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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..

Community
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David Carchedi
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  • I remember I read in the "red book" that the notion of variety and scheme is the same over an algebraically closed field. – google Sep 10 '15 at 19:16
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    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. – Donu Arapura Sep 10 '15 at 21:50
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    How do you define the topological space associated to a scheme which is not of finite type? – Sam Gunningham Sep 11 '15 at 00:24
  • @SamGunningham Spec? Or, since David's only looking at homotopy types, the Artin-Mazur etale homotopy type (http://ncatlab.org/nlab/show/%C3%A9tale+homotopy)? – David Roberts Sep 11 '15 at 00:48
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    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. – Sam Gunningham Sep 11 '15 at 01:10
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    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 $B\widehat{\mathbb Z}$. – Sam Gunningham Sep 11 '15 at 01:14
  • @SamGunningham I wasn't quite thinking about the difference between the analytic and Zariski topology :-S. – David Roberts Sep 11 '15 at 01:22
  • @David Remark 3 of http://arxiv.org/abs/1310.2784 might be of relevance, it mentions the difference between A^1 homotopy and homotopy of the analytic topological space of schemes over C. – David Roberts Sep 11 '15 at 01:22
  • @SamGunningham I think locally of finite type is enough. For the general case, I don't think this question makes sense since GAGA does not apply. – user40276 Sep 11 '15 at 01:54
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    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 http://arxiv.org/abs/math/0609665. In this case, we have that $H^2(M) = 0$ by Mayer-Vietoris, but the Hausdorffification is contractible. – Sam Gunningham Sep 11 '15 at 02:25
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    @Sam: Thanks for pointing this out. I had even glanced at that paper earlier this year but had forgotten. It's reasonable to expect from this that non-seperated schemes over $\mathbb{C}$, even if of finite type, my fail to have the homotopy type of a finite CW-complex. However, the example you give turns out to be a $K(\mathbb{Z},1),$ so the question still remains if 2.) holds... – David Carchedi Sep 11 '15 at 02:53
  • The answer and comments to the MO question you pointed to only explain why the complex points of a separated scheme of finite type has a CW-structure, but it does not explain why we get homotopy types of finite CW-complexes. By virtue of a well known theorem of Whitehead, for such schemes, having the homotopy type of a finite CW-complex is thus equivalent to having the weak homotopy type of a finite CW-complex. Such a finiteness property is provided by Morse theory, resolution of singularities, and proper descent. – D.-C. Cisinski Oct 14 '15 at 14:55

2 Answers2

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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.

D.-C. Cisinski
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  • do you really need all of the complicated machinery to see finiteness? –  May 30 '19 at 08:43
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    @user141225 this complicated machinery is called mathematics. And yes, in order to have the level of precision expressed above, you need sophisticated tools. – D.-C. Cisinski May 31 '19 at 07:38
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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.

darx
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    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"? – Qfwfq Oct 11 '15 at 16:22
  • @Qfwfq 'comes from' means under pullback by the map $\mathbf A^{\infty} \to \mathbf A^1$. – R. van Dobben de Bruyn Aug 02 '19 at 03:36