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 $h(B) = h(A) + h(C)$?



Faltings height in short exact sequences
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](http://library.msri.org/books/Book49/files/09yang.pdf) by Tonghai Yang.


