Diagonalizing quaternionic unitary matrices

Question

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.

Answer 1

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.