Hello everyone

My question is the following:

Given a locally compact space $X$, assume it is also connected for simplicity, its K-theory ring is defined as $\tilde{K}(X^+)$, where $X^+$ is the one point compactification and $\tilde{K}$ denotes reduced K-theory (with respect to the point at infinity). Alternatively, we can take differences $[E]-[F]$, where $E$ and $F$ are isomorphism classes of vector bundles over $X$, each one trivial outside a compact set and both with the same rank. Notice that, since both bundles are trivial outside a compact set, they can be extended to $X^+$.

So far so good. There is another definition consisting of equivalence classes of triples $(E,F,\alpha)$, where $E$ and $F$ are now bundles over $X$ and $\alpha:E\to F$ is an isomorphism outside a compact set in $X$. Two triples $(E,F,\alpha)$ and $(E',F',\beta)$ are equivalent if there are ''degenerate'' triples $(G,G,id)$, $(H,H,id)$ and isomorphisms $u:E\oplus G\to E'\oplus H$ and $ v:F\oplus G\to F'\oplus H$ commuting with the maps $\alpha\oplus id$ and $\beta\oplus id$. Notice that in this definition, the bundles are not trivial outside a compact set so there is, a priori, no way to extend them to $X^+$.

Well, both definitions are equivalent. This is easy to prove for compact finite dimensional manifolds (for example Atiyah's book contains a nice argument). For non compact finite dimensional manifolds, it turns out that every vector bundle is a direct summand of a trivial vector bundle (it is even more surprising: for a given bundle, there is a finite covering of the manifold such that the bundle restricted to each set of the covering is trivial). A proof of this is very difficult to find and relies on Dimension theory (Wells, Differential Analysis on Complex Manifolds prop 4.1). Using this property, the proof in this case is immediate.

Looking for the general proof, I realized that every single place where this equivalence is mentioned simply quotes Segal's paper on equivariant K-theory. I have to admit that I don't understand the argument explained there. It is clear that every locally compact space is open and dense in its one point compactification. Take $U$ open in a compact space $Y$, using the first definition of $K(U)$, we should be able to prove that $K(U)\cong K(Y,B)$, where $B=Y-U$, but that would require ''chopping off'' the bundles outside a compact set and gluing something to make the triple be represented by bundles trivial at infinity and, therefore, with a natural extension to $X^+$. This seems to me as a highly invasive process and I don't really see why it is correct.

Could someone please help me with this???


Let $(E,F,\alpha)$ represent an element of the second group. Wlog $\alpha:E\to F$ is defined globally even though it's only an iso outside $K$. Choose global compactly supported sections $s_1,\dots ,s_n$ of the dual of $E$ such that on $K$ they span the bundle; for large enough $n$ this is possible. Let $E'$ be the trivial rank $n$ bundle and interpret these sections as an injective map $s:E\to E'$. We now have a bundle map $(\alpha,s):E\to F\times E'$ which is injective everywhere. Split off the resulting subbundle of $F\times E'$ and call the other part $F'$. This leads to an isomorphism $E\times F'\cong F\times E'$. Replace $(E,F,\alpha)$ by $(E\times E',F\times E',\alpha\times 1)$. Near infinity, where $\alpha\times 1$ is an iso, the composed iso $E\times E'\to F\times E'\to E\times F'$ looks likes $(1, 0)$ on the $E$ part and therefore looks like $(?,\alpha')$ on the $E'$ part where $\alpha'$ is an iso. Make a little adjustment so that it looks like $(0,\alpha')$ on the $E'$ part. This iso $\alpha':E'\to F'$ near infinity yields $(E',F',\alpha')$ equivalent to $(E,F,\alpha)$, but with the bundles trivial near infinity. (In fact $E'$ is trivial globally.)

  • $\begingroup$ That's the sneaky bit I was missing. It's not necessary to embed $E$ as a subbundle of something trivial globally but only over the (compact) bit where $\alpha$ is not an isomorphism. $\endgroup$ – Loop Space May 13 '11 at 11:36
  • $\begingroup$ @Andrew: Yes, I was thinking along the same lines as you at first, making up counterexamples, until I realized that they were counterexamples to the wrong statement. $\endgroup$ – Tom Goodwillie May 13 '11 at 13:48
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    $\begingroup$ Dear professors Goodwillie and Stacey. Thank you very much for your interest and your answers to my question. The argument is very clever and much simpler than Segal's. This equivalence is fundamental in the proof of the index theorem but this is the first time that I see it written. Thanks again Best $\endgroup$ – user15072 May 13 '11 at 18:46
  • $\begingroup$ You're very welcome. I'm glad you raised the question. I had never really had occasion to think about vector bundles over arbitrary locally compact spaces until now. $\endgroup$ – Tom Goodwillie May 13 '11 at 22:08

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