When is the normalizer of the maximal torus maximal? 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 $.
 A: 
$
\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.
A: $\def\fg{\mathfrak{g}}\def\ft{\mathfrak{t}}\def\long{\text{long}}\DeclareMathOperator\SO{SO}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SU{SU}$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.)
