## The Question :

*109 people think this question is useful*

I was reading Barry Mazur’s biography and come across this part:

*Grothendieck was exceptionally patient with me, for when we first met I knew next to nothing about algebra. In one of his first conversations with me, he raised the question (asked of him by Washnitzer) of whether a smooth proper algebraic variety defined over a real quadratic field could yield topologically different differentiable manifolds realized by the two possible imbeddings of the number field into the reals. What a perfect question, at least for me! Not that I answered it. But it was surely one of the very few algebro-geometric questions that I then had the background to appreciate. … the question provided quite an incentive for a topologist to look at algebraic geometry. I began to learn the elements of algebraic geometry working with Mike Artin.*

Is the problem still open? I am an algebraic topology student so it feels very surprising someone will come up with a question like this. But I am at a loss how to experimentally find some toy examples one can work by hand.

*The Question Comments :*

## The Answer 1

*4 people think this answer is useful*

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…

## The Answer 2

*10 people think this answer is useful*

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I received the email about one year ago and forget to post it, as I am no longer active at the site.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.

## The Answer 3

*2 people think this answer is useful*

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.