Let $X$ be a projective, smooth variety over $\mathbb{C}$, and let $D$ be a irreducible, big Cartier divisor(notice, I do not assume nefness). Then is it true that $${\rm{H}}^1(X, K_X + D) = 0\quad?$$

On the one hand, I doubt it because the result seems to be written nowhere; on the other hand, I feel the conditions here are much nicer than many vanishing theorem, and I was wondering if some modification would reduce it to one of the vanishing theorem.

Besides, any references/results/comments on vanishing theorem for big divisors are great welcome!


Let $X\to \mathbb P ^1$ be a family of smooth quadric surfaces degenerating to $X_0=\mathbb F _2=\mathbb P (\mathcal O _{\mathbb P ^1}\oplus \mathcal O _{\mathbb P ^1} (2))$ where $0\in \mathbb P ^1$. We denote by $E$ the $-2$ curve in $X_0$, $F$ the fiber of $X_0\to \mathbb P ^1$ and by $f:X\to Z$ the corresponding small contraction (following the notation of arXiv:0901.0389 remark 4.4 and remark 4.9). For $b>>0$ consider $L$ a line bundle whose restriction to the general fiber is $\mathcal O _{\mathbb P ^1} (b+1)\boxtimes \mathcal O _{\mathbb P ^1} (b-1)$ and whose restriction to the special fiber is $\mathcal O _{X_0}(b(E+2F)+E)$ and $D\in |L|$ a general member.

We claim that $R^1f_*\omega _X(D)\ne 0$. To see this consider the short exact sequence $0\to \omega _X(D-X_0)\to \omega _X(D)\to \omega _{X}(D)|_{X_0}\to 0$. Since $R^if_*\omega _X(D-X_0)\cong R^if_*\omega _X(D)\otimes \mathcal O_{Z}(-Z_0)$ (by the projection formula), then it follows that $R^1f_* \omega _X(D)\to R^1f_*\omega _{X}(D)|_{X_0}\cong R^1f_* \omega _{X_0}(D|_{X_0})$ is surjective. Consider now the short exact sequence $0\to \omega _{X_0}\to \omega _{X_0}(D|_{X_0})\to \omega _{D|_{X_0}}\to 0$. Since $R^1f_*\omega _{X_0}=0$ (by Grauert-Riemanschneider), it follows that $R^1f_* \omega _{X_0}(D|_{X_0})\to R^1f_*\omega _{D|_{X_0}}$ is surjective. Since $|L|_{X_0}|=|b(E+2F)|+E$ where $E+2F$ is the pull back of an ample divisor $H$ on $X_0$, we have $R^1f_*\omega _{D|_{X_0}}\cong R^1f_*\omega _E\otimes \mathcal O _{Z_0}(bH)$. Since $H^1(\omega _E)\cong H^0(\mathcal O _E)^\vee\ne 0$, we have $R^1f_*\omega _{D|_{X_0}}\ne 0$ and the claim follows.

It should be easy to see that $D$ is big, it is not nef because $D\cdot E =E^2=-2$, moreover we can arrange that $H^1(f_*\omega _X(D))=0$ (if necessary replace $D$ by $D$ plus the pull-back of a sufficiently ample divisor on $Z$ and apply Serre vanishing). Then, by the Leray spectral sequence $H^1(\omega _X (D))\cong H^0(R^1f_*\omega _X (D))\ne 0$.

Maybe there are easier examples, but this is the first one that came to mind. I do not think there can be surface examples, because if $D$ is irreducible and not nef, then $D^2<0$, but as $D$ is big, $D\sim _{\mathbb Q}D'\neq D$ subtracting common components of $D$ we obtain $\lambda D\sim _{\mathbb Q}D''$ with $0<\lambda$ and the support of $D''$ does not contain $D$. But then $D^2=\frac 1 {\lambda ^2} D\cdot D''\geq 0$ a contradiction.

  • $\begingroup$ Thank you for your terrific answer!! I should look the above mentioned paper in more details! $\endgroup$
    – Li Yutong
    Feb 4 '15 at 1:18

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