K Group Determined by Chern Classes

Question

Why is it that the complex reduced K group of $\mathbb{CP}^2$ is determined by Chern classes $c_1$ and $c_2$?

I am aware of the fact that the cohomology ring of complex Grassmannians is generated by Chern classes up to its dimension. How does this relate to my question? Or am I heading the wrong way?

Answer 1

Complex vector bundles over a CW complex $X$ with $\dim X \leq 4$ (e.g. $X = \mathbb{CP}^2$) are determined up to isomorphism by their rank and Chern classes. More precisely, if $\operatorname{Vect}_r^{\mathbb{C}}(X)$ denotes isomorphism classes of rank $r$ complex vector bundles on $X$, then

$$c_1 : \operatorname{Vect}^{\mathbb{C}}_1(X) \to H^2(X; \mathbb{Z})$$

is a bijection, and for every $r > 1$,

$$(c_1, c_2) : \operatorname{Vect}_r^{\mathbb{C}}(X) \to H^2(X; \mathbb{Z})\times H^4(X; \mathbb{Z})$$

is a bijection. See this answer for a proof.

In particular, two complex vector bundles over $X$ are stably isomorphic if and only if they have the same Chern classes. Therefore, the map $(c_1, c_2) : \widetilde{K}(X) \to H^2(X; \mathbb{Z})\times H^4(X;\mathbb{Z})$ is a bijection.