\Gamma(Rp_{1*}\mathcal{RHom}^*(\Delta_*E,\Delta_*E))=R\Gamma(Sym(T[-1])\otimes \mathcal{End}(E))$$. The k-th hypercohomology space thus is given by $\oplus_{i+j=k}H^i(\wedge^j T\otimes \mathcal{End}(E))$.

(and yes, people use the notation $HH^{\bullet}(X,E)$ to $RHom^{\bullet}_{X\times X}(\mathcal{O}_X,E)$.)

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  • $\begingroup$ I'm a little bit nervous that the HKR isomorphism, as far as I know, is on the level of cohomology, not on the sheaves. Hence we cannot say $\Delta^*\Delta_*\mathcal{O}\cong\text{Sym}(\Omega[1])$ as sheaves. In fact, I expect that the "HKR map with coefficients" intertwine with $\mathcal{E}nd E$. $\endgroup$ – Zhaoting Wei Sep 22 '13 at 3:47
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    $\begingroup$ They are isomorphic as algebra objects in the derived category of $X$. This phenomenon is not unique for the diagonal embedding. If you have a closed embedding $i:X\rightarrow Y$, and $\mathcal{F}$ is any vector bundle, then $i^*i_*\mathcal{F}$ and $F\otimes Sym(N^{\vee}[1])$ are isomorphic, if $X$ splits in $Y$ (so the normal sequence splits). (see When is the self-intersection of a subvariety a fubration? by Arinkin and Caldararu) $\endgroup$ – Marci Sep 22 '13 at 13:13
  • $\begingroup$ Oh, yes, I think you're right and I misunderstood the HKR map. By the way, do you have any idea of how to compute $H^i(\wedge^j T\otimes \mathcal{E}nd (E))$? Could we express it in terms of $H^k(\wedge^j T)$ and $H^l(\mathcal{E}nd (E))$? $\endgroup$ – Zhaoting Wei Sep 22 '13 at 21:59
  • $\begingroup$ I do not think one should expect a formula involving the $H^k(\wedge ^j)$ and $H^l(\mathcal{End}(E))$. $\endgroup$ – Marci Sep 22 '13 at 23:40

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