# Given a vector bundle over a compact space, how can we determine the smallest Grassmannian classifying it?

It is a standard result that for $$\mathbb{F}=\mathbb{R},\mathbb{C},\mathbb{H}$$, the Grassmannian $$G_n(\mathbb{F}^{\infty})$$ is a homotopical classifying space for $$n$$-plane bundles over any paracompact Hausdorff space. However, if we assume the base is compact, we also find that an $$n$$-plane bundle can be realized as the pullback of the tautological bundle over any sufficiently large finite-dimensional Grassmannian $$G_n(\mathbb{F}^m)$$ ($$m\gg 1$$). Is there a way to determine what the smallest such $$m$$ is? Moreover, is the map in this case unique up to homotopy? If this is the case, then the finite-dimensional Grassmannians would also be classifying spaces, but for a specific subclass of $$n$$-plane bundles, which is what I'm trying to figure out.

The finite-dimensional Grassmannians do classify a subclass of vector bundles. $$\text{Gr}_n(F^m)$$ classifies exactly the rank $$n$$ subbundles of the trivial bundle $$F^m$$; it follows that the classifying map of a vector bundle $$V$$ of rank $$n$$ factors through $$\text{Gr}_n(F^m)$$ iff there exists another vector bundle $$W$$ of rank $$m-n$$ such that $$V \oplus W \cong F^m$$ is trivial(izable).
It's known that if $$X$$ is a $$d$$-dimensional CW complex then such a $$W$$ always exists of rank at most $$d$$, so the classifying map of a vector bundle of rank $$n$$ factors through $$\text{Gr}_n(F^{n+d})$$.
Finding the smallest $$W$$ is delicate. There are obstructions coming from characteristic classes. For $$F = \mathbb{R}$$, $$X$$ a smooth manifold of dimension $$n$$, and $$V$$ the tangent bundle of $$X$$, it's known that in the worst case ($$X$$ a suitable product of real projective spaces) the smallest $$W$$ has rank $$n - \alpha(n)$$ where $$\alpha(n)$$ is the number of $$1$$s in the binary expansion of $$n$$. This is closely related to the question of the minimal dimension of a smooth immersion of $$X$$ into $$\mathbb{R}^m$$; see, for example, these notes (which contain a proof of the claim in the second paragraph).
• Thank you! Now that you mention it, I see why the subbundles of the rank $n$ trivial bundles are precisely those which come from the $m$th Grassmannian. It's less clear to me, however, why the map defining the pullback must be unique up to homotopy. The proof for the infinite-dimensional case makes significant use of the infinite-dimensionality. Oct 6, 2020 at 19:28
• @Doron: I didn't make any claim about uniqueness. A priori there may be inequivalent choices of an embedding of $V$ into a trivial bundle. Oct 6, 2020 at 19:30