Let $G$ be a complex connected reductive group, and let $P \subseteq G$ be a parabolic subgroup. My question is the following: if $g$ and $h$ are elements of $P$ which are conjugate as elements of $G$, are they necessarily also conjugate in $P$?

Edit: In order to rule out the cases pointed out in the comments, I'm refining my question. Let $U \subset P$ be the unipotent radical, and $L = P/U$ the reductive quotient. Let $g$ and $h$ be elements of $P$, which project to the same element in $L$, and which are conjugate as elements of $G$. Are they then necessarily also conjugate in $P$?

Further Edit: If $g$ and $h$ are elements of $P$ which project to the same element of $L$, and if $g$ and $h$ both lie in some (but not necessarily the same) Levi subgroup, then they are conjugate in $P$. This is because all Levi subgroups are conjugate by an element of $U$, and so we can conjugate $g$ and $h$ into the same Levi (without changing their projection), and then they must be equal. This implies that without loss of generality, we can assume that $g$ and $h$ have the same semisimple component.

Now there are counter-examples if we do not require $h$ and $g$ to lie in some Levi. For example, let $G = GL(3,\mathbb{C})$, and $P$ to be the subgroup of upper triangular matrices. Then the following two elements are conjugate in $G$ but not in $P$, and both project to the same element of the Levi:

$\begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \ \ \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$.

To rule out this case, we should then also require that $h$ lies in some Levi. The question is then whether the conjugate element $g$ must also lie in some (possibly different) Levi. By the above discussion, this would imply that $h$ and $g$ are conjugate in $P$.

  • 4
    $\begingroup$ No: if $P$ is Borel and $T$ a maximal torus, two elements of $T$ are conjugate in $P$ iff they're equal. But could be conjugate in $G$ (Weyl group action). $\endgroup$
    – YCor
    May 24, 2020 at 2:23
  • $\begingroup$ @YCor, that is certainly true for regular elements of $T$, but is it true for any two elements? $\endgroup$
    – LSpice
    May 24, 2020 at 2:31
  • 5
    $\begingroup$ @LSpice in a Borel $B$, $T$ is a retract, so two elements of $T$ are conjugate in $B$ iff they're conjugate in $T$. Since $T$ is abelian, this means being equal. $\endgroup$
    – YCor
    May 24, 2020 at 2:41
  • $\begingroup$ @YCor okay what if we also assume that g and h project to the same element in the reductive quotient of P. $\endgroup$ May 24, 2020 at 10:00
  • $\begingroup$ @unknownymous it might be a reasonable; please edit your question accordingly? if you know some particular cases (e.g., when $g,h$ belong to the unipotent radical of $P$) could you mention it too? $\endgroup$
    – YCor
    May 24, 2020 at 10:10

1 Answer 1


I think your question has a negative answer.

For reductive $G$, there are only finitely many conjugacy classes of unipotent elements in $G$. However, for a parabolic subgroup $P < G$ with unipotent radical $U$, usually the action of $P$ on $U$ by conjugation has an infinite number of orbits.

Here is an explicit example from [1]. Let $G = \operatorname{GL}_6(K)$ with $K$ an infinite field and let $P$ be the parabolic (Borel) subgroup formed by the lower triangular matrices. (This also gives you examples for $G = \operatorname{GL}_n(K)$ for any $n > 6$). For $\alpha \in K$ define $$u_{\alpha} = \begin{pmatrix} 1 & & & & & \\ 1 & 1 & & & & \\ 0 & 0& 1 & & & \\ 0 & 1 & 1 & 1 & & \\ 0 & \alpha & 0 & 0 & 1 & \\ 0 & 0 & 0 & 1 & 1 & 1 \end{pmatrix}$$ A calculation shows that $u_{\alpha}$ and $u_{\beta}$ are $P$-conjugate if and only if $\alpha = \beta$. On the other hand if $\alpha, \beta \not\in \{ 0,-1 \}$, then $u_{\alpha}$ and $u_{\beta}$ are $G$-conjugate.

For the question of when a parabolic subgroup $P$ has finitely many orbits on $U$, see for example [2], [3], [4] below and references therein.

[1] Djoković, Dragimir Ž.; Malzan, J. Orbits of nilpotent matrices. Linear Algebra Appl. 32 (1980), 157–158.

[2] Hille, L.; Röhrle, G. A classification of parabolic subgroups of classical groups with a finite number of orbits on the unipotent radical. Transform. Groups 4 (1999), no. 1, 35–52.

[3] Jürgens, U.; Röhrle, G. MOP—algorithmic modality analysis for parabolic group actions. Experiment. Math. 11 (2002), no. 1, 57–67.

[4] Goodwin, S.; Röhrle, G. Finite orbit modules for parabolic subgroups of exceptional groups. Indag. Math. (N.S.) 15 (2004), no. 2, 189–207.

  • $\begingroup$ Very interesting example! I give another counter example in my question. But I've refined my question once again to rule out these counter examples. $\endgroup$ May 24, 2020 at 15:05

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