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I'm really new to derived categories, so i hope this isn't a stupid question.

I'm trying to understand how the duality described as for example in Residues and Duality of R. Hartshorne, using the language of derived categories, gives back the "classical" Serre-duality $$\mathrm{Ext}^{n-i}(\mathcal{F},\omega _X)\cong H^{i}(X,\mathcal{F})^{\vee}$$ which i know from algebraic geometry if $X$ is a smooth variety of dimension $n$ over a field $k$, $f:X\rightarrow\mathrm{Spec}(k)$ its structure morphism, $\omega _X$ its canonical bundle and $\mathcal{F}$ a coherent sheaf on $X$ (and $(-)^{\vee}$ is denoting the dual). This brought me to this question and further to the text Derived categories of sheaves: a skimming recommended there, which really helped me a lot, but there is still one point i don't see (on page 16 of this text) -- in the above notation:

Why is it that $$\mathrm{Hom}_{k}(\mathbf{R}f_{\ast}\mathcal{F}[i],k)\cong H^{i}(X,\mathcal{F})^{\vee}$$ where $\mathcal{F}[i]$ is the complex consisting only of $\mathcal{F}$ at the $(-i)$'th place (and $0$ everywhere else)?

(Obviously $f_{\ast}$ is the global-sections functor in this case, so the statement definitely seems reasonable -- but unfortunately i just don't see how to prove it.) $\mathrm{Hom}_k$ here shall correspond to the $\mathrm{Hom}_{\mathrm{pt}}$ of the text referred to above.

EDIT: I feel that my first guess on what is actually meant by $\mathrm{Hom}_k$ (resp. $\mathrm{Hom}_{\mathrm{pt}}$) was wrong -- it now seems to me as if it is supposed to denote $\mathrm{Hom}_{D(k)}$ where $D(k)$ shall denote the "derived category of coherent sheaves on $\mathrm{Spec} (k)$". So the left hand side of the above were maps of complexes of $k$ vector spaces, corresponding -- if we were without loss of generality allowed to consider everything as living in the derived category of quasicoherent (instead of coherent) sheaves, with enough injectives -- to maps $f_{\ast}\mathcal{I}^{i}\rightarrow k$ if $\mathcal{I}^{\bullet}$ is an injective resolution of $\mathcal{F}$, and such a map again of course induced a map on cohomologies, $H^{i}(X,\mathcal{F})\rightarrow k$ ... In case this works up to here, why would this be reversible, yielding the desired isomorphism (as i still don't see why $\mathbf{R}f_{\ast}\mathcal{F}[i]$ would be concentrated in a single degree)?

Thanks in advance!

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I'm assuming you believe that $\hom_{\mathrm D(k)}(\mathrm R f_\ast \mathcal F[i],k)=\hom_{\mathrm D(k)}(k,\mathrm R f_\ast \mathcal F[i])^\vee$. Now let $f:\mathcal A\to \mathcal B$ be a left-exact functor on abelian categories. It is well-known (see Corollary 10.5.7 of Weibel, for example) that there are natural isomorphisms $\mathrm H^i(\mathrm R f(A))\simeq \mathrm R^i f(A)$ for $A\in \mathcal A$. Moving back to the category $\mathrm{D}(k)$, it should be clear that $\mathrm H^0(C^\bullet[i])=\mathrm H^i(C^\bullet)$ for any $C^\bullet$. Thus, we have $$ \hom_k(k,\mathsf R f_\ast \mathcal F[i]) = \mathrm H^0(\mathrm R f_\ast \mathcal F[i]) = \mathrm H^i(\mathrm R f_\ast \mathcal F) = \mathrm H^i(X,\mathcal F) $$

Edit: Work for the moment in the derived category $\mathrm{D}(k)$ of chain complexes of $k$-vector spaces. Since vetor spaces are all injective modules, this is actually just the category of complexes of $k$-vector spaces up to homotopy. Let $V^\bullet$ be a complex. A morphism $f:k\to V^\bullet$ (where $k$ is concentrated in degree zero) consists of a morphism $f:k\to V^0$ such that $d^0\circ f=0$. So such a morphism just picks out an element $c\in \ker(d^0)$. You can check that two morphisms are homotopic precisely if the corresponding elements of $\ker(d^0)$ are the same in $\mathrm H^0(V^\bullet)$. To sum it up, $\hom_{\mathrm D(k)}(k,V^\bullet)=\mathrm H^0(VC^\bullet)$.

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  • $\begingroup$ Thank you very much for your answer! I'm sorry i have to ask, but: Why $\mathrm{Hom}_{D(k)}(k,\mathbf{R}f_{\ast}\mathcal{F}[i])=H^0(\mathbf{R}f_{\ast} \mathcal{F} [i])$ in the last line? Why not $\ldots =\mathbf{R}f_{\ast}\mathcal{F}[i]^0$, the degree $0$-part of the complex $\mathbf{R}f_{\ast}\mathcal{F}[i]^{\bullet}$? (My thought with this: as $k$ is concentrated in degree $0$ the maps of complexes should be the maps of degree $0$ parts) I must be missing some basic fact on derived categories... $\endgroup$ – user178979 Oct 20 '14 at 13:28
  • $\begingroup$ Thanks again! It's all clear to me now. $\endgroup$ – user178979 Oct 20 '14 at 14:51

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