Given a topological ring $R$, under what conditions and in what way, can one induce a topology on the $R$-points of a scheme $X$? For example, if $X$ is $P^n$ or $A^n$, one has natural topology on the $R$-points.


If $G$ is a group scheme/A and $R$ is $A$-algebra (still a topological ring), will the induced topology on $G\left(R\right)$ (as above) automatically make $G$ into a topological group.


For number theorists, if $G$ is an algebraic group/Q, we can consider the adelic points $G\left(A\_{K}\right)$ for any number field $K$. Is the induced topology on $G\left(A\_{K}\right)$ that of a restricted direct product? Under what conditions will $G\left(A\_{K}\right)$ be locally compact or satisfy other nice properties? 



How to topologize X(R) when R is a topological ring?
Brian Conrad has some notes on this on his website ("[Some notes on topologizing the adelic points of schemes, unifying the viewpoints of Grothendieck and Weil](http://math.stanford.edu/~conrad/papers/adelictop.pdf)"). The short version is that if X is affine, you can topologize X(R) in a natural, functorial way (specifically, the weakest topology such that the functions X(R)-->R induced by elements of the structure sheaf are continuous). If X isn't affine, you have to be more careful, because the units of R might not be open in X and might not be a topological group wrt the subspace topology. But those are the only problems, and if your ring doesn't have those problems, you can glue the naturally topologized affines and everything is functorial. Happily for number theorists, the adeles are fairly nice, and for a finite-type separated K-scheme X, X(A\_K) can be naturally topologized, and it is locally compact and Hausdorff.


