# 3 Topological $K$ -theory

We will end by saying a bit more about the functor $KU$ defined above. Pullback of vector bundle makes it a contravariant functor on the category of finite CW complexes. We can extend this to a functor on all CW complexes by defining it to be the functor represented by $B\mathrm{U}\times \mathbb {Z}$, but its values on infinite complexes have less straightforward descriptions.
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For technical reasons, we actually want a reduced version of this — on a based space $X$, we have

This corresponds to virtual vector bundles that are rank zero on the base point component. This is really not that important and not worth worrying about.

This functor $\widetilde{KU}$ behaves like the degree $0$ part of a (reduced) cohomology theory. For example, it satisfies an appropriate form of Mayer–Vietoris. So we will write it as $KU^0$ instead. The goal is the manufacture a (generalized) cohomology $KU$ whose degree $0$ part is this $KU^0$ we already have. This is called (complex) topological $K$-theory, and is of utmost importance in algebraic topology.

We first do it for *negative* degrees, which is easy. If $h^*$ is a (reduced) cohomology theory, then Mayer–Vietoris implies we always have

So for $n \geq 0$, we can simply define

$KU^{-n}(X) = KU^0(\Sigma ^ n X).$The functor $KU^{-n}(X)$ is then represented by $\Omega ^ n (B\mathrm{U}\times \mathbb {Z})$.

The key fact is that Bott periodicity tells us $\Omega ^2 (B\mathrm{U}\times \mathbb {Z}) \cong B\mathrm{U}\times \mathbb {Z}$. So another way to state Bott periodicity is that

There is a canonical isomorphism

$KU^ k(X) \cong KU^{k - 2}(X)$whenever both are defined.

Once we know this, we can simply define the remaining groups by

$KU^ n(X) = \begin{cases} KU^0(X) & n\text{ even}\\ KU^{-1}(X) & n\text{ odd}. \end{cases}$We then know automatically that this satisfies properties like Mayer–Vietoris, and hence is a generalized cohomology theory.

For completeness, we state the corresponding real result as well.

There is a canonical isomorphism

$KO^0(\Sigma ^8 X) \cong KO^0(X).$