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For a topological manifold M with $KO$-the grothendieck group (actually a ring) of its vector bundles.

Whether the ring $KO$ is generated by vector bundles of rank $\leq N$,where $N$ is large sufficiently,or just finitely generated (true for spheres at least I guess)?

If not,what is a counter example? And when is it true?

Tyrone
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wsh
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1 Answers1

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If $M$ is any space homotopy equivalent to a finite-dimensional CW complex, then there is an integer $N$ for which any bundle over $M$ is of the form $\xi\oplus\text{triv}$, where $\xi$ is a bundle of rank $\leq N$. If $M$ is homotopy equivalent to a compact CW complex, then $KO(M)$ will be finitely generated.

The first part follows from the fact that $\widetilde {KO}(M)\cong[M,BO(\infty)]$. The map $BO(n)\rightarrow BO(\infty)$ is $N$-connected, so if $M$ is (homotopy equivalent to a) CW complex of dimension $\leq n$, there is a surjection $[M,BO(n)]\rightarrow\widetilde{KO}(M)$. This shows that every virtual vector bundle over $M$ is represented by vector bundle of rank $\leq n$.

On the other hand, suppose that $M$ is homotopy equivalent to a finite CW complex. By homotopy invariance of the homology functor we may as well assume that $M$ is a CW complex. If $\dim M\leq 0$, then $M$ is a finite set of points and of course $KO(M)$ is finitely generated.

Now suppose that we can show that $KO(M)$ is finitely generated whenever $\dim M\leq n$, and that $\dim M=n+1$. Then we have a long exact sequence $$\dots\leftarrow KO(\bigsqcup^k_{i=1}S^{n})\leftarrow\ KO(N)\leftarrow KO(M)\leftarrow KO(\bigsqcup^k_{i=1}S^{n+1})\leftarrow\dots$$ where $N$ is a CW complex of dimension $\leq n$ (namely the $n$-skeleton of $M$). Note that all the groups here are abelian.

The group $KO(\bigsqcup^k_{i=1}S^{n+1})\cong\bigoplus^k_{i=1}KO(S^{n+1})$ is known to be finitely generated. The group $KO(N)$ is finitely generated by induction. As $KO(M)$ is sandwiched between these two finitely generated groups in the exact sequence above, we see that $KO(M)$ is finitely generated (for instance, see here).

Any (compact) topological manifold $M$ is homotopy equivalent to a (compact) CW complex of finite dimension. In fact we can assume the dimension of this CW complex to be equal to the manifold dimension of $M$.

Tyrone
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  • thank you！but why $BO(n)$simply connected?I think its fundamental group is $Z/2Z$.And could you tell me the details about your last claim on $[M,K]$ finitely generated? – wsh Oct 13 '22 at 02:21
  • @wsh sorry it took me so long to get back to you. I've had a pretty busy week. What I wrote originally was for complex K-theory, and I changed it to KO-theory before posting without checking it thoroughly. The argument with finite complexes still goes through, but it is not worth pursuing (We can assume $M$ is simply connected by using Bott periodicity to replace it with $\Sigma^8M$. Now that $M$ is simply connected we can replace $BO(n)$ with its universal cover $BSO(n)$. Argue as before) – Tyrone Oct 18 '22 at 21:07
  • my question is almostly solved,but why there is an exact sequence of these $KO$ groups? – wsh Oct 25 '22 at 05:30
  • @wsh Because $KO$ is a cohomology theory. The $(n+1)$-dimensional CW complex $M$ is being formed by attaching $(n+1)$-cells to its $n$-skeleton $N$ (an $n$-dimensional CW complex). I changed the wedge symbol to a disjoint union (technically more correct, as I am not assuming $M$ is connected). I hope that wasn't confusing you. – Tyrone Oct 25 '22 at 10:03
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    great！my question has been totally solved and I have learnt a lot！ – wsh Oct 25 '22 at 16:43