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Let $\pi:E\to M$ be a rank $k$ vector bundle over a compact manifold $M$. The usual method to associate a sphere bundle to $E$ is by considering only vectors of length 1 in each fiber of $E$ (after choosing a metric on the bundle). This yields a bundle $S(E)\to M$ with fiber $S^{k-1}$.

My question is: Can we construct a $k$-sphere bundle $C(E)\to M$ from $E$ by looking at the one-point compactification of each fiber of $E$?

If this is indeed possible some details to the construction and references would be appreciated.

I suppose that the zero section of $E\to M$ would induce a section of $C(E)\to M$. This construction is probably related to the construction of the Thom-space, where the one-point compactification of the total space $E$ is considered.

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Yes you can. Look at local trivializations for the vector bundle to make the local trivializations for the new sphere bundle... at some point you might have to use the fact that a linear map between vector spaces extends to a continuous map between the one-point compactifications, but that is easy (a linear map is proper!) –  Dylan Wilson Nov 29 '12 at 16:02
Not only does the zero section of $E \to M$ lift to a section of $C(E) \to M$, there is also an $\infty$-section into $C(E) \to M$. –  Alexander Thumm Nov 29 '12 at 16:04
@Alexander Thanks for this observation. –  Dave Hartman Nov 29 '12 at 23:40
I'm pretty sure this gives the same answer as if you take the fiberwise quotient of the disk bundle by the sphere bundle. But as long as you're thinking about this, you should look up the Thom space and the Thom isomorphism, if you haven't seen it before. It's very cool -- you can think of a Thom space as a "twisted suspension" of the base space, and the Thom class is then a generalized version of the fundamental class of the sphere, which measures the extent to which your co/homology theory can "see" the twistedness of the original vector bundle. –  Aaron Mazel-Gee Dec 10 '12 at 23:42

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