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I´m looking for information about the intersection of two vector bundles (principally trivial bundles, but no necessarily). I´m trying to make a picture (literally) of reflexive finite generated modules.


Another related topic is sub-budles of a vector bundles.


All suggestions are wellcome!


Edit: I try to be more specific.

Suppose two sub-bundles of a bundle over a topological space X. We can do the intersection of their vector fibers in each point of base space and collect this intersection vector fibers along the base space. Then we can give it a topology, restriction of the bigger vector bundle.

What´s about of this "submodule of sections"?

Is it another vector bundle?

Has another interesting structure?

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What do you mean the "intersection" of two vector bundles? –  Charles Siegel Dec 5 '09 at 23:59
    
Literally intersection: Suppose two sub-bundles of a bundle over a topological space X. We can do the intersection of their vector fibers in each point of base space and collect this intersection vector fibers along the base space. Then we can give it a topology, restriction of the bigger vector bundle. What´s about of this "submodule of sections"? Is it another vector bundle? Has another interesting structure? – –  Francisco Perdomo Dec 6 '09 at 0:36
    
I might be able to help more if you could tell me what a reflexive module is? I did a quick google search and couldn't find the definition, and I'm not familiar with it (though might have run across it by a different name, or as an unnamed property, perhaps) –  Charles Siegel Dec 6 '09 at 14:41
    
A module M is said reflexive iff (R:(R:M))=M. Here (R:M)=Hom(M,R). For example, projective modules are reflexives. When R is noetherian integrally closed domain, all prime ideals with height 1 are reflexives modules. It is known (Bourbaki, chapter 7) all reflexives finitely generated modules are the intersection of two free modules. I´m trying to make a picture of that. Vector bundle is the corresponding topological concept to projective finitely generated module and that´s the reason of my question. –  Francisco Perdomo Dec 6 '09 at 16:15
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3 Answers

up vote 2 down vote accepted

You need to refine the question to get better answers, but here are some thoughts:

1) Over a normal variety, you can think of line bundles as divisors, and "intersect" them.

2) A vector bundle can be represented by a reflexive sheaf, but being reflexive is a lot weaker.

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Where can I request about reflexive sheaves? –  Francisco Perdomo Dec 6 '09 at 16:20
    
    
thanks for reference! –  Francisco Perdomo Dec 8 '09 at 1:27
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More specific than Ilya's answer. To see that the intersection of two subbundles need not be a bundle, take, on your space $M$, the rank two trivial bundle $M\times \mathbb{R}^2$. Then take two line bundles $L=(m,x,0)|m\in M, x\in\mathbb{R}$ and let $K$ be a nontrivial line bundle contained in $M\times\mathbb{R}^2$ such that the fiber over some point $m_0$ is the $y=0$ line. Then away from that point, the intersection will be just the origin, however, at $m_0$, the intersection is a rank 1 vector space. Thus, the dimension can jump, and so you don't get an actual vector bundle, merely a family of vector subspaces of $\mathbb{R}^2$ parameterized by $M$. And for this, all we need is some manifold which has a nontrivial subbundle of $\mathbb{R}^2$.

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Yes, I think with you the intersection of two bundle need not be a bundle and the dimension can jump. I haven´t trouble with this. My problem is with the Ilya´s assertion "the dimension can go only up" Why can dimension go down? I think I´ve lost something implicit in his comment. –  Francisco Perdomo Dec 6 '09 at 3:46
    
The idea is roughly that the generic situation should be minimal intersection, because the condition that they intersect more will be a closed condition, and the condition that they intersect less will be open. This is by analogy with just vector spaces. Take a vector space and then take two subspaces. Their intersection will be some minimum dimension, unless there are some extra conditions that happen, and then the condition jumps up, but it can never jump down. –  Charles Siegel Dec 6 '09 at 4:06
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The intersection of two subbundles K, L of a bundle M doesn't have to be bundle itself: its dimension can jump up on some algebraic submanifold. The reason why the dimension can go only up is quite simple to visualize by thinking about the intersection of two manifolds EK, EL (total spaces of bundles K, L) in a manifold EM.

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I don´t understand why you say about dimension go up. For example if the bundle M has dimension 4 and the two subbundles have dimension 2, I think almost always dimensión is constant,say 1. But the vector space intersection of vector fibers may be 0 if the subbundles are complemtary or 2 where the fibers in K and L are equal. Why don´t it can occur? I don´t understand –  Francisco Perdomo Dec 6 '09 at 2:19
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