# Cohomology of quadric in $\mathbb{C}^4$ at infinity

How to prove that the singular cohomology of $$X=V(xy-zw)\setminus 0 \subset \mathbb{C}^4$$ is $$H^*(X,\mathbb{Z})=\{\mathbb{Z},0,\mathbb{Z},\mathbb{Z},0,\mathbb{Z}\}?$$ Preferably, I would like to know whether the stronger result $$X \simeq S^2 \times S^3$$ holds?

• I have to go and don't have time to see whether this approach will work or not, but one thing to try is working with the projectivization of $X$ and use Meyer-Vietoris. The projectivization if the Grassmannian G(2, 4), for which you know the cohomology (or you can appeal to various complex-analytic theorems for a hypersurfaces). Commented May 1, 2019 at 14:49
• The map $f : X \mapsto \Bbb P^1, (x,y,z,w) \mapsto x/z$ has fiber $f^{-1}(a) = (az,y,z,ay)\cong \Bbb C^2 \backslash \{0\}$, so your space is homotopy equivalent to a $S^3$-bundle over $S^2$. Commented May 1, 2019 at 15:19
• @hunter The (complex) projectivisation of X is known to be isomorphic to $\mathbb{C} P^1 \times \mathbb{C} P^1$ and I am not sure where the Grassmannian G(2,4) (real or complex?) appears here? Commented May 1, 2019 at 15:36
• @Filip92 I don't think that's true -- the dimensions are wrong. The projectivization of $X$ is a quadric in $\mathbb{P}^4$, not $\mathbb{P}^3$. Commented May 1, 2019 at 15:42
• Ah, I see what you are saying - I thought that by ''projectivisation'' you mean projection of X in $\mathbb{C} P^3$, whereas you mean the same curve in $\mathbb{C}P^4 \supset \mathbb{C}^4$ Commented May 1, 2019 at 15:45