Bott Periodicity — The spaces BUB\mathrm{U} and BOB\mathrm{O}

2 The spaces BUB\mathrm{U} and BOB\mathrm{O}

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

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

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

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

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

:BU(n)×BU(m)BU(n+m):BU(n)×BU(m)BU(nm). \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 BU(n)B\mathrm{U}(n) relate to U(n)\mathrm{U}(n)? Fix any base point of BU(n)B\mathrm{U}(n), and consider the space of based loops in BU(n)B\mathrm{U}(n), written ΩBU(n)\Omega B\mathrm{U}(n).

Proposition 3

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

Proof
The core content of the statement is that clutching functions work. Indeed, suppose XX is a connected based space. Then we have [X,ΩBU(n)]=[ΣX,BU(n)]=[ΣX,BU(n)], [X, \Omega B\mathrm{U}(n)]_* = [\Sigma X, B\mathrm{U}(n)]_* = [\Sigma X, B\mathrm{U}(n)], where the last equality comes from π0BU(n)=π1BU(n)=0\pi _0 B\mathrm{U}(n) = \pi _1 B\mathrm{U}(n) = 0 (e.g. by inspecting the construction of BU(n)B\mathrm{U}(n) as a Grassmannian to see it only has cells of dimension 2\geq 2). So the proposition is equivalent to saying that vector bundles on ΣX\Sigma X are the same as (based) maps XU(n)X \to \mathrm{U}(n), which is exactly the clutching construction. 1

Remark

BU(n)B\mathrm{U}(n) is not characterized (up to homotopy) by the above property. Note, however, that U(n)\mathrm{U}(n) is in particular a topological monoid, and ΩBU(n)\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, A\mathbb {A}_\infty -spaces). This property does characterize BU(n)B\mathrm{U}(n).

The importance of this proposition is that it allows us to read off the homotopy groups of BU(n)B\mathrm{U}(n) from those of U(n)\mathrm{U}(n). Of course, this is not too useful until we pass on to the limit nn \to \infty . There is a map BU(n)BU(n+1)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 U(n)U(n+1)\mathrm{U}(n) \hookrightarrow \mathrm{U}(n + 1) we had previously. We then let

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

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

Corollary 4

We have

πkBU={Zk0 even0otherwise. \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 BU(n)BU(n+1)B\mathrm{U}(n) \hookrightarrow B\mathrm{U}(n + 1), and so gives rise to a map

:BU×BUBU. \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

(E1)(F1)EF1. (E \oplus 1) \otimes (F \oplus 1) \neq E \otimes F \oplus 1.

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

Definition 5

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

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

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

Lemma 6

For any vector bundle EE over XX, there is some other vector bundle FF such that EFE \oplus F is trivial.

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

Theorem 7

If XX is a finite CW complex, then [X,BU][X, B\mathrm{U}] is the group of rank 00 virtual vector bundles, where the group structure comes from the direct sum map :BU×BUBU\oplus : B\mathrm{U}\times B\mathrm{U}\to B\mathrm{U}.

Proof
Since XX is finite, we have [X,BU]=colimn[X,BU(n)]. [X, B\mathrm{U}] = \operatorname*{colim}_{n \to \infty } [X, B\mathrm{U}(n)]. If a map f:XBU(n)f: X \to B\mathrm{U}(n) classifies a vector bundle EE, then the correspondence sends this to [E]nK(X)[E] - n \in K(X).

Corollary 8

If XX is a finite CW complex, then

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

Proof
Use the Z\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)×KU(X)KU(X)KU(X) \times KU(X) \to KU(X), classified by a map

:(BU×Z)×(BU×Z)BU×Z. \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 BU×ZB\mathrm{U}\times \mathbb {Z}, corresponding to the trivial rank 0 vector bundle, kills everything under \otimes , so this factors to give a map

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

where as always, XY=X×Y/XYX \wedge Y = X \times Y/X \vee Y.

This is important, since it induces a ring structure on π(BU×Z)\pi _* (B\mathrm{U}\times \mathbb {Z}) — if fi:SkiBU×Zf_ i: S^{k_ i} \to B\mathrm{U}\times \mathbb {Z}, then smashing them together gives

f1f2:Sk1+k2Sk1Sk2(BU×Z)(BU×Z)BU×Z. 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 π(BU×Z)\pi _*(B\mathrm{U}\times \mathbb {Z}) is Z\mathbb {Z} in every even degree, and is zero otherwise. The ring structure is the best you can hope for.

Theorem 9 (Complex Bott periodicity)
π(BU×Z)Z[u],degu=2. \pi _*(B\mathrm{U}\times \mathbb {Z}) \cong \mathbb {Z}[u],\quad \deg u = 2.

This has some nice geometric consequences. Observe that π(BU×Z)π(Ω2(BU×Z))\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 BU×ZΩ2(BU×Z)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:S2BU×Zu: S^2 \to B\mathrm{U}\times \mathbb {Z} be a generator of π2(BU×Z)\pi _2(B\mathrm{U}\times \mathbb {Z}). Then we get a map

S2(BU×Z)f1(BU×Z)(BU×Z)(BU×Z). 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 (BU×Z)Ω2(BU×Z)(B\mathrm{U}\times \mathbb {Z}) \to \Omega ^2 (B\mathrm{U}\times \mathbb {Z}) is then multiplication by uu, which is an isomorphism. So

Corollary 10

The map above gives a homotopy equivalence

BU×ZΩ2(BU×Z). B\mathrm{U}\times \mathbb {Z}\simeq \Omega ^2 (B\mathrm{U}\times \mathbb {Z}).
This is a geometric incarnation of the Bott periodicity theorem, which says two spaces are homotopy equivalent.

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

Theorem 11

The class uu can be chosen to be represented by the map

S2CP1CPBU(1)BUBU×Z. 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 [γ]1K(CP1)[\gamma ] - 1 \in K(\mathbb {CP}^1), where γ\gamma is the tautological bundle over CP1\mathbb {CP}^1.

The real version of these results is slightly less pretty.

Theorem 12 (Real Bott periodicity)

We have

π(BO×Z)Z[η,α,β]/(2η,η3,α24β) \pi _*(B\mathrm{O}\times \mathbb {Z}) \cong \mathbb {Z}[\eta , \alpha , \beta ]/(2\eta , \eta ^3, \alpha ^2 - 4 \beta )

where degη=1,degα=4,degβ=8\deg \eta = 1, \deg \alpha = 4, \deg \beta = 8. Therefore,

BO×ZΩ8(BO×Z). B\mathrm{O}\times \mathbb {Z}\cong \Omega ^8(B\mathrm{O}\times \mathbb {Z}).