The MO question https://mathoverflow.net/questions/331549/compactness-of-the-automorphic-quotient-and-genericity made me realise that I don't really understand the topology on the adèlic points of an algebraic group $G$. Indeed, the way I realised this is by making the wrong comment https://mathoverflow.net/questions/331549/compactness-of-the-automorphic-quotient-and-genericity#comment826950_331549. To save your following the link, the question is about consequences of compactness of $G(F)\backslash G(\mathbb A)$. Now I thought that, for each place $v$,

  1. the group $G(F_v)$ is a closed subgroup of $G(\mathbb A)$; and,

  2. since $G(F)$ is embedded diagonally in $G(\mathbb A)$, it intersects $G(F_v)$ trivially; so

  3. $G(F_v)$ sits as a closed subset of $G(F)\backslash G(\mathbb A)$.

Since further discussion (for example, paulgarrett's answer) showed that the conclusion is wrong, obviously one of the steps is wrong—but I'm embarrassed to say I don't know which one. Is it the first? If so, is there any easier way to see why than "because there's a definition of closed set that $G(F_v)$ doesn't satisfy"?

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    $\begingroup$ Why is step 3 true? Is the quotient map closed? $\endgroup$ – rj7k8 May 18 at 2:37
  • $\begingroup$ @rj7k8, well, one (or more) of the steps is false; maybe that's the one! $\endgroup$ – LSpice May 18 at 19:34

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