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Let $(T,E)$ be a polarized abelian variety ($T=V/L$, $\dim_\mathbb{C} V=g$, $E:V\times V\to\mathbb{R}$ a nondegenerate real alternating bilinear form, with $E(L\times L)\subseteq\mathbb{Z}$ and $E(iv,iw)=E(v,w)$). I'm trying to prove the existence of a basis for $L$ such that the matrix of $E$ has the form $$E=\left(\begin{array}{cc}0&D\\-D&0\end{array}\right),$$ where $D$ is a diagonal matrix with diagonal $d_1,\ldots,d_g$, and $d_i\mid d_{i+1}$.

Now what I've tried so far is considering $\min\{E(v,w):v,w\in L,E(v,w)>0\}$, and taking $e_1,e_{g+1}$ that achieve this minimum. Then I would like to decompose $V$ into $V=\langle e_1,e_{g+1}\rangle\oplus\langle e_1,e_{g+1}\rangle^\perp$ (this can be done since $E$ is nondegenerate). I would like to repeat this same process in $\langle e_1,e_{g+1}\rangle^\perp$, but I have no way of knowing "how many" lattice points are in $\langle e_1,e_{g+1}\rangle^\perp$. Am I attacking this problem the wrong way, or is there something I'm missing?

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See mathoverflow.net/questions/5108/… –  David Speyer Sep 6 '11 at 18:45
    
I looked at it, but I still have some questions. How is it possible to find $e$ and $f$ such that the lattice spanned by $e$ and $f$ is the intersection of the whole lattice with the vector space spanned by $e$ and $f$? It also says to complete the basis $\{e,f\}$ to a basis for the whole lattice, but I'm not sure you can always complete a basis for a lattice... –  Robert Auffarth Sep 6 '11 at 18:51
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In what form are you given your lattice? For most presentations, this should be standard linear algebra. As far as the last point goes, if $L \cap \mathrm{Span}_{\mathbb{R}}(e,f) = \mathrm{Span}_{\mathbb{Z}}(e,f)$, then $L / \langle e,f \rangle$ is torsion free, so it is free of rank $\dim L-2$. Take any basis for this free module and lift it to $L$; the lifted vectors, together with $e$ and $f$, are a basis for $L$. –  David Speyer Sep 6 '11 at 19:09
    
Thanks, you made it a lot clearer. I'm able to complete the proof now. –  Robert Auffarth Sep 6 '11 at 20:23
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