# Normalizers of subsystem subgroups of Lie groups

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.

• 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). – Mikko Korhonen Dec 18 '18 at 16:28
• 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. – Jim Humphreys Dec 18 '18 at 19:04
• $D_6\times A_1$ 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 $A_1$). – Victor Petrov Dec 20 '18 at 16:00