# Topology of the automorphic quotient

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"?

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