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I have read in some lecture notes on Lie theory that any compact Lie algebra $\mathfrak{g}$ can be factored as a direct sum of of irreducible ideals for the $\mathrm{ad}$ representation. That is, $$\mathfrak{g}=\mathfrak{a}_1\oplus\dots\oplus\mathfrak{a}_k$$ where $\mathfrak{a}_i$ is a stable subspace of $\mathfrak{g}$ under $\mathrm{ad}$ and has no other stable subspaces except for $0$ and itself.

The author explains that this follows from the fact that any compact Lie group can be equipped (via averaging) with an $\mathrm{Ad}$ invariant inner product which implies that $\mathrm{ad}$ is skew-symmetric.

I can't see how these two facts are related, because the second one implies that $\mathrm{ad}$ is diagonalizable with imaginary eigenvalues over $\mathfrak{g}_\mathbb{C}$, and I wonder if the author has any way of combining those one dimensional complex eigenspaces in mind.

I consider the above factorization as a consequence of the fact that any ideal of a compact Lie algebra has an ideal complement, and thus by an inductive process one can factor $\mathfrak{g}$ into a direct sum of irreducible ideals.

Any thoughts on this are welcome.

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Don't worry about eigenvalues, just use the fact that if $W$ is a subspace of a space with an inner product and $W$ is invariant under a family of symmetric or skew symmetric operators then its complement is again invariant. Indeed, we have $$\langle Tx, y\rangle = \pm \langle x, Ty \rangle $$ and use $(W^{\perp})^{\perp}= W$

The ideals of $\mathfrak{g}$ are subspaces of $\mathfrak{g}$ invariant under $\text{ad}$. Assume that $\mathfrak{g}$ has an invariant inner product. Then the operators $\text{ad} X$ are skew-symmetric. Take an ideal $\mathfrak{a}$. The orthogonal complement is again an ideal of $\mathfrak{g}$ because it is an invariant subspace. Moreover, $$\mathfrak{a}\oplus \mathfrak{a}^{\perp} = \mathfrak{g}$$ (here we use that the inner product is definite ).

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