Topological type $x_0^2+x_1^2+x_2^2+ x_3^2+x_4^2 = 0$ in $\mathbb P^4$ Let's consider projective variety $V$ given by th equation $x_0^2+x_1^2+x_2^2+ x_3^2+x_4^2 = 0 \ $ in $\mathbb CP^4$.
I was wondering what is the Picard group of $V$ ? Or cohomology ring of $V$ ? 
 A: If you $2$-uply embed $\mathbb{P}^4$ into $\mathbb{P}^{14}$, then this $3$-fold is a hyperplane section of $\mathbb{P}^4$. By the Lefschetz hyperplane theorem, we deduce that $\mathrm{Pic}(V) \cong \mathbb{Z}$ and the betti numbers of $V$ are
$$1,\ 0,\ 1,\ ?,\ 1,\ 0,\ 1.$$
In fact, the middle term is $0$. The reason I know this is that your space is $SO(5)(\mathbb{C})/P$ for an appropriate parabolic $P$, and the complex homogenous spaces have no odd cohomology. There is probably a better way to see this.
The Lefschetz hyperplane theorem tells us that $H^2(\mathbb{P}^4) \to H^2(V)$ is an isomorphism. Letting $\zeta$ be a generator of $H^2$, we see that $\zeta^3$ is twice the fundamental class because $V$ has degree $2$. Poincare duality tells us that $\zeta^2$ must be twice the generator of $H^4$. So integer generators of the cohomology groups are
$$1,\ \zeta,\ \frac{\zeta^2}{2},\ \frac{\zeta^3}{2}$$
and the ring structure is obvious. 
A: More generally, the cohomology of any smooth hypersurface in $\mathbb{C}P^n$ is described in the paper 'Topology of nonsingular complex hypersurfaces' by Kulkarni and Wood.  It depends only on $n$ and the degree of the hypersurface.  (In fact, any two smooth hypersurfaces of the same degree are even diffeomorphic.)  The proof is basically Morse theory.
