Nature of Invertible Sheaves in which there are no global sections. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T19:45:39Z http://mathoverflow.net/feeds/question/8242 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8242/nature-of-invertible-sheaves-in-which-there-are-no-global-sections Nature of Invertible Sheaves in which there are no global sections. Csar Lozano Huerta 2009-12-08T20:10:23Z 2009-12-09T06:28:43Z <p>EDIT: Let me try to make the question clearer.</p> <p>Consider the invertible sheaves $\mathcal{O}(d)$ over the projective space $\mathbb{P}^n$ where $d\in \mathbb{Z}$. Now, if $d>0$, among many properties, it is known that the space of global sections of such sheaves $V=\Gamma({\mathbb{P}^n},\mathcal{O}(d))$ encode information about embeddings between the projective spaces $\mathbb{P}^n$ and $\mathbb{P}(V)=\mathbb{P}^N$.</p> <p>My question is the following. What kind of information do the sheaves $\mathcal{O}(d)$ encode when $d&lt;0$?. By "information" I mean, in which context do they appear and seem to be helpful?</p> <p>References are very welcome.</p> http://mathoverflow.net/questions/8242/nature-of-invertible-sheaves-in-which-there-are-no-global-sections/8246#8246 Answer by Alberto García-Raboso for Nature of Invertible Sheaves in which there are no global sections. Alberto García-Raboso 2009-12-08T20:46:17Z 2009-12-08T20:46:17Z <p>I'm not sure what if this is what you are looking for, but here goes. All the information that you are associating to sheaves $\mathcal{O}(d)$ for positive $d$ seems to be essentially attached to their global sections. By Serre duality, $H^0(\mathcal{O}(-d)) \cong H^n(\mathcal{O}(d-n-1))$, and so these global sections talk about the other nonzero cohomology group of the invertible sheaves of positive degree.</p> http://mathoverflow.net/questions/8242/nature-of-invertible-sheaves-in-which-there-are-no-global-sections/8273#8273 Answer by Andrew Critch for Nature of Invertible Sheaves in which there are no global sections. Andrew Critch 2009-12-08T23:45:07Z 2009-12-08T23:45:07Z <p>They can still give you (non-canonical) rational maps to $\mathbb{P}^n$:</p> <p>Even when an invertible sheaf $L$ on $X$ has no global sections, one can still find open subsets $U$ of $X$ such that $L|_U$ is globally generated, for example when $U$ is affine. This isn't so interesting, but if you can find such a $U$ that is not contained in an affine, then $L|_U$ might not be trivial, and then you might get morphisms from $U$ to $\mathbb{P}^n$ (by choosing generators) not coming from $\mathcal{O}_U$. If $X$ was integral, then $U$ will be automatically dense, so you get a rational map from $X$ to $\mathbb{P}^n$.</p> <p>One way you could look for such a $U$ is to pick $n$ elements of the stalk $L_x$ at a point $x$, then intersect $n$ neighborhoods on which these elements extend, remove their common zero locus, and let the result be <br>$U$. If $U$ isn't contained in an affine, maybe you've found something cool. If you really wanted you could try working out a nice description for the rational map you've defined (though I've never done this). Even if $U$ was affine, maybe you've found a more interesting description of a less interesting map. </p> <p>From a very different perspective, since $\mathcal{O}(-d)$ is the inverse of $\mathcal{O}(d)$, in some sense any "information" it contains is obtainable from inverting the transition functions of $\mathcal{O}(d)$... so with this restrictive view of "information", perhaps it would be interesting to study what sort of morphisms/maps to $\mathbb{P}^n$ one can get from an invertible sheaf $L$ on $X$ when neither $L$ nor $L^\vee$ is ample/very ample.</p> http://mathoverflow.net/questions/8242/nature-of-invertible-sheaves-in-which-there-are-no-global-sections/8278#8278 Answer by Ben Webster for Nature of Invertible Sheaves in which there are no global sections. Ben Webster 2009-12-09T00:24:22Z 2009-12-09T00:24:22Z <p>You can still get a geometric realization of the Chern class: you look at a meromorphic section of your bundle, and you take minus the sum of the divisors on which it has a pole (weighted by order) plus the sum of the divisors where it has a 0 (also weighted by order).</p>