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Suppose I have a smooth manifold $M$, and want to consider the $K$-theory $K^0(M)$. I could define this in the usual way (by taking the Grothendieck group of the monoid of equivalence classes of vector bundles) or in a 'smooth' way (considering only smooth vector bundles, and taking the Grothendieck group as usual).

I haven't seen this discussed anywhere. Is there any difference between the two approaches? And are there any references in which this question is discussed?

(Even better: if $M$ is a $G$-space for a compact Lie group $G$, is the equivariant $K$-theory affected by taking only smooth bundles?)

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1 Answer 1

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The $K^0(M)$ based on continuous vector bundles is the same as that based on smooth vector bundles because of the fundamental (and perhaps not sufficiently advertised) result:

Theorem Every continuous vector bundle on a $C^\infty$ manifold has a compatible $C^\infty$ vector bundle structure. Such a structure is unique up to $C^\infty$ isomorphism.

You can find a proof in Hirsch's Differential Topology , Chapter 4, Theorem 3.5, page 101.

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  • $\begingroup$ Cheers, you're right about the result being insufficiently advertised. I'll spread the word to anyone who will listen. I'm not sure if this answers my question about equivariant $K$-theory (although the literature seems to suggest that the answer is the same) since this proof doesn't see the $G$-action at all... $\endgroup$
    – user24452
    May 13, 2012 at 12:20
  • $\begingroup$ Dear Alexander, I have indeed not addressed the issue of $G$-action in my answer. $\endgroup$ May 13, 2012 at 12:55
  • $\begingroup$ That's fine, I didn't mean to imply otherwise! Thanks again for the answer. $\endgroup$
    – user24452
    May 13, 2012 at 13:10

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