6
$\begingroup$

This is a cross post from MSE of

https://math.stackexchange.com/questions/4562196/normalizer-of-maximal-torus-is-maximal

Let $ T $ be a maximal torus in a compact connected simple Lie group $ K $. For which groups $ K $ is the normalizer $ N(T) $ maximal among the proper closed subgroups of $ K $?

I know this is true for the infinite families of compact connected simple groups $ SU_n, n \geq 2 $ and $ SO_{2n}, n\geq 3 $, see

https://arxiv.org/abs/math/0605784

table 5 fourth row $ p=1 $ case for $ SU_n, n \geq 2 $ and table 7 first row $ p=2 $ case for $ SO_{2n}, n \geq 3 $.

$\endgroup$
5
  • 8
    $\begingroup$ I don't think that that $N(T) $ is maximal for the compact group ${\rm Sp}(n)$. I think that there is a larger subgroup: $N(T)\cdot {\rm Sp}(1)^n$ where $${\rm Sp}(1)^n={\rm Sp}(1)\times\dots\times {\rm Sp}(1)$$ embedded diagonally. The same refers to ${\sf G}_2$. However, I think that I can prove the assertion for the irreducible root systems in which all roots have the same length, for example ${\sf E}_6$, ${\sf E}_7$, $\sf{E}_8$. $\endgroup$ Nov 28, 2022 at 18:39
  • $\begingroup$ @MikhailBorovoi good point. Going back and checking the reference shows that $ N(T) $ is only maximal for $ SU_n $ ($ A_{n-1} $ root system) and $ SO_{2n} $ ($ D_n $ root system). Which are exactly the root systems in which all roots have the same length as you predicted. I've updated my question to reflect this. Also Are you claiming that $ N(T)=T^2 \rtimes D_6 $ is not maximal in $ G_2 $? That is consistent with the root system but just seems odd because I don't know what subgroup it would be contained in? $\endgroup$ Nov 28, 2022 at 19:34
  • 2
    $\begingroup$ @IanGershonTeixeira I haven't checked this carefully, but ... Let $\mathfrak{g}$ be a simple Lie algebra with root decomposition $\mathfrak{h} \oplus \bigoplus \mathfrak{g}_{\beta}$. Suppose that there are two lengths of roots, and put $\mathfrak{g}_{\text{long}} = \mathfrak{h} \oplus \bigoplus_{\beta \ \text{long}} \mathfrak{g}_{\beta}$. Then $\mathfrak{g}_{\text{long}}$ is a subalgebra with corresponding connected subgroup $G_{\text{long}}$. I think that $N(T)$ normalizes $G_{\text{long}}$. If so, then $G_{\text{long}} N(T)$ is a larger subgroup than $N(T)$. $\endgroup$ Nov 28, 2022 at 19:44
  • 2
    $\begingroup$ In the compact $G_2$ case, the $G_{\text{long}}$ group is an $SU(3)$, which intersects $N(T)$ in $T \rtimes D_3$. So the claim is that the product of $SU(3)$ and $N(T)$ over the common intersection $T \rtimes D_3$ is the larger subgroup you want. We can also describe this as $SU(3) \rtimes C_2$, where $C_2$ acts by an outer automorphism of $SU(3)$. $\endgroup$ Nov 28, 2022 at 19:45
  • $\begingroup$ Yup about 30 seconds ago I looked back at my question math.stackexchange.com/questions/4557593/… and realized $ N(T) $ was in $ SU_3:2 $ since $ T^2:S_3 $ is already in $ SU_3 $ then you just extend and get $ T^2:S_3:2=T^2:D_6 $ so that indeed $ N(T) $ in $ SU_3:2 $ $\endgroup$ Nov 28, 2022 at 19:47

2 Answers 2

10
$\begingroup$

$\def\fg{\mathfrak{g}}\def\ft{\mathfrak{t}}\def\long{\text{long}}\DeclareMathOperator\SO{SO}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SU{SU}$The point of this answer, as discussed in comments, is to note that this is NOT true in types BCFG. In comments, Mikhail Borovoi said he could prove the result is true in types ADE (now an answer), so collectively this resolves the claim.

Let $G$ be a compact connected simple Lie group with Lie algebra $\fg$. Choose a torus $T$ and with corresponding Lie algebra $\ft$, and let $$\fg = \ft \oplus \bigoplus_{\beta} \fg_{\pm \beta}$$ be the root decomposition, where $\beta$ runs over positive roots and each $\fg_{\pm \beta}$ is two dimensional. (Since I'm not tensoring by $\mathbb{C}$, I don't get to go all the way down to one dimensional subspaces.)

Now, suppose that $\fg$ has two lengths of roots, "long" and "short". A case by case check of the root systems shows that the sum of two long roots is always either $0$, long or not a root. Therefore, $$\fg_{\long} = \ft \oplus \bigoplus_{\beta\ \long} \fg_{\pm \beta}$$ is a subalgebra; let $G_{\long}$ be the corresponding subgroup.

Then $T \subsetneq G_{\long}$ and, because the only choice in defining $G_{\long}$ is the choice of $T$, the normalizer $N(T)$ also normalizes $G_{\long}$. So $G_{\long} N(T)$ is a subgroup containing $N(T)$ of dimension $\dim G_{\long} > \dim T$.

Explicitly, we have:

  • In type $B_n = \SO(2n+1)$, the group $G_{\long}$ is $\SO(2n)$ and $G_{\long} N(T) = O(2n)$.

  • In type $C_n = \Sp(2n)$, the group $G_{\long}$ is $\Sp(2)^n$ and I think that $G_{\long} N(T) = \Sp(2)^n \rtimes S_n$.

  • In type $F_4$, the group $G_{\long}$ is $\SO(8)$, or possibly one of the other compact real forms of $D_4$. I'm not sure what $G_{\long} N(T)$ is explicitly; it might depend on which compact form of $F_4$ we are using.

  • In $G_2$, the group $G_{\long}$ is $\SU(3)$. I think that $G_{\long} N(T)$ is $SU(3) \rtimes C_2$, with $C_2$ acting by an outer automorphism.

These examples, and a little bit of thought about the $F_4$ case, lead me to conjecture: $G_{\long} N(T)$ sits in a short exact sequence $1 \to G_{\long} \to G_{\long} N(T) \to \operatorname{Out}(G_{\long}) \to 1$. (I'm not ready to guess whether or not the sequence is always semidirect.)

$\endgroup$
13
  • 6
    $\begingroup$ I will type my proof for $ADE$ tomorrow. $\endgroup$ Nov 28, 2022 at 20:16
  • $\begingroup$ Is that $ Sp(2n) $ supposed to be $ Sp(n) $? So that for example when $ n=2 $ we have the group $ G_{long}N(T)=Sp(1)\times Sp(1) \rtimes C_2 $ which passes through the double cover $ Sp(2) \to SO(5) $ to the group $ G_{long}N(T)=O(4)=SO_4 \rtimes C_2 $. $\endgroup$ Nov 28, 2022 at 21:07
  • 2
    $\begingroup$ @LSpice You're right. My usual convention is that $Sp(2n)$ is of type $C_n$, so I should write $Sp(2)^n$, not $Sp(1)^n$. $\endgroup$ Nov 28, 2022 at 23:46
  • 1
    $\begingroup$ Does $F_4$ have multiple compact forms? I believe the simply connected version is centerless. In the simply connected form, the $SO(8)$ is really a $Spin(8)$. This follows because there is (famously?) a $Spin(9)\subseteq F_4$ and Borel and de Siebenthals' work implies the "$SO(8)$" must be conjugate to a subgroup of this $Spin(9)$. $\endgroup$ Nov 29, 2022 at 16:26
  • 1
    $\begingroup$ The homogeneous space $F_4/Spin(9)$ is diffeomorphic to the Cayley plane, so appears a lot when talking about Riemannian manifolds of positive sectional curvature or Riemannian symmetric spaces. @Ian: I think Jyrki's comment's math.stackexchange.com/questions/88330/the-weyl-group-of-f-4 support your guess for $G_{long}N(T)$. $\endgroup$ Nov 30, 2022 at 15:28
7
$\begingroup$

$ \newcommand{\g}{{\mathfrak g}} \newcommand{\h}{{\mathfrak h}} \newcommand{\t}{{\mathfrak t}} \newcommand{\C}{{\mathbb C}} \newcommand{\Ad}{{\rm Ad}} \newcommand{\ad}{{\rm ad}} $Theorem. Let $G$ be a connected compact Lie group, $T\subset G$ be a maximal torus, and $H\subseteq G$ be a Lie subgroup such that $H\supseteq N(T)$. We write $\g$, $\t$, $\h$ for the corresponding Lie algebras. Assume that $G$ is almost simple and that all roots $\alpha\in R=R(\g_\C,\t_\C)$ have the same length. Then either $H=N(T)$ or $H=G$.

Proof. Assume that $H\neq N(T)$. Let $h\in H$, $h\notin N(T)$. Then $h T h^{-1}\neq T$, whence $\Ad_h(\t)\neq\t$. Since $\h\supseteq\t$, $\,\h\supseteq \Ad_h(\t)$, we see that $\h\neq \t$.

We pass to the complexifications $\g_\C$, $\t_\C$, $\h_\C$. We have the root decomposition $$\g_\C=\t_\C\oplus\bigoplus_{\alpha\in R} \g_\alpha$$ where $R=R(\g_\C,\t_\C)$ is the root system. Since $\h_\C\supset \t_\C$, we see that $\h_\C$ is an $\ad_{\,\t_\C}$-invariant subspace of $\g_\C$, and therefore we have $$\h_\C=\t_\C\oplus\bigoplus_{\alpha\in S} \g_\alpha$$ for some subset $S\subseteq R$. Since $\h\neq \t$, the set $S$ is non-empty. Since $\h$ is a Lie subalgebra, the subset $S\subseteq R$ is closed, but we shall not use this.

Since $N(T)\subseteq H$, we see that $\h$ is $N(T)$-invariant, whence $S$ is $W$-invariant where $W=N(T)/T$ is the Weyl group. Since $R$ is irreducible and all roots in $R$ have the same length, the only non-empty $W$-invariant subset of $R$ is $R$ itself. Thus $S=R$ and $\h=\g$. Since $G$ is connected, it follows that $H=G$, as required.

$\endgroup$
7
  • $\begingroup$ +1. What is the implication of assuming here that $ G $ is almost simple? Since $ G $ is already assumed compact and connected isn't assuming $ G $ almost simple equivalent to just assuming $ G $ simple? On another note, If we assumed $ G $ complex instead of $ G $ compact would an essentially identical proof show that the normalizer of a maximal algebraic torus is a maximal subgroup (when all roots have equal length)? $\endgroup$ Nov 29, 2022 at 14:13
  • $\begingroup$ (1) "Almost simple "= simple modulo finite center. For example, ${\rm SU}(2)$ is almost simple. Some people use this fancy term, others write just "simple". $\endgroup$ Nov 29, 2022 at 14:54
  • 1
    $\begingroup$ (2) Yes, exactly. When writing about compact groups, I actually had in mind algebraic groups over an algebraically closed field of characteristic 0. $\endgroup$ Nov 29, 2022 at 14:57
  • $\begingroup$ Also what is the etiquette here? Whose answer should I accept? $\endgroup$ Nov 30, 2022 at 2:04
  • 1
    $\begingroup$ @LSpice: Thank you for editing. $\endgroup$ Nov 30, 2022 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.