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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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    $\begingroup$ 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!) $\endgroup$ – Dylan Wilson Nov 29 '12 at 16:02
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    $\begingroup$ 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 $C(E) \to M$. $\endgroup$ – Alexander Thumm Nov 29 '12 at 16:04
  • $\begingroup$ @Alexander Thanks for this observation. $\endgroup$ – Dave Nov 29 '12 at 23:40
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    $\begingroup$ 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. $\endgroup$ – Aaron Mazel-Gee Dec 10 '12 at 23:42
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A fiber bundle consists of two things. The action of a group $G$ on a fiber $F$, as well as the transition functions. Given an open cover $U_i$ of the base manifold, there are local trivializations $\phi_i:\pi^{-1}(U_i)\cong U_i\times F$. On the overlap there are maps $\phi_i\phi_j^{-1}:U_i\cap U_j\times F\rightarrow U_i\cap U_j\times F$, which have the form $\phi_i\phi_j^{-1}(x,v)=(x,\tau_{ij}(x)\cdot v)$ (the dot is the group action specified above. The $\tau_{ij}$ are maps $\tau_ij:U_i\cap U_j\rightarrow G$ and satisfy the cocycle condition. Now it is a fact that if we have a continuous group action $G\times F\rightarrow F$ and $F$ is Hausdorff and locally compact, the group action extends to a continuous action $G\times \widehat{F}\rightarrow \widehat{F}$ of the one-point compactification $\widehat F$ of $F$. (I found proof of this fact in this generality in a paper by J. de Vries). Thus we can use the transition functions we had before, but the new action to construct a new fiber bundle: "the fiberwise one-point compactification".

For vector bundles the story is easier. The bundle I constructed above is bundle isomorphic to the unit sphere bundle of $E\times \mathbb R\rightarrow M$ (with respect to some auxiliary metric). To prove this you should think of the fiberwise stereographic projection. From this it is also easy to see that there are indeed two sections of this sphere bundle namely $x\mapsto (0,1)$ and $x\mapsto (0,-1)$.

Or you can think of unit disc bundle and collapse every sphere separately as Aaron Mazel-Gee suggested.

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