When is the normalizer of maximal torus 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: $\def\fg{\mathfrak{g}}\def\ft{\mathfrak{t}}\def\long{\text{long}}$This question was crossposted to MO
https://mathoverflow.net/questions/435457/normalizer-of-maximal-torus-is-maximal-in-compact-simple-group/435501#435501
Where Mikhail Borovoi gave a short conceptual proof
https://mathoverflow.net/a/435501/387190
that if all the roots have the same length (types $ A_n,D_n,E_6,E_7,E_8 $) then $ N(T) $ is maximal.
And David E Speyer gave a nice conceptual proof
https://mathoverflow.net/a/435474/387190
that if the roots are not all the same length (i.e. there are long roots and short roots i.e. types $ B_n,C_n,F_4,G_2 $) then $ N(T) $ is not maximal since it is contained in the proper closed subgroup of strictly larger dimension
$$
G_{long}N(T)=N(G_{long})
$$
whose Lie algebra is spanned by the lie algebra of $ T $ together with the roots spaces for all the long roots.
In particular let me highlight this part of the answer from David E Speyer:

*

*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 $G_{\long} N(T) = Sp(2)^n \rtimes S_n$ [this group is table 8 fifth row $ p=1 $ case in https://arxiv.org/abs/math/0605784].


*In type $F_4$, the group $G_{\long}$ is $Spin(8)$ and $ G_{\long} N(T)=Spin(8)\rtimes S_3 $ [see comments on https://math.stackexchange.com/questions/88330/the-weyl-group-of-f-4 for some details about this subgroup]


*In $G_2$, the group $G_{\long}$ is $SU(3)$. $G_{\long} N(T)$ is $SU(3) \rtimes 2$, with $2$ acting by the unique outer automorphism of $ SU_3 $ (complex conjugation).
In general it seems that
$$
G_{long}N(T)=G_{long}\rtimes Out(G_{long})
$$
