Take the 2-minute tour ×

I'm looking for an example in the literature where $\mbox{Pic}^0(X)$, $\mbox{Pic}(X)$, and $NS(X)$ of a projective surface $X$ over a field are calculated. I want them for an example I'm trying to work out, so ideally $X$ would be relatively simple, perhaps a cubic hypersurface in $\mathbb{P}^3$, or something along those lines. I know it's out there, but googling and browsing arXiv and MathSciNet haven't quite panned out.

share|improve this question
Other easy examples might include things like toric surfaces and ruled surfaces. –  Karl Schwede Sep 30 '10 at 4:44
Sorry for the dumb question, but what is $Pic^0$? –  Martin Brandenburg Sep 30 '10 at 12:24
It the connected component of the identity in the Picard Scheme in general, see for example en.wikipedia.org/wiki/Picard_variety See pages 21 and 74 of Mumford's Abelian Varieties for other descriptions when X is an abelian variety (in that case, in char 0 at least, Pic^0(X) is the dual abelian variety). –  Karl Schwede Sep 30 '10 at 15:30

5 Answers 5

up vote 16 down vote accepted

Manin's book "Cubic forms" contains the calculations of these groups when $X$ is a smooth projective cubic surface. In particular, $\operatorname{Pic}^0(X)=\{0\}$ and $\operatorname{Pic}(X)=\operatorname{NS}(X)$ is a free commutative group of rank 7.

Another class of examples is provided by products $X=E \times E'$ of two elliptic curves. In particular, if $E=E'$ has no complex multiplication then $NS(E\times E)$ is a free commutative group of rank 3 generated by the classes of $E \times \{0\}$, $\{0\}\times E$ and the diagonal while $\operatorname{Pic}^0(E\times E)=E \times E$. See Mumford's Abelian Varieties.

share|improve this answer
Welcome to MO, Professor Zarhin. –  Pete L. Clark Sep 30 '10 at 3:31
The case of cubic surfaces is also explained in Chapter 5 of Hartshorne. (More precisely, the Pic is computed there; I forget whether Hartshorne says anything about NS, although of course in this case they are equal.) –  Emerton Sep 30 '10 at 9:08
Just for completeness, let me mention that these groups are easy to compute for all blowups of P^2, which includes the case of cubic surfaces. Result: Pic^0 is always trivial, and Pic = NS is always free abelian, with rank increasing by 1 each time you blow up. –  user5117 Sep 30 '10 at 9:53

For surfaces in $\mathbb{P}^3$ of degree at most 3 the calculation of $Pic(X)$ is relatively easy: In this case $X$ is rational, hence $NS(X)$ modulo torsion equals $H^2(X,\mathbb{Z})$. (If you work over the complex numbers you might also apply Lefschetz (1,1)-Theorem.)

Starting from degree 4 the calculation of $NS(X)$ is much more involved. The difficulty depends on how you present $X$. In case you give $X$ just by an equation it is not so easy to calculate $NS(X)$, at least if you work in characteristic 0. See e.g., http://pjm.math.berkeley.edu/ant/2007/1-1/p01.xhtml where an example is given of a quartic surfaces with $\mathrm{rank} Pic(X)=\mathrm{rank} NS(X)=1$.

Over a finite fields (or over $\overline{\mathbb{F}_p}$) you can find an upper bound for the rank of $NS(X)$ in terms of the zeta function of $X$ (see loc. cit.). If you believe the Tate conjecture then this upper is the actual rank of $NS(X)$. In concrete examples you might try to use this upper bound, try to find sufficiently many curves on $X$ and then use the intersection pairing to prove that the classes of these curves are independent in $NS(X)$.

share|improve this answer

In general, if $X$ is a smooth complex projective variety which is simply connected, then we have $\rm{Pic}^0(X)=0$. Indeed we have $H^1(X,\mathbb{Z})=0$, and then Hodge theory implies that $H^1(X,\mathcal{O}_X)=0$. The exponential sheaf sequence http://en.wikipedia.org/wiki/Exponential_sheaf_sequence then implies that the natural map $\rm{Pic}(X) \to H^2(X, \mathbb{Z})$ is injective.

In particular, any hypersuface of dimension greater than $1$ is simply connected (by the Lefschetz hyperplane section theorem), and so $\rm{Pic}^0(X)$ is always trivial in this case.

share|improve this answer
Over any field, $H^1(X,\mathcal{O}_X)$ is the tangent space to $\mathrm{Pic}(X)$ (or equivalently $\mathrm{Pic}^0(X)$) at the origin. Hence, in char. zero, $h^1(X,\mathcal{O}_X)$ is the dimension of $\mathrm{Pic}^0(X)$. –  Laurent Moret-Bailly Sep 30 '10 at 16:01

There are also some computations for conic bundles by Sansuc in

MR0695346 (85d:14014) Sansuc, Jean-Jacques À propos d'une conjecture arithmétique sur le groupe de Chow d'une surface rationnelle.

share|improve this answer

For Reid's list of 95 K3 surfaces Picard lattices have been computed by Belcastro. Her paper can be downloaded from the arXiv at http://arxiv.org/PS_cache/math/pdf/9809/9809008v2.pdf . She has also made her thesis available, which can be downloaded at http://www.toroidalsnark.net/sm_thesis.pdf .

share|improve this answer
In the above paper the Picard lattice of a very general member of the family is computed. Although this is highly non-trivial, it is definitely easier then calculating the Picard lattice of a specific member in the family. (The OP asked for specific examples.) I.e., for a very general surface in $\mathbb{P}^3$ of degree at least 4 one has $Pic(X)=NS(X)=\mathbb{Z}$ (theorem of Noether and Lefschetz); for a concrete surface in $\mathbb{P}^3$ the calculation of $Pic(X)$ is more complicated. –  Remke Kloosterman Sep 30 '10 at 16:27

Your Answer


