Calculations of Pic^0, Pic, NS of surfaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T09:08:02Z http://mathoverflow.net/feeds/question/40555 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40555/calculations-of-pic0-pic-ns-of-surfaces Calculations of Pic^0, Pic, NS of surfaces Justin Shih 2010-09-29T23:37:46Z 2010-09-30T15:09:02Z <p>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.</p> http://mathoverflow.net/questions/40555/calculations-of-pic0-pic-ns-of-surfaces/40563#40563 Answer by Yuri Zarhin for Calculations of Pic^0, Pic, NS of surfaces Yuri Zarhin 2010-09-30T01:12:37Z 2010-09-30T03:31:04Z <p>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.</p> <p>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 <em>Abelian Varieties</em>.</p> http://mathoverflow.net/questions/40555/calculations-of-pic0-pic-ns-of-surfaces/40573#40573 Answer by Chandan Singh Dalawat for Calculations of Pic^0, Pic, NS of surfaces Chandan Singh Dalawat 2010-09-30T04:42:51Z 2010-09-30T04:42:51Z <p>There are also some computations for conic bundles by Sansuc in </p> <p>MR0695346 (85d:14014) Sansuc, Jean-Jacques <a href="http://www.digizeitschriften.de/main/dms/img/?PPN=GDZPPN002544741" rel="nofollow">À propos d'une conjecture arithmétique sur le groupe de Chow d'une surface rationnelle.</a></p> http://mathoverflow.net/questions/40555/calculations-of-pic0-pic-ns-of-surfaces/40580#40580 Answer by Remke Kloosterman for Calculations of Pic^0, Pic, NS of surfaces Remke Kloosterman 2010-09-30T07:37:33Z 2010-09-30T07:37:33Z <p>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.)</p> <p>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., <a href="http://pjm.math.berkeley.edu/ant/2007/1-1/p01.xhtml" rel="nofollow">http://pjm.math.berkeley.edu/ant/2007/1-1/p01.xhtml</a> where an example is given of a quartic surfaces with $\mathrm{rank} Pic(X)=\mathrm{rank} NS(X)=1$.</p> <p>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 $NS(X)$. </p> http://mathoverflow.net/questions/40555/calculations-of-pic0-pic-ns-of-surfaces/40598#40598 Answer by Daniel Loughran for Calculations of Pic^0, Pic, NS of surfaces Daniel Loughran 2010-09-30T10:48:57Z 2010-09-30T10:48:57Z <p>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 <a href="http://en.wikipedia.org/wiki/Exponential_sheaf_sequence" rel="nofollow">http://en.wikipedia.org/wiki/Exponential_sheaf_sequence</a> then implies that the natural map $\rm{Pic}(X) \to H^2(X, \mathbb{Z})$ is injective.</p> <p>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.</p> http://mathoverflow.net/questions/40555/calculations-of-pic0-pic-ns-of-surfaces/40627#40627 Answer by Laie for Calculations of Pic^0, Pic, NS of surfaces Laie 2010-09-30T15:09:02Z 2010-09-30T15:09:02Z <p>For Reid's list of 95 K3 surfaces Picard lattices have been computed by Belcastro. Her paper can be downloaded from the arXiv at <a href="http://arxiv.org/PS_cache/math/pdf/9809/9809008v2.pdf" rel="nofollow">http://arxiv.org/PS_cache/math/pdf/9809/9809008v2.pdf</a> . She has also made her thesis available, which can be downloaded at <a href="http://www.toroidalsnark.net/sm_thesis.pdf" rel="nofollow">http://www.toroidalsnark.net/sm_thesis.pdf</a> .</p>