Proposition 1

$a)$ Let $\mathfrak{g}$ be a real form of $\mathfrak{a}$ and $\sigma$ the conjugation of $\mathfrak{a}$ relative to $\mathfrak{g}$. Then:

$\sigma^2=$ Id$_{\mathfrak{a}}, \sigma (\lambda x+\mu y) = \overline{\lambda \sigma}(x) + \overline{\mu\sigma}(y),[\sigma (x), \sigma (y)] =\sigma [x, y]$ (2)

for $\lambda , \mu\in \mathbf{C},x, y\in \mathfrak{a}$. An element $x$ of $\mathfrak{a}$ belongs to $\mathfrak{g}$ if and only if $\sigma (x) =x$.

b) Let $\sigma :\mathfrak{a}\rightarrow \mathfrak{a}$ be a map satisfying (2). Then the set $\mathfrak{g}$ of fixed points of $\sigma$ is a real form of $\mathfrak{a}$, and $\sigma$ is the conjugation of $\mathfrak{a}$ relative to $\mathfrak{g}$.

The proof is immediate.

Note that if B denotes the Killing form of $\mathfrak{a}$, and if $\mathfrak{g}$ is a real form of $\mathfrak{a}$, the restriction of B to $\mathfrak{g}$ is the Killing form of $\mathfrak{g}$; in particular, B is real-valued on $\mathfrak{g}\times \mathfrak{g}$. Assume that $\mathfrak{a}$ is reductive; then the real Lie algebra $\mathfrak{g}$ is compact if and only if the restriction of B to $\mathfrak{g}$ is negative (§1, no. 3). In that case we say that $\mathfrak{g}$ is a compact real form of $\mathfrak{a}$.