@senti_today : The first statement is true, because the number of conjugacy classes in a finite group cannot remain bounded as the size of the group increases. The second statement is false: Reflections inside a dihedral group are a conjugacy class of measure $\frac{1}{2}$, and you can convert this to an infinite example.

Quick ad hoc calculation (not checked): standard permutation representation of D_8 on a square gives a 4-dimensional linear representation, which is the sum of a trivial representation, the representation you want, and a quadratic character. The discriminant of the associated quartic field is (at 2) of valuation 10, so you just need to figure out what the associated quadratic field is; I think (without checking) it's $\mathbb{Q}(\sqrt{2})$, so that gives 7. Did that very hastily, though, don't trust i

Maybe I'm misunderstanding. Firstly if X is projective, the C-points of X((t)) are the same as X(C[[t]]), right? I don't know what the "usual topology" on this is, but surely is homotopy equivalent to X(C), not to the loops on X(C). By taking the affine cone over this you get a similarly funny situation with affine varieties.

No: I think that if $V$ is $2$-dimensional in characteristic $p$, then $Sym^p(V)$ (considered as a $GL(V)$-module) has a two-dimensional submodule, but no two-dimensional quotient.

I think that, if $SO(q), SO(q')$ are isomorphic, then $q$ is isometric to $\lambda q'$ for some $\lambda \in K$. We can intrinsically construct the isomorphism class of the pair $(K q,V)$ from the group: When the dimension of $\mathrm{SO}(q)$ is greater than $1$, it is the line of invariant quadratic forms on the smallest-dimensional representation of the group.

You mean "surjective" in the first paragraph, if your coefficients are $\mathbb{Q}_{\ell}(r)$. The last line of your first paragraph is incorrect.

I deleted my answer (it is $k=43$, which I found by fairly routine computation) because of this excellent reference.

I do not follow: surely there are infinitely many non-neat arithmetic subgroups?

So is Tits' group just the $\mathbf{Z}$-points of the normalizer, after one chooses a model over $\mathbf{Z}$?

Use the splitting of $A \rightarrow G \rightarrow B$ whenever the orders of $A$ and $B$ are relatively prime.

A standard analysis result of this type is the Schwartz kernel theorem.

Nice example! Stupid generalization: this is a flag variety of a reductive group over $\mathbf{Z}$; one gets a similar example from any such that is compact over $\mathbf{R}$, see Gross' paper for a list.

I don't know of any examples of smooth proper varieties over $\mathbf{Z}$, except those constructed in some simple way from the flag varieties of Chevalley groups. Does anyone know other interesting examples? Also, the answer to your question is certainly NO if one replaces the integers by the ring of integers in a suitable number field. Do you know some variant of the question that might remain true?

Maybe I am misunderstanding the question, but you can write down (many) nonconstant bounded harmonic functions on a trivalent regular tree.

Nope. I have no idea about it at all. I don't even have interesting examples of projective smooth varieties with such torsion. Maybe it's a good topic for a question here...

There's a distinction between the mod 5 etale cohomoology and the reduction of the $\mathbb{Q}_5$ cohomology. What you discuss above would in principle detect (the semisimplification of the) latter, not the former; they will differ if there is torsion in the $\mathbb{Z}_5$-cohomology. (I also missed this in my comment below.)

Note this is really hard to do in practice (although I don't have a better idea). Even if you tried to compute H^2 of a surface fibered in genus 2 curves over a base curve X, then (to compute the cohomology of X with coefficients in the relevant local system) you have to pass to a 125-fold covering of X and compute the Jacobian of that beast; I think it's very hard even to write down the Jacobian of a curve of genus three! In principle one can count points on X over various finite fields and this determines K, but this is also wildly impractical.