# Do $\mathbb{R}^n$ bundles have unit sphere bundles?

Recall there are multiple ways to define the unit sphere bundle of a vector bundle. One is by constructing a fiberwise vector space metric and declaring the sphere bundle to have fibers the unit spheres in each of the vector space fibers. The other way is to use the equivalence of vector bundles and principal $$O(n)$$ bundles and then since $$O(n)$$ acts faithfully on $$S^{n-1}$$ we may replace the fiber to obtain a sphere bundle.

There is, however, a distinction between vector bundles and $$\mathbb{R}^n$$ bundles, that is fiber bundles with fiber $$\mathbb{R}^n$$ and structure group $$Homeo(\mathbb{R}^n)$$. It is not always possible to assign a coherent vector space structure to the fibers to make it a vector bundle. Is there a notion of an associated sphere bundle in this context?

Now I tend to just believe fiberwise constructions are always possible, so I would believe someone if they said we could pick a metric on the fibers of any $$\mathbb{R}^n$$ bundle and define its unit sphere bundle in the same way as for vector bundles. But on the principal bundle side of things, this seems to me to be asserting that $$Homeo(\mathbb{R}^n)$$ has a subgroup $$H$$ so that the inclusion $$H \rightarrow Homeo(\mathbb{R}^n)$$ is a weak equivalence, and $$H$$ preserves the unit sphere $$S^{n-1}$$ while acting faithfully on it. This seems very difficult to believe.

I will add that this is an unstable question. We can always form the fiberwise one point compactification of a $$\mathbb{R}^n$$ bundle since homeomorphisms extend to the one point compactification. If the unit sphere bundle exists, it's fiberwise suspension should coincide with this sphere bundle.

• Why do you need a faithful action? I think you need it only to recover the $\mathbb{R}^n$-bundle from the sphere bundle, which is not something you're asking for... – Denis Nardin Jan 7 at 15:48
• @DenisNardin I guess this would only be necessary if the bundle was obtained by replacing the fiber which apriori it doesn't need to be. – Connor Malin Jan 7 at 16:28
• Thinking more about it, I don't think there should be any issue with a choice of fiberwise metric. I suspect the space of metrics on $\mathbb{R}^n$ inducing the standard topology is contractible, so we should have a section of the bundle of metrics over our bundle. I guess what I am after is a description of the map $Homeo(\mathbb{R}^n) \rightarrow Homeo(S^{n-1})$ that classifies it. – Connor Malin Jan 7 at 16:32
• The "unit sphere" in a metric on R^n does not need to be a topological sphere, so your strategy does not work. (Use a metric on R^5 = S^5 \ p coming from the homeomorphism $\Sigma^2 P \cong S^5$, where $P$ is the Poincare homology sphere. Then there is a point whose unit sphere is $\Sigma P$.) What properties are you looking for your construction to have? Can't prove this is impossible unless we know what "this" is. – mme Jan 7 at 16:36
• @MikeMiller It should lift the vector bundle unit sphere bundle in the cases when the $\mathbb{R}^n$ bundle lifts to a vector bundle structure. The isomorphism class should be natural with respect to pullbacks. There should also be geometric constraints coming from normal bundles to topological submanifolds and the boundary of their regular neighborhoods (when they exist), but I don't know how well I could form these. – Connor Malin Jan 7 at 16:46

It is not true in general that there is a subgroup $$H$$ of $$Homeo(\mathbb{R}^n)$$ such that the inclusion is a homotopy equivalence and $$H$$ preserves the unit sphere. If there was, then $$H$$ would also preserve the unit disk and thus every topological $$\mathbb{R}^n$$-bundle would contain a $$D^n$$-bundle. But it is not the case by a result of Browder, Theorem 1 of Open and Closed Disk Bundles.