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If X is a variety over the complex numbers, one reasonable thing to do is to consider the associated analytic space $X_{an}$ and to take the topological Euler characteristic of that.

Is there a purely algebraic way to obtain this number?

If X is non-singular then one might define it as the integral of the top Chern class of its tangent bundle.

The reason I ask is that I'm currently reading Joyce's survey on Donaldson-Thomas invariants and I wanted to know if by any chance he were using some more sophisticated notion.

On related note: if X is a non-proper scheme over C, why is its Euler characteristic well-defined?

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If X is e.g. smooth projective you can use Hodge decomposition. –  Kevin H. Lin Aug 10 '10 at 18:46
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1) You can use étale cohomology to get a purely algebraic definition. (You can also use algebraic de Rham cohomology as Kevin suggests which has an extension to the singular non-proper case). 2) On way of seeing that the Euler characteristic is well-defined is to use the fact that $X$ is homotopy equivalent to a finite complex. (You can also use the finiteness theorem for cohomology of constructible sheaves.) –  Torsten Ekedahl Aug 10 '10 at 19:07
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Yet another approach to the finiteness of the Betti numbers (which proves much more). Complex algebraic varieties can be compactified in the following way : let $X$ be such a variety, then there exists $\bar{X}$ smooth projective in which $X$ is a dense Zariski open and $D=\bar{X}-X$ is a strict normal crossing divisor in $\bar{X}$. You can then relate the cohomology groups of $X$ to those of $\bar{X}$ and $D$ (which are finite dimensional) in De Rham theory by using logarithmic differential forms : see e.g. the treatment of this in Claire Voisin's book on Hodge theory. –  Simon Pepin Lehalleur Aug 11 '10 at 7:58
    
@Torsten Ekedahl and Kevin Lin: Is using the algebraic de Rham complex to obtain the Euler characteristic just an application of GAGA or is there something I'm missing? –  babubba Aug 11 '10 at 23:25
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@angoleirovero: There is a little bit more than (classical) GAGA. First there is Grothendieck's result that for any (proper or not) smooth variety the algebraic de Rham cohomology is equal to the analytic one. Second for singular ones one embeds the variety (when it is possible) into a smooth and takes the de Rham cohomology of the formal completion of the variety in the ambient one. –  Torsten Ekedahl Aug 12 '10 at 4:21
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3 Answers

up vote 14 down vote accepted

If $X$ is a smooth variety over the complex numbers, you can use the topological Euler characteristic of the support $X(\mathbb{C})$ of the scheme. The most efficient way to compute it is to use Chern classes and the Poincaré-Hopf theorem: the Euler characteristic is the degree of the top Chern class. Chern classes can be defined purely algebraically (in homology) as it is done for example in Chapter 3 of Fulton's book "Intersection Theory".

When $X$ is singular, there are many nonequivalent generalizations of the Euler characteristic. One efficient approach is again to work in the spirit of the Poincaré-Hopf theorem by first defining Chern classes for singular varieties. This provides a purely algebraic treatment. I will give a tour of the 4 most famous examples:

-The Chern-MacPherson class (also known as the Chern-Schwartz-MacPherson class ) is probably the most important. It can be defined on the Chow group of a variety using constructible functions. It enjoys beautiful functorial properties under proper maps and it is compatible with specializations. The existence of the Chern-MacPherson class realizes a conjecture of Deligne and Grothendieck of 1969 as the unique natural transformation between the covariant functor of constructible functions and the usual covariant functor of $\mathbb{Z}$-homology such that it maps the characteristic function of a manifold to its (homological) Chern class. The Chern-MacPherson class computes the topological Euler characteristic of a (possibly singular) variety.

-The Chern-Mather-class enters the definition of the Chern-MacPherson class. It is relevant to compute stringy invariants. It has nice properties under proper birational maps and can be simply defined using a Nash-blow-up.

These two classes (MacPherson and Mather) can be generalized to scheme by considering the support of the scheme. This makes calculations much more easy, but obviously one looses a lot of the information carried by the scheme. The next two definitions are much more "scheme friendly" but they have less understood functorial properties.

-The Chern-Fulton and the Chern-Fulton-Johnson classes are both extremely sensitive to the scheme structure. They coincide in the case of local proper intersections. In general they don't satisfy the "inclusion-exclusion principle" which is satisfies by the topological Euler characteristic.

References

For a friendly introduction with references to the appropriate literature, I would recommend this short lecture notes by Paolo Aluffi. The general treatment is discussed in Fulton's book Intersection Theory (specially section s 4.2.6, 4.2.9 and 19.1.7). For applications to motivic integration and stringy invariants see for example this review or this one.

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While all the above are arguably "better" Euler characteristics, all that Joyce needs to get started is actually the naive Euler characteristic of the topological space $X(\mathbb{C})$. –  Arend Bayer Aug 12 '10 at 13:09
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@Arend Actually, I was really answering the question "what is the Euler characteristic of a scheme?" so I had to consider as well the singular case. –  JME Aug 12 '10 at 13:35
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Yes, my point was even in the singular case, Joyce just needs the naive Euler characteristic. To be a little more constructive: Joyce absolutely needs excision: $\chi(X) = \chi(Z) + \chi(X \setminus Z)$ for $Z \subset X$ closed. This basically forces $\chi$ to depend only on the reduced scheme structure, also it is determined by its values on smooth varieties (using a stratification and generic smoothness). Also, resolution of singularities implies that it is determined by its values on smooth proper varieties. Why is this is well-defined? This is indicated in Torsten Ekedahl's comments. –  Arend Bayer Aug 18 '10 at 23:54
    
Again, I was answering the question "what is the Euler characteristic of a scheme?". So I consider the singular case and I do not restrict myself to the case of Joyce since in that case the scheme structure is irrelevant. As I mention in my answer, the Chern-MacPherson class computes the topological Euler characteristic (what you call the "naive Euler characteristic"). It relies on the Borel-Moore homology (homology with compact support) and it is therefore well defined even in the non-proper case. Chern classes give a purely algebraic definition which is also computationally powerful. –  JME Aug 19 '10 at 9:57
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One comment on the "excision" property for Euler characteristics of complex algebraic varieties: $\chi(X) = \chi(Z) + \chi(X\backslash Z)$. There is a short proof of this in Fulton's Introduction to Toric Varieties, p. 142, which avoids the sticky triangulation issue and doesn't seem to be so well-known. In fact, it's equivalent to the statement that $\chi(X) = \chi_c(X)$ for any algebraic variety (the latter being the Euler characteristic for compactly supported cohomology).

This doesn't exactly help with the first question (about an algebraic definition), but it does show why the Euler characteristic is well-defined for any variety, per Arend's comment.

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Hi Dave, I just want to mention that your definition using excision actually also helps with the algebraic definition since the Euler characteristic you refer to matches the topological Euler characteristic and therefore can be defined purely algebraically using the Chern-MacPherson class. I discuss it in my answer. –  JME Aug 19 '10 at 11:01
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Here are some comments on the questions of the posting.

Every complex algebraic variety has the homotopy type of a finite CW-complex, so the Betti numbers are finite and the Euler characteristic is well defined. To simplify consider the quasi-projective case. Let $X\subset \mathbf{P}^n(\mathbf{C})$ be a projective variety and let $Y$ be a closed subvariety, then we can consider both $X$ and $Y$ as real algebraic varieties in $\mathbf{P}^{2n}(\mathbf{R})$. Now real projective spaces can be embedded in affine spaces, as opposed to the complex case: we can associate to a line $l$ in $\mathbf{R}^{k+1}$ the unique orthogonal projection with image $l$. A similar trick exists over an arbitrary field (the Jouanolou trick), but it is the uniqueness of the projection that gives us an embedding in the real case. In coordinates we take a point $(x_0:\cdots: x_k)$ to the $k+1$ by $k+1$ matrix $(\frac{x_i x_j}{\sum x_i^2})$.

This implies that real projective varieties are affine. Then one can use the triangulation theorem for real affine varieties, as presented e.g. in Hironaka's Arcata 1974 lectures to triangulate $X$ so that $Y$ is a subcomplex; this gives a triangulation of $X\setminus Y$ (infinite if $Y$ is nonempty) and a finite CW complex homotopy equivalent to $X\setminus Y$.

I'm pretty sure a similar result should hold for arbitrary (not necessarily projective) complex algebraic varieties and also for algebraic spaces, but I've never seen the details worked out in the literature.

The other question (whether or not the algebraic structure determines the cohomology) is not completely trivial either (and the answer depends on what exactly one means by cohomology). If $X$ is defined over a finite extension $F$ of $\mathbf{Q}$ and $X'=X\times_F\mathbf{C}$ for some embedding $F\subset\mathbf{C}$, then the cohomology ring of $X'(\mathbf{C})$ with finite coefficients does not depend on the embedding (and hence neither does the complex cohomology ring), see Freitag-Kiehl, \'Etale cohomology, theprem 11.6. It was an old question of Grothendieck whether the rational cohomology ring can depend on the embedding. It turned out even the real cohomology ring can, as recently shown by F. Charles. See www.math.ens.fr/~charles/crll5855.pdf

upd: here is an explicit procedure to obtain the Euler characteristic from the algebraic data: as mentioned in the comments, if $X$ is smooth and complete, we can take the alternating sum of the Euler characteristics of $\underline{\Omega^i}_X$'s. If $X$ the complement of a simple normal crossing divisor $D_1\cup\cdots \cup D_k$ in a smooth complete $D_{\varnothing}$, then $$\chi(X)=\sum_{I\subset\{1,\ldots,k\}}(-1)^{|I|}\chi(\cap_{i\in I} D_i).$$ Here we use the fact that the differentials in a spectral sequence do not change the Euler characteristic. In general one stratifies $X$ so that the difference of any two consecutive strata is smooth and takes the sum of the Euler characteristics of the strata. Using a similar procedure one can compute the Serre characteristic, which is a 2 variable analog; it can be seen as the image of the compactly supported cohomology in the Grothendieck group not of $\mathbf{Q}$-vector spaces, but of the mixed Hodge structures.

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For non-projective varieties, why should one expect them to be homotopy equivalent to a finite CW complex? A theorem of Whitney from the 60s says that they always have finitely many connected components, which is a start, but he has to do some work to prove that. –  Dan Ramras Aug 19 '10 at 2:56
    
Dan -- this seems to be one of those things that everyone believes in but that are difficult to locate in the literature. I would say one can prove this by first completing (Nagata's theorem) and then triangulating the resulting complete variety (and then using the fact that the complement of a finite subpolyhedron in a finite polyhedron is again homotopy equivalent to a finite polyhedron). I've just asked a similar question mathoverflow.net/questions/36050/…. If the answer to that question is positive, than the strategy works. –  algori Aug 19 '10 at 3:32
    
Whitney's paper is from 1957 (jstor.org/stable/1969908?origin=crossref) so I guess it's quite possible there's a more modern proof. I don't know Nagata's theorem. I did find a paper of his which constructs complete varieties which are not projective - projecteuclid.org/…. I'm not sure how that affects the argument you're sketching. Are you claiming that every complex variety embeds in a complete projective variety? –  Dan Ramras Aug 20 '10 at 16:56
    
Dan -- re: Are you claiming that every complex variety embeds in a complete projective variety?: no, but I'm pretty sure it embeds as a real analytic subvariety, which suffices for the triangulation purposes. –  algori Aug 22 '10 at 3:42
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