2
$\begingroup$

I've been thinking recently about moduli spaces defined over $\mathbb{Z}$, and this led me to the following question:

Question

Riemann existence says that if we have a variety over $\mathbb{C}$, $X_{\mathbb{C}}$, then $\widehat{\pi_1^{top}(X_{\mathbb{C}}(\mathbb{C}))}\cong\pi_1^{et}(X_{\mathbb{C}})$. For which (the first, the second, or both) of the following interpretations of $X_{\mathbb{C}}(\mathbb{C})$ does this theorem work: a. It is the set of all morphisms $Spec(\mathbb{C})\rightarrow X_{\mathbb{C}}$.

or

b. It is the set of all sections of the structure morphism $X_{\mathbb{C}}\rightarrow Spec(\mathbb{C})$.

Motivation

The motivation to the question, as I said before, is because I've been thinking about moduli spaces over $\mathbb{Z}$. Let's say $X$ is a priori a scheme of finite type over $\mathbb{Z}$. Here $X(\mathbb{C})$ has only the first interpretation, since the structure morphism is not going to $\mathbb{C}$, but to $\mathbb{Z}$. I want to say something about the geometry of $X(\mathbb{C})$, but how could I if this is not what Riemann Existence is talking about?

Remark

Let me give you a sense of why the two interpretations are so different. If $X_{\mathbb{C}}$ is a variety over $\mathbb{C}$, the second interpretation would mean that $X_{\mathbb{C}}(\mathbb{C})$ is in correspondence with the maximal points of $X_{\mathbb{C}}$. However, in the first interpretation we would also have phenomena like the following: If $Y_{\mathbb{C}}$ is a subvariety of $X_{\mathbb{C}}$, then $\kappa(Y_{\mathbb{C}})$ (the function field of $Y_{\mathbb{C}}$) is going to be a field of cardinality $2^{\aleph_0}$ and so its algebraic closure is isomorphic to $\mathbb{C}$. This means that we have a $Spec(\mathbb{C})\rightarrow X_{\mathbb{C}}$ such that its image is at the generic point of $Y$. Obviously, the first interpretation makes more geometric sense, but as I mentioned in the motivation, I am very interested in how to deal with the second. And so, in particular, I'm asking if we have a Riemann Existence theorem for the second interpretation.

$\endgroup$
5
  • 1
    $\begingroup$ It should be the second (b) with the analytic topology. In the first case, you would get the union of $\sigma(X(\mathbb{C}))$ as $\sigma$ varies over $Aut(\mathbb{C})$, which would be quite strange $\endgroup$ Jul 19, 2011 at 19:04
  • $\begingroup$ Then, if I have a scheme $X$ over $\mathbb{Z}$, what can I say about $X(\mathbb{C})$? Can I say anything constructive? $\endgroup$ Jul 19, 2011 at 19:07
  • $\begingroup$ Let me give a rephrasing of what I'm perplexed about. Say $X$ is the (coarse, fine, it doesn't matter) moduli space over $\mathbb{Z}$ of some moduli problem. Say $W_{\mathbb{C}}$ is the moduli space over $\mathbb{C}$. My intuition says that $X\times Spec(\mathbb{C})\cong W_{\mathbb{C}}$. But $X(\mathbb{C})$ has a different interpretation from $W(\mathbb{C})$? This seems weird, since they need to classify the same thing! $\endgroup$ Jul 19, 2011 at 19:15
  • $\begingroup$ It seems that the main question here was settled. However, I remain confused. So I will start a new question with my new source of confusion (basically summed up in the above comment). $\endgroup$ Jul 19, 2011 at 19:24
  • $\begingroup$ It can be found here: mathoverflow.net/questions/70773/… $\endgroup$ Jul 19, 2011 at 19:38

1 Answer 1

2
$\begingroup$

The set of points should be given by the second choice, i.e., the set of $\operatorname{Spec} \mathbb{C}$-points, over $\operatorname{Spec} \mathbb{C}$. However, there is an additional step you need to do before defining $\pi_1$ (besides choosing a basepoint), which is applying an analytification functor to endow the set with a suitable topology. This functor takes locally finite type schemes over $\operatorname{Spec} \mathbb{C}$ to complex analytic spaces. There is a brief exposition of analytification in SGA 1, Exp. 12, and in Serre's GAGA.

Regarding your comment about making a topological space from a variety defined over $\mathbb{Z}$, a $\mathbb{C}$-point of the base change to $\mathbb{C}$ over $\mathbb{C}$ is the same as a $\mathbb{C}$-point of the original scheme. This is the universal property of fiber product.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.