Classification of rank $\geq 2$ vector bundles over Grassmannians 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.
 A: 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.
