# Reference for Arakelov's theorem: $K^2_f=0$ iff $f$ is locally trivial

Let $f:X\longrightarrow B$ be a family of curves, with $f$ relatively minimal, over a fixed curve $B$ ($B$ is projective, irreducible and smooth). The fibration $f$ is said locally trivial if all fibers are smooth and isomorphic, and moreover the number $K^2_f$ is defined as $$K^2_f=K^2_X-8(g-1)(g(B)-1)$$ where $g$ is the genus of a fiber.

Edit: $g\ge2$

I'd like to have a reference for the proof of the following theorem:

$K^2_f=0$ if and only if $f$ is locally trivial.

All authors attribute this theorem to Arakelov and cite the article "S.Ju. Arakelov - family of algberaic curves with fixed degeneracy". In this article I can't find the above theorem! Can you suggest me an alternative reference?

• You must assume $g\geq 2$, otherwise this is false. Then this Theorem 2' in "exposé no. 3" by Szpiro, Asterisque 86. – abx Feb 28 '15 at 12:52
• Many thanks. Unfortunately this article is not free on the web. – Dubious Feb 28 '15 at 13:33
• $K_{f}^2 = \omega_{X/B}^2$, the self-intersection of the relative dualizing sheaf considered in Arakelov's paper. If $X/B$ is locally trivial, it is obvious that $\omega_{X/B}^2 = 0$, for then $\omega_{X/B}$ is a pull-back after a base change. Otherwise, Arakelov proves in Proposition 3.2 that $\omega_{X/B}$ is an amples sheaf: this is his famous rigidity theorem. Hence, $\omega_{X/B}^2 > 0$ when $f$ is not locally trivial. – Vesselin Dimitrov Feb 28 '15 at 17:02
• Please don't write $K^2=\omega^2$!! One of them is a divisor, the other is a sheaf. There is a reason these notions are distinguished. $\omega^2$ is the line bundle corresponding to the divisor $2K$ and not the number $K^2$. Of course, what you have in mind is different, but then why not write down what you have in mind? It's not that hard. You could have written for instance that $K^2=c_1(\omega)^2$... – Sándor Kovács Mar 2 '15 at 19:56
• @SándorKovács: OK, I see your point. – Vesselin Dimitrov Mar 2 '15 at 20:51

## 1 Answer

What you are saying is not entirely correct.

1. That equation for $K_f^2$ is not a definition. It's a consequence of the definition of $K_f$ as $K_X-f^*K_B$: If $F$ denotes a fiber of $f$, then since $K_X\cdot F=\deg K_F = 2g-2$, $$K_f^2 = K_X^2 - 2 K_X\cdot f^*K_B = K_X^2 - 2 K_X\cdot (2g(b)-2) F = K_X^2 - 2(2g-2)(2g(B)-2).$$
2. The statement you are looking for is not exactly like that. $K_f^2=0$ is not equivalent to $f$ being locally trivial, but being generically locally trivial. That is, for instance it might have some singular fibers, but the smooth fibers are isomorphic. The right word actually is isotrivial, which is defined as two general fibers being isomorphic.

With that said, Vesselin already gave you the correct reference, but at the same time there is a whole lot of newer results that deal with these kind of questions.

The modern approach is that first of all regardless of any assumption on isotriviality, the sheaf $f_*\mathscr O_X(mK_{f})$ is semipositive for large and divisible enough $m$. Then using the fact that the fibers are canonically polarized varieties (that's the real reason you need $g\geq 2$), the natural map $$f^*f_*\mathscr O_X(mK_{f})\to \mathscr O_X(mK_{f})$$ is generically surjective, so with a bit of work you can prove that $K_f$ is nef and then $K_f^2\geq 0$ always and then $K_f^2> 0$ if and only if $K_f=K_{X/B}$ is big. That last condition follows from 3.2 in Arakelov's paper as Vesselin already mentioned. (By the way, that result is for a stable model of $f$ which is not necessarily smooth, but everything works out at the end).

The advantage of this approach is that this works in higher dimensions as well.

• I only add that a stable isotrivial fibration of genus $g\ge 2$ is locally trivial. Therefore in the stable case my statement should be correct. – Dubious Mar 2 '15 at 20:57