All manifolds in this post are hausdorff and second-countable.

Is it true that two smooth fiber bundles with same fiber, base manifold and structure group (that is a Lie group $G$ of diffeomorphisms of the fiber) are equivalent if and only if are equivalent as continuous fiber bundle (so the equivalence can be only continuous)? If it is true, can you give me a reference?

Thank you.

  • $\begingroup$ This is far from my expertise, but are exotic spheres (seen as fiber bundles over a point) counter examples? $\endgroup$
    – Thomas Rot
    Dec 15, 2015 at 11:05
  • $\begingroup$ @ThomasRot I edited the post: I was assuming same fiber, base manifold and structure group. I think your counterexample fails because those bundles does not have the same fibre. $\endgroup$
    – Ervin
    Dec 15, 2015 at 11:19

1 Answer 1


As long as the Lie groups are finite dimensional, yes, this is true. The key is that you can make finite-dimensional approximations to the classifying space $BG$: that is, there are finite dimensional smooth manifolds $B_iG$ with inclusion maps $B_iG \hookrightarrow B_{i+1}G$ such that, as topological spaces, $BG = \lim B_iG$. I don't have a good reference for this or a sketch of the proof. For your favorite groups, it's obvious: $BO(n) = \text{Gr}(n,\infty) = \lim_k \text{Gr}(n,n+k)$, for instance, or $BU(1) = \lim_k \Bbb{CP}^k$. The general case, then, is the same idea: you find a sequence of spaces $E_nG$ that $G$ acts freely on whose limit is contractible. If I remember a reference or a proof for this I'll edit it in.

Now that we have this: let $M$ be a finite dimensional smooth manifold. Two $G$-bundles being smoothly equivalent is the same as saying that the smooth maps $M \to BG$ are smoothly homotopic. (To make sense of this, one either makes a Hilbert manifold out of $BG$, in which case the first paragraph wasn't necessary, as we'll see, or you assume $M$ is compact so the image lies in some $B_iG$.) The proof of this is essentially the same as the proof that the same is true continuously of continuous maps.

Now we're in a setting in which we can apply the various smooth map approximation theorems. This is the reason I tried to make things finite dimensional; I don't know how to prove the same approximation theorems in the infinite-dimensional setting, though they're almost certainly true. Anyway, let's get going.

First, any continuous map is homotopic to a smooth map. So for any continuous vector bundle on $M$, this must be isomorphic to a smooth vector bundle. Next, suppose I have smooth vector bundles $f_i: M \to BG$. Suppose they are continuously homotopic: that is, there is a continuous map $f_t: M \times I \to BG$. Now (again, either because we can restrict the codomain to a finite-dimensional manifold or because you're willing to use infinite-dimensional approximation theorems) we know that we can approximate this by a smooth map without changing the values on the boundary, since they're already smooth. So this means that if vector bundles $f_0$, $f_1$ are continuously isomorphic, they are smoothly isomorphic.

This becomes false if you allow the Lie group to become infinite-dimensional (say, $\text{Diff}(M)$.) The simplest examples I know are 4-manifold bundles over the circle which are topologically trivial but not smoothly trivial. But because $M$-bundles over the circle are in bijection with $\pi_0 \text{Homeo}(M)$ topologically or $\pi_0 \text{Diff}(M)$ smoothly, you just need a diffeomorphism that's continuously isotopic to the identity but not smoothly. One is provided, eg, in Ruberman, "An obstruction to smooth isotopy in dimension 4".

  • 2
    $\begingroup$ I would be cautious about invoking the claims in this answer for noncompact manifolds $M$. The reason being that I'm not confident about these so-called approximation results for Hilbert manifold codomains. In particular, what's stopping you from using them on $B\text{Diff}(M)$? It must not be a Hilbert manifold, or something like that. At the very least, it doesn't have finite-dimensional approximations, which is why the other argument goes through fine. $\endgroup$
    – user98602
    Dec 15, 2015 at 15:03
  • $\begingroup$ Dear Mike, I have a question about the last paragraph of your answer. It seems that you are changing the "Lie group" from Diff(M) to Homeo(M). But what if, as in the original question, you don't change the group but you change whether the transition maps from the base into the group are continuous vs smooth? For Diff(M) this perhaps does not make sense directly, but we could speak instead of bundles that are "totally smooth" vs "fiberwise smooth." (I.e. the total space is a manifold and the trivializations are smooth, vs. a continuous fiber bundle with structure group Diff(M).) $\endgroup$
    – Cary
    Jul 17, 2018 at 8:48
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    $\begingroup$ @Cary I'm confused about the last paragraph of this answer. It does seem that I am changing the structure group, and I do believe that continuous maps $X \to B\text{Diff}(M)$ still classify smooth $M$-bundles, which is contradictory to my last paragraph here. There should be a smooth approximation argument of some sort (you could model $B\text{Diff}(M)$ as the space of submanifolds of $\Bbb R^\infty$ that are diffeomorphic to $M$; then for a sufficiently small chart in $X$, our map is given by a continuous map $U \times M \to \Bbb R^\infty$ which is an embedding on each $x \times M$) $\endgroup$
    – user98602
    Jul 18, 2018 at 21:00
  • $\begingroup$ Thanks for clarifying. This has been confirmed by others I've asked about this -- both notions of "smooth bundle" are classified by BDiff(M), by careful use of smooth-approximation arguments. $\endgroup$
    – Cary
    Jul 27, 2018 at 11:16
  • $\begingroup$ @Cary I really appreciate your comment correcting my mistake. $\endgroup$
    – user98602
    Jul 27, 2018 at 11:35

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