I've been studying isometric quotients (by compact Lie groups) of compact simply connected homogeneous spaces $G/H$ and their inherited curvature. One of the issues that continually arises is the following problem:

Let $\mathfrak{g} = \mathfrak{h}\oplus\mathfrak{p}$ be an orthogonal decomposition with respect to a Ad G invariant inner product on $\mathfrak{g}$. When is there a $g\in G$ such that $Ad(g)\mathfrak{p}\cap \mathfrak{p} = \{0\}$?

Of course, if the dimension of $\mathfrak{p}$ is too large (equivalently, if the dimension of $\mathfrak{h}$ is too small), then they must intersect nontrivially for dimension reasons. This leads us to the general form of the question:

Let $G$ be a compact Lie group and let $V\subseteq \mathfrak{g} = Lie(G)$ be a vector subspace (but not neccesarily a subalgebra). Assume dim($V)\leq \frac{1}{2}$ dim($\mathfrak{g})$. Is there a $g\in G$ such that $Ad(g)V\cap V = \{0\}$?

Of course, if there is one $g$, an entire neighborhood around $g$ will have this property. But

when is there an open and dense set of $g$ such that $Ad(g)V\cap V=\{0\}$?

(If it helps, feel free to assume that $G$ is simple. I'm pretty sure (but haven't been able to prove) that one can reduce the question to simple $G$ anyway.)

Note that for $G=SU(2)$ or $G=SO(3)$, there is an open and dense set of such $g$. This follows because in this case, $Ad:G\rightarrow SO(\mathfrak{g})$ is surjective and, of course the usual $SO(n)$ action on $\mathbb{R}^n$ acts transitively on k-subspaces. Further, the set of all elements of $SO(3)$ that fix a given line in $\mathbb{R}^3$ is isomorphic to an $SO(2)$, which is codimension 2 in $SO(3)$.

However, for any other $G$, this trick doesn't work as $Ad$ will fail to be surjective.

Thank you in advance for your time and feel free to add more tags as appropriate!