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Let $G$ be a semisimple complex Lie group, and let $H$ be a subgroup corresponding to a subset of the extended Dynkin diagram of $G$ (à la Borel - de Siebenthal). I would like to know if there is a recipe for computing the normalizer of $H$. My feeling is that this must be known, but I could not find anything.

For concreteness, consider this example. Let $G=E_7$ (simply connected, say). There is a subgroup of type $A_7$, which has index $2$ inside its normalizer. This corresponds to the involution of the Dynkin diagram of type $A_7$, that one can check respects the highest root of $E_7$ and so respects the extended Dynkin diagram of type $E_7$. On the other hand, consider the subgroup of type $D_6\times A_1$. The index of the subgroup in its normalizer is at most $2$, because the Dynkin diagram has automorphism group $2$. It is clear that the obvious involution of the Dynkin diagram does not extend to an involution of the extended Dynkin diagram of $E_7$, because the highest root is sent to some other root. Does this imply that the subgroup is self-normalizing? I would expect that this subgroup is not self-normalizing.

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    $\begingroup$ What you are asking for is equivalent to computing the normalizer of the Weyl group of $H$ in the Weyl group of $G$. For this computation in the Weyl group, see Proposition 28 in Carter, "Conjugacy classes in the Weyl group" (1972). Looking at Table 10 in Carter's paper, you see that $D_6 \times A_1$ is self-normalizing in $E_7$. Another relevant reference is Howlett, "Normalizers of parabolic subgroups of reflection groups" (1980). $\endgroup$
    – Mikko Korhonen
    Commented Dec 18, 2018 at 16:28
  • $\begingroup$ As you say, "My feeling is that this must be known, but I could not find anything." It may be "known" to someone, but it doesn't seem to be written down. $\endgroup$
    – Jim Humphreys
    Commented Dec 18, 2018 at 19:04
  • $\begingroup$ $D_6\times A_1$ seems to be self-normalizing, because it is the centralizer of $\mu_2$, the center of $A_1$ (for any $g$ normalizing $D_6\times A_1$ leaves $A_1$ invariant and therefore commutes with the center of $A_1$). $\endgroup$
    – Victor Petrov
    Commented Dec 20, 2018 at 16:00

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