Let $K$ be a number field and $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ a short exact sequence of abelian varieties over $K$. Let $h(A)$ denote the logarithmic Faltings height (normalized so that it is invariant upon base change to any finite extension $K'/K$; thus, due to this normalization, one may assume that $A$, $B$, and $C$ have semiabelian reduction everywhere). Is $h(B) = h(A) + h(C)$?
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$\begingroup$ Does you know already if this equality holds when the short exact sequence splits? $\endgroup$– Daniel LoughranCommented Nov 1, 2014 at 20:59
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$\begingroup$ I think so: this boils down to additivity under tensor product of Arakelov height of metrized line bundles. $\endgroup$– Question MarkCommented Nov 1, 2014 at 21:33
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$\begingroup$ The following is very naive: Does not every short exact sequence of abelian varieties split after passing to an isogeny? So can you not perform an isogeny to reduce to the split case, then use the formula for how the Faltings height changes under isogeny? I seem to recall that the Faltings height is preserved under isogeny, unless the degree of the isogeny is divisible by a given collection of bad primes. $\endgroup$– Daniel LoughranCommented Nov 2, 2014 at 6:51
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$\begingroup$ For splitting up to an isogeny one doesn't even need to pass to a field extension, but I am not sure how to proceed afterwards: in Faltings' Mordell paper there is an explicit formula for how isogeny affects $h(B)$, but I can't see how to use it to settle the question. If you could explain this in detail, that would be very welcome. $\endgroup$– Question MarkCommented Nov 2, 2014 at 16:28
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$\begingroup$ Well as I said (or at least tried to say), it was my first naive idea on how to approach the problem. If you have already considered this approach, then you probably understand better than me what is going on. Do you believe that the stated equality holds? How hard have you tried to construct counter-examples? $\endgroup$– Daniel LoughranCommented Nov 2, 2014 at 17:21
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3 Answers
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I think the following should give a counterexample. Let $\mathcal{O}$ be an order in an imaginary quadratic field $K$ and $\mathcal{O}_K$, the ring of integers. Then it's not too hard to find a (non-split) short exact sequence of $\mathcal{O}$-modules: $$0 \to \mathcal{O}_K \to \mathcal{O} \oplus \mathcal{O} \to \mathcal{O}_K \to 0,$$
e.g. if $1, \omega$ is a basis of $\mathcal{O}_K$, with $\omega^2 \in \mathbb{Z}$, then send $(a,b)$ to $\omega a - b$. If $A$, $B$, and $C$ are the abelian varieties (over $\mathbb{C}$) corresponding to these lattices (so $A = \mathbb{C}/\mathcal{O}_K$, etc.), then $$0 \to A \to B \to C \to 0.$$
Indeed, the maps on lattices induce $\mathbb{C}$-linear maps on complex vector spaces which preserve the lattices, so you get maps $A \to B \to C$. $B \to C$ is clearly surjective, and $A \to B$ is injective because $\mathcal{O}_K$ (the cokernel of the map of lattices) is torsion-free. Exactness in the middle you can check by hand.
If the Faltings height is additive then this exact sequence of abelian varieties gives that $h(\mathbb{C}/\mathcal{O}) = h(\mathbb{C}/\mathcal{O}_K)$, where I really mean to take the heights of models over a number field. But in general $h(\mathbb{C}/\mathcal{O}) \neq h(\mathbb{C}/\mathcal{O}_K)$, as can be seen from the formulas on pages 273-274 of this paper by Tonghai Yang.
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1$\begingroup$ In the short exact sequence, how is the injection $\mathcal{O}_K \hookrightarrow \mathcal{O} \oplus \mathcal{O}$ defined? $\endgroup$ Commented Nov 4, 2014 at 6:43
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$\begingroup$ If $\mathcal{O} = \mathbb{Z} + f \mathcal{O}_K$, then this map sends $\alpha$ to $(f\alpha, f\omega\alpha)$. $\endgroup$ Commented Nov 4, 2014 at 14:03
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$\begingroup$ Then why is your sequence exact at $\mathcal{O} \oplus \mathcal{O}$? For instance, if it happens that $\omega \in \mathcal{O}$, why is $(1, \omega)$ in the image of $\mathcal{O}_K$? $\endgroup$ Commented Nov 4, 2014 at 21:27
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$\begingroup$ If $\omega \in \mathcal{O}$, then $f$ must be 1, and $(1,\omega)$ is the image of $\alpha = 1$. $\endgroup$ Commented Nov 5, 2014 at 2:49
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$\begingroup$ Thanks, I should've figured out that one. Could you clarify though how you get the short exact sequence of abelian varieties from the one of lattices? I presume, e.g., $A = \mathbb{C}/\mathcal{O}_K$, but then it's not clear to me how the first exact sequence leads to the second one. $\endgroup$ Commented Nov 5, 2014 at 5:18
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Proposition 3.3 of Ullmo's paper "Hauteur de Faltings de quotients de J_0(N) " (American Journal of Math., 2000) seems to answer your question.
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4$\begingroup$ When $K=\mathbf Q$ and $B$ has semi-stable reduction, Ullmo shows that the difference $h(A)+h(C)-h(B)$ is of the form $n\log(2)$, for some integer $n\in[0,\dim(C)]$. It's hard to have something much better in general. Indeed, the immersion $A\to B$ does not extend to an immersion of the Neron models in general, there is some finite kernel (which is controlled by a theorem of Raynaud, though). This happens at places $v$ of $K$ such that $e_v\geq p_v-1$; when $K=\mathbf Q$, the only problematic prime is $2$. $\endgroup$– ACLCommented Nov 6, 2014 at 21:07
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I am sorry for answering this question nine years later, but I would like to add that a recent theorem of Rémond shows that the Faltings height is sub-additive in short exact sequences. In other words, $h(A) + h(C) \leq h(B)$ for every short exact sequence of abelian varieties $0 \to A \to B \to C \to 0$. This implies in particular that every isogeny class of abelian varieties contains a member of minimal Faltings height and maximal endomorphism ring.
The paper of Rémond containing the aforementioned results is published here, and its preprint version can be found here.
