I have a flat morphism $p: X \to Y$ from a smooth projective $X$ to a smooth projective $Y$. I have a line bundle $L$ on $X$ whose restriction to every fiber of $p$ is big and nef. I need the vanishing of $Rp_*(\Omega^b X\otimes L^{-1})$ for b small compared to $\dim X-\dim Y$. I could not find this in Esnault-Viehweg. Is it true? If yes, what is the reference?


2 Answers 2


First of all I suppose you meant $R^ip_*$ for $i>0$ and not $Rp_*$. Actually, neither is true, but the latter is obviously false while the former may seem believable first. Second, I suppose you are working over an algebraically closed field of characteristic zero.

In any case, unfortunately this fails even over an algebraically closed field of characteristic zero and already if $Y$ is a point. (This is not surprising as any statement you might hope for would be enough to prove in this case).

Let $X_0$ be an arbitrary smooth projective variety over an algebraically closed field of characteristic zero and $\mathscr L$ an arbitrary ample line bundle on $X_0$. Let $\pi:X\to X_0$ be the blowing up of $X_0$ at an arbitrary smooth point with exceptional divisor $E\simeq\mathbb P^{r-1}$ (i.e., $\dim X=r$). Then for any $b\in \mathbb N$, $0<b<r$ we have the following non-vanishing: $$ H^b(X,\Omega_X^b\otimes\pi^*\mathscr L)\neq 0 $$

Proof: Observe that by Bott's formula for cohomology on $\mathbb P^{r-1}$ applied to $E$, we have that $R^i\pi_*\Omega_X^b=0$ if and only if $i\neq 0, b$. By Kodaira-Akizuki-Nakano applied to $\mathscr L$ on $X_0$ we have that $H^i(X_0,\pi_*\Omega_X^b\otimes\pi^*\mathscr L)=0$ for $i>0$ (this is because $\pi_*\Omega_X^b/\Omega_{X_0}^b$ is supported at the point that was blown up). So, the Leray spectral sequence computing $H^i(X,\Omega_X^b\otimes\pi^*\mathscr L)$ degenerates and gives that for $i>0$, $$ H^i(X,\Omega_X^b\otimes\pi^*\mathscr L)=H^{i-b}(X,R^b\pi_*\Omega_X^b\otimes\mathscr L) $$ For $i=b$ the latter is clearly non-zero. $\square$

Corollary: $H^{b}(X, \Omega_X^b\otimes \pi^*\mathscr L^{-1})\neq 0$ for $0<b<r$. In particular, if $\dim X\geq 3$, then $H^{1}(X, \Omega_X\otimes \pi^*\mathscr L^{-1})\neq 0$

The main reason vanishing fails here is that there is this big blob of a divisor, $E$, where the chosen line bundle, $\pi^*\mathscr L$, is trivial.

There are actually still some results that do give you vanishing, but they need more assumption. Sommese proved various versions with assumptions on the fiber of the morphism induced by some high power of the line bundle, so in particular these results assume that the line bundle is actually semi-ample, however it does not always need big once you assume the restriction on the fibers. For more on this see the book by Shiffman and Sommese: Vanishing theorems on complex manifolds.

Another couple of interesting papers were written by Arapura: Check out this and this.

My result Karl mentioned in his remark is actually for singular varieties. The main idea is that since you can't expect general vanishing for not necessarily ample but big and nef line bundles, you might be able to do it on the model where they are ample, however, this usually requires working on singular varieties, which makes dealing with $\Omega$ a bit tougher. For more details see this paper.

More recent similar vanishing results were obtained by Greb-Kebekus-Peternell and myself. The most recent relevant paper is this, but you could also look at this and this.


I guess it depends on exactly what you need. Take $Y=Spec\ \mathbb{C}$. Then here's a cautionary example that $L$ nef and big is usually inadequate for vanishing for $\Omega_X^b\otimes L^{\pm 1}$ when $0 <b<n=\dim X$. Let us suppose that $ X$ is obtained by blowing up a smooth variety $Z$ at a point $p$. Let $f:X\to Z$ be the projection, and suppose that $L= f^*M$ with $M$ ample.

Suppose also that $n>2$. Then $H^1(Z,\Omega_Z^1\otimes M^{-1})=0$ by Kodaira-Nakano. One might ask if $H^1(X,\Omega_X^1\otimes L^{-1})=0$? Pick $n>3$ and use Kodaira-Nakano on $Z$ and Leray to get $$H^1(X,\Omega_X^1\otimes L^{-1})=H^0(Z, R^1f_*\Omega_X^1\otimes M^{-1})=\mathbb{C}$$ because $R^1$ is the sky-scraper sheaf $\mathbb{C}_p$.

  • $\begingroup$ I guess I should occasionally reload my browser... This answer was not here when I started mine, then I was interrupted and it took a while before I got back to finishing it... Sorry for the duplication, but it might be worth leaving my answer there as it has some references. $\endgroup$ Mar 16, 2012 at 18:32
  • $\begingroup$ No problem. Your answer is certainly more thorough. $\endgroup$ Mar 16, 2012 at 18:35

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