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If an algebraic variety $X$ is proper, then the (naive) hodge numbers $h^{p,q}:= dim\, H^p(X, \Omega^q)$ are finite.

To what extent is the converse true?

E.g., finiteness of $h^{0,0}$ tells you that at least the variety is not (affine and not proper).

I am happy to assume X is smooth.

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  • $\begingroup$ @WillSawin Is what you're saying consistent with e.g. the fact that $H^1(A^2 \setminus 0; \mathcal{O})$ is nonzero (in fact infinite dimensional)? $\endgroup$ Commented Nov 22, 2016 at 19:55
  • $\begingroup$ Interesting question. My guess is that the answer should be no in general. Do we know a non-proper example with $h^{p, 0} < \infty$ for all $p$? $\endgroup$ Commented Nov 22, 2016 at 19:59
  • $\begingroup$ @VivekShende No, I calculated that case wrong and generalized incorrectly. $\endgroup$
    – Will Sawin
    Commented Nov 22, 2016 at 20:15
  • $\begingroup$ Are there smooth counterexamples with trivial canonical bundle? $\endgroup$ Commented Nov 23, 2016 at 0:35
  • $\begingroup$ I would like to know this as well $\endgroup$ Commented Nov 23, 2016 at 0:37

1 Answer 1

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Let $L$ be a degree $0$, nontorsion line bundle on an elliptic curve $E$, viewed as an affine scheme $X$ over $E$ with $\pi: X \to E$ the structure morphism. I claim all the Hodge numbers of $X$ are finite.

We have a canonical isomorphism $$\omega_X = \pi^* (\omega_E \otimes L^{-1})$$ and short exact sequence $$0 \to \pi^* \omega_E \to \Omega^1_X \to \pi^* L^{-1}\to 0$$ that, together with the non-canonical isomorphism $\omega_E=\mathcal O_E$, reduce us to showing $H^p(X, \mathcal O_X)$ and $H^p(X, \pi^* L^{-1})$ are finite.

Because $\pi$ is affine, we have $$H^p(X, \mathcal O_X) = H^p (E, \pi_* \mathcal O_X)= H^p (E, \sum_{n=0}^\infty L^{-\otimes n}) = \sum_{n=0}^{\infty} H^p(E, L^{- \otimes n}) = H^p(E,\mathcal O_X)$$ because as $L$ is nontorsion of degree $0$, all its powers have vanishing cohomology in every degree. Similarly $$H^p(X, \pi^* L^{-1}) = H^p (E, \pi_* \pi^* L^{-1}) = H^p (E, \sum_{n=0}^{\infty} L^{-\otimes (n+1)}) = \sum_{n=0}^\infty H^p(E, L^{- \otimes (n+1)}) = 0$$ by the same logic.


In general, let $X$ be a variety with compactification $\overline{X}$ whose boundary is a simple normal crossings union of smooth divisors $D_1,\dots D_k$. If I calculated correctly, a sufficient condition for finiteness of Hodge numbers is that finitely many of the cohomology groups indexed by numbers $p,q$ and tuples $m_1,\dots,m_k$

$$H^p \left( \bigcap_{ i | m_i>0} D_i, \Omega^q_X \otimes N_1^{\otimes m_1} \otimes N_2^{\otimes m_2} \otimes \dots \otimes N_k^{\otimes m_k}\right)$$

are nonvanishing.

These should be the cohomology groups of the associated graded components of the pole order filtration of $\Omega^q_X$, viewed as a quasicoherent sheaf on $\overline{X}$.

This condition explains my example when applied to the obvious compactification as a $\mathbb P^1$ bundle. However, it is not necessary, as can be seen by simply choosing another compactification.

This vanishing might be an interesting condition to study. There are some explicit constructions generalizing my example (for $A$ an abelian variety and $X$ any variety, $L_1$ a very general degree $0$ line bundle on $A$ and $L_2$ any line bundle on $X$, a smooth divisor $X \times A$ with normal bundle $L_1 \otimes L_2$ does the job.)

The only necessary conditions I found were a bunch of vanishing characteristic class integrals (I think the integral of any nonconstant monomial in $c_1(D_1),\dots,c_1(D_k)$ times the Chern character of $\Omega^q$ times the Todd class must vanish) which don't really say much about the structure of the $D_i$.

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  • $\begingroup$ Do your necessary conditions shed any light on the question of whether this can happen when X is calabi-yau, in the sense of having trivial canonical bundle? $\endgroup$ Commented Nov 24, 2016 at 22:04
  • $\begingroup$ @VivekShende I doubt I will be able to say much. In my example, $K$ is numerically trivial on $X$, but the necessary conditions are purely numerical, so I don't know how to distinguish $K$ numerically trivial from $K$ trivial. $\endgroup$
    – Will Sawin
    Commented Nov 25, 2016 at 8:31

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