Classification of rank $\geq 2$ vector bundles over Grassmannians

Question

Are there classification results of higher rank (complex) vector bundles over (complex) Grassmannian manifolds? For example, we know that line bundles are in correspondence with the $H^2(G)$, the second cohomology group.

Answer 1

If $X$ is a paracompact topological space, then we have $\operatorname{Vect}_k^{\mathbb{C}}(X) = [X, \operatorname{Gr}_k(\mathbb{C}^{\infty})]$ where the left hand side is the set of isomorphism classes of rank $k$ complex vector bundles on $X$ and the right hand side is the set of free homotopy classes of maps $X \to \operatorname{Gr}_k(\mathbb{C}^{\infty})$.

In particular, as $\operatorname{Gr}_m(\mathbb{C}^n)$ is paracompact, $\operatorname{Vect}_k^{\mathbb{C}}(\operatorname{Gr}_m(\mathbb{C}^n)) = [\operatorname{Gr}_m(\mathbb{C}^n), \operatorname{Gr}_k(\mathbb{C}^{\infty})]$. I am not aware of a simpler description than this.

---

The reason why complex line bundles offer a more useful description is that $\operatorname{Gr}_1(\mathbb{C}^{\infty}) = \mathbb{CP}^{\infty}$ is an Eilenberg-Maclane space, namely a $K(\mathbb{Z}, 2)$; i.e. $\pi_2(\mathbb{CP}^{\infty}) = \mathbb{Z}$ and $\pi_n(\mathbb{CP}^{\infty}) = 0$ for $n \neq 2$. Using the general fact that $[X, K(G, n)] = H^n(X; G)$, we see that

$$\operatorname{Vect}_1^{\mathbb{C}}(X) = [X, \mathbb{CP}^{\infty}] = [X, K(\mathbb{Z}, 2)] = H^2(X; \mathbb{Z}).$$

One might hope that $\operatorname{Gr}_k(\mathbb{C}^{\infty})$ is an Eilenberg-Maclane space for other values of $k$ so that we can obtain a similar description of $\operatorname{Vect}^{\mathbb{C}}_k(X)$ as a cohomology group, but this is not the case. For every $k > 1$, $\operatorname{Gr}_k(\mathbb{C}^{\infty})$ has infinitely many non-zero homotopy groups and is therefore not an Eilenberg-Maclane space.