Let $V$ be a proper singular complex algebraic variety, possibly nonprojective ($dim(V)=n>0$). I would like to know:


1) if its second Betti number is non zero, 


2) same question but now $V$ is a compact Moishezon space. 


Edit: (as written by Francesco Polizzi) $b\_2\geq 1$ in the case of normal projective varieties (we even have a Hard Lefschetz theorem in intersection cohomology).
We also have $b\_2\geq 1$ in the case of smooth complete algebraic varieties, this gives a topological criterion to prove "non-algebraicity" of some smooth complex manifolds.
Thus my question is really about:


1') complete but singular algebraic,


2') compact Moishezon spaces (because they are analytification of complete algebraic spaces). 



Second betti number of compact analytic spaces
$\def\ZZ{\mathbb{Z}}$ Let $X$ be a complete toric variety. I will show that $H^2(X, \mathbb{Z}) \cong A^1(X, \mathbb{Z}) \cong \mathrm{Pic}(X)$. Since there are examples of complete toric $3$-folds with trivial $\mathrm{Pic}$, this shows that these toric varieties have trivial $H^2$. I rely heavily on the theory of Minkowski weights by [Fulton and Sturmfels](http://arxiv.org/abs/alg-geom/9403002). See corollary 2.4 of that paper for the isomorphism $\mathrm{Pic}(X) \cong A^1(X)$ and example 5.6 for a toric 3-fold with trivial $\mathrm{Pic}$.


It is possible that if I understood mixed Hodge theory better, I could read this off immediately from the Fulton-Sturmfels paper. The last sentence of Section 1 says that, for $X$ a complete toric variety, $A^{\ast}(X) \otimes \mathbb{Q}$ is isomorphic to the appropriate weight space of $H^{2 \ast}(X)$, and maybe it is obvious to an expert that the other weight spaces of $H^2(X)$ vanish in our setting. But it isn't clear to me, so I will compute $H^2(X)$ explicitly and show that it matches with the description of $A^1(X)$ by Fulton and Sturmfels. 


Let $\Sigma$ be a complete fan and $X$ the corresponding toric variety. I'll write $\Sigma^k$ for the set of cones in $\Sigma$ of codimension $k$. For $\rho$ a full dimensional cone of $\Sigma$, and $\sigma$ a codimension $1$ wall bounding $\sigma$, write $m\_{\rho \sigma}$ for the minimal vector of the dual lattice which is perpendicular to $\sigma$ and positive on $\rho$.


Fulton and Sturmfels (Lemma 2.3) show that $A^1(X)$ is isomorphic to the functions $c: \Sigma^1 \to \ZZ$ which obey the following condition: Let $\tau$ be any cone of codimension $2$ and let $\sigma\_1 \subset \rho\_1 \supset \sigma\_2 \supset \cdots \supset \sigma\_t \subset \rho\_t \supset \sigma\_{t+1} = \sigma\_1$ be the cones surrounding it, with $\rho\_i$ of codimension $0$ and $\sigma\_i$ of codimension $1$. Then we must have
$$\sum\_i c(\sigma\_i) m\_{\rho\_i \sigma\_i} = 0.$$


I must now realize this result in cohomology. For any cone $\kappa$, let $U\_{\kappa}$ be the corresponding affine open. So, if $\kappa \subset \lambda$ then $U\_{\kappa}$ is an open subset of $U\_{\lambda}$. We have an open cover $X = \bigcup\_{\rho \in \Sigma\_0} U\_{\rho}$, indexed by the maximal cones of $\Sigma$. We now use what is basically the Meyer-Vietores spectral sequence (chapter 15 of Bott and Tu). I say basically, because I only want to use the intersection $U\_{\rho\_1} \cap U\_{\rho\_2}$ if $\rho\_1$ and $\rho\_2$ meet on a common wall. Since this is an MO answer not a paper, I'm going to be sloppy about this (but advertise two relevant questions [1](https://mathoverflow.net/questions/91237/calculating-mayer-vietoris-efficiently) [2](https://mathoverflow.net/questions/23175/mathematically-mature-way-to-think-about-mayer-vietoris).) We should get a spectral sequence which starts with $\bigoplus\_{\alpha \in \Sigma^p} H^q(U\_{\alpha\_p})$, with the first page coming from the maps $H^q(U\_{\alpha}) \to H^q(U\_{\beta})$ whenever $\beta$ is a codimension one cone in $\alpha$.


For any $\alpha \in \Sigma^p$, we have $U\_{\alpha} \cong (\mathbb{C}^{\ast})^p \times V\_{\alpha}$ where $V\_{\alpha}$ is Spec of a pointed semi-group ring. We can contract $V\_{\alpha}$ onto its torus fixed point, so $H^q(U\_{\alpha}) \cong H^q((S^1)^p) = \bigwedge^q \ZZ^p$. Explicitly, the $\ZZ^p$ in question is lattice vectors orthogonal to $\alpha$. So our first page has the form
$$\begin{matrix}
 & & & & & & & \ddots \\
 & & & & & & \ZZ^{\Sigma^3} & \cdots \\
 & & & & \ZZ^{\Sigma^2} & \longrightarrow & \ZZ^{3 \times \Sigma^3} & \cdots \\
 & & \ZZ^{\Sigma^1} & \longrightarrow & \ZZ^{2 \times \Sigma^2} & \longrightarrow & \ZZ^{3 \times \Sigma^3} & \cdots \\
\ZZ^{\Sigma^0} & \longrightarrow & \ZZ^{\Sigma^1} & \longrightarrow & \ZZ^{\Sigma^2} & \longrightarrow & \ZZ^{\Sigma^3} & \cdots \\
\end{matrix}$$
(You'll have to imagine those diagonal dots go the right way; MathJax doesn't seem to have the `\iddots` command.)


The bottom row is the cohomology of a ball shaped $CW$-complex (the dual to the fan), so it wipes out on the next page. The map $\ZZ^{\Sigma^1} \to \ZZ^{2 \Sigma^2}$ exactly sends $\sigma \mapsto c(\sigma)$ to $\tau \mapsto \sum\_i c(\sigma\_i) m\_{\rho\_i \sigma\_i}$ in our earlier notation. So the $(1,1)$ element on the next page is $A^1(X)$, and in general I believe the $(p,p)$ element is $A^p(X)$. (There is an issue I haven't checked carefully: Let $N$ be the lattice in which $\Sigma$ lives and $M$ the dual lattice. Let $\ZZ \alpha$ be the sublattice of $N$ spanned by $\alpha$ and let $\ZZ \alpha^{\perp}$ be the orthogonal sublattice of $M$. The spectral sequence naturally wants the map from the $(p,p)$ term to the $(p+1, p)$ term to go $\bigoplus\_{\alpha \in \Sigma^p} \bigwedge^p \ZZ \alpha^{\perp} \to \bigoplus\_{\beta \in \Sigma^{p+1}} \bigwedge^p \ZZ \beta^{\perp}.$ Fulton and Sturmfels work with $\bigoplus\_{\alpha \in \Sigma^p} \ZZ \to \bigoplus\_{\beta \in \Sigma^{p+1}} N/ \ZZ \beta$. That's the same thing up to choosing orientations of all the cones. For $p=1$, Fulton and Sturmfels work out the translation for you and you can see that it gives the right answer; I haven't done the analogous computation for $p>1$.)


So the next page has the form
$$\begin{matrix}
 & & & & & \ddots \\
 & & & & A^3(X) & \cdots \\
 & & & A^2(X) & ? & \cdots \\
 & & A^1(X) & ? & ? & \cdots \\
\ZZ & & & & & \cdots \\
\end{matrix}$$
with arrows going right $2$ and down $1$. We see that $H^2$ has stabilized at $A^1$, as promised.


