Does there exist a reductive group $G$ of type $E_7$ over a given totally real field $L^+$ such that for every embedding $\tau:L^+\to \mathbb{R}$ except one , $G_{\tau,\mathbb{R}}(\mathbb{R})$ is compact? Is there a way to construct a group with a given number of compact forms?
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2$\begingroup$ Yes such a group should exist. For example it can be constructed as an inner form of the split adjoint group $G$ of type $E_7$ using the short exact sequence of pointed sets $H^1(L^+, G) \to \bigoplus_{v} H^1(L^+_{v},G) \to \pi_1(G)_{\Gamma_{L^+}}$. Here $\pi_1(G)$ is Borovoi's algebraic fundamental group and $\Gamma_{L^+}$ is the absolute Galois group of $L^+$.... $\endgroup$– Pol van HoftenCommented 8 hours ago
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2$\begingroup$ ... We need to show that there is a class in $H^1(L^+,G)$ whose image in $H^1(L^+_{v},G)$ for infinite places $v$ of $L^+$ corresponds to the compact inner form of $G_{\mathbb{R}}$ for all but one $v$. This can always be done using the short exact sequence abpve, by choosing nontrivial classes at some auxiliary finite places. In fact, a single finite place should be enough since I think that $\pi_1(G)=\mathbb{Z}/2\mathbb{Z}$ with trivial Galois action. $\endgroup$– Pol van HoftenCommented 8 hours ago
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$\begingroup$ Thank you very much. $\endgroup$– ALi1373Commented 5 hours ago
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1$\begingroup$ @PolvanHoften, of course it would be more stable, here, if you wrote your excellent "comments" as "an answer". :) $\endgroup$– paul garrettCommented 58 mins ago
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