User emmy - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T20:22:38Z http://mathoverflow.net/feeds/user/24415 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99428/connections-between-grothendiecks-and-serres-duality connections between Grothendieck's and Serre's duality emmy 2012-06-13T10:55:17Z 2013-02-08T15:49:30Z <p>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 <code>$f^! \mathcal{O}_Y$</code>, appearing in Grothendieck's duality, is the dualizing sheaf for $X$. Let <code>$F$</code> be a coherent sheaf on <code>$X$</code>. Starting from <code>$Hom_{\mathcal{O}_X}(F, f^! \mathcal{O}_Y) \simeq Hom_{{O}_Y}(Rf_* F, \mathcal{O}_Y)$</code> applying the cohomology functor $H^i$ we obtain <code>$Ext^i(F, f^! \mathcal{O}_Y) \simeq Ext^i(Rf_* F, \mathcal{O}_Y).$</code> Using Yoneda's Formula, the right term becomes <code>$Hom^i_{D(Y)}(Rf_* F, \mathcal{O}_Y) \cong Hom_{D(Y)}(Rf_* F, \mathcal{O}_Y[i]) \cong Hom(\widetilde{H^1(X,F)}[-i], \mathcal{O}_Y),$</code> 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 <code>$\Gamma(Y, \mathcal{O}_Y)= \mathbb{C}$</code>, the last term is equal to <code>$Hom(H^1(X,F)[-i], \mathbb{C}) = H^{1-i}(X,F)'$</code>. Now, we have <code>$Hom^i_{D(X)}(F, f^! \mathcal{O}_Y) \cong H^{1-i}(X,F)'$</code> and, shifting by <code>$(-n+1-i)$</code>, <code>$Hom(F, f^! \mathcal{O_Y}[-n+1]) \cong H^n(X,F)'$</code>. So, <code>$f^!(\mathcal{O}_Y)[-n+1]$</code> is a dualizing sheaf for <code>$X$</code>. But we know that, for a nonsingular projective variety the (unique) dualizing sheaf is the canonical sheaf <code>$\omega$</code>. Thus, we must have <code>$f^!\mathcal{O}_Y[-n+1] \cong \omega$</code>, then <code>$f^! \mathcal{O}_Y = \omega[n-1]$</code>.</p> <p>I should obtain $f^! \mathcal{O}_Y = \omega[n]$... what's wrong? Thank you</p> http://mathoverflow.net/questions/100664/finite-homological-dimension finite homological dimension emmy 2012-06-26T08:23:46Z 2012-06-26T08:23:46Z <p>Hi, I found the following in the proof of a theorem: <code>$Z \subset Y \times M$</code> 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 <code>$\mathcal{O}_Z$</code> is of finite homological dimension (is quasi-isomorphic to a bounded complex of locally free sheaves), as an <code>$\mathcal{O}_{ Y \times M}$</code>-module. Why? (all schemes are of finite type over $\mathbb{C}$) Thank you.</p> http://mathoverflow.net/questions/99984/line-bundles-and-rational-singularities Line bundles and rational singularities emmy 2012-06-19T09:50:49Z 2012-06-19T11:57:10Z <p>Hi, I have some problem to understand the proof of lemma 3.2 of this article: <a href="http://www.ams.org/journals/jams/2001-14-03/S0894-0347-01-00368-X/" rel="nofollow">http://www.ams.org/journals/jams/2001-14-03/S0894-0347-01-00368-X/</a>.</p> <p>The lemma states the following: Let <code>$X$</code> be a variety and <code>$f: Y \rightarrow X$</code> a resolution of singularities. Assume that <code>$X$</code> has rational singularities. Then a line bundle <code>$L$</code> on <code>$Y$</code> is the pullback <code>$f^*M$</code> of some linebundle <code>$M$</code> on <code>$X$</code> if and only if the restriction of $L$ to each formal fibre of $f$ is trivial. Moreover, when this holds, <code>$M=f_*L$</code>.</p> <p>For the proof of the "if" part, suppose that the restriction of <code>$L$</code> to each formal fibre is trivial. The teorem on formal functions shows that the completions of the stalks of the sheaves <code>$R^if_* \mathcal{O}_Y$</code> and <code>$R^if_*L$</code> at any point <code>$x \in X$</code> are isomorphic for each $i$. Since $X$ has rational singularities, <code>$R^if_*L=0$</code> for all <code>$i&gt;0$</code> and <code>$M=f_*L$</code> is a linebundle on $X$.</p> <p>Since <code>$f^*M$</code> is torsion free, the natural adjunction map <code>$\eta: f^*f_*L \rightarrow L$</code> is injective, so there is a short exat sequence <code>$$0 \rightarrow f^*f_*L \stackrel{\eta}{\rightarrow} L \rightarrow Q \rightarrow 0.$$</code> By tha projection formula and the fact that $X$ has rational singularities, <code>$R^if_*(f^*M)=M \otimes R^if_* \mathcal{O}_Y=0$</code> for all <code>$i&gt;0$</code>. The fact that <code>$\eta$</code> is the unit of adjunction implies that <code>$f_* \eta$</code> has a left inverse, and in particular is surjective. Applying <code>$f_*$</code> to the exact triple we conclude that <code>$f_*Q=0$</code>, and, by the theorem on formal functions <code>$f_*(Q \otimes L^{-1})=0$</code>, in particular <code>$Q \otimes L^{-1}$</code> has no nonzero global sections. Tensoring the exact triple with <code>$L^{-1}$</code> gives a contradiction, unless $Q=0$. Hence $\eta$ is an isomorphism and we are done.</p> <p>I did not understand this last step. Tensoring the short exact sequence with <code>$L^{-1}$</code> and then taking global sections, we get <code>$\Gamma(Y, f^*M \otimes L^{-1}) \cong \Gamma(Y, \mathcal{O}_Y)$</code>, because the last term is zero. How can I deduce from this that <code>$f^*M \otimes L^{-1} \simeq \mathcal{O}_Y$</code> and then <code>$f^*M \cong L$</code>? Where is the contradiction? Thank you</p> http://mathoverflow.net/questions/100664/finite-homological-dimension Comment by emmy emmy 2012-07-14T08:49:57Z 2012-07-14T08:49:57Z thank you, Jason :) http://mathoverflow.net/questions/99428/connections-between-grothendiecks-and-serres-duality Comment by emmy emmy 2012-06-19T17:53:48Z 2012-06-19T17:53:48Z In &quot;Residues and duality&quot; 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 <code>$f^!\mathcal{O}&#95;Y$</code> 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 <code>$f^!\mathcal{O}&#95;Y \cong \omega[n]$</code> as a special case, without proving that the shifted canonical sheaf satisfies the duality? I hope my question is clear... http://mathoverflow.net/questions/99984/line-bundles-and-rational-singularities/99992#99992 Comment by emmy emmy 2012-06-19T15:35:30Z 2012-06-19T15:35:30Z Thank you, Francesco. But I don't know how to use Chern classes and Picard groups. http://mathoverflow.net/questions/99428/connections-between-grothendiecks-and-serres-duality Comment by emmy emmy 2012-06-15T21:31:42Z 2012-06-15T21:31:42Z Ok, thank you to you all. :) I will try to solve the problem in another way. http://mathoverflow.net/questions/99428/connections-between-grothendiecks-and-serres-duality Comment by emmy emmy 2012-06-14T06:54:14Z 2012-06-14T06:54:14Z Maybe I didn't understand this point. The cohomology functors <code>$H^i(X,-)$</code> are, by definition, the right derived functors of the global section <code>$\Gamma(X,-)$</code>. So, for <code>$i=1$</code> don't I get <code>$R \Gamma(X,F) \cong H^1(X,F)$</code>? http://mathoverflow.net/questions/99428/connections-between-grothendiecks-and-serres-duality Comment by emmy emmy 2012-06-13T21:46:32Z 2012-06-13T21:46:32Z The identity <code>$R^if&#95;&#42;(F) \cong \widetilde{H^i(X,F)}$</code> is right for a morphism <code>$f: X \rightarrow Y$</code> 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) http://mathoverflow.net/questions/99428/connections-between-grothendiecks-and-serres-duality Comment by emmy emmy 2012-06-13T19:23:18Z 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 <a href="http://www.ams.org/journals/jams/1996-9-01/S0894-0347-96-00174-9/" rel="nofollow">ams.org/journals/jams/1996-9-01/&hellip;</a> and I was trying to compute $f^! \mathcal{O}_Y$ in that particular case, just using this and classical Serre's duality. Thank you