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Let $X$ be a smooth projective variety over an algebraically closed characteristic $0$ field $\mathbb{K}$, and let $\Omega^\bullet_X$ be the de Rham sheaf of complexes of $\mathcal{O}_X$-modules of algebraic differential forms on $X$. Let $\mathbf{R}\Gamma\Omega^\bullet_X$ be (a complex representing) the derived gloval sections of $\Omega^\bullet_X$, and let $End(\mathbf{R}\Gamma\Omega^\bullet_X)$ be the total complex of hom-bicomplex $Hom(\mathbf{R}\Gamma\Omega^\bullet_X,\mathbf{R}\Gamma\Omega^\bullet_X)$.

I have two questions.

i) Is $End(\mathbf{R}\Gamma\Omega^\bullet_X)$ quasi-isomorphic to $End(H^\bullet(X;\mathbb{K}))$as differential graded Lie algebras? (where the latter has trivial differential). (edit: this should be easy and true in general. thanks Torsten).

ii) Now consider a sheaf endomorphism $f:\Omega^\bullet_X\to \Omega^\bullet_X$ (as sheaves of complexes of vector spaces) such that $f:ker(d)\to Im(d)$, so that $f$ will induce the zero morphism on the cohomology sheaf $\mathcal{H}^\bullet(\Omega_X^\bullet)$. And consider the induced morphism $\mathbf{R}\Gamma f: \mathbf{R}\Gamma \Omega^\bullet_X \to \mathbf{R}\Gamma \Omega^\bullet_X$. Is it true that $\mathbf{R}\Gamma f: ker(d)\to Im(d)$? (I suspect the answer is yes in this case: a morphism of sheaves wich is trivial on local cohomology should be trivial on global cohomology, since there is a spectral sequence with coefficients in local cohomology abutting to global cohomology)

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Quasi-isomorphic as what? As complexes of vector spaces it is trivially true (true for all complexes). –  Torsten Ekedahl Nov 25 '10 at 22:02
    
Sorry, I was too inaccurate. I meant quasi-isomorphic as differential graded Lie algebras. I guess the answer is yes in this case, too, again for any complex of vector spaces $C^\bullet$, by considering the span $End(C^\bullet)\leftarrow L\rightarrow End(H^\bullet(C^\bullet))$, where $L=\{\varphi\in End(C^\bullet)\,|\, \varphi(Z^\bullet)\subseteq Z^\bullet\,,\, \varphi(B^\bullet)\subseteq B^\bullet\}$. But since I wasn't able to find this trivial remark anywhere, I was unsure. –  domenico fiorenza Nov 26 '10 at 6:38
    
I think it depends on if you mean End of dg vector spaces or End of dg algebras... –  Kevin H. Lin Nov 26 '10 at 7:34
    
I am interpreting it to mean End of dg algebras... –  Kevin H. Lin Nov 26 '10 at 7:35
    
I was intending End of dg vector spaces. Yet going in the direction of dg algebras is a very nice cue, thanks! –  domenico fiorenza Nov 26 '10 at 8:02

1 Answer 1

Here is an extended version of a comment above.

First, one can consider $R\Gamma\Omega^\bullet_X$ as a complex of vector spaces. Any complex of vector spaces is quasi-isomorphic to its cohomology. Taking this complex to be $End(R\Gamma\Omega^\bullet_X)$ we get $$End(R\Gamma\Omega^\bullet_X)\sim H^\bullet (End(R\Gamma\Omega^\bullet_X))\cong End (H^\bullet(X)).$$

Here $\sim$ stands for quasi-isomorphic and we use the fact that if $K^\bullet$ and $L^\bullet$ are two complexes of vector spaces, then (as graded vector spaces) $$H^\bullet (Hom(K^\bullet, L^\bullet))\cong Hom (H^\bullet(K^\bullet), H^\bullet(K^\bullet)).$$

There is one slightly tricky point here: both $End(R\Gamma\Omega^\bullet_X)$ and $ End (H^\bullet(X))$ are dg-algebras (the latter with zero differential) and one can ask if they are quasi-isomorphic as such. The answer is yes; this can be shown as follows: if $K^\bullet$ and $L^\bullet$ are homotopy equivalent complexes (in fact, in any abelian category), there is an $A_\infty$ quasi-isomorphism between $End(K^\bullet)$ and $End(L^\bullet)$ and hence $End(K^\bullet)$ and $End(L^\bullet)$ are quasi-isomorphic as dg-algebras, meaning they can be connected by a chain of quasi-isomorphisms. (This is probably an overkill, but I can't think of a more straightforward argument at the moment.)

Second, e.g. by taking the Dolbeault resolution one can think of $R\Gamma\Omega^\bullet_X$ as a commutative dg-algebra (cdga) and so one can consider $End_{R\Gamma\Omega^\bullet_X}(R\Gamma\Omega^\bullet_X)$, the endomorphisms of $R\Gamma\Omega^\bullet_X$ as a module over itself. The result will be again $R\Gamma\Omega^\bullet_X$: we use the fact that the natural identification of algebras $Hom_A(A,A)\cong A$ is compatible with the differentials if $A$ is a dg-algebra. Now, again from the Dolbeault resolution, $R\Gamma\Omega^\bullet_X$ is just the algebra of smooth complex-valued forms on $X$, which, as shown by Deligne, Griffiths, Morgan, SUllivan, Real homotopy theory... is quasi-isomorphic to $H^\bullet (X)$ with zero differential.

So we get $$End_{R\Gamma\Omega^\bullet_X}(R\Gamma\Omega^\bullet_X)\cong R\Gamma\Omega^\bullet_X\sim H^\bullet (X)\cong End_{H^\bullet(X)}(H^\bullet (X)).$$

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