# Simple Lie algebras: making subspaces 'very transversal'

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?

• You can ask more generally for irreducible finite-dimensional representations of semisimple Lie algebras. Of course it's more general, but it at least yields more checkable test-cases (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 $\dim(V)\le\dim(G)/2$. – YCor Oct 11 at 21:56
• 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. – YCor Oct 12 at 9:06
• 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 4-dimensional 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 $e_{i-2}$. 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 $gV+V=W_3$. – YCor Oct 12 at 9:28
• 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 (7-dimensional) hyperplane of $sl_3$ with all basis elements (including the Cartan), except $e_{-w}$. Then $V+[g,V]\subset H$ for every $g\in sl_3$. – YCor Oct 12 at 9:48
• 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). – YCor Oct 12 at 14:59

This is not true. Namely, in the 8-dimensional $$\mathfrak{sl}_3$$, consider the 4-dimensional 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 "south-east"-triangular. Right-multiplying $$g$$ by a suitable element of $$[\mathfrak{b},\mathfrak{b}]$$, we can suppose that $$g$$ is an anti-diagonal matrix. Then we see that $$g$$ centralizes $$\mathfrak{d}\cap \mathfrak{v}$$, hence conjugates $$\mathfrak{d}\cap \mathfrak{v}$$ into itself and this contradicts $$\mathfrak{v}\cap g\mathfrak{v}g^{-1}=0$$.
• 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. – H A Helfgott Oct 15 at 13:57