2 The spaces $B\mathrm{U}$ and $B\mathrm{O}$

A better way to think about Bott periodicity is to not look at $\mathrm{U}$, but $B\mathrm{U}$. To describe $B\mathrm{U}$, we again start with the “unstable” versions $B\mathrm{U}(n)$.

$B\mathrm{U}(n)$ is defined to be a space such that for any CW complex $X$, there is a canonical bijection

$$[X, B\mathrm{U}(n)] \leftrightarrow \Big\{ \text{$n$ dimensional (complex) vector bundles on $X$}\Big\} .$$

An explicit model of $B\mathrm{U}(n)$ can be described as the Grassmannian of $n$-planes in $\mathbb {C}^\infty$, the countable dimension complex vector space.

This universal property of $B\mathrm{U}(n)$ is very useful because it gives us a very geometric handle on the spaces $B\mathrm{U}(n)$. For example, the direct sum and tensor product of vector bundles are classified by maps

$$\begin{aligned} \oplus : B\mathrm{U}(n) \times B\mathrm{U}(m) & \to B\mathrm{U}(n + m)\\ \otimes : B\mathrm{U}(n) \times B\mathrm{U}(m) & \to B\mathrm{U}(nm). \end{aligned}$$

The first question to ask is — how does $B\mathrm{U}(n)$ relate to $\mathrm{U}(n)$? Fix any base point of $B\mathrm{U}(n)$, and consider the space of based loops in $B\mathrm{U}(n)$, written $\Omega B\mathrm{U}(n)$.

The core content of the statement is that clutching functions work. Indeed, suppose $X$ is a connected based space. Then we have

$$[X, \Omega B\mathrm{U}(n)]_* = [\Sigma X, B\mathrm{U}(n)]_* = [\Sigma X, B\mathrm{U}(n)],$$

where the last equality comes from $\pi _0 B\mathrm{U}(n) = \pi _1 B\mathrm{U}(n) = 0$ (e.g. by inspecting the construction of $B\mathrm{U}(n)$ as a Grassmannian to see it only has cells of dimension $\geq 2$). So the proposition is equivalent to saying that vector bundles on $\Sigma X$ are the same as (based) maps $X \to \mathrm{U}(n)$, which is exactly the clutching construction. ¹

The importance of this proposition is that it allows us to read off the homotopy groups of $B\mathrm{U}(n)$ from those of $\mathrm{U}(n)$. Of course, this is not too useful until we pass on to the limit $n \to \infty$. There is a map $B\mathrm{U}(n) \hookrightarrow B\mathrm{U}(n + 1)$ given by adding a trivial line bundle. Under the clutching construction, this corresponds to the map $\mathrm{U}(n) \hookrightarrow \mathrm{U}(n + 1)$ we had previously. We then let

$$B\mathrm{U}= \operatorname*{colim}_{n \to \infty } B\mathrm{U}(n).$$

In particular, there is a map $* = B\mathrm{U}(0) \to B\mathrm{U}$ which we will choose to be our canonical basepoint of $B\mathrm{U}$.

The direct sum and of vector bundles is compatible with the inclusion $B\mathrm{U}(n) \hookrightarrow B\mathrm{U}(n + 1)$, and so gives rise to a map

$$\oplus : B\mathrm{U}\times B\mathrm{U}\to B\mathrm{U}.$$

We would like a map that comes from tensor products as well, but that is not compatible with the inclusion, since

$$(E \oplus 1) \otimes (F \oplus 1) \neq E \otimes F \oplus 1.$$

To fix this, we need to think about what $B\mathrm{U}$ represents.

If $E$ is a vector bundle, we write $[E]$ for its image in $KU(X)$. Then every element in $KU(X)$ is of the form $[E] - [F]$, and its rank is $\dim E - \dim F$. ² We also write $n$ for the $n$-dimensional trivial vector bundle.

Since $X$ is finite, we have

$$[X, B\mathrm{U}] = \operatorname*{colim}_{n \to \infty } [X, B\mathrm{U}(n)].$$

If a map $f: X \to B\mathrm{U}(n)$ classifies a vector bundle $E$, then the correspondence sends this to $[E] - n \in K(X)$.

Use the $\mathbb {Z}$ factor to keep track of the rank, since every virtual vector bundle is the sum of a rank zero virtual vector bundle plus a trivial bundle.

Now since $\otimes$ is linear, it induces a map $KU(X) \times KU(X) \to KU(X)$, classified by a map

$$\otimes : (B\mathrm{U}\times \mathbb {Z}) \times (B\mathrm{U}\times \mathbb {Z}) \to B\mathrm{U}\times \mathbb {Z}.$$

In fact, we get something even better, since the basepoint of $B\mathrm{U}\times \mathbb {Z}$, corresponding to the trivial rank 0 vector bundle, kills everything under $\otimes$, so this factors to give a map

$$\otimes : (B\mathrm{U}\times \mathbb {Z}) \wedge (B\mathrm{U}\times \mathbb {Z}) \to B\mathrm{U}\times \mathbb {Z},$$

where as always, $X \wedge Y = X \times Y/X \vee Y$.

This is important, since it induces a ring structure on $\pi _* (B\mathrm{U}\times \mathbb {Z})$ — if $f_i: S^{k_i} \to B\mathrm{U}\times \mathbb {Z}$, then smashing them together gives

$$f_1 \wedge f_2: S^{k_1 + k_2} \cong S^{k_1} \wedge S^{k_2} \to (B\mathrm{U}\times \mathbb {Z}) \wedge (B\mathrm{U}\times \mathbb {Z}) \overset {\otimes }{\to } B\mathrm{U}\times \mathbb {Z}.$$

As a group, the ring $\pi _*(B\mathrm{U}\times \mathbb {Z})$ is $\mathbb {Z}$ in every even degree, and is zero otherwise. The ring structure is the best you can hope for.

This has some nice geometric consequences. Observe that $\pi _*(B\mathrm{U}\times \mathbb {Z}) \cong \pi _*(\Omega ^2(B\mathrm{U}\times \mathbb {Z}))$ abstractly as groups, and we know this without using the ring structure. This does not automatically imply $B\mathrm{U}\times \mathbb {Z}\cong \Omega ^2(B\mathrm{U}\times \mathbb {Z})$, since we need a map that realizes this isomorphism of groups in order to apply Whitehead's theorem. The ring structure provides exactly this.

Indeed, let $u: S^2 \to B\mathrm{U}\times \mathbb {Z}$ be a generator of $\pi _2(B\mathrm{U}\times \mathbb {Z})$. Then we get a map

$$S^2 \wedge (B\mathrm{U}\times \mathbb {Z}) \overset {f \wedge 1}\to (B\mathrm{U}\times \mathbb {Z}) \wedge (B\mathrm{U}\times \mathbb {Z}) \overset {\otimes }\to (B\mathrm{U}\times \mathbb {Z}).$$

The adjoint map $(B\mathrm{U}\times \mathbb {Z}) \to \Omega ^2 (B\mathrm{U}\times \mathbb {Z})$ is then multiplication by $u$, which is an isomorphism. So

This is a geometric incarnation of the Bott periodicity theorem, which says two spaces are homotopy equivalent.

Given the importance of the map $u$, it is reassuring to know there is a very concrete description of it:

The real version of these results is slightly less pretty.