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 $B$?

  • 1
    $\begingroup$ 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. $\endgroup$ Aug 20, 2021 at 15:33
  • $\begingroup$ I think that this is actually an answer @ConnorMalin, at least as soon as you invoke Brown-representability. Nice! $\endgroup$ Aug 20, 2021 at 18:41
  • $\begingroup$ Both your comments helped a lot, thanks! @ConnorMalin If you write an answer I will accept it. $\endgroup$ Aug 21, 2021 at 8:17

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .