# 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 $\Omega B\mathrm{U}(n)$.

$\Omega B\mathrm{U}(n) \cong \mathrm{U}(n)$.

^{1}□

$B\mathrm{U}(n)$ is not characterized (up to homotopy) by the above property. Note, however, that $\mathrm{U}(n)$ is in particular a topological monoid, and $\Omega B\mathrm{U}(n)$ can be made one by considering loops of all lengths so that composition of loops is strictly associative. The above homotopy equivalence is then one of topological monoids (or rather, $\mathbb {A}_\infty$-spaces). This property *does* characterize $B\mathrm{U}(n)$.

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 $B\mathrm{U}$.

We have

$\pi _ k B\mathrm{U}= \begin{cases} \mathbb {Z}& \text{$k \neq 0$ even}\\ 0 & \text{otherwise} \end{cases}.$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.

A *virtual vector bundle* is a formal difference of two vector bundles.

More precisely, if $X$ is a finite CW complex, write $\operatorname{Vect}_\mathbb {C}(X)$ for the monoid of vector bundles over $X$ (up to isomorphism) under direct sum. Write $KU(X)$ to be the group completion of $\operatorname{Vect}_\mathbb {C}(X)$. A virtual vector bundle is then an element of $KU(X)$.

*rank*is $\dim E - \dim F$.

^{2}We also write $n$ for the $n$-dimensional trivial vector bundle.

For any vector bundle $E$ over $X$, there is some other vector bundle $F$ such that $E \oplus F$ is trivial.

Hence, any virtual vector bundle can be written as $[E] - n$ for some vector bundle $E$.□

If $X$ is a finite CW complex, then $[X, B\mathrm{U}]$ is the group of rank $0$ virtual vector bundles, where the group structure comes from the direct sum map $\oplus : B\mathrm{U}\times B\mathrm{U}\to B\mathrm{U}$.

If $X$ is a finite CW complex, then

$KU(X) \cong [X, B\mathrm{U}\times \mathbb {Z}].$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

The map above gives a homotopy equivalence

$B\mathrm{U}\times \mathbb {Z}\simeq \Omega ^2 (B\mathrm{U}\times \mathbb {Z}).$*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 class $u$ can be chosen to be represented by the map

$S^2 \cong \mathbb {CP}^1 \hookrightarrow \mathbb {CP}^\infty \cong B\mathrm{U}(1) \hookrightarrow B\mathrm{U}\hookrightarrow B\mathrm{U}\times \mathbb {Z}.$Equivalently, it is $[\gamma ] - 1 \in K(\mathbb {CP}^1)$, where $\gamma$ is the tautological bundle over $\mathbb {CP}^1$.

The real version of these results is slightly less pretty.

We have

$\pi _*(B\mathrm{O}\times \mathbb {Z}) \cong \mathbb {Z}[\eta , \alpha , \beta ]/(2\eta , \eta ^3, \alpha ^2 - 4 \beta )$where $\deg \eta = 1, \deg \alpha = 4, \deg \beta = 8$. Therefore,

$B\mathrm{O}\times \mathbb {Z}\cong \Omega ^8(B\mathrm{O}\times \mathbb {Z}).$