Hypersurface of complex projective space - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T00:26:32Z http://mathoverflow.net/feeds/question/42038 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42038/hypersurface-of-complex-projective-space Hypersurface of complex projective space YCho 2010-10-13T17:46:26Z 2010-10-15T07:39:30Z <p>Is there example of hypersurface $X \subset \mathbb{P}^n$ satisfying </p> <ol> <li>$X$ is of degree 2. (I mean, the Poincare dual PD(X) is 2u, where u is a generator of $H^2(\mathbb{P}^n, \mathbb{Z})$.</li> <li>Some odd betti number is non-zero. </li> </ol> <p>Thank you for any comment. </p> http://mathoverflow.net/questions/42038/hypersurface-of-complex-projective-space/42048#42048 Answer by Francesco Polizzi for Hypersurface of complex projective space Francesco Polizzi 2010-10-13T18:49:49Z 2010-10-15T07:39:30Z <p>As noticed by Sasha in his comment, the answer is <strong>no</strong>. </p> <p>The following proof also shows that this result cannot be generalized for higher values of the degree.</p> <p>For any smooth complex hypersurface of degree $d$, say $X_d \subset \mathbb{P}^{n+1}$, by standard arguments involving Lefschetz theorem we have </p> <p>$H^k(X_d)= H^k(\mathbb{P}^n)$ for $k \neq n$.</p> <p>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</p> <p>$\chi(X_d)= \langle c_n(T_{X_d}), [X_d] \rangle =\frac{1}{d}[(1-d)^{n+2}-1]+n+2$,</p> <p>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 $b_3(X)=10$. </p> <p>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. </p>