What is the geometric point of view of an algebraic line bundle compared to a analytic line bundle? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T22:53:09Z http://mathoverflow.net/feeds/question/92588 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92588/what-is-the-geometric-point-of-view-of-an-algebraic-line-bundle-compared-to-a-ana What is the geometric point of view of an algebraic line bundle compared to a analytic line bundle? Johannes 2012-03-29T17:19:58Z 2012-04-04T10:18:39Z <p>Hi folkz,</p> <p>I'm trying to learn more about line bundles, invertible sheaves and divisors on schemes. I understand the connection beweteen Cartier and Weil Divisors and the connection between Cartier Divisors and invertible sheaves and how to get from one to another (as far as possible).</p> <p>But compared to my analytic imagination of a line bundle I don't see how to come from an invertible sheaf to the line bundle (apart from the fact, that these two terms coincide). Where is '<i>the line</i>' in my locally free of rank one $\mathcal{O}_X$-module?</p> <p>greatz Johannes</p> http://mathoverflow.net/questions/92588/what-is-the-geometric-point-of-view-of-an-algebraic-line-bundle-compared-to-a-ana/92591#92591 Answer by Simon Rose for What is the geometric point of view of an algebraic line bundle compared to a analytic line bundle? Simon Rose 2012-03-29T17:36:07Z 2012-03-29T17:36:07Z <p>Perhaps this might help as some intuition. Instead of looking for "the line" in a locally free sheaf, let's look in the other direction. Let's start with a line bundle, and move back towards sheaves.</p> <p>So take a line bundle $\pi : L \to X$. This bundle has a sheaf of sections $\mathcal{O}_L$ defined by</p> <p>$$\mathcal{O}_L(U) = \{s : U \to L \mid \pi \circ s = id_U\}$$</p> <p>i.e. over an open set $U$ in $X$, $\mathcal{O}_L(U)$ is the collection of all sections of $L$ over $U$. It can be shown that this is a locally free sheaf of rank one.</p> <p>Now, for a vector bundle of rank $n$, all of this is true, but the locally free sheaf is now or rank $n$.</p> <p>Hopefully this provides at least a little intuition for the relation between the two.</p> http://mathoverflow.net/questions/92588/what-is-the-geometric-point-of-view-of-an-algebraic-line-bundle-compared-to-a-ana/93038#93038 Answer by Allen Knutson for What is the geometric point of view of an algebraic line bundle compared to a analytic line bundle? Allen Knutson 2012-04-03T20:43:48Z 2012-04-04T10:18:39Z <p>Let $\mathcal F$ be a locally free $\mathcal O_X$-module. Then $\mathcal R := Sym_{\mathcal O_X}(\mathcal F)$, the tensor products being over $\mathcal O_X$, is a sheaf of rings, and we can take its $\bf Spec$ to get a space over $X$. That space is the corresponding vector bundle.</p> <p>$\mathcal R$'s grading is what gives the dilation action on the fibers. The map $\mathcal F \to (\mathcal F \otimes \mathcal O_X) \oplus (\mathcal O_X \otimes \mathcal F)$, $f \mapsto (f\otimes 1) + (1\otimes f)$ induces a cocommutative comultiplication $\mathcal R \to \mathcal R \otimes \mathcal R$, which gives the vector addition on ${\bf Spec}\ \mathcal R$, I think.</p>