# Conjugacy of elements in a parabolic subgroup

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 $$P$$.

• 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).
– YCor
May 24, 2020 at 2:23
• @YCor, that is certainly true for regular elements of $T$, but is it true for any two elements? May 24, 2020 at 2:31
• @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.
– YCor
May 24, 2020 at 2:41
• @YCor okay what if we also assume that g and h project to the same element in the reductive quotient of P. May 24, 2020 at 10:00
• @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?
– YCor
May 24, 2020 at 10:10

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.

• 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. May 24, 2020 at 15:05