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Columbus1
Columbus1
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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.

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). 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.

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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Columbus1
Columbus1
  • 71
  • 2

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). 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.