# Does Bott periodicity imply homotopy equivalences?

1. Aug 13, 2015

### 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 $K^n(X) = [X, B^n Gr^\infty(m)]$ 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 $\Omega (B X) = X$ where $\Omega$ takes the loop space).

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

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

$$\boxed{ B^n Gr^\infty \simeq B^{n+2} Gr^\infty} \;?$$

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 $B^2 Gr^\infty \simeq \mathbb Z \times Gr^\infty$. This seems weird, as I wouldn't expect $[X, Gr^\infty ] \cong [X, \mathbb Z \times Gr^\infty ]$ ... 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?

2. Aug 14, 2015

### nonequilibrium

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