Let $S$ be a smooth projective curve over $\mathbf C$ and let $f:X\to S$ be a projective flat morphism with "semi-stable" fibres (i.e., the fibres are reduced and strict normal crossings divisors) and normal generic fibre. In particular, $X$ is normal and has only rational singularities. Moreover, the morphism $f:X\to S$ has only finitely many singular fibres. Assume $X\to S$ to be polarized. Let $h$ be the Hilbert polynomial of the generic fibre $X_\eta\to$ Spec $\mathbf C$.

Let $m=m(X/S)$ be the smallest positive integer such that $\deg \det f_\ast \omega_{X/S}^{\otimes m} $ is non-zero. (Who showed that this implies $\deg \det f_\ast \omega_{X/S}^{\otimes m} >0$?)

Do we have that, for all $n\geq m$, the real number $\deg \det f_\ast \omega_{X/S}^{\otimes n}$ is non-zero? If not, does there exist an integer $m$ for which this property holds?

Is there an explicit upper bound on $m$ which depends only on $h$?

If the generic fibre is a curve of genus at least two one can take $m=1$ by Arakelov's theorem.

  • 1
    $\begingroup$ Hilbert polynomial for what embedding to projective space? $\endgroup$
    – Will Sawin
    Dec 22, 2012 at 0:21
  • $\begingroup$ Let's say the canonical sheaf is ample. $\endgroup$
    – Masse
    Dec 22, 2012 at 10:30

1 Answer 1


You need to assume also that $f$ is non-isotrivial. (Think of $X=S\times F$). On the other hand, you don't need the semi-stable assumption for any of these, except that the explicit bound you are asking about is better if you know semi-stability. Also, if there are only finitely many singular fibres, then the general fibre is smooth, right? I also assume that you take $h$ to be the Hilbert polynomial of a fixed power of $\omega_{X_\eta}$.

For your first question in parentheses: The answer to "who showed this" depends on the context. One should probably mention the names Fujita, Kawamata, Viehweg, Kollár. Perhaps more.

For the second question the answer is "yes with $m=2$, as long as $f_*\omega_{X/S}^{\otimes n}\neq 0$". In fact, in that case $f_*\omega_{X/S}^{\otimes n}$ is ample. This was proved by Esnault and Viehweg in Ample sheaves on moduli schemes. Algebraic geometry and analytic geometry (Tokyo, 1990), 53–80, ICM-90 Satell. Conf. Proc., Springer, Tokyo, 1991.

[EDIT]: I've just realized that you asked an explicit upper bound on $m$. The previous paragraph answers that. Here I "answered" a question I thought you were asking about an upper bound for the degree:

There exists an upper bound on the degree, which slightly depends on your assumptions, but they all come down to be something like $(const)\cdot \mathrm{rank}(f_*\omega_{X/S}^{\otimes n})\cdot (2g(S)-2+\delta)$, where $\delta$ is the number of singular fibres. Results of this kind have been obtained by Bedulev-Viehweg, Viehweg-Zuo$^1$ (J. Algebraic Geom. 10 (2001), no. 4, 781–799), Kovács.

In fact, there have been a lot of results related to these questions and various generalizations. You may start with the survey Kovács, Sándor J. Subvarieties of moduli stacks of canonically polarized varieties: generalizations of Shafarevich's conjecture, Algebraic geometry—Seattle 2005. Part 2, 685–709, Proc. Sympos. Pure Math., 80, Part 2, Amer. Math. Soc., Providence, RI, 2009, and look at the references.

These results have strong implications to hyperbolicity of moduli spaces and generalizations of the Shafarevich conjecture to higher dimensional varieties. See for instance a paper by Kovács-Lieblich.

  • $\begingroup$ Thanks for your answer and sorry for the late reply! I think I'm not understanding something completely; if $f_\ast \omega^{\otimes n} =0$, it can happen that $f_\ast \omega^{\otimes (n+1)}$ is non-zero? I find this very strange...The question reuces to whether $f_\ast \omega^{\otimes n}$ is non-zero for ALL $n>>0$. (And here I really mean for ALL $n$ big enough. Thus, without exceptions.) $\endgroup$
    – Masse
    Jan 4, 2013 at 13:42
  • $\begingroup$ Also, I find the condition $f_\ast omega^{\otimes n}$ non-zero in Arakelov's inequality a bit strange. Isn't the degree of the trivial line bundle zero? Thus, isn't Arakelov's inequality trivial in this case? (See for example Theorem 1.1 in the survey of Viehweg on Arakelov inequalities.) $\endgroup$
    – Masse
    Jan 4, 2013 at 13:44

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