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The quaternionic unitary group $\mathrm{U}(n,\mathbb{H})$, also called the compact symplectic group $\mathrm{Sp}(n)$, consists of $n \times n$ quaternionic matrices $g$ such that $gg^\ast = 1$, where

$$ (g^\ast)_{ij} = \overline{g}_{ji}$$

and the overline denotes quaternionic conjugation.

My question: is every quaternionic unitary matrix conjugate to a diagonal one?

That is: given a quaternionic unitary matrix $g$, is there a quaternionic unitary matrix $h$ such that $h g h^{-1}$ is diagonal, meaning that $(h g h^{-1})_{ij} = 0$ when $i \ne j$?

I feel pretty sure this is true. If it is, what I really want is a reference to a proof, or a proof.

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1 Answer 1

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This follows from the general fact that, in a compact connected Lie group, every element is conjugate to an element in a maximal torus (and all maximal tori are conjugate). This result is proved in just about every book that treats compact Lie groups. For example, see Helgason's "Differential Geometry, Lie Groups, and Symmetric Spaces" or Bröcker and tom Dieck's "Representations of Compact Lie Groups".

The diagonal elements of $\mathrm{U}(n)\subset\mathrm{Sp}(n)$ form a maximal torus in $\mathrm{Sp}(n)$, so every element in $\mathrm{Sp}(n)$ is conjugate in $\mathrm{Sp}(n)$ to a diagonal unitary matrix.

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  • $\begingroup$ Thanks! The maximal torus idea occurred to me, but for some reason I hadn't thought $\mathrm{U}(n) \subset \mathrm{Sp}(n)$ was big enough to provide a maximal torus, even though I knew it for $n = 1$. $\endgroup$
    – John Baez
    Feb 11, 2021 at 18:36
  • $\begingroup$ @JohnBaez: If you are satisfied with my answer, you can go ahead and accept it so that it will stop coming up as 'open'. $\endgroup$ Feb 17, 2021 at 14:51
  • $\begingroup$ Okay. I still need to figure out why the diagonal elements of U(n) form a maximal torus of Sp(n), but it's probably easy now that I know to try. $\endgroup$
    – John Baez
    Feb 17, 2021 at 22:23
  • $\begingroup$ @JohnBaez: Well, the diagonal elements of $\mathrm{U}(n)$ are a connected abelian subgroup $\mathbb{T}^n$ and, if you look at the centralizer of this subgroup in $\mathrm{Sp}(n)$, you'll see easily that it is $\mathbb{T}^n$ itself, so it's maximal. Thus, it's a maximal abelian subgroup and hence is a maximal torus. $\endgroup$ Feb 17, 2021 at 23:52
  • $\begingroup$ Okay, that's nice and quick. I indeed easily see that. $\endgroup$
    – John Baez
    Feb 18, 2021 at 1:58

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