Does anyone know if the following holds?


Conjecture: Any two commuting elements in a reductive algebraic group G over C of rank>1 lie in a proper parabolic subgroup of G.


To make things easier, you can assume that these elements are semi-simple.


Note that if G is simply-connected then the centralizer of any semi-simple element is connected. This implies that if both are semi-simple then they lie in a torus and hence the conjecture holds.


On the other hand, the rank assumption is important since the diagonal orthogonal matrices D(-1,1,-1) and D(1,-1,-1) commute in SO(3,C) but one can prove that they don't belong to the same parabolic subgroup.



commuting elements in a reductive group
Consider images in $\mathrm{PGL}\_n$ of the matrices $A$ and $B$, where $A$ is the diagonal matrix whose $i^{\rm th}$ diagonal entry is $\omega^i$, where $\omega$ is a primitive $n^{\rm th}$ root of $1$, and $B$ is the permutation matrix associated with the cycle $(1\, 2\, \ldots\, n)$. These two elements are semisimple and commute, but are not contained in a common parabolic subgroup.


