Let $\bf g$ be a finite-dimensional real simple Lie algebra of compact type and let $\left<-,-\right>$ denote the positive-definite inner product induced from the negative of the Killing form. Let $\Omega$ denote the trilinear map defined by
$$\Omega(X,Y,Z) = \left<[X,Y],Z\right> .$$
It is easy to see that it is alternating, because of the ad-invariance of the Killing form. Let us call a subspace $S\subset{\bf g}$ **isotropic** if $\Omega$ vanishes identically when restricted to $S$; that is, if
$$\Omega(X,Y,Z) = 0, \forall X,Y,Z \in S.$$

In other words, $S$ is isotropic iff $[S,S] \subset S^\perp$, where ${}^\perp$ means the perpendicular complement relative to the Killing form.

Furthermore we say that an isotropic subspace is **maximal** if it is not properly contained in an isotropic subspace. It is not hard to show that $S$ is maximal isotropic if and only if $[S,S] = S^\perp$.

The question is how to characterise the maximal isotropic subspaces of $\bf g$.

It is easy to see that the maximally isotropic *subalgebras* are precisely the Cartan subalgebras, but I am interested in subspaces which are not necessarily subalgebras.

The only examples I know are those for which $S = {\bf k}^\perp$ and ${\bf k} < {\bf g}$ a subalgebra, whence $${\bf g} = {\bf k} \oplus S$$ is a symmetric decomposition corresponding to the compact riemannian symmetric space $G/K$.

Question: *Are there any other maximal isotropic subspaces?*

subalgebras, but not maximally isotropic subspaces. In other words, they are properly contained in a maximally isotropic subspace which is not a subalgebra. Does this make sense? – José Figueroa-O'Farrill Nov 4 '09 at 13:41