Let $G$ be a Lie group or group of Lie type whose Lie algebra $\mathfrak{g}$ is simple. Because the Lie algebra is simple, for any proper subspace $V\subset \mathfrak{g}$, there is a $g\in G$ such that $g V g^{1} \ne V$. Is it the case that there is a $g\in G$ such that $V$ and $g V g^{1}$ are "as transversal as possible", meaning that $\dim(V + g V g^{1}) = \min(2 \dim(V), \dim(G))$? Is this the case, at least, for $G$ a classical group?

2$\begingroup$ You can ask more generally for irreducible finitedimensional representations of semisimple Lie algebras. Of course it's more general, but it at least yields more checkable testcases (in your setting the case of $sl_2$ is trivial, and the case of $sl_3$ already seems hard). Note: if we restrict to the case of representation with a nonzero invariant bilinear form (as in the adjoint representation case), we can restrict to $\dim(V)\le\dim(G)/2$. $\endgroup$ – YCor Oct 11 '18 at 21:56

1$\begingroup$ Working more, I can handle the case of $sl_3$ with the exception of a subspace $V$ of dimension 4 on which the determinant vanishes. $\endgroup$ – YCor Oct 12 '18 at 9:06

2$\begingroup$ Actually I tried this approach in the case of representations and it does not work. Namely, the infinitesimal condition you require is much stronger. More precisely, consider the 4dimensional irrep $W_3$ of $sl_2$. It has a basis $(e_3,e_1,e_{1},e_{3})$ with $he_i=ie_i$, $xe_i$ proportional to $e_{i+2}$, $ye_i$ proportional to $e_{i2}$. Let $V$ be the subspace with basis $(e_3,e_1)$. Then $gV+ V$ has dimension $\le 3$ for every $g$ in the Lie algebra, since it's contained in the subspace with basis $(e_3,e_1,e_{1})$. While there exists $g\in SL_2$ such that $gV+V=W_3$. $\endgroup$ – YCor Oct 12 '18 at 9:28

1$\begingroup$ I looked again: the infinitesimal version is false for the adjoint representation of $sl_3$. Fix a choice of positive roots, and let $u,v,w=u+v$ be the positive roots. Choose $h$ in the Cartan subalgebra such that $w(h)=0$. Let $V$ be the subspace with basis $(h,e_u,e_v,e_w)$ (it is a solvable subalgebra). Let $H$ be the (7dimensional) hyperplane of $sl_3$ with all basis elements (including the Cartan), except $e_{w}$. Then $V+[g,V]\subset H$ for every $g\in sl_3$. $\endgroup$ – YCor Oct 12 '18 at 9:48

1$\begingroup$ My answer below contradicts my previous claim (that a counterexample $V$ should have determinant identically zero); I've detected the error in my draft of proof... sorry for the mess. I now think the only two potential counterex. have $\dim(V)=4$, and either (a) $V$ has vanishing det, and does not only consist of nilpotent matrices (b) $V$ has a codim $\le 1$ subset of matrices that are conjugate to a nonzero scalar multiple of the diagonal matrix $(1,1,2)$. In both cases I have no classification of such subspaces $V$ (there are many of type (a)). The counterexample below has type (b). $\endgroup$ – YCor Oct 12 '18 at 14:59
This is not true. Namely, in the 8dimensional $\mathfrak{sl}_3$, consider the 4dimensional Lie subalgebra $\mathfrak{v}$ consisting of matrices of the form $$\begin{pmatrix} a & x & z\\ 0 & 2a & y\\ 0 & 0 & a\end{pmatrix}.$$ (This is the centralizer of $E_{13}$ in $\mathfrak{sl}_3$.) I claim that $g\mathfrak{v}g^{1}\cap \mathfrak{v}\neq 0$ for every $g\in \mathrm{GL}_3$.
Indeed, let $\mathfrak{b}$ be the normalizer of $\mathfrak{v}$, namely the Lie subalgebra of upper triangular matrices of trace zero. Let $\mathfrak{d}$ be the subalgebra of diagonal matrices: it is a Cartan subalgebra in both $\mathfrak{b}$ and $\mathfrak{sl}_3$. It is known that the intersection of any two Borel subalgebras contains a Cartan subalgebra $\mathfrak{d}'$. This applies to the intersection $\mathfrak{b}\cap g\mathfrak{b}g^{1}$. Conjugating by some upper triangular matrix conjugating $\mathfrak{d}'$ into $\mathfrak{d}$, we can suppose that $\mathfrak{d}'=\mathfrak{d}$. So $g\mathfrak{b}g^{1}$ is one of the 6 Borel subalgebras containing $\mathfrak{d}$ (images of $\mathfrak{b}$ by the Weyl group $\mathfrak{S}_3$). The condition $[\mathfrak{b},\mathfrak{b}]\cap g[\mathfrak{b},\mathfrak{b}]g^{1}=0$ forces $g\mathfrak{b}g^{1}$ to be the opposite Borel subalgebra $\mathfrak{b}_$, that is, the Lie subalgebra of lower triangular matrices. So $g$ maps the unique flag preserved by $\mathfrak{b}$ to the unique flag preserved by $\mathfrak{b}_$. So $g$ is "southeast"triangular. Rightmultiplying $g$ by a suitable element of $[\mathfrak{b},\mathfrak{b}]$, we can suppose that $g$ is an antidiagonal matrix. Then we see that $g$ centralizes $\mathfrak{d}\cap \mathfrak{v}$, hence conjugates $\mathfrak{d}\cap \mathfrak{v}$ into itself and this contradicts $\mathfrak{v}\cap g\mathfrak{v}g^{1}=0$.

$\begingroup$ Nice. The idea here can also be used to give an example of a variety $W\subset G=\SL_3(K)$ of dimension $\dim(G)/2=4$ such that $\dim(WgW)=7<\dim(G)$ for $g$ generic. $\endgroup$ – H A Helfgott Oct 15 '18 at 13:57