Diagonalizing elements of compact lie groups

Question

Chapter 5 of Sepanski's Compact Lie Groups starts with this paragraph:

"Since a compact Lie group $G$ can be thought of as a Lie subgroup of $U(n)$, it is possible to diagonalize each $g\in G$ using conjugation in $U(n)$.

In fact, the main theorem of this chapter shows that it is possible to diagonalize each $g\in G$ using conjugation in $G$."

The first part of this paragraph, I believe, is due to the fact that $G$ has a faithful unitary representation, and also the spectral theorem which guarantees unitary diagonalizability of unitary matrices.

The second part of this paragraph is something that I have a little difficulty following. Consider for instance the compact abelian group $SO(2)$ as sitting inside $U(2)$. Then according to the above statement, every element of $SO(2)$, that is every rotation, can be diagonalized by a conjugation by a rotation!

Is the above quotation correct, or perhaps I am misinterpreting it?

Answer 1

"Diagonalize" has to be interpreted appropriately: in particular, it's not clear a priori what the "diagonal" subgroup of an arbitrary compact (connected) Lie group $G$ is. There turns out to be a very good answer to this: $G$ has a maximal torus $T$, and every element of $G$ is conjugate, in $G$, to an element of $T$.

When $G = U(n)$, the maximal torus is $T = U(1)^n$, the diagonal matrices, but in general it can be more complicated. For example, when $G = SO(n)$, the diagonal matrices do not form a maximal torus inside $G$ (since, as you correctly observe, it's just not true that you can diagonalize every element of $SO(n)$ in this sense). Instead, for a maximal torus you can choose $SO(2)^{\lfloor \frac{n}{2} \rfloor}$ embedded block-diagonally.

In particular, $SO(2)$ is its own maximal torus.