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Are virtual Betti numbers of a projective algebraic variety obtained from a cell decomposition the same as the usual Betti numbers?
To be more precise, let $X$ be an algebraic variety (not necessary irreducible) over the complex numbers, we define the $i$-th Betti number to be the dimension of the $i$-th singular cohomology group of $X$ with coefficients in $\mathbb{C}$.
Assume that $X$ has a cell decommposition, i.e. there is a filtration of closed subvarieties $ F^1\subset \cdots \subset F^n =X$ such that the complements $F^i \setminus F^{i-1}$ are disjoint union of affine spaces, which we call cells.
Then, the virtual Betti numbers (following Byalinicki-Birula) are
$ b_{2i+1}=0$, $b_{2i}=$ number of $i$-dim cells.

Here are two remarks on it.
1) If one takes Borel-Moore homology instead of singular cohomology the equality of virtual Betti numbers with Betti numbers is a direct consequence of the long exact localization sequence (for an open subset and its closed complement) and the knowledge of the BM-homology on affine spaces. This does not use the assumption of $X$ being projective. So, for spaces with cell decomposition is the BM-homology isomorphic to singular cohomology? But there must be assumptions on $X$ to get an isomorphism, else $X=\mathbb{C}^n$ would be a counterexample.

2) In Fresse: Betti numbers of Springer fibre in type A, p.12 they state the equality of virtual Betti numbers with the dimensions of the usual sheaf cohomology with coefficients in $\mathbb{Q}$. He gives as source Kashiwara, Schapira (Sheaves on Manifolds), section 4.6 which does not seem to exist.

Of course, when you assume $X$ to be smooth (and projective), or having an operation of a torus with only finitely many fixed points, there are much more methods available.

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If you're only concerned with smooth projective varieties over C (which in particular are compact spaces) then BM homology coincides with singular homology. –  Faisal Oct 24 '11 at 7:18
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2 Answers

Sure, it is true. Let's check once again the definition of virtual Betti numbers:

  1. For $X$ smooth and compact $b_i(X)=\dim_{\mathbb{Q}} H^i(X,\mathbb{Q})$.
  2. For any closed $Y\subset X$ one has $b_i(X) = b_i(Y)+b_i(X-Y)$.

Now, whenever you have an algebraic cell decomposition $F^1\subset\ldots\subset F^{n-1}\subset F^n=X$, by (2) you have $b_i(X)=b_i(F^n-F^{n-1})+\ldots+b_i(F^2-F^1)+b_i(F^1)$, where each $F^i-F^{i-1}$ is a disjoint union of some affine spaces. It is left to see that $b_i(\mathbb{A}^j)=b_i(\mathbb{P}^j)-b_i(\mathbb{P}^{j-1})=\delta^i_{2j}$.

Now combine with the first Proposition in [Fresse, 4.1] and get that whenever $X$ is projective $b_i(X)=\dim_{\mathbb{Q}}H^i(X,\mathbb{Q})$. This holds because $X$ compact implies $H^i_c(X,\mathbb{Q})=H^i(X,\mathbb{Q})$.

Ok, let's see why this Proposition holds in our particular situation. Take the first step of the stratification: $F^{n-1}\subset F^n=X$. Denote $U=F^n-F^{n-1}$. Then there is a long exact sequence associated to a closed subset: $\ldots\to H^i_c(U)\to H^i_c(X)\to H^i_c(F^{n-1})\to H^{i+1}_c(U)\to\ldots$. All the coefficients are $\mathbb{Q}$. Now $U$ is a disjoint union of affine cells of dimension equal to $\dim X$ and $\dim F^{n-1}<\dim X$ (you can always achieve this by a proper choice of stratification). Note that $F^{n-1}$ also admits a cell decomposition. So we may assume that the statement holds for $F^{n-1}$ by an inductive argument. In Particular, $H^i_c(F^{n-1})=0$ for all $i>2\dim F^{n-1}$, thus, for all $i > 2(\dim X-1)$. Combine with the long exact sequence and the fact that $\dim H^i_c(\mathbb{A}^n)=\delta^i_{2n}$. This finishes the proof.

The reference for the long exact sequence associated to a closed subset will be Iversen, Cohomology of sheaves.

However, as far as I understand, having an algebraic cell decomposition is somewhat strong. Note that in general virtual Betti numbers can be negative.

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Thanks a lot, that is much easier than I expected. –  Julia Sauter Oct 24 '11 at 17:48
    
But singular cohomology does not fulfill property (2). For example, $H^0_{sing}(\mathbb{C})\neq H^0_{sing} (pt) \oplus H^0_{sing}(\mathbb{C}\setminus pt)$. –  Julia Sauter Oct 24 '11 at 17:54
    
You just proved that an alternative definition of virtual Betti numbers gives the definition of virtual Betti numbers that I gave. –  Julia Sauter Oct 24 '11 at 17:56
    
Julia, exactly. It's well known that if you want, say, additive Euler characteristics, you should take cohomology with compact support. –  Anton Fonarev Oct 24 '11 at 18:06
    
Julia, I've updated my answer to clarify the situation. –  Anton Fonarev Oct 24 '11 at 18:22
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Julia,

This more of an extended series of comments.

  1. Rational Borel-Moore homology is dual to compactly supported cohomology. So if the underlying space is compact then the dimensions are the usual Betti numbers but not in general (e.g. as you point out, not for $\mathbb{C}^n$). [Added: To see the duality, take a look at item 19.1.1 of Fulton's Intersection Theory. You'll see an isomorphism of Borel-Moore with homology of the one point compactification modulo infinity. The previous statement follows easily.]

  2. As Anton points out, most varieties do not have cell decompositions. Among smooth curves, the affine and projective lines are the only examples. So a definition that relies on this is pretty limited.

  3. I'm not quite sure what exactly you mean by virtual Betti number in general, but again I'll follow Anton here (I hope he doesn't mind). There is subtlety, however, and that is to prove that an invariant exists satisfying his properties 1. and 2. It is not a priori clear. One case use $$b'_i(X) = \pm\sum_j (-1)^j Gr_W^i H_c^j(X)$$ where $W$ is Deligne's weight filtration and I use $b_i'$ to distinguish it from the usual Betti number. I think this invariant goes back to Danilov and Khovanskii, but I'm not completely sure.

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Of course, I don't mind. Let me also point out that usually one proves this kind of existence by means of the weak factorization conjecture (which is theorem nowadays). –  Anton Fonarev Oct 25 '11 at 6:27
    
Thank you Donu, I did not know that rational BM-homology is dual to compactly supported cohomology ($H_c^*(X, \mathbb{Q} )$), that would give answer to me, where can I find this duality? I do not know how vBN are defined in general, my definition for varieties with cell decompositions is from an old article of Bialynicki-Birula. –  Julia Sauter Oct 25 '11 at 6:57
    
Dear Donu, I think one needs to use $H^j_c$ instead of $H^j$ in the definition of virtual Betti numbers. –  ulrich Oct 25 '11 at 7:45
    
Julia: I'll try to dig up a reference when I have time. Thanks Ulrich, I fixed it. –  Donu Arapura Oct 25 '11 at 11:15
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