Vector bundle of dimension $\leqslant n$ on $n$-connected space is trivial

Question

I wonder whether any vector bundle of dimension $\leqslant n$ on an $n$-connected CW-complex is trivial? It seems that,

1. the complex can be given cellular structure with exactly one 0-dimensional cell and no other cells of dimension $\leqslant n$,
2. there is a inner product on the vector bundle by the paracompactness of CW-complexes,
3. the corresponding $S^{n-1}$-fiber bundle, by some obstruction theory argument (I am not sure which one, but it should use the fact that all maps $S^{\leqslant (n-1)} \to S^n$ are homotopic), has a section,
4. and its orthogonal complement has smaller dimension, so allowing the proof by induction.

For $n=1$ it is true: any covering (with two sheets) over simply-connected space is trivial. But I have not found the general statement anywhere, so it there an error?

Answer 1

Your proposed argument fails at the third step. It is just not true that if $X$ is $n$-connected then every $S^{n-1}$-bundle over it is trivial.

The desired statement is not even true rationally: the rational homotopy groups of $BO(n)$ are not hard to calculate and in particular for $n \ge 3$,

$$\pi_{4n-4}(BO(2n)) \otimes \mathbb{Q} \cong \mathbb{Q}$$

so it follows that there are countably many nontrivial $2n$-dimensional real vector bundles over $S^{4n-4}$, which moreover can be distinguished by their Pontryagin classes $p_{n-1}$, and this gives an infinite family of counterexamples to the desired statement.

Answer 2

No, this is not the case. As Najib points out in the comments, this is the same as asking "if $X$ is $n$-connected, is every map $X \to BO(n)$ null-homotopic?"

Instead, let's show this isn't true for oriented bundles; if every bundle is (unoriented) trivial, then there are 2 oriented bundles up to isomorphism.

Take $X = S^4$ and $n = 3$. $SO(3)$ has a double cover isomorphic to the group of unit quaternions $S^3$, which fits into a fibration $\Bbb Z/2\Bbb Z\to BSO(3) \to BS^3$, and $BS^3 = \Bbb{HP}^\infty$ (this coming from a more general fibration $G/H \to BH \to BG$). Passing to the long exact sequence of homotopy groups we see $\pi_k(BSO(3)) = \pi_k(\Bbb{HP}^\infty)$. By cellular approximation, $\pi_4(\Bbb{HP}^\infty) = \pi_4(S^4) \cong \Bbb Z$, which gives far more oriented 3-bundles over $S^4$ than possible if they were all (unoriented) trivial.

Answer 3

This is more of a comment on the other answers.

We can interpret the case $n = 0$ in terms of classifying spaces as well. The point is that $BO(1) = \mathbb{RP^{\infty}}$, and any map $f$ from a simply connected space $X$ to $BO(1)$ will lift to the universal cover of $BO(1)$, which is $S^{\infty}$. In particular, $f$ will factor through a contractible space, and thus is null homotopic.

Of course, one can also prove this more directly, also using the lifting lemma on the simply connected space itself, and without referring to classifying spaces. But phrasing it this way makes it natural that one would want to a "lifting" lemma for universal principle bundles (in particular for $G = O(n)$). However, there doesn't seem to be one, as the answers to this question show. (Or if there is one, it is more subtle than just checking induced maps on homotopy groups.) (Note that the map corresponding to a vector bundle lifts to the universal bundle, which is contractible, iff the map is homotopic, iff the vector bundle is trivial.)