Isn't your function just the sum of squares of an orthonormal basis of V? (So it doesn't even depend on the G-action.)

Say you have a binary cubic form of discriminant $D$. Your 3-torsion ideal class is, considered as a binary quadratic form, basically its Hessian; it has discriminant $−3D$ (up to squares, maybe). So what you are seeing is a actually correspondence between 3-torsion in $C_{-3D}$ and index 3 subgroups of $C_D$. For that, look up ``Scholz reflection.'' Underlying that is actually a duality which explains this.

I guess this might be the motivation of your question, but for a curve over a finite field this is a consequence of reduction theory, i.e. I think it is precisely the translation of the fact that a "Siegel set" contains a fundamental domain. For that, there is an article of Springer "Reduction theory over global fields."

As of right now, this article is available from AMS books online here: ams.org/books/pspum/017. (Also as of now, Google doesn't seem to know this.)

Yes, our proofs seem to be basically the same, but your treatment of the degenerate cases is cleaner than mine. Thanks.

Yes, the symmetric spaces were classified by E. Cartan. See en.wikipedia.org/wiki/Symmetric_space#Classification_result for the list. However, I haven't seen anywhere a list of their homologies. Probably it was done in the 1950s, but I couldn't find it in e.g. the Borel-Hirzebruch papers.

Search for "SL_2 hashing" for an example where such ideas are proposed.

But the proof -- at least, Grothendieck's proof -- that de Rham and topological cohomology are isomorphic uses resolution of singularities. How to avoid this?

This is not a reference as you are looking for, but see BCnrd's comment at mathoverflow.net/questions/19830/….

@DagOskarMadsen, at least the way I am thinking about it, it is important that everything is finite-dimensional over $K$. I don't know what happens in the general case. Also, it is a bit annoying that one has to treat the cases of K finite and infinite separately. Surely there is a uniform way of proceeding...

Sorry, here it is: Modules that become isomorphic to $M$ over the algebraic closure of our finite field $K$ are classified by $H^1(G, \mathrm{Aut}(M'))$, where $M'$ is $M$ base-changed to the algebraic closure, and $G$ is the absolute Galois group of $K$. Now $\mathrm{Aut}(M')$ is the set of $\bar{K}$-points of a connected algebraic $K$-group, namely, the automorphism group of $M$ (considered as a $K$-variety). There is a theorem of Lang and Steinberg that says $H^1$ always vanishes in this setting.

I don't think so. It comes down to this: Take a polynomial $f \in K[x_1, \dots, x_n]$. If there exists $(a_1, \dots, a_n) \in L^n$ such that $f(a_1, \dots, a_n) \neq 0$, then also there exists $(b_1, \dots, b_n) \in K^n$ with $f(b_1, \dots, b_n) \neq 0$. The existence of $a_i$ means that $f$ is not identically vanishing.