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If $( A B | C D)$ is a symplectic matrix with entries in the finite field with two elements, is it necessarily the case that $\sum_{i,j,k} a_{ij}b_{ij}c_{ik}d_{ik} = 0$?

This arose in connection with some calculations involving theta functions. There seems to be some indication that it might be true, and I could not come up with a counterexample; but experts in the area may well know the answer right off.

Thanks for any help.

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  • $\begingroup$ I fixed the math display to look like the one you put into the duplicate post. I hope it is what you want. $\endgroup$
    – Willie Wong
    Commented Sep 30, 2010 at 20:03
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    $\begingroup$ Does $(AB|CD)$ mean $\begin{pmatrix} A & B \\ C & D \end{pmatrix}$? $\endgroup$
    – Steven Sam
    Commented Sep 30, 2010 at 22:19
  • $\begingroup$ I don't understand the question. I assume that $(AB|CD)=\begin{pmatrix}A&B\\C&D\end{pmatrix}$ is a block matrix over $\mathbb F_2$. Your notation suggests moreover that $A,B,C,D$ are all squares. "symplectic" presumably means that it fixes (infinitesimally) some particular symplectic form, which (if I had to guess) you are taking to be $(AB|CD)=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, where by "$1$" I mean the identity square matrix of whatever size you're working with. If this is all correct, fine, but you should edit the question to make precise the notation. $\endgroup$
    – Theo Johnson-Freyd
    Commented Oct 1, 2010 at 3:45
  • $\begingroup$ (and, I guess, since you are over $\mathbb F_2$, $1=-1$ is fine in the matrix.) $\endgroup$
    – Theo Johnson-Freyd
    Commented Oct 1, 2010 at 3:46
  • $\begingroup$ (and maybe, looking at wikipedia, you don't want it to fix the form infinitesimally, but rather honestly). $\endgroup$
    – Theo Johnson-Freyd
    Commented Oct 1, 2010 at 3:50

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