Picard/cohomology lattice of surfaces of low degree in $\mathbb P^3$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:49:13Z http://mathoverflow.net/feeds/question/92072 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92072/picard-cohomology-lattice-of-surfaces-of-low-degree-in-mathbb-p3 Picard/cohomology lattice of surfaces of low degree in $\mathbb P^3$ Dmitry Kerner 2012-03-24T11:19:15Z 2012-03-29T16:44:29Z <p>Let $S_{d>3}\subset\mathbb{P}^3_{\mathbb{C}}$ be a smooth surface of degree $d$. What is known (where to read?) about the Picard/cohomology lattice for small d?</p> <p>e.g. for $d=4$ the cohomology lattice is $E_8^{\oplus2}\oplus H^{\oplus3}$, but what is the class of a hyperplane section in this lattice? Probably many things are known for $d=5,6$? (I need the results for any smooth $S_d$, not necessarily generic.)</p> <p>And what is known for "not-too-singular" $S_d$? (e.g. with nodes only)</p> <p>upd. I guess any two smooth surfaces of the same degree in $\mathbb{P}^3$ have the same cohomology lattice? I want to know this lattice and in particular what is the plane section in this lattice, i.e. $\mathcal{O}(1)$.<br> \ In my actual problem I have a curve lying on such a surface, want to identify its equivalence class. (If you wish: the $\mathbb{P}^3$-degree of $C$ is $\frac{d(d-1)}{2}$, the self-intersection on $S_d$ is: $C^2=\frac{d(d-1)(d-2)}{6}$, so the arithmetic genus is: $p_a=\frac{(2d+1)(d-2)(d-3)}{6}$. Such a curve is quite special, lying on the surface it forces the surface to be rather special too.)</p> <p>So, I'd need as much information as possible about $S_{d>3}\subset\mathbb{P}^3$. Such a very-classical geometry.</p> http://mathoverflow.net/questions/92072/picard-cohomology-lattice-of-surfaces-of-low-degree-in-mathbb-p3/92582#92582 Answer by Johannes Nordström for Picard/cohomology lattice of surfaces of low degree in $\mathbb P^3$ Johannes Nordström 2012-03-29T16:44:29Z 2012-03-29T16:44:29Z <p>You can compute the invariants necessary to determine the lattice structure of $H^2(S_d)$ from the characteristic classes. $(1+d\sigma)(1+c_1+c_2) = (1+4\sigma+6\sigma^2)$, so $c_1 = (4-d)\sigma$ and $c_2 = (d^2 -4d + 6)\sigma^2$ (where $\sigma$ is the hyperplane class on $S_d$). Identifying top classes with their integrals $\sigma^2 = d$, so the Euler characteristic is $\chi = c_2 = d^3 -4d^2 +6d$ and the signature is $\frac{1}{3}p_1 = \frac{1}{3}(-c_2 + c_1^2) = -\frac{d^3 - 4d}{3}$. ($H^1(S_d) = 0$, so the rank of $H^2(S_d)$ is $\chi -2$.) The Stiefel-Whitney class $w_2 = c_1$ mod 2, so vanishes if and only if $d$ is even. </p> <p>So for $d = 5$ we get the odd unimodular lattice of rank $53$ and signature $-35$, i.e. $H^2(S_5) = 9\langle 1 \rangle \oplus 44\langle -1 \rangle$. For $d = 6$ we get the even unimodular lattice of rank $106$ and signature $-64$, i.e. $H^2(S_6) = 8E_8 \oplus 19H$. (Sanity check: for $d = 3$ the argument gives that $H^2(S_3)$ is odd of rank $7$ and signature $-5$, as it should be since $S_3$ is the blow-up of $\mathbb{P}^2$ at 6 points.)</p> <p>There's not much to do to identify the hyperplane class $\sigma$ in the cohomology lattice; the lattice has plenty of automorphisms, so any two primitive elements of the same norm are equivalent. Similarly, if you want to know the relation between $\sigma$ and the cohomology class of $C$, then you just need to know the intersection form on the sublattice spanned by $\sigma$ and $C$; the fact that $H^2(S_d)$ is indefinite implies that any two (primitive) isometric sublattices of rank $&lt; \frac{rk \; H^2(S_d)}{2} - 1$ are equivalent under automorphisms of $H^2(S_d)$ (see Theorem 1.4.8 of Dolgachev, "Integral quadratic forms: Applications to algebraic geometry [After V. Nikulin]" for the even case (where the inequality does not have to be strict); the odd case can be deduced from Theorem 1.16.10 in Nikulin, "Integral symmetric bilinear forms and some of their applications").</p> <p>I'm not sure this is quite what you were asking for, but perhaps it's a start.</p>