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.

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    $\begingroup$ Other easy examples might include things like toric surfaces and ruled surfaces. $\endgroup$ – Karl Schwede Sep 30 '10 at 4:44
  • $\begingroup$ Sorry for the dumb question, but what is $Pic^0$? $\endgroup$ – Martin Brandenburg Sep 30 '10 at 12:24
  • $\begingroup$ 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). $\endgroup$ – Karl Schwede Sep 30 '10 at 15:30

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.

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    $\begingroup$ Welcome to MO, Professor Zarhin. $\endgroup$ – Pete L. Clark Sep 30 '10 at 3:31
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    $\begingroup$ 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.) $\endgroup$ – Emerton Sep 30 '10 at 9:08
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    $\begingroup$ 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. $\endgroup$ – 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 $NS(X)$.


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.

  • $\begingroup$ 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 $\mathrm{Pic}^0(X)$. $\endgroup$ – 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.


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 .

  • $\begingroup$ 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. $\endgroup$ – Remke Kloosterman Sep 30 '10 at 16:27

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