Does Bott periodicity imply homotopy equivalences?

[nonequilibrium]

Hello!

Trying to learn some basics of (topological) K-theory and came up with the following question:

From what I can gather, we can define (complex, topological) K-theory as [itex] K^n(X) = [X, B^n Gr^\infty(m)] [/itex] with m going to infinity (indeed for m large enough, the answer is independent of it) and where B is taking the classifying space (this means: for any topological space X, we define BX such that [itex] \Omega (B X) = X[/itex] where [itex]\Omega[/itex] takes the loop space).

Let me know if so far I have made a mistake. As an illustration this tells us [itex]K^0(X) = [X,Gr^\infty] [/itex] (where I have implicitly taken the limit [itex]m \to \infty[/itex]) which indeed classifies vector bundles on X up to (stable) isomorphism/equivalence.

So Bott periodicity tells us [itex]K^n(X) \cong K^{n+2}(X)[/itex]. In other words it tells us [itex] [X, B^n Gr^\infty] \cong [X, B^{n+2} Gr^\infty] [/itex]. My question is: have I made a mistake, or does Bott periodicity imply

[tex]\boxed{ B^n Gr^\infty \simeq B^{n+2} Gr^\infty} \;?[/tex]

To focus on a specific example, let's take n = -1. Then this would tell us that the classifying space of the infinite Grassmannian is homotopic to U(n) (for n large enough). Is this true?...

EDIT: It seems wikipedia agrees with the above bold statements. However, not completely. For example it implies that in fact [itex] B^2 Gr^\infty \simeq \mathbb Z \times Gr^\infty [/itex]. This seems weird, as I wouldn't expect [itex] [X, Gr^\infty ] \cong [X, \mathbb Z \times Gr^\infty ][/itex] ... Do we have to take reduced cohomology/mod out by something to make it all work, or does it work like this anyway and do I just fail to see it?

 

[nonequilibrium]

Aha, I have realized the answer. The key point is that [itex] [X,Y] \cong [X,Y \times \mathbb Z] [/itex] since in these contexts we are looking at *base point preserving* maps, which are insensitive to the number of disconnected components.

 

[math771]

Hi there! I'm not an expert in K-theory, but from what I understand, you have the right idea. K-theory is indeed defined as K^n(X) = [X, B^n Gr^\infty(m)] with m going to infinity, and this does classify vector bundles on X up to stable isomorphism. And yes, Bott periodicity does imply that B^n Gr^\infty \simeq B^{n+2} Gr^\infty, so your statement is correct.

As for your example with n=-1, I believe you are correct in saying that B^2 Gr^\infty \simeq \mathbb Z \times Gr^\infty. This is because the classifying space of the infinite Grassmannian is homotopy equivalent to the classifying space of the group U(n), which is homotopy equivalent to the product of the classifying space of the unitary group U(n) and the classifying space of the integers, which is just \mathbb Z. So it is not an error, it just follows from the definition of the classifying space.

I hope this helps clarify things for you! K-theory can definitely be a tricky subject, so don't worry if it takes some time to fully grasp it. Keep up the good work!