Topological K-theory

Remarks

Using real vector bundles instead of complex ones gives real K-theory $KO(X)$. The tensor product of bundles makes $K^0(X)$ a commutative ring. By the Bott periodicity theorem, $K^{-n-2}(X) \cong K^{-n}(X)$, and this defines $K^n(X)$ for all integers $n$.

Example: for a point, a vector bundle is a finite-dimensional complex vector space, so $K^0(\text{pt}) \cong \Z$ through the dimension. For the $2$-sphere $S^2$, $K^0(S^2) \cong \Z \oplus \Z$. It is generated by the trivial line bundle $1$ and the class $H$ of the tautological line bundle on $\mathbb{CP}^1 \cong S^2$, with the relation $(H - 1)^2 = 0$ in the ring $K^0(S^2)$.

Non-example: the monoid of isomorphism classes is not itself $K^0(X)$. In it, the trivial line bundle has no inverse, because the rank of a direct sum is the sum of the ranks. Also, two bundles $E$ and $F$ with $E \oplus 1 \cong F \oplus 1$ have the same class in $K^0(X)$ even when $E \not\cong F$. For real bundles and $KO$, the tangent bundle of $S^2$ and the trivial bundle of rank $2$ are such a pair.