Timeline for In how many ways does a Lie algebra decompose as an orthogonal direct sum of Cartans?

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Mar 19 at 15:22	history	edited	Theo Johnson-Freyd	CC BY-SA 4.0	Fixed the link
Mar 19 at 12:41	comment	added	David E Speyer		It turns out that $\mathbb{F}_p^{2m}$ is equipped with a natural symplectic form, and the natural condition is that each $L \cup \{ 0 \}$ is Lagrangian for this form. There are many different such partitions for $m>1$. This is Section 1.2 in the book, although I understand that you don't have access to this. (I have the whole book, but it's long!)
Mar 19 at 12:39	comment	added	David E Speyer		One caveat to "the answer is yes" is that Kostrikin and Tiep's construction of OD's in $A(p^m)$ is a bit more complicated. There is a copy of $\mathbb{Z}/(p \mathbb{Z})^{2m}$ in $PSU(p^m)$, for which $\mathfrak{su}(p^m)$ breaks into $p^{2m-1}$ one dimensional orthogonal eigenspaces indexed by the nonzero vectors in $\mathbb{F}_p^{2m}$. In order to group them into Cartans, one must partition the nonzero vectors of $\mathbb{F}_p^{2m}$ into sets $L$ of size $p^m-1$. (continued)
Mar 19 at 3:20	history	asked	Theo Johnson-Freyd	CC BY-SA 4.0