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An endomorphism of a vector space is said to be semisimple if every invariant subspace admits an invariant complement.

If $\mathfrak g$ is a complex semisimple Lie algebra that is the complexification of a compact Lie algebra $\mathfrak u$, why are the endomorphisms $\mathrm {ad}_X : \mathfrak g \rightarrow \mathfrak g$ semisimple for $X\in \mathfrak u$ but not necessarily for $X \in \mathfrak g$?

Callum
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Rodrigo
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1 Answers1

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This won't be a rigorous proof, but an intuitive answer (hopefully you can fill in the missing details if you wish). Any finite-dimensional complex semisimple $\mathfrak{g}$ can be embedded in $ \mathfrak{gl} (V) $ for some finite-dimensional vector space $V$; and we can introduce an inner product on $V$ such that the image of $ \mathfrak{u} $ w.r.t. this embedding is a subalgebra of the skew-Hermitian matrices $ \mathfrak{u} (V) $. Now, for an element $X \in \mathfrak{gl} (V)$ we have that $ \mathrm{ad}_X $ is semisimple iff $X$ is diagonalizable as a matrix. While not every matrix is diagonalizable, it is true that skew-Hermitian matrices are always diagonalizable.

I am fairly sure one can prove this without using the embedding into $ \mathfrak{gl} (V) $ (probably by directly using the construction of the compact real form), but I think my way intuitively targets the "why" part better.

smitke6
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    This can be easily made into a rigourous answer by considering that the map $\mathrm{ad}:\mathfrak{g}\to\mathfrak{gl}(\mathfrak{g})$ (so we don't need to embed in some arbitrary $\mathfrak{gl}(V)$, we can use $\mathfrak{gl}(\mathfrak{g})$) sends each $X$ to a skew-symmetric endomorphism for the Killing form on $\mathfrak{g}$. If $\mathfrak{g}$ is compact then the negative of the Killing form defines an inner product. Then $\mathrm{ad}_X$ is skew-symmetric for this inner product and thus diagonalisable over $\mathbb{C}$. – Callum May 01 '24 at 11:32