# Hypersurface of complex projective space

Is there example of hypersurface $X \subset \mathbb{P}^n$ satisfying

1. $X$ is of degree 2. (I mean, the Poincare dual PD(X) is 2u, where u is a generator of $H^2(\mathbb{P}^n, \mathbb{Z})$.
2. Some odd betti number is non-zero.

Thank you for any comment.

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All quadrics are cellular. In particular, it implies all odd Betti numbers are zero. –  Sasha Oct 13 '10 at 18:15

As noticed by Sasha in his comment, the answer is no.

The following proof also shows that this result cannot be generalized for higher values of the degree.

For any smooth complex hypersurface of degree $d$, say $X_d \subset \mathbb{P}^{n+1}$, by standard arguments involving Lefschetz theorem we have

$H^k(X_d)= H^k(\mathbb{P}^n)$ for $k \neq n$.

In particular, all the odd Betti numbers are zero, except possibly the middle Betti number when $n$ is odd. On the other hand, the Euler-Poincare characteristic of $X_d$ is equal to

$\chi(X_d)= \langle c_n(T_{X_d}), [X_d] \rangle =\frac{1}{d}[(1-d)^{n+2}-1]+n+2$,

so for $n$ odd and $d=2$ the middle cohomology group must be zero too. Notice that for $n$ odd and $d >2$ one always has a non-zero middle Betti number. For instence, if $X \subset \mathbb{P}^4$ is a smooth cubic hypersurface, then $b_3(X)=10$.

A good reference for these results is Dimca's book "Singularities and topology of hypersurfaces", Chapter 5, which also considers the case of hypersurfaces in $\mathbb{P}^{n+1}$ with isolated singularities.

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I don't know whether the question was meant to include non-smooth quadric hypersurfaces, but the answer is the same for them: their (co)homology is free and concentrated in even degrees. In fact, every singular quadric $X$ in $\mathbb P^{n+1}$ is the projective cone on some quadric $Y$ in $\mathbb P^{n}$, making $H_k(X)$ isomorphic to $H_{k-2}(Y)$ for every $k>0$. –  Tom Goodwillie Oct 13 '10 at 19:22
Yes, that's true. However, I find this proof interesting since its provides the Betti numbers for hypersurfaces of any degree d, not only for quadrics, showing that in higher degrees, when n is odd, one always has a non-zero odd Betti number. I have edited the answer to make this more clear. –  Francesco Polizzi Oct 13 '10 at 21:31