TITLE: Is a space with no nontrival vector bundles contractible?
QUESTION [10 upvotes]: Let $X$ be a "nice" space, say having the homotopy type of a CW complex. Suppose also that $X$ is connected. Suppose that all real vector bundles on $X$ are trivial. Does it follow that $X$ is contractible?
If not, is there an example where $X$ is very nice, say a compact smooth manifold?
Here is an idea I had for the construction of a counter example. Let $G$ be a group having no nontrivial representations into any real finite-dimensional vector space. For instance, one can take $G = \varinjlim A_n$, the "alternating group on $\mathbf N$" consisting of those permutations of $\mathbf N$ fixing almost all elements pointwise and having sign $+1$. Let $X$ be a $K(G,1)$. Then I believe that $X$ has no nontrivial vector bundles, though I am not certain of my argument (the essential point of which is the equality $$\check{H^1}(X, \text{GL}_n(\mathbf R)) \overset{?}{=} \text{Hom}(G, \text{GL}_n(\mathbf R).)$$
I also thought of using the Atiyah-Hirzebruch spectral sequence to produce a counter-example, but I didn't come up with anything concrete. Anyways, I think it's an interesting question.

REPLY [3 votes]: It is not the case that $X$ must be contractible. For example, take $X = S^3$. By the clutching construction, the isomorphism classes of real rank $n$ vector bundles on $S^3$ are given by $[S^2, O(n)] = \pi_2(O(n)) = 0$ as $O(n)$ is a Lie group. More generally, every principal $G$-bundle on $S^3$ is trivial for any Lie group $G$.