Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Two connections on a smooth vector bundle are called equivalent if they can be mapped to each other by a self-diffeomorphism of the total space that covers identity on the base? Is there a classification of connections up to equivalence?

EDIT: Recall that if $\xi$ is a vector bundle with total space $E$, and if $V$ is the subbundle of the tangent bundle $TE$ whose fibers are the tangent spaces to the fibers of $\xi$, then a connection on $\xi$ is a scale-invariant subbundle $H$ of $TE$ such that $TE=V\oplus H$. Here scale-invariant means that $H$ is preserved by each diffeomorphisms $r:E\to E$ that multiplies a vector by the non-zero real number $r$. From this definition one might suspect that any two connections on $\xi$ are equivalent in the sense of the previous paragraph. Is this true?

UPDATE:

  1. The answer to the last question in the edit is NO, as was explained to me by John Etnyre. For example, consider two connections with subbundles $H_1$, $H_2$ such that $H_1$ is integrable everywhere, and $H_2$ is not. Since diffeomorphisms preserve integrability, the two connections aren't equivalent in my sense.

  2. I was hoping to use one of Gromov's h-principles to see when $H_1$ can be deformed to $H_2$ by an ambient isotopy of $E$. Indeed, it is easy to see that any two $H_1$, $H_2$ are homotopic subbundles, e.g. choose Riemannian metric $g_i$ on $E$ such that $H_i$ is $g_i$-orthogonal to the fibers of $E\to B$, and then $g_t=tg_1+(1-t)g_2$ is a Riemannian metric on $E$ and we can let $H_t$ be the orthogonal complement to the fibers; this $H_t$ is the desired homotopy. So one hopes that h-principle can upgrade the homotopy to an isotopy but a simple dimension count shows that we are in the wrong dimension range so h-principle does not apply, as confirmed by the example in part 1.

share|improve this question
    
Are you allowing self-diffeomorphisms of the total space that are not vector bundle maps, i.e. diffeomorphisms that are not fiber-wise linear? Bundle isomorphisms should preserve the conjugacy class (in $GL_n$) of the holonomy along any given loop in the base, so I guess you must be allowing more general maps? –  Dan Ramras May 4 '10 at 19:33
    
Also, I'm confused by your notion of scale-invariance. If H is a sub-vector bundle of E, then isn't it automatically invariant under multiplication by any real number? –  Dan Ramras May 4 '10 at 19:35
1  
@Dan, I do not require the self-diffeomorphisms to be bundle automorphisms. The definition in the edit is one of many standard definitions of the connection (see e.g. Spivak, volume 2). As for scale invariance, the differential of the map $r$ takes takes $H_p$ to a subspace of $T_{rp}E$, so $r_*(H_p)=H_{rp}$ is a non-trivial restriction. –  Igor Belegradek May 4 '10 at 20:10
    
Right, I see what you're saying now about scale invariance. Interesting question. Is there a way to translate this question to a question about connections on principal bundles, where the scale invariance condition just becomes an equivariance condition? It's not clear to me how diffeomorphisms of the total space of a vector bundle are related to diffeomorphisms of the total space of the associated principal bundle, so these two versions of the question may be different. –  Dan Ramras May 4 '10 at 20:50
    
Where did this question come from? More precisely: why would you (or anyone) want to forget the linear structure on the fibers when forming the equivalence between the two connections? -R Montgomery –  Richard Montgomery Jan 22 '12 at 4:15
show 1 more comment

1 Answer

Maybe this is not what you are after, but at least for line bundles, there is a classification of line bundles with connection up to isomorphism in terms of smooth Deligne cohomology. This is explained in Brylinski's Loop spaces, characteristic classes and geometric quantization -- particularly Theorem 2.2.12.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.