Ah right, didn't think you can just quantify over automorphisms. Checks out though.

How do you define $\mathbb R$ in $\mathbb C$ in SOL assuming AD?

It would be enough to prove $\mathbb Q(\gamma)=\mathbb Q(10^{-1/k})$. You should be able to do it with Galois theory - look at the subgroups of $Gal(\mathbb Q(\theta,\zeta_m),\mathbb Q)\cong(\mathbb Z/m)\rtimes(\mathbb Z/m)^\times$ containing $Gal(\mathbb Q(\theta,\zeta_m),\mathbb Q(\theta)\cong(\mathbb Z/m)^\times$, and check there is only one of each index dividing $m$.

As a field, $\mathbb R$ is immensely complicated (infinite transcendence degree over the prime field, etc.). I think the question is unapproachable even in much simpler cases, like for valuations on $\mathbb Q(x,y)$. Can we give a classification for this ring, at least for rank 1 valuations?

@LSpice It is a nontrivial fact that any nonarchimedean norm on any field can be extended to any extension, algebraic or not. It's very noncanonical but always exists.

@ToddTrimble Missed it, thank you

I don't see how the result of the first paragraph precludes existence of a (homotopy-invariant) invariant for a CW structure. It may be that the obstruction is only applicable for spaces which are closed manifolds. So unless we have that any closed manifold has homotopy type of a CW complex which is also a closed manifold, I fail to see the conclusion.

Neither the paper nor the MO post you link seem to mention the $\dim\geq 6$ result you mention. Would you happen to have a reference for it?

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0 - yes, some examples are $GL_n$ for $n>2$ or $GL_{2,K}$ for $K$ a number field which is not totally real. 3 - if you find a good answer to this question, you may be able to help propel Langlands program quite considerably. AFAIK the best idea we have is to relate the $G$ and its locally symmetric spaces to some other $G'$ which do have Shimura varieties, like unitary groups.

You can very easily do this in any CAS as well. In Sage, I can generate the list of all such primes below $10^7$ in about 5 seconds with [p for p in prime_range(10^7) if primitive_root(p)==2].

There is also some nuance as to whether this actually is an example of nonclassical logic. There is a subtle distinction between a logic being two-valued (for every $A$, either $A$ is true or $\neg A$ is true) and law of excluded middle (for every $A$, $A\lor\neg A$ is true). The latter still holds for logics valued in nontrivial Boolean algebras, which need not be two valued. Internal logic of topos on a discrete space will satisfy axioms of classical logic. For truly intuitionistic behavior, you'd need to introduce properties which hold on some open whose complement is not open.

@SpicetheBird You can see that it's not topologically finitely generated quite directly - if it was, there'd only be finitely many continuous maps $G_{\mathbb Q}\to\mathbb Z/2$, that is finitely many open index $2$ subgroups, that is finitely many degree $2$ extensions of $\mathbb Q$. But adjoining square roots of different primes gives you infinitely many quadratic extensions.

It's been a long while since I've read that proof, what sort of non-algebraic inputs does it use?

@Turbo Given some set $P$ of $k$ primes, the set $S$ of numbers divisible only by primes in $P$ satisfies $|S\cap [n]|=O((\log n)^k)$. Thus if you have an increasing sequence $a_m$ whose elements are in $S$, then necessarily $a_m\gg 2^{O(\sqrt[k]{m})}$

@FedorPetrov In the sense that no one has produced one, at least not in full generality. I don't think one is known even for something like primes of the form $7n+3$, say.

@JamesEHanson Someone has done the work of algebraically establishing the properties of the class of Archimedean fields to show there is a maximal one and has the usual completeness proeprty: drive.google.com/file/d/1fb1pNJVMuROmpLq-kDwXXeY2QxzCV0hy/vi‌​ew