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Let $G$ be a complex algebraic group, and write $Z(g)$ for the centralizer of a semisimple element $g$ in $G$. I will assume $G$ is simply connected, in which case $Z(g)$ is connected. Let $G^\vee$ and $Z(g)^\vee$ be the Langlands dual complex algebraic groups. If I understand, $Z(g)^\vee$ is called an "endoscopic group" for $G^\vee$, though I've taken all of the arithmetic out of it.

Most of the time, $Z(g)$ is a Levi subgroup of $G$, in which case $Z(g)^\vee$ is naturally a Levi subgroup of $G^\vee$. The handful of centralizers that are not Levis were classified by Kac. It's often emphasized that in these more interesting cases the associated endoscopic group is not a subgroup of $G^\vee$. For instance, $Z(g) = SL(2) \times SL(2)$ is a centralizer of an element of order 2 in $G = Sp(4)$, but there is no way to include $Z(g)^\vee = PGL(2) \times PGL(2)$ into $G^\vee = SO(5)$.

I just noticed that the endoscopic groups of $G^\vee$ do include into $G^\vee$ whenever $G$ is simply laced. This works out in type A because all centralizers are Levis, in type D essentially because of the self-duality of the groups $SO(2n)$, and in types $E_6,E_7,E_8$ by checking the centralizers one by one.

  1. There are 2 interesting centralizers in $E_6$, 4 in $E_7$, and 8 in $E_8$, so plenty of opportunities for me to have made a mistake. Is my claim correct?

  2. If correct, is there a simpler explanation for it than a case-by-case check?

  3. Is the existence or nonexistence of an endoscopic sub group relevant in the kinds of endoscopic things people study in the Langlands program?

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"The handful of centralizers that are not Levis were classified by Kac." -- I would've thought to credit rather [Borel-de Siebenthal]. mathoverflow.net/questions/28878/… –  Allen Knutson Jul 1 '11 at 22:15
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