Let $X$ be a non-compact metric space (though if the answer to the question is positive, then it probably also holds for more general spaces like, e.g., paracompact Hausdorff) and $E \to X$ a vector bundle over it.

Suppose that over every compact subset $K \subset X$ the restricted bundle $E|_K$ is trivial. Can we conclude that $E$ is globally trivial?

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    $\begingroup$ Nice question! Wouldn't the natural assumption be that $X$ is locally compact? I also suppose a natural first start is if we have an exhaustion of $X$ by compact subspaces. $\endgroup$
    – David Roberts
    Aug 18, 2013 at 1:36
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    $\begingroup$ @DavidRoberts: The long line is a locally compact space (not metrizable though), whose tangent bundle is not trivial (whereas it is trivial over the closed subintervals). $\endgroup$ Aug 18, 2013 at 2:33
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    $\begingroup$ If $X$ is a CW-complex, then the questions asks whether there is a non-nullhomotopic phanom map from $X$ to $BO(n)$, where a map is phantom if its retriction to every finite subcomplex is null-homotopic. There is some literature on phantoms maps. I do not know enough about phantom maps into $BO(n)$ but I suspect not all of them are null-homotopic. $\endgroup$ Aug 18, 2013 at 2:38
  • $\begingroup$ @ChrisGerig I don't know enough about the long line: are all compact subsets of it finite unions of closed intervals so that its tangent bundle is a counterexample to my question (if formulated for non-metric spaces)? $\endgroup$
    – AlexE
    Aug 18, 2013 at 2:51
  • $\begingroup$ @DavidRoberts Why should "locally compact" be a natural assumption here? I would think that "paracompact" is much more natural since this has direct consequences for vector bundles over such spaces (e.g., existence of metrics). Note that the long line is not paracompact, but metric spaces are. But if you find a proof which needs local compactness, I would be also satisfied with this. You can also assume other things like separability or second countability or even X being a manifold, if needed. $\endgroup$
    – AlexE
    Aug 18, 2013 at 2:59

2 Answers 2


As Igor Belegradek showed in the comments, one could find an example by finding a CW-complex $X$ and a map $X \to BO(n)$ which is not nullhomotopic, but where the restriction to every finite subcomplex is nullhomotopic. Such a map is called a phantom map. The question "is this map nullhomotopic?" has the same answer whether or not we are asking our maps to preserve the basepoint, and so I will take some steps that are casual about basepoints.

For our example, we're going to take $n = 3$ and $X = \Sigma \mathbb{CP}^\infty$, the suspension of $\mathbb{CP}^\infty$. This is a CW-complex whose finite subcomplexes are $\Sigma \mathbb{CP}^n$.

These spaces are simply connected, so $[\Sigma \mathbb{CP}^n, BO(3)] = [\Sigma \mathbb{CP}^n, BSO(3)]$ for all $n \leq \infty$.

Then $[\Sigma \mathbb{CP}^n,BSO(3)] = [\mathbb{CP}^n, SO(3)]$ for all $n \leq \infty$ by the loop-suspension adjunction. ("A vector bundle on a suspension is determined by a clutching function.")

We can also identify $SO(3)$ with $\mathbb{RP}^3$, which has $S^3$ as a double cover. Again because $\mathbb{CP}^n$ is simply connected, $[\mathbb{CP}^n,SO(3)] = [\mathbb{CP}^n, S^3]$ for all $n \leq \infty$.

One of the famous examples of phantom maps is a map constructed by Brayton Gray: a map $\mathbb{CP}^\infty \to S^3$ which is not nullhomotopic, but where the restriction to $\mathbb{CP}^n$ is nullhomotopic for any $n$. (I believe that this is in his paper "Spaces of the same $n$-type, for all $n$", and that a proof can be given using Milnor's $\lim^1$ sequence.) Pushing this back, we get a vector bundle on $\Sigma \mathbb{CP}^\infty$ whose restriction to any finite subcomplex is trivial.

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    $\begingroup$ In the last line in should be $\Sigma \mathbb CP^\infty$ ;) In fact $n$-dimensional vector bundles on $\mathbb CP^\infty$ identifies with representations. More generally it is a consequence of the work of Jackowski-McClure-Oliver the question IS true over classifying spaces of compact Lie groups. $\endgroup$ Aug 21, 2013 at 21:30
  • $\begingroup$ @JesperGrodal: Oops. Fixed. $\endgroup$ Aug 21, 2013 at 21:32
  • $\begingroup$ This is awesome! Can one explicitly write down clutching functions? Akhil Matthew has a nice writeup of the Gray example here: amathew.wordpress.com/2012/06/13/an-example-of-a-phantom-map but I haven't yet tried to make it explicit enough to do this. $\endgroup$ Aug 21, 2013 at 22:15
  • $\begingroup$ @TylerLawson This is a great answer, thanks! Just one quick question regarding that X: is it metrizable? (Are there metrization theorems for CW-complexes?) $\endgroup$
    – AlexE
    Aug 22, 2013 at 0:11
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    $\begingroup$ Nice example! For CW complexes metrizability is equivalent to local finiteness, so $\Sigma\mathbb{CP}^\infty$ is not metrizable but one can use mapping telescope construction (explained e.g. in Hatcher's "Algebraic Topology" textbook) to replace $\Sigma\mathbb{CP}^\infty$ with a homotopy equivalent locally finite complex which will then be metrizable, and still have the same properties (by construction, I think). $\endgroup$ Aug 22, 2013 at 1:45

If you let $T=S^1 \times D^2$ be the solid torus and pick an embedding $i: T \to \mathrm{int}(T)$ which multiplies by 2 in $\pi_1$, the direct limit $X = \varinjlim(T \xrightarrow{i} T \xrightarrow{i} \dots)$ is a smooth 3-dimensional manifold (non-compact, but admitting a proper embedding into $\mathbb{R}^4$). Its homotopy type is $K(\mathbb{Z}[\frac12],1)$, so by the universal coefficient theorem $[X,\mathbb{C}P^\infty] = \mathrm{Ext}(\mathbb{Z}[\frac12],\mathbb{Z}) \neq 0$, so there exists a non-trivial complex line bundle $L \to X$. (We can even take $L \subset X \times \mathbb{C}^2$ since $[X,\mathbb{C}P^1] = [X,\mathbb{C}P^\infty]$.) Any compact $K \subset X$ is contained in a submanifold diffeomorphic to $T \simeq S^1$, so $L \vert_{K}$ is trivial.

EDIT: in fact, pick any homomorphism $\pi_1(X) = \mathbb{Z}[\frac12] \to \mathbb{C}^\times$ which doesn't factor through the exponential map $\mathbb{C} \to \mathbb{C}^\times$ and use that to define $L$.

  • $\begingroup$ And welcome to MO Anders And! $\endgroup$ Aug 26, 2013 at 21:40

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