I am interested in whether I can find non-trivial bundles in the case of a fiber bundle with base space $S^3$ and fiber either $\mathbb{R}$ or $S^1$. I know that in the case of a principal bundle triviality is equivalent to the existence of a global section. In the vector bundle case (with fiber $\mathbb{R}$) I'm not really sure how to know about triviality/non-triviality of a bundle (there always exists a global zero section). I know that if the base space is contractible (e.g. $\mathbb{R}^n$) all bundles over it are trivial, but this is not the case with $S^3$. As far as I understand, $S^3$ is simply connected but not contractible. Obviously I have noted that both $S^1$ and $\mathbb{R}$ are one-dimensional so that it is a rank 1 bundle, but I'm not sure if this is important.

Question: On a bundle with base space $S^3$ and fiber either a line $\mathbb{R}$ or $S^1$, how can I show if the bundle is trivial or non-trivial? (I.e. in the case of a principal bundle - how do I show the existence or non-existence of a global section?)

If you have any references for me I would be delighted. I have mostly used Nakahara's book "Geometry, topology and physics" in my studies. Personally I would say that I know the basics of manifolds and bundles but I am realistic about my knowledge on these topics, and it's all quite complicated to me. So please be gentle with me as I don't have a strict mathematical background. =)


Over spheres, you actually have a nice way to distinguish bundles.

Every sphere $S^n$ can be written as a union of two contractible spaces, $U \cup V$, whose intersection is a small neighborhood of the equator, $\mathbb{R} \times S^{n-1}$. For instance if $x,y$ are the north and south poles, respectively, you can take $U = S^n - \{x\}, V = X^n - \{y\}$.

Over $U$ and $V$, any bundle is trivial, as you noted. The bundle on the sphere is glued together from these two trivial bundles.

If you're interested in a vector bundle, the transition function for the bundle is given by a map $U \cap V \to GL_m$, for a vector bundle of rank $m$. If this map is null-homotopic, meaning that it can be homotoped to the constant map, then your bundle is trivial, as any trivialization over $U$ can then be glued to a trivialization over $V$. So in the end you are reduced to the study of maps from $U \cap V \simeq S^{n-1}$ to your structure group -- i.e., $\pi_{n-1}GL_m$.

For $m=1$, this amounts to asking whether any map $S^{n-1} \to GL_1$ is null-homotopic. In your case for $S^3$, since $GL_1(\mathbb{R})$ is a disjoint union of two contractible spaces, and $S^2$ is connected, any map is null-homotopic.

In the fiber bundle case with fiber $S^1$ and base $S^3$, you are looking at maps $S^2 \to S^1$. But $\pi_2(S^1) = *$ so any circle bundle is contractible over $S^3$.

Note that both facts hold for higher spheres as well.

  • $\begingroup$ Thanks! I am still not sure I follow what you write on homotopy. I get that the intersection $U \cup V$ is basically $S^{n-1}$ and that the transition function $t_{UV}$ go from $U\cap V$ to $G=S^1$ (for a circle bundle), but why does this mean that one should look at $\pi_{n-1}(S^1)$? And in the case of a line bundle I don't really understand the following: italic"Since GL1(R) is a disjoint union of two contractible spaces, and S2 is connected, any map is null-homotopic, and hence contractible."italic $\endgroup$
    – kloptok
    Jan 4 '13 at 21:45
  • $\begingroup$ And by the way, I think there was an $n$ too much in your section on vector bundles -- both the rank of the bundle and the base space sphere use the letter $n$. I guess what you meant is that one should study $\pi_{n-1}(GL_m(\mathbb{R}))$ for a rank $m$ bundle? $\endgroup$
    – kloptok
    Jan 4 '13 at 21:49
  • $\begingroup$ In a principle fiber bundle with fiber $G$, the transition functions are given by a map from $U_i \cap U_j \to G$; that's why you look at maps form $U \cap V \to S^1$ in your example of a principle circle bundle. As it turns out, if you're interested in any old circle bundle, the group of automorphisms of a circle is homotopy equivalent to a disjoint union of two circles -- the components being given by whether or not one preserves orientations. So you're right that to consider $S^1$ bundles (not necessarily principle) you should consider maps of $U \cap V$ into $S^1 \sqcup S^1$. $\endgroup$
    – user54535
    Jan 6 '13 at 1:53
  • $\begingroup$ The reason that the study of maps $U \cap V \to B$ can be replaced by studying maps $S^{n-1} \to B$ is that $U \cap V$ is homotopy equivalent to $S^{n-1}$ -- the homotopy type of a mapping space is unchanged by homotopy equivalences. Finally, the $GL_1(R)$ comment. The word "contractible" was redundant and even confusing; I removed it. The reason any map is null-homotopic: $GL^_1(\mathbb{R})$ is homotopy equivalent to two disjoint points, so two maps $f,g$ aren't homotopic if and only if they map components of the domain into different components of the target. But your domain is connected. $\endgroup$
    – user54535
    Jan 6 '13 at 1:58

Every vector bundle or sphere bundle over $S^{3}$ is trivial by the following reason, (respectively):

1)As it is said in the answer by user54535, the structure of vector bundle on $S^{3}$ is determined by the homotopy type of the clutching function from $S^{2}$ to the Lie group $GL(n)$ but every lie group has trivial second fundamental group: $https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups

2)The following shows that every sphere bundle on 3- sphere is trivial.

The cllasification of sphere bundles,



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