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Let $X$ be a scheme over $\mathbb{C}$.

  1. When does the topological space $X\left(\mathbb{C}\right)$ of $\mathbb{C}$-points have the homotopy type of a finite CW-complex?

  2. When does the topological space $X\left(\mathbb{C}\right)$ of $\mathbb{C}$-points have the weak homotopy type of a finite CW-complex, (i.e. when is it a finite space)?

By "when", I mean what adjectives do I have to add to make this true, e.g. finite type, separated, smooth...

  1. I'm also interested in question 1.) for when $X$ is affine.

If you happen to know a reference also, that would be fantastic. Thanks!

P.S. I'm aware of this mathoverflow question:

How to prove that a projective variety is a finite CW complex?

However, it addresses only the case of varieties, unless I am missing something..

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    $\begingroup$ Google: no, a schemes can be pretty wild even over $\mathbb{C}$. You need it to be reduced separated and of finite type to qualify as a variety. David: I expect that separated and of finite type would ensure 1, but I have not really thought about carefully. $\endgroup$ Sep 10, 2015 at 21:50
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    $\begingroup$ How do you define the topological space associated to a scheme which is not of finite type? $\endgroup$ Sep 11, 2015 at 0:24
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    $\begingroup$ What is Spec of a scheme? I assumed the question was referring to the complex topology, not the Zariski. For a finite type scheme, locally it is cut out by an ideal in $\mathbb C^n$, so I know what to do. $\endgroup$ Sep 11, 2015 at 1:10
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    $\begingroup$ By the way, I expect the etale homotopy type would give a different answer to the analytic homotopy type in general. For example, for $X= \mathbb C^\times$, the analytic homotopy type is that of $S^1 = B\mathbb Z$, but the etale homotopy type (I guess) is $B\widehat{\mathbb Z}$. $\endgroup$ Sep 11, 2015 at 1:14
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    $\begingroup$ The topological space associated to the non-separated ``line with a doubled origin'' is the non-Hausdorff manifold $M=\mathbb C \sqcup_{\mathbb C^\times} \mathbb C$. I think this does not have the weak homotopy type of a finite CW complex. The proof can be adapted from Prop 5.1 in arxiv.org/abs/math/0609665. In this case, we have that $H^2(M) = 0$ by Mayer-Vietoris, but the Hausdorffification is contractible. $\endgroup$ Sep 11, 2015 at 2:25

2 Answers 2

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Any scheme which is separated of finite type, has at least a triangulation, hence is, in particular, a CW-complex. In fact, by a theorem of Lojasiewicz, this is true for any semi-algebraic set (one can even get this for subanalytic sets, by a result of Hironaka, in Triangulation of algebraic sets, Proc. Amer. Math. Soc. Inst. Algebra Geom. Arcata(1974)); however, the case of (possibly singular) algebraic varieties goes back to the early times of Algebraic Topology: e.g. these papers of van der Waerden and of Lefschetz and Whitehead).

If you only are interested in weak homotopy types, it follows from Lurie's proper base change theorem that considering complex points satisfies proper (hyper)descent (this is Prop. 3.21 in this paper of A. Blanc, which is now published in Compositio Math.). Using Hironaka's resolution of singularities theorem, this implies that, for any scheme of finite type $X$, the space $X(\mathbf{C})$ is a finite homotopy colimit of spaces of the form $Y(\mathbf{C})$ with $Y$ affine and smooth (using Mayer-Vietoris-like homotopy pushouts associated to blow-ups and to coverings by Zariski open subschemes). A smooth affine algebraic variety has the homotopy type of a finite CW-complex: this follows from Morse theory, as can be seen from (the proof of) Theorem 7.2, page 39 in Milnor's book Morse Theory.

From all this, we get that a sufficient condition for $X(\mathbf{C})$ to have the weak homotopy type of a finite CW-complex is to be of finite type, while a sufficient condition to get the homotopy type of a finite CW-complex is to be separated of finite type. A sufficient condition to get an actual finite triangulated space is to be proper.

If we drop the assumption that the scheme is of finite type, I don't see how we can control/define what happens unless we work with pro-homotopy types of some sort.

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  • $\begingroup$ do you really need all of the complicated machinery to see finiteness? $\endgroup$
    – user141225
    May 30, 2019 at 8:43
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    $\begingroup$ @user141225 this complicated machinery is called mathematics. And yes, in order to have the level of precision expressed above, you need sophisticated tools. $\endgroup$ May 31, 2019 at 7:38
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Let $X$ be a scheme over $\mathbf{C}$. I think we should endow $X(\mathbf{C})$ with a topology as follows. If $X = \text{Spec}(A)$ is affine, then we write $A = \text{colim}\ A_i$ with $A_i$ of finite type over $\mathbf{C}$, so $X = \lim X_i$ with $X_i = \text{Spec}(A_i)$. Then $X(\mathbf{C}) = \lim X_i(\mathbf{C})$ and we endow the left hand side with the limit topology where each $X_i(\mathbf{C})$ is endowed with the usual one. This just means that a set is open if it comes from an open in one of the $X_i(\mathbf{C})$. In general we glue these topologies; I think this obviously works but I didn't check the details.

Some weird things can happen here, for example it can happen that $X(\mathbf{C})$ is empty even though $X$ is not empty. Also, you can get $X(\mathbf{C})$ to be homeomorphic to any profinite space you like for affine $X$. If $A = S^{-1}\mathbf{C}[x, y]$ then you get $\mathbf{C}^2$ where you remove (possibly infinitely many) plane curves; so you can get $U = \{(x, y) \in \mathbf{C}^2 \mid x \not \in \mathbf{Z}\}$ for example.

All I am trying to say here is that we should probably require $X$ to be (at least) locally of finite type over $\mathbf{C}$. Not an answer.

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    $\begingroup$ I'm a bit confused by the statement "a set is open if it comes from an open in one of the $X_i(\mathbb{C})$": in the case of $\mathbb{A}^{\infty}_{\mathbb{C}}=\lim_i Spec \mathbb{C}[x_1,\cdots, x_i]$, then $\mathbb{A}^1=Spec \mathbb{C}[x_1]$ would be open in $\mathbb{A}^{\infty}$ because open in $X_1$? Shouldn't the phrase read "closed if closed in one $X_i$ instead"? $\endgroup$
    – Qfwfq
    Oct 11, 2015 at 16:22
  • $\begingroup$ @Qfwfq 'comes from' means under pullback by the map $\mathbf A^{\infty} \to \mathbf A^1$. $\endgroup$ Aug 2, 2019 at 3:36

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