Explicit Hermitian scalar product on a simple complex Lie algebra

Question

Every simple complex Lie algebra $\mathfrak g$ has a unique (up to isomorphism) compact real form $\mathfrak k$. That is, $\mathfrak k$ is a real Lie algebra with negative definite Killing form, and $\mathfrak k\otimes_{\mathbb R}\mathbb C$ is isomorphic to $\mathfrak g$.

If $\mathfrak k$ is the Lie algebra of a compact Lie group $K$, then the representation $\mathfrak g$ of $K$ has a $K$-invariant Hermitian scalar product given by $$ (x,y)=\int_K\left\langle gx,gy\right\rangle dg $$ where $\left\langle-,-\right\rangle$ is the Killing form of $\mathfrak g$ and $dg$ is the Haar measure on $K$.

Question 0: are the above statements correct?

Question 1: is there a way to express the Hermitian scalar product $(-,-)$ on $\mathfrak g$ purely in Lie algebra terms, without referring to any Lie groups or Haar integration?

Answer 1

Your statements are not quite correct, since the Killing form on $\mathfrak g$ is complex bilinear rather than Hermitian, so what you produce is not a Hermitian form. But you can run the averaging-argument starting with any positive definite Hermitian form on $\mathfrak g$, there are no invariance properties required at this stage. However, in the algebraic picture, things can be obtained in a much simpler way.

Since $\mathfrak k$ is a real form of $\mathfrak g$, you can write any element of $\mathfrak g$ as $X+iY$ for $X,Y\in\mathfrak k$. Now you can simply use the negative of the real Killing form $B$ on $\mathfrak k$ (which is negative definite since $\mathfrak k$ is compact and semisimple and $K$-invariant) and extend it to a Hermitian form on $\mathfrak g$. This means that you define $\langle X_1+iY_1,X_2+iY_2\rangle:=B(X_1,X_2)+B(Y_1,Y_2)+i(B(X_2,Y_1)-B(X_1,Y_2))$.

The standard description for this is via a Cartan-involution $\theta$ on $\mathfrak g$, which by definition has the property that evaluating the negative of the Killing form of $\mathfrak g$ on $X$ and $\theta(Y)$ defines a positive definite inner product. Then $\mathfrak k$ is obtained as the fixed point set of $\theta$.