# Normalizer of Levi subgroup

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 $$nxn^{-1} = gxg^{-1}$$.

Remarks:

• 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.

• 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 $L$. Apr 24 at 2:02
• 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 $\text{diag}(2,1,1)$.
– a_g
Apr 24 at 16:57
• 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? Apr 24 at 17:24
• 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).
– a_g
Apr 24 at 22:27
• Aren't minimal Levis of a reductive group over an algebraically closed field just maximal tori? Apr 24 at 23:28

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.

• 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!
– a_g
May 11 at 15:01
• 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.) May 14 at 21:52
• 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". May 14 at 21:59
• @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 $\mathrm{GL}_8$). May 14 at 23:48
• Re, ah, yes, that'd do it! May 14 at 23:50