Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $V$ ?

share|improve this question
add comment

2 Answers

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.

share|improve this answer
3  
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
2  
@Will, the cohomology is not that of $\mathbb P^3$, though, no? –  Mariano Suárez-Alvarez Mar 9 '12 at 20:09
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.