If $\mathfrak{g}$ is a semisimple $\Rightarrow$ $\mathfrak{h} \subset \mathfrak{g} $ imply $\mathfrak{h} \cap \mathfrak{h}^\perp = \{0\}$

Question

Let $\mathfrak{g}$ be a semisimple Lie Algebra, $\langle \cdot,\cdot\rangle$ the Killing form, and $\mathfrak{h}\subset \mathfrak{g}$ a subalgebra of $\mathfrak{g}$.

I'm trying to find a counterexample (or prove) the following affirmation:

Affirmation: $\mathfrak{h} \cap \mathfrak{h}^\perp = \{0\}$, where $\mathfrak{h}^\perp = \{X \in \mathfrak{g};\ \langle X, Y\rangle =0, \ \forall \ Y\in \mathfrak{h} \}$.

Does anyone know how to prove this or have a nice counterexample?

NB: In my opinion this affirmation seems false, however all spaces that I tried $\mathfrak{h} \cap \mathfrak{h}^\perp = \{0\}$ held.

Answer 1

Take any semi-simple algebra $g$ which is not compact, the killing form is not a scalar product, there exists $x\in g$ such that $\langle x,x\rangle =0$, $Vect(x)$ is a subalgebra of $g$ and $x$ is in the orthogonal of $Vect(x)$.

Answer 2

A good example for this is parabolic subalgebras. Almost by definition they contain their Killing polars. That is let $\mathfrak{g}$ be a non-compact semisimple Lie algebra and let $\mathfrak{p} \leq \mathfrak{g}$ be a parabolic subalgebra (containing a Borel subalgebra) then $\mathfrak{p}^\perp \subset \mathfrak{p}$. For a concrete example of a parabolic subalgebra just think of your favourite matrix Lie algebra and the subalgebra of upper triangular matrices.