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Let $X$ be a smooth rationally connected variety (say in characteristic zero). Then it is straightforward to prove that $H^0(X,(\Omega_X^1)^{\otimes m}) = 0$ for all $m > 0$ since there is a very free rational curve $f: \mathbb{P}^1 \rightarrow X$ for which $f^* T_X$ is ample (see for example Rational Curves on Algebraic Varieties by J. Kollár, chapter IV). In fact, the vanishing of these cohomology groups conjecturally characterizes rationally connected varieties.

However, this doesn't seem to immediately rule out the existence of a nonzero global $m$-form for some $m > 1$, since $\Omega^m_X$ is a quotient of $(\Omega_X^1)^{\otimes m}$.

Question. Are there any rationally connected (or even Fano) varieties with nonzero global differential $m$-forms?

For $X$ smooth Fano of dimension $n$, we would necessarily have $1 < m < n$.

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    $\begingroup$ In characteristic $0$, $\Omega_X^m$ is a summand of $(\Omega^1_X)^{\otimes m}$, not just a quotient, so the vanishing argument goes through. $\endgroup$
    – Will Sawin
    May 19, 2021 at 21:18
  • $\begingroup$ Thanks Will! I forgot about that fact. $\endgroup$
    – Irwin
    May 19, 2021 at 21:42

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For Fano manifolds in characteristic $0$, we can also use the following argument.

By Dolbeault theorem and Hodge symmetry we get, for any compact Kähler manifold $X$ and for any $m>0$, the isomorphism $$H^0(X, \, \Omega^m_X)\simeq H^m(X, \, \mathcal{O}_X) =H^m(X, \, K_X \otimes K_X^{-1}).$$ If $X$ is Fano then $K_X^{-1}$ is ample and so the last group is zero by Kodaira vanishing theorem.

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    $\begingroup$ A variant of this argument also shows that $H^2(X,T_X)$ vanishes, so that deformations of complex Fano manifolds are unobstructed. In positive characteristic, we have neither of these. $\endgroup$ May 19, 2021 at 22:22

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