User emmy - MathOverflow

connections between Grothendieck's and Serre's duality

2012-06-13T10:55:17Z

Hi, I would like to show that if $f: X \rightarrow Y=Spec \, \mathbb{C}$, where $X$ is a nonsingular complex projective variety, is the projection to a point, then the complex $f^! \mathcal{O}_Y$ , appearing in Grothendieck's duality, is the dualizing sheaf for $X$. Let $F$ be a coherent sheaf on $X$ . Starting from $ Hom_{\mathcal{O}_X}(F, f^! \mathcal{O}_Y) \simeq Hom_{{O}_Y}(Rf_* F, \mathcal{O}_Y) $ applying the cohomology functor $ H^i $ we obtain $Ext^i(F, f^! \mathcal{O}_Y) \simeq Ext^i(Rf_* F, \mathcal{O}_Y).$ Using Yoneda's Formula, the right term becomes $Hom^i_{D(Y)}(Rf_* F, \mathcal{O}_Y) \cong Hom_{D(Y)}(Rf_* F, \mathcal{O}_Y[i]) \cong Hom(\widetilde{H^1(X,F)}[-i], \mathcal{O}_Y),$ where, for the last isomorphism, I use Theorem 8.5 from Hartshorne'a Algebraic Geometry, p.251. Now, by Corollary 5.5 pag.151 from Hartshorne, and remembering that $\Gamma(Y, \mathcal{O}_Y)= \mathbb{C}$ , the last term is equal to $Hom(H^1(X,F)[-i], \mathbb{C}) = H^{1-i}(X,F)'$ . Now, we have $Hom^i_{D(X)}(F, f^! \mathcal{O}_Y) \cong H^{1-i}(X,F)'$ and, shifting by $(-n+1-i)$ , $Hom(F, f^! \mathcal{O_Y}[-n+1]) \cong H^n(X,F)'$ . So, $f^!(\mathcal{O}_Y)[-n+1]$ is a dualizing sheaf for $X$ . But we know that, for a nonsingular projective variety the (unique) dualizing sheaf is the canonical sheaf $\omega$ . Thus, we must have $f^!\mathcal{O}_Y[-n+1] \cong \omega$ , then $f^! \mathcal{O}_Y = \omega[n-1]$ .

I should obtain $f^! \mathcal{O}_Y = \omega[n]$... what's wrong? Thank you

finite homological dimension

2012-06-26T08:23:46Z

Hi, I found the following in the proof of a theorem: $ Z \subset Y \times M$ where $M$ is a smooth projective variety over $\mathbb{C}$, $Y$ is a scheme and $Z$ is a subscheme of the product, flat over $Y$. Then $\mathcal{O}_Z$ is of finite homological dimension (is quasi-isomorphic to a bounded complex of locally free sheaves), as an $\mathcal{O}_{ Y \times M}$ -module. Why? (all schemes are of finite type over $\mathbb{C}$) Thank you.

Line bundles and rational singularities

2012-06-19T09:50:49Z

Hi, I have some problem to understand the proof of lemma 3.2 of this article: http://www.ams.org/journals/jams/2001-14-03/S0894-0347-01-00368-X/.

The lemma states the following: Let $X$ be a variety and $f: Y \rightarrow X$ a resolution of singularities. Assume that $X$ has rational singularities. Then a line bundle $L$ on $Y$ is the pullback $f^*M$ of some linebundle $M$ on $X$ if and only if the restriction of $L$ to each formal fibre of $f$ is trivial. Moreover, when this holds, $M=f_*L$ .

For the proof of the "if" part, suppose that the restriction of $L$ to each formal fibre is trivial. The teorem on formal functions shows that the completions of the stalks of the sheaves $R^if_* \mathcal{O}_Y$ and $R^if_*L$ at any point $x \in X$ are isomorphic for each $i$. Since $X$ has rational singularities, $R^if_*L=0$ for all $i>0$ and $M=f_*L$ is a linebundle on $X$.

Since $f^*M$ is torsion free, the natural adjunction map $\eta: f^*f_*L \rightarrow L$ is injective, so there is a short exat sequence $$ 0 \rightarrow f^*f_*L \stackrel{\eta}{\rightarrow} L \rightarrow Q \rightarrow 0.$$ By tha projection formula and the fact that $X$ has rational singularities, $R^if_*(f^*M)=M \otimes R^if_* \mathcal{O}_Y=0$ for all $i>0$ . The fact that $\eta$ is the unit of adjunction implies that $f_* \eta$ has a left inverse, and in particular is surjective. Applying $f_*$ to the exact triple we conclude that $f_*Q=0$ , and, by the theorem on formal functions $f_*(Q \otimes L^{-1})=0$ , in particular $Q \otimes L^{-1}$ has no nonzero global sections. Tensoring the exact triple with $L^{-1}$ gives a contradiction, unless $Q=0$. Hence $\eta$ is an isomorphism and we are done.

I did not understand this last step. Tensoring the short exact sequence with $L^{-1}$ and then taking global sections, we get $ \Gamma(Y, f^*M \otimes L^{-1}) \cong \Gamma(Y, \mathcal{O}_Y)$ , because the last term is zero. How can I deduce from this that $f^*M \otimes L^{-1} \simeq \mathcal{O}_Y$ and then $f^*M \cong L$ ? Where is the contradiction? Thank you

Comment by emmy

2012-07-14T08:49:57Z

thank you, Jason :)

Comment by emmy

2012-06-19T17:53:48Z

In "Residues and duality" one proves Grothendieck's duality by showing some particular cases (smooth, finite morphism...) and then reducing the general case to them. Here is simple to show that $f^!\mathcal{O}_Y$ is the shifted canonical sheaf. But, if one proves Grothendieck's duality for a proper morphism in a more general way, without using this special cases (see for example Neeman's article I cited), is there a way to recover that $f^!\mathcal{O}_Y \cong \omega[n]$ as a special case, without proving that the shifted canonical sheaf satisfies the duality? I hope my question is clear...

Comment by emmy

2012-06-19T15:35:30Z

Thank you, Francesco. But I don't know how to use Chern classes and Picard groups.

Comment by emmy

2012-06-15T21:31:42Z

Ok, thank you to you all. :) I will try to solve the problem in another way.

Comment by emmy

2012-06-14T06:54:14Z

Maybe I didn't understand this point. The cohomology functors $H^i(X,-)$ are, by definition, the right derived functors of the global section $ \Gamma(X,-)$ . So, for $i=1$ don't I get $R \Gamma(X,F) \cong H^1(X,F)$ ?

Comment by emmy

2012-06-13T21:46:32Z

The identity $R^if_*(F) \cong \widetilde{H^i(X,F)}$ is right for a morphism $ f: X \rightarrow Y$ from a noetherian to an affine scheme, where $F$ is a quasi-coherent sheaf on $X$. (See Hartshorne's Algebraic Geometry, page 251, Thm. 8.5)

Comment by emmy

2012-06-13T19:23:18Z

By dualizing sheaf I mean a sheaf that satisfies the usual Serre's duality (definition p.241 Hartshorne Algebraic Geometry). I read the proof of duality for smooth morphisms from Residues and Duality, but I need the proof of Grothendieck's duality from this article ams.org/journals/jams/1996-9-01/… and I was trying to compute $f^! \mathcal{O}_Y $ in that particular case, just using this and classical Serre's duality. Thank you