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I was wondering if any vector bundles on a manifold other than the tangent bundle give topological invariants. I guess stiefel Whitney classes also come from the inverse bundle - but other than that.

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somewhat related question (in the algebraic context): mathoverflow.net/questions/57682/… –  Liran Shaul Mar 22 '11 at 13:33
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On a spin manifold, the associated bundle to the spin representation is a natural home for some topological invariants - is that the sort of thing you're looking for? –  Paul Siegel Mar 22 '11 at 15:34
    
The grothendieck group of all vector bundles on $X$ modulo exact sequences is the K-theory of $X$, and hence a topological invariant. But in general, only the tangent bundle and bundles constructed from it are naturally associated to a "naked" manifold. –  Paul Mar 22 '11 at 16:46
    
sure do you have a reference? the more I think about it the more I realize that this is a vague question. The thought was that if one looks at all vector bundles and their Euler classes that one might pick up something other than the Euler characteristic and maybe even get some important submanifolds through tranverse self intersection of the zero section of the bundle. –  marc Mar 22 '11 at 20:26
    
thanks Paul I am amazed by this –  marc Mar 23 '11 at 3:56
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1 Answer

up vote 5 down vote accepted

The top exterior power of the tangent bundle determines orientablility (the most basic of topological invariants, after dimension) in the sense that $\wedge^nTM$ is a trivial line bundle iff $M^n$ is orientable.

The complexified tangent bundle $TM\otimes_{\mathbf{R}} \mathbf{C}$ is used to define the Pontryagin classes, which are themselves smooth invariants (although the Pontryagin numbers and the rational Pontryagin classes turn out to be topological invariants).

Both of these examples are built from the tangent bundle, so I suspect are not what you are looking for. Aside from this you could look at the stable normal bundle (as you suggest) or put some additional structure on your manifold (as Paul Siegel suggest in his comment).

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