# Does a finite-dimensional Grassmannian classify the subbundles of a trivial vector bundle?

It is known that the universal vector bundle over the infinite-dimensional Grassmannian, $$E \longrightarrow Gr_n(\mathbb{R}^{\infty}),$$ classifies the rank $$n$$ vector bundles in the sense that any such vector bundle (let me assume that $$B$$ is a compact CW complex) $$E' \longrightarrow B$$

is isomorphic to the pullback $$f^{*}E \longrightarrow B$$

for some $$f: B \rightarrow Gr_n(\mathbb{R}^{\infty})$$. Furthermore, two bundles $$f^{*}E$$ and $$g^{*}E$$ are isomorphic if and only if $$f$$ and $$g$$ are homotopic.

Is there a version of this correspondence for the subbundles of a fixed trivial bundle? To be precise,

1. Let $$F_n^{D}$$ be the canonical vector bundle over $$Gr_n(\mathbb{R}^D)$$. For fixed $$n, d$$, can we find $$D$$ with the following property: for every rank $$n$$ subbundle $$F'$$ of the trivial bundle $$B\times \mathbb{R}^d$$, there is a $$f:B \rightarrow Gr_n(\mathbb{R}^D)$$ such that $$F'\approx f^{*}F_n^D$$ ?

2. Can we assume $$D=d$$?

3. For two $$f,g: B \rightarrow Gr_n(\mathbb{R}^D)$$, does $$f^{*}F_n^D \approx g^{*}F_n^D$$ hold if and only if $$f,g$$ homotopic? How does the answer depend on the dimension of $$B$$?

• I think the issue is that to be representable there is a pushout condition which says if we glue two vector bundles over $X$ and $Y$ along an isomorphism between the restriction to a subcomplex of $X$ and $Y$, then the result must be a sub bundle of the trivial rank n bundle. This does not happen even in the case where both vector bundles are trivial to begin with. Aug 20, 2021 at 15:33
• I think that this is actually an answer @ConnorMalin, at least as soon as you invoke Brown-representability. Nice! Aug 20, 2021 at 18:41
• Both your comments helped a lot, thanks! @ConnorMalin If you write an answer I will accept it. Aug 21, 2021 at 8:17

Connor Malin's comment pointed out that there is not likely to be a representability theorem of subbundles of a trivial bundle as in the infinite-dimensional Grassmannian case. This resolved most of my initial concerns.

Aside from that, I think I found a direct counter-example of 1&2&3 in the case $$B=S^1\times S^1\times S^1$$.

For simplicity, I will take $$F=\mathbb{C}$$ as the base field. The following construction is used in condensed matter physics to build a model of Hopf Insulator.

$$f: B \longrightarrow Gr_1(\mathbb{C}^2)=\mathbb{C}P^1, \\ (k_1,k_2,k_3) \mapsto [\sin{k_1}+i\sin{k_2}:\sin{k_3}+i(\cos{k_1}+\cos{k_2}+\cos{k_3}-3/2)],$$

where $$k_i$$'s are the periodic coordinates on $$B=T^3$$.

$$f$$ defines the pullback line bundle $$f^{*}E$$ of the canonical bundle $$E$$ over $$\mathbb{C}P^1$$, which is trivial since it has a nonvanishing section

$$\sigma:B\longrightarrow f^{*}E \subseteq B\times \mathbb{C}^2, \\ \mathbf{k} \mapsto \bigg(\mathbf{k},\big(\sin{k_1}+i\sin{k_2},\sin{k_3}+i(\cos{k_1}+\cos{k_2}+\cos{k_3}-3/2) \big)\bigg).$$

Hence hypotheses 1&2&3 in the question cannot be true simultaneously in the complex case unless $$f$$ is nulhomotopic.

In fact, $$f$$ restricts to the nulhomotopic maps on the surfaces $${1}\times S^1\times S^1, S^1\times {1}\times S^1, S^1\times S^1\times {1}$$ and homotopy extension property of CW pairs enables us to see $$f$$ as $$\widetilde{f}: S^3 \longrightarrow \mathbb{C}P^1\approx S^2.$$

$$\widetilde{f}$$ has nonzero Hopf invariant,

$$\chi=-\frac{1}{4\pi^2}\int_{T^3} d\mathbf{k}\;\mathbf{F}\cdot\mathbf{A}=1, \quad\text{where} \\ \mathbf{A}(\mathbf{k})=i \overline{f(\mathbf{k})} \cdot \nabla_{\mathbf{k}} f(\mathbf{k}) \;(\text{here f maps into \mathbb{C}^2}),\; \mathbf{F}(\mathbf{k})= \nabla_{\mathbf{k}} \times \mathbf{A}(\mathbf{k}),$$ which implies that $$\widetilde{f}$$ cannot be homotoped to a constant map.