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The question is here.

For convenience:

Let $S=M_n(\mathbb{Z}_2)$. Let $D^{(n)}_2$ be the matrices in the rank $2$ $\mathcal{D}$-class of $S$. Find

$$N_n=\left\lvert\left\{\begin{array} \, e & \mathcal{L} & f \\ \mathcal{R} & \, & \mathcal{R} \\ h & \mathcal{L} & g \end{array}\mid e, f, g, h\in E\left(D^{(n)}_2\right)\right\}\right\rvert;$$ that is, find $N_n$, the number of quadruples $(e, f, g,h)\in E\left(D_2^{(n)}\right)^4$ such that $e\mathcal{L}f\mathcal{R}g\mathcal{L}h\mathcal{R}e$.

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    $\begingroup$ As written, no. If you define terminology and instead ask how to find it, it may become suitable. Gerhard "Not Following External Links Presently" Paseman, 2017.04.04. $\endgroup$
    – Gerhard Paseman
    Apr 4, 2017 at 21:01
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    $\begingroup$ This question is completely understandable to experts (although the MSE question gives the definitoins). This can be reformulated entirely in terms of linear algebra, which would make it easier to get an answer. If I am not mistaken, you want the number of $2$-dimensional subspaces $U,U'$ of $\mathbb Z_2^n$ and $n-2$-dimensional subspaces $V,V'$ of $\mathbb Z_2^n$ such that $U\cap V=U\cap V'=U'\cap V=U'\cap V'=0$. $\endgroup$
    – Benjamin Steinberg
    Apr 7, 2017 at 18:46

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