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.


1 Answer 1


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).

  • $\begingroup$ 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. $\endgroup$ Oct 6, 2020 at 19:28
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    $\begingroup$ @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. $\endgroup$ Oct 6, 2020 at 19:30

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