Grothendieck 's question - any update?
Since this question has been hanging around for a few years, I think I should post Prof. Mazur's official answer below:
Hello Bombyx mori,
Take a look at
J.-P. Serre, Exemples de variétés projectives conjuguées non homéomorphes, C. R. Acad. Sci. Paris 258 (1964), 4194-4196.
which actually gives two conjugate algebraic varieties with different fundamental groups, and therefore different homotopy types.
Best wishes,
Barry Mazur
I received the email about one year ago and forget to post it, as I am no longer active at the site. This is in fact the source of David Speyer's blog post, so it is well known already. Whether the real quadratic case is closed I am still not sure (as I am not an expert on number fields), but hopefully this gives fellow mathematicans some encouragement to read Serre's original paper. What Prof. Vasiu was refering to should be something similar to this paper.
Just noticed this hanging (?) question: depending on what is wanting, maybe the following is worthwhile...
How about the variety that is a quaternion division algebra $B$ over a real quadratic extension $k$ of $\mathbb Q$ which splits at one real place and not the other. Then the norm-one group at one of the real places is compact, and the other isn't.
Now, yes, there is a clear causal mechanism for such an example, and maybe that's not what's desired...
One interpretation of the question is the following. Let $X$ be a smooth proper algebraic variety over a real quadratic number field $K$. Let $\sigma_1$ and $\sigma_2$ be the two embeddings of $K$ into $\mathbf R$. Denote by $X(\mathbf{R}_i)$ the set of real points of $X$ with respect to the embedding $\sigma_i$ of $K$ into $\mathbf R$. Can it happen that $X(\mathbf{R}_1)$ and $X(\mathbf{R}_2)$ are not homeomorphic?
It is quite easy to construct elementary examples of such varieties. Consider quadrics in projective $n$-space $\mathrm{P}^n$, for example. For $n=2$ you have the nonsingular conic $C$ defined by the equation $x^2+y^2+z^2\sqrt2=0$ over the real quadratic number field $K=\mathbf{Q}(\sqrt2)$. If you map $\sqrt2\in K$ to the positive square root of $2$ in $\mathbf R$ then the set of real points of $C$ will be the empty differentiable manifold of dimension $1$. If you map $\sqrt2\in K$ to the negative square root of $2$ in $\mathbf R$ then the set of real points of $C$ will be diffeomorphic to the circle $S^1$. They are certainly not homeomorphic.
In higher dimension one can obtain more interesting examples. For $n=3$ one can consider the nonsingular quadric $Q$ defined by the equation $x^2+y^2+z^2\sqrt2-w^2=0$ in $\mathbf{P}^3$ over $K$. With respect to the first embedding of $K$ into $\mathbf R$, the set of real points of $Q$ is diffeomorphic to the sphere $S^2$. With respect to the second, the set of real points of $Q$ is diffeomorphic to the torus $S^1\times S^1$. Again, they are not homeomorphic.