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This question arose from an unsuccessful attempt to settle another question of mine: Vector fields on complete intersections

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 $R^i f_* \mathcal{F}$ 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?

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.

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.

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Can we assume anything special about $\mathcal{F}$. For example, $\mathcal{F} = \Omega_{X/Y}^i$. –  Karl Schwede Aug 1 '11 at 13:07
    
Karl -- yes, that's the example I had in mind. –  algori Aug 1 '11 at 14:07
    
algori, maybe you could say exactly what you had in mind then. There are a lot of results about local-freeness for twists of these guys, especially if $i$ in $\Omega_{X/Y}^i$ is the relative dimension. For example, see Theorem 7.9(2) in arxiv.org/PS_cache/arxiv/pdf/0902/0902.0648v4.pdf I think there are also some results along these lines in arxiv.org/PS_cache/arxiv/pdf/1003/1003.2913v3.pdf Finally, since you said you can assume your base is a DVR, you just need these pushdowns to be torsion free see Kollar's papers Higher direct images... for a starting place. –  Karl Schwede Aug 1 '11 at 16:54
    
Karl -- thanks for the references. More specifically, I'm interested in the dual of $\Omega^1_{X/Y}$, which happens to be $\Omega^{r-1}_{X/Y}(k)$. Here r is the relative dimension of $X$ and $k$ is, in the case I'm interested in, a positive number between $1$ and $n−2$ where $n$ is the relative dimension of the projective space over $Y$ where $X$ lives. I know that all direct images of any $\Omega^i_{X/Y}$ are locally free. –  algori Aug 2 '11 at 3:10

2 Answers 2

Dear algori,

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.

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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 not a full-fledged answer.

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) cyclic cover $\rho : Z \to X$ (as discussed in that paper, or see Kollar-Mori, the book).

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?)

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?

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Karl -- thanks. My hope was that there is some general magic that I'm unaware of and that would do the trick (such a thing did happen to me once). But this was, of course, probably foolish, as Jason Starr points out. –  algori Aug 3 '11 at 4:32

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