Local freeness of direct images - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:39:37Z http://mathoverflow.net/feeds/question/71791 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71791/local-freeness-of-direct-images Local freeness of direct images algori 2011-08-01T10:23:29Z 2011-08-02T18:09:07Z <p>This question arose from an unsuccessful attempt to settle another question of mine: <a href="http://mathoverflow.net/questions/41813/vector-fields-on-complete-intersections" rel="nofollow">http://mathoverflow.net/questions/41813/vector-fields-on-complete-intersections</a></p> <p>Let $X\to Y$ be a smooth projective morphism of noetherian schemes and let $\mathcal{F}$ be a locally free (coherent) sheaf on $X$ such that all direct images <code>$R^i f_* \mathcal{F}$</code> are free. Are there any reasonable conditions which would imply that the direct images of the twisted sheaf $\mathcal{F}(1)$ and/or of the dual $\mathcal{H}om_{\mathcal{O_X}}(\mathcal{F},\mathcal{O}_X)$ are locally free?</p> <p>I can't see any such conditions, but I may be missing something. If this helps, one can assume that $Y$ is the spectrum of a discrete valuation ring.</p> <p>upd: counter-examples would be welcome as well, i.e. an example of $X,Y,f,\mathcal{F}$ as above such that all direct images of $\mathcal{F}$ are locally free, but some direct image of the dual is not.</p> http://mathoverflow.net/questions/71791/local-freeness-of-direct-images/71799#71799 Answer by Jason Starr for Local freeness of direct images Jason Starr 2011-08-01T13:28:11Z 2011-08-01T13:28:11Z <p>Dear algori,</p> <p>There are many counterexamples. For instance, let $R$ be a DVR with uniformizing parameter $t$, let $Y$ equal $\text{Spec} R$, and let $X$ equal $\mathbb{P}^1_R = \text{Proj} R[X_0,X_1]$. Let $\mathcal{H}$ be the invertible sheaf $\mathcal{O}_X$, let $\mathcal{G}$ be the locally free sheaf $\mathcal{O}_X \oplus \mathcal{O}_X(1)\oplus \mathcal{O}_X(1)$. Let $\phi:\mathcal{H}\to \mathcal{G}$ be the $\mathcal{O}_X$-homomoprhism with coordinates $(t \text{Id},X_0,X_1)$. Let $\mathcal{F}$ be $\text{Coker}(\phi)$. Then $\mathcal{F}$ is $\mathcal{O}(1)\oplus\mathcal{O}(1)$ specializing to $\mathcal{O}\oplus \mathcal{O}(2)$. It is easy to see that $R^q\pi_*\mathcal{F}$ is locally free for every $q$. But when you dualize $\mathcal{F}$, this is definitely not true. </p> http://mathoverflow.net/questions/71791/local-freeness-of-direct-images/71906#71906 Answer by Karl Schwede for Local freeness of direct images Karl Schwede 2011-08-02T18:09:07Z 2011-08-02T18:09:07Z <p>algori, my suggestion would be to try to use methods similar to those in the Kollar-Kovacs paper above. The only reason that this is not a comment, is that I couldn't make it fit. It is certainly <em>not</em> a full-fledged answer.</p> <p>In particular, you could try the following: working in characteristic zero, choose a general section of $\mathcal{O}_X(k)$ for some $k \gg 0$ (I'll assume my base is local, you said a DVR was fine). Use this to form a (finite) <em>cyclic</em> cover $\rho : Z \to X$ (as discussed in that paper, or see Kollar-Mori, the book). </p> <p>Now try to study $\rho_* \Omega_{Z/Y}^{r-i}$ as a $O_X$-module. In particular, maybe it can be expressed as a direct sum of the $\Omega_{X/Y}^j(n)$ for various $n$ and $j$. Perhaps someone who knows what this is off the top of their heads will chime in and say there is no way this can work... (Sandor?)</p> <p>Thus, if you can show that the images of the $\Omega_{Z/Y}^{r-i}$ are locally free, then you can also conclude that the higher direct images of the $\Omega_{X/Y}^j(n)$ are locally free as well (at least for those $n$ that appear in the direct sum). Try varying the choice of $k$ and perhaps that will give you what you want?</p>