The question is a very little more than what's in the title. It is easy (for some values of ‘easy’) to produce examples of endoscopic groups that are not subgroups. When I asked a colleague, he mentioned $\mathrm{PGL}_3$ as an endoscopic group of $\mathrm G_2$. However, in this example, $\mathrm{PGL}_3$ is isogenous to $\mathrm{SL}_3$, which is an actual subgroup of $\mathrm G_2$, and similarly for the other examples that I know. Is this kind of ‘central failure’ as bad as it can get, or is there an example of an endoscopic group that is not even isogenous to a subgroup?
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2$\begingroup$ You may like check mathoverflow.net/questions/69292/… $\endgroup$– T. AmdeberhanCommented May 24, 2017 at 5:41
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$\begingroup$ @T.Amdeberhan, I agree, that looks like a very interesting question, but it doesn't seem to have been answered! $\endgroup$– LSpiceCommented May 24, 2017 at 7:09
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1 Answer
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If I understand the definition correctly, a connected reductive group $H$ is an endoscopic group for a connected reductive group $G$ if its Langlands dual $H^\vee$ is a connected centralizer in $G^\vee$.
So $H=SO(2p+1)\times SO(2q+1)$ is endoscopic in $G=SO(2p+2q+1)$ since $H^\vee=Sp(2p)\times Sp(2q)$ is a centralizer in $G^\vee=Sp(2p+2q)$. But clearly $H$ is not isogenous to a subgroup of $G$. Another example is $H=SO(2p)\times Sp(2q)$ for $G=SO(2p+2q)$.
Using Borel-de Siebenthal the problem comes from the fact that for finite Dynkin diagrams one may have $$ \text{extended dual}\ne\text{dual extended}. $$
answered May 24, 2017 at 13:39
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1$\begingroup$ This is perfect: the examples answer the question that I asked, but the slogan answers the question that I meant ("how / why does this happen?"). Thank you! $\endgroup$– LSpiceCommented May 24, 2017 at 14:12