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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 $V$ ?

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up vote 9 down vote accepted

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.

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The hyperplane theorem follows from the fact that the complement of the variety in $\mathbb{C}P^n$ is homotopy equivalent to a cell complex whose (real) dimension is $n$. In this case, one can see explicitly that the complement deformation retracts to $\mathbb{R}P^3$, and deduce that $b_3(V) = 0$. – Johannes Nordström Mar 9 '12 at 17:37
You can chance coordinates to $a=x_0+ix_1$, $b=x_0-ix_1$, $c=x_2+ix_3$< $d=x_2-ix_3$ to get the equation $ab+cd+x_4^2=0$, which shows that the surface is birational to $\mathb P^3$. I don't know enough about 3-folds to tell if this all implies it is isomorphic to $\mathbb P^3$. – Will Sawin Mar 9 '12 at 19:28
@Will, the cohomology is not that of $\mathbb P^3$, though, no? – Mariano Suárez-Alvarez Mar 9 '12 at 20:09

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.

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The fact you mention is a direct consequence of Ehresmann's fibration theorem (which seems more elementary to me than Morse theory). – Henri Mar 10 '12 at 9:44

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