Timeline for A lie Subgroup of SO(4n)

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Jan 2, 2013 at 16:48	comment	added	Robert Bryant		(correction) Of course, I meant to type $\mathrm{Sp}(m)\times\mathrm{Sp}(n)$ in my comment above instead of $\mathrm{Sp}(n)\times\mathrm{Sp}(n)$.
Jan 2, 2013 at 13:00	comment	added	Robert Bryant		More precisely, it's the 'quaternionic tensor product'. In the usual representation, one has $\mathrm{Sp}(n)\subset\mathrm{SO}(4n)$, so the naïve tensor product would map $\mathrm{Sp}(n)\times\mathrm{Sp}(n)$ into $\mathrm{SO}(4m\cdot 4n)=\mathrm{SO}(16mn)$, which is not what you want. The point is that the commuting ring of $\mathrm{Sp}(n)\subset\mathrm{SO}(4n)$ acting on $\mathbb{R}^{4n}$ is the quaternions $\mathbb{H}$, and one can use this to construct an invariant subspace of dimension $4mn$ in the tensor product $\mathbb{R}^{4m}\otimes \mathbb{R}^{4n}\simeq \mathbb{R}^{16mn}$.
Jan 2, 2013 at 9:23	history	answered	Bruce Westbury	CC BY-SA 3.0