Let $G$ be a reductive group (we can work on an algebraically closed field if needed) and let $L$ be a parabolic subgroup, i.e. the centralizer of a certain torus $T \subseteq G$.

Associated with this situation is the normalizer subgroup $N=N_G(L)$ of the Levi subgroup, which acts on $L$ by conjugation. Fix a finite subset $X \in L$. We will suppose that $L$ is minimal for $X$ in the sense that there exists no proper Levi $L' \subsetneq L$ containing $S$.

Now, suppose that we take $g \in G$ satisfying that $gxg^{-1} \in L$ for all $x \in X$.

Question: Does there exist an element $n \in N$ such that $nxn^{-1} = gxg^{-1}$ for all $x \in X$?

Example: Take $G = \text{GL}_4$ and consider the torus consisting of matrices of the form

\begin{pmatrix}\lambda_1 & 0 & 0 & 0 \\ 0 & \lambda_2 & 0 & 0 \\ 0 & 0 & \lambda_3 & 0 \\ 0 & 0 & 0 & \lambda_3 \end{pmatrix}

whose associated Levi subgroup is made of matrices of the form

$$ \begin{pmatrix}\star & 0 & 0 & 0 \\ 0 & \star & 0 & 0 \\ 0 & 0 & \star & \star \\ 0 & 0 & \star & \star \end{pmatrix}. $$

In particular, we can take $$ x = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & a_1 & a_2 \\ 0 & 0 & a_3 & a_4 \end{pmatrix} \in L, \qquad g = \begin{pmatrix} g_1 & g_2 & 0 & 0 \\ g_3 & g_4 & 0 & 0 \\ 0 & 0 & h_1 & h_2 \\ 0 & 0 & h_3 & h_4 \end{pmatrix} \in G. $$ We have that $gxg^{-1} \in L$ but in general $g \not\in N$ (it does not normalize elements with different eigenvalues in the first two columns). However, we can take $$ n = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & h_1 & h_2 \\ 0 & 0 & h_3 & h_4 \end{pmatrix} $$ that belongs to $N$ (actually, it's an element of $L$) and satisfies that $nxn^{-1} = gxg^{-1}$.


  • I think it is not hard to show that this fact is true in $\text{GL}_n$ (and probably in $\text{SL}_n$ and $\text{PGL}_n$ too). Sketch: $N = N_G(L)$ is a semidirect product of $L$ and a certain subgroup $W_L$ of the Weyl group $W = S_n$. Furthermore, without loss of generality, we can suppose that $L$ is a standard Levi subgroup, so $L$ is a product of general linear groups of lower order (the blocks of the matrix). Then, $gxg^{-1}$ must have the same block structure as $x$. Even though $g$ might not be an element of $L$ if some block of $x$ lies in the center, the same effect can be obtained by applying an 'intra-block conjugation' (an element of $L$) followed by an 'inter-block conjugation' (an element of $W_L$).

  • A naive attempt to prove it would be to consider $Y = \{g \in G \mid gXg^{-1} \subseteq L\}$ and analyze the action of $N$ on $Y$ by left multiplication. Then, the question reduces to show that on each $N$-orbit there is an element fixing $X$. However, I don't know how to prove this fact either.

  • As @LSpice pointed out in a comment, the hypothesis that $L$ is minimal is needed to guarantee that you cannot mix different blocks.

Extra: Is this true for general subgroups? More precisely, given $H < G$ and $x \in H$, is it true that $G \cdot x \cap H = N_G(H) \cdot x$? Here, $K \cdot x$ denotes the $K$-orbit of $x$ by conjugation, and $N_G(H)$ is the normalizer of $H$. I don't think this is true, but I couldn't find a simple counterexample.

Edit: New hypotheses added to consider only minimal Levi subgroups.

  • 5
    $\begingroup$ For a random element $x$, no. You can conjugate $\operatorname{diag}(2, 1, 1) \in L \mathrel{:=} \operatorname{GL}_1 \times \operatorname{GL}_2$ in $\operatorname{GL}_3$ to $\operatorname{diag}(1, 2, 1)$, but not in a way that normalises $L$. $\endgroup$
    – LSpice
    Apr 24 at 2:02
  • $\begingroup$ Absolutely! You need to suppose that $L$ is the minimal Levi subgroup (question edited to add this hypothesis). In this case, the minimal Levi subgroup containing $\text{diag}(2,1,1)$ is the standard maximal torus, and there is an element of the Weyl group conjugating $\text{diag}(1,2,1)$ to $\text{diag}(2,1,1)$. $\endgroup$
    – a_g
    Apr 24 at 16:57
  • $\begingroup$ If you do not require that the field be algebraically closed, then you can take non-conjugate, elliptic elements $g_1$, $g_2$, and $g_3$ of $\operatorname{SL}_2$, regard $(g_1, g_3), (g_2, g_3) \in \operatorname{SL}_2 \times \operatorname{SL}_2$ as elliptic elements of $\operatorname{Sp}_4$, put $G = \operatorname{Sp}_8$ and $L = \operatorname{GL}_2 \times \operatorname{Sp}_4$, and then conjugate $\operatorname{diag}(g_1, (g_2, g_3)) \in L$ to $\operatorname{diag}(g_2, (g_1, g_3)) \in L$ in $G$. Does this count as an answer, or do you want to require that the field is algebraically closed? $\endgroup$
    – LSpice
    Apr 24 at 17:24
  • $\begingroup$ Good point! But yes, I want to consider the case when the underlying field is algebraically closed (I suspected that this assumption was needed, but I wasn't sure). $\endgroup$
    – a_g
    Apr 24 at 22:27
  • $\begingroup$ Aren't minimal Levis of a reductive group over an algebraically closed field just maximal tori? $\endgroup$ Apr 24 at 23:28

1 Answer 1


If one assumes the ground field $k$ is algebraically closed (this is unnecessary; see the edit), the answer to the main question seems to be yes. More generally, let $H$ be a closed $k$-subgroup of $G$, let $L$ be a Levi of $G$ containing $H$ which is minimal with this property, and let $g \in G(k)$ be such that $gHg^{-1} \subset L$. Then I claim that there exists $n \in N_G(L)(k)$ such that the maps $h \mapsto ghg^{-1}$ and $h \mapsto nhn^{-1}$ coincide. This recovers the answer to your question by taking $H$ to be the Zariski closure of the subgroup of $G(k)$ generated by $X$, but it is a bit stronger in positive characteristic: for instance, we don't even assume that $H$ is smooth.

Let $S$ be the maximal central torus of $L$, so $S$ is contained in the centralizer $Z_G(H)$. By minimality of $L$, note that $S$ is a maximal torus of $Z_G(H)$. (In fact, all such $L$ arise as centralizers of maximal tori of $Z_G(H)$.) Moreover, since $gHg^{-1} \subset L$, we have $S \subset gZ_G(H)g^{-1}$, or in other words $g^{-1}Sg \subset Z_G(H)$. By conjugacy of maximal tori, it follows that there is some $x \in Z_G(H)(k)$ such that $x^{-1}g^{-1}Sgx = S$. In particular, since $L = Z_G(S)$, the element $n := gx$ lies in $N_G(L)(k)$, and it satisfies the desired property since conjugation by $x$ has no effect on $H$.

The answer to the extra question is no. Let $G = \mathrm{GL}_2$, and let $B$ be the upper-triangular Borel, so $N_G(B) = B$. Let $x = \mathrm{diag}(a, b)$ for $a \neq b$, and note that $B \cdot x$ consists of those upper-triangular matrices with diagonal entries $(a, b)$ (in that order), while $G \cdot x \cap B$ consists of those upper-triangular matrices with diagonal entries either $(a, b)$ or $(b, a)$.

EDIT: As LSpice pointed out in a comment, this argument doesn't seriously use that $k$ is algebraically closed -- to deal with a general ground field, let $S$ be the maximal split central torus of $L$ above, and use the fact that maximal split tori are rationally conjugate in reductive groups over any field.

  • 1
    $\begingroup$ Great and crystal-clear answer! Now I realize that the key point is precisely the fact that both $S$ and $g^{-1}Sg$ are maximal tori of $Z_G(H)$ and thus $Z_G(H)$-conjugated, as you mentioned. Thanks a lot! $\endgroup$
    – a_g
    May 11 at 15:01
  • 2
    $\begingroup$ Agreed, exactly the right answer. I was trying to muck about with details of Bala–Carter theory, whereas, as your answer makes clear and @a_g emphasises, it is really about maximal tori in $Z_G(H)$. The answer uses reductivity of $G$ in the claim that $L$ equals $Z_G(S)$. Upon replacing "central torus" by "split …", I think that I don't see where this uses that $k$ is algebraically closed; does it really? (If it doesn't, then my proposed counterexample is certainly wrong.) $\endgroup$
    – LSpice
    May 14 at 21:52
  • 1
    $\begingroup$ It may also be worth mentioning explicitly, as you know but might not be completely clear to @a_g, that it should be regarded as a miracle (stating something interesting about the way that the subgroup $H$—of the question, not of the answer—sits inside $G$) when the answer to the extra question is "yes", rather than as an unfortunate pathology when it is "no". $\endgroup$
    – LSpice
    May 14 at 21:59
  • 1
    $\begingroup$ @LSpice, I agree with both comments. I guess the problem with your proposed counterexample is that $\mathrm{diag}(g_1, (g_2, g_3))$ is not actually conjugate to $\mathrm{diag}(g_2, (g_1, g_3))$ in $\mathrm{Sp}_8$ (although of course they are conjugate in $\mathrm{GL}_8$). $\endgroup$
    – SeanC
    May 14 at 23:48
  • $\begingroup$ Re, ah, yes, that'd do it! $\endgroup$
    – LSpice
    May 14 at 23:50

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