Complexification of a Lie subalgebra of a compact real form

Question

I'm currently reading the paper Lie algebra Cohomology and the Generalized Borel–Weil theorem written by B. Kostant, and I have a question about Remark 3.9 he made.

In this paper, $\mathfrak{g}$ is a complex semisimple Lie algebra and $\mathfrak{t}$ is its compact real form. Using the decomposition $\mathfrak{g}=\mathfrak{t}\oplus i\mathfrak{t}$ the author defines the complex conjugate $\ast$ on $\mathfrak{g}$ by defining $(a+ib)^{\ast}=-a+ib$ for $a,b\in\mathfrak{t}$. (He uses the subspace $i\mathfrak{t}$ as a real part.) In Remark 3.9 of the paper, the author mentions that

In case $\mathfrak{a}$ is a Lie subalgebra and $\mathfrak{a}=\mathfrak{a}^{\ast}$ it follows that $\mathfrak{a}$ is necessarily reductive in $\mathfrak{g}$ since it clearly arises as the complexification of a Lie subalgebra of $\mathfrak{t}$.

First of all, it was not clear to me what does it mean by saying that $\mathfrak{a}$ is reductive ‘in’ $\mathfrak{g}$. Does it simply mean that $\mathfrak{a}$ is a reductive Lie algebra? Secondly, if it means that $\mathfrak{a}$ is a reductive Lie algebra, then I am curious why the fact that $\mathfrak{a}$ arises as a complexification of a Lie subalgebra of $\mathfrak{t}$ implies that $\mathfrak{a}$ is reductive.

Answer 1

Historically, I think the paper may simply be early enough that it wasn't clear whether the various notions of reductivity in characteristic $0$ were equivalent. “Reductive in $\mathfrak g$” could mean, for example, that the adjoint representation on $\mathfrak g$ was completely reducible.

A Lie subalgebra of $\mathfrak k$ is the Lie algebra of a compact group, so its complexification is the Lie algebra of the complexification of a compact group, which is reductive.

(By the way, the letter is a Fraktur k, $\mathfrak k$ \mathfrak k, not a Fraktur t, $\mathfrak t$ \mathfrak t (setting aside the usual issues of Fraktur vs. Sütterlin, for which you can read elsewhere on MO (e.g.)). This matters to the extent that it is very hard for a Lie theorist, or at least for me, to read $\mathfrak t$ as being anything other than the Lie algebra of a torus.)