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Let $X$ be a (singular) projective variety, in other words something given by a collection of polynomial equations in $\mathbb CP^n$ or $\mathbb RP^n$. How can one prove it is a finite $CW$ complex?

Similar question: Suppose that $X$ affine (i.e. given by polynomial equations in $\mathbb C^n$, or $\mathbb R^n$). How can one prove its one point compactification is a finite $CW$ complex?

These questions are sequel to the discussions here:

For which classes of topological spaces Euler characteristics is defined?

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    $\begingroup$ See mathoverflow.net/questions/15087/… $\endgroup$ Jun 3, 2010 at 14:40
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    $\begingroup$ Particularly my answer :) $\endgroup$ Jun 3, 2010 at 14:41
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    $\begingroup$ You may also want to look at the paper by Hironaka: "Triangulations of Algebraic Sets", p. 165--185, in the proceedings from the 1974 AMS Arcata conference in Algebraic Algebraic geometry. The purpose of the article is exactly to give a simple demonstration of a fact which "everyone knows", but which is reputed to be difficult. $\endgroup$
    – Mike Roth
    Jun 3, 2010 at 14:51

2 Answers 2

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The Lojasiewicz theorem says that every semi-algebraic subset of $\mathbf{R}^n$ can be triangulated. Moreover, there is a similar statement for pairs of the form (a semi-algebraic set, a closed subset). See e.g. Hironaka, Triangulations of algebraic sets, Arcata proceedings 1974 and references therein (including the original paper by Lojasiewicz).

The case of an arbitrary (not necessarily quasi-projective) complex algebraic variety follows from Nagata's theorem (every variety can be completed) and Chow's lemma (every complete variety can be blown up to a projective one).

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  • $\begingroup$ Algori, thanks! What about one-point compactification? How do you show that this is a finit CW-complex? $\endgroup$ Jun 3, 2010 at 17:02
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    $\begingroup$ This is not too difficult - take a completion and then trianguate the resulting couple. If $K$ is a subpolyhedron of a polyhedron $K$, then $K/L$ can also be triangulated. $\endgroup$
    – algori
    Jun 3, 2010 at 17:33
  • $\begingroup$ That is, if $L$ is a subpolyhedron... $\endgroup$
    – algori
    Jun 3, 2010 at 17:34
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The following thesis seems to be more general and worth referencing at this post:

Triangulation of Locally Semi-Algebraic Spaces. by K.R. Hofmann.

I quote from the abstract:

"We give necessary and sufficient conditions for a locally semi-algebraic space to be homeomorphic to a simplicial complex. Our proof does not require the space to be embedded anywhere, and it requires neither compactness nor projectivity of the space. A corollary is that every real or complex algebraic variety is triangulable, a result which does not seem to be available in the literature when the variety is neither projective nor real and compact."

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