Let $\mathfrak{g}$ be a (finite-dimensional, real) Lie algebra equipped with a nondegenerate symmetric bilinear form $\langle \cdot{,}\cdot \rangle$ that is ad-invariant, i.e., such that
\begin{equation*}
\langle [x,y],z\rangle = \langle x,[y,z]\rangle \quad \text{for all $x,y,z \in \mathfrak{g}$};
\end{equation*}
let $S$ be a nondegenerate subspace.
Question. Suppose that $S$ is closed under the Lie bracket, i.e., it is a Lie subalgebra. Does it then follow that its orthogonal complement $S^{\perp}$ (which is automatically nondegenerate) is also a Lie subalgebra?
I believe that, in general, the answer is no. However, it would be great if somebody could provide a counterexample.
Best Answer
Take $\mathfrak{g} = \mathbb{R}^3$ with $[x,y] = x \times y$ the cross product and as symmetric bilinear form $\langle \cdot, \cdot \rangle$ the usual scalar product. Then the space $S = \langle e_1 \rangle$ is closed under the Lie bracket and $S^\perp = \langle e_2, e_3 \rangle$ but $e_2 \times e_3 = e_1$.