how to see whether a bundle is trivial or not? Let $Z_2$ be the group with $2$ elements. Let $a\in Z_2$ be the nontrivial element. 
Let $S^n$ be the $n$-sphere. Let $C(S^n,2)=\{(x,y)\in S^n\times S^n\mid x\neq y\}$.
Let $a$ act on $C(S^n,2)$ by
$
a(x,y)=(y,x).
$
And $a$ act on $S^1$ by antipodal map $a(e^{i\theta})=e^{i(\theta+\pi)}$.
Consider a fibre bundle with fibre $S^1$ and base space $ C(S^n,2)/Z_2$
$$
\xi: S^1\to C(S^n,2)\times_{Z_2}S^1\to C(S^n,2)/Z_2. 
$$
How to detect whether the bundle is  trivial or not?
 A: I believe that it's trivial. Since the automorphism of $S^1$ is oriented, your question is equivalent to asking whether or not the (complex) line bundle determined by the bundle is trivial. This is classified by a map
$$
C(S^n,2)/(\mathbb{Z}/2) \to \mathbb{CP}^\infty
$$
or more neatly, an element of $H^2(C(S^n,2), \mathbb{Z})$. 
In order to compute this, define $U$ to be the complement of the diagonal in $S^n \times S^n$ (this is the ordered configuration space). We can look at the long exact sequence of the pair $(S^n \times S^n, U)$, from which we obtain (assuming at least that $n > 2$)
$$
0 = H^2(S^n \times S^n) \to H^2(U) \to H^3(S^n \times S^n, U)
$$
which tells us that the cohomology of the ordered  configuration space injects into $H^3(S^n \times S^n, U)$. Since the cohomology of the quotient is just the invariant cohomology, it suffices to show that  $H^2(U)$ is zero, and it of course then suffices to show that $H^3(S^n \times S^n , U) = 0$.
This should actually be the same (due to Excision) as the Thom space of a rank $n$ bundle over $S^n$, which means that it only has cohomology in degrees 0, $n, 2n$. So if $n > 3$ then you're done.
This is a pretty ugly proof, I admit. It doesn't obviously work for $n = 1, 2, 3$, and it uses a lot of pretty heavy machinery. I'm sure there is a simpler proof out there, but this is the first argument that came to mind...
