Picard/cohomology lattice of surfaces of low degree in $\mathbb P^3$

2012-03-24T11:19:15Z

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?

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.)

And what is known for "not-too-singular" $S_d$? (e.g. with nodes only)

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. $\mathcal{O}(1)$.
\ 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.)

So, I'd need as much information as possible about $S_{d>3}\subset\mathbb{P}^3$. Such a very-classical geometry.

Answer by Johannes Nordström for Picard/cohomology lattice of surfaces of low degree in $\mathbb P^3$

2012-03-29T16:44:29Z

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.

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.)

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 $< \frac{rk \; H^2(S_d)}{2} - 1$ 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").

I'm not sure this is quite what you were asking for, but perhaps it's a start.

Picard/cohomology lattice of surfaces of low degree in $\mathbb P^3$ - MathOverflow

Picard/cohomology lattice of surfaces of low degree in $\mathbb P^3$

Answer by Johannes Nordström for Picard/cohomology lattice of surfaces of low degree in $\mathbb P^3$