Triviality/non-triviality of line/circle bundle over $S^3$ 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. =)
 A: 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.
A: 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, 
www.maths.ed.ac.uk/~aar/papers/steen4.pd
