I've been getting started with fiber bundles and I'm really confused up to now. The definition I have is clumsy and incorporates three things: local trivializations, transition maps and structure groups. It is in short like that:

A fiber bundle is a tuple $\xi=(E,B,\pi,F,G)$ where $E,B,F$ are topological spaces respectively called the total space, base space and typical fiber, $\pi : E\to B$ is a continuous surjection such that for each $x\in B$, $\pi^{-1}(x) =F_x$ is a space homeomorphic to $F$ called the fiber at $x$ and $G$ is a topological group of homeomorphisms of $F$. Those items should satisfy:

  1. There is an open cover $\{U_i\}$ of $B$ such that for each $U_i$ there's $\phi_i : \pi^{-1}(U_i)\to U_i\times F$ homeomorphism with $\pi_1\cdot \phi_i = \pi$ where $\pi_1$ is the projection onto the first factor. From this follows that $\pi_2 \cdot \phi_i = \tilde{\phi}_i$ is a homeomorphism between $F_x$ and $F$.

  2. Whenever $x \in U_i$ we can consider the restriction of $\tilde{\phi}_i$ to $F_x$, the map $\tilde{\phi}_{i, x}$. Given $x\in U_i\cap U_j$ we can form the composition $\tilde{\phi}_{i, x}\circ \tilde{\phi}^{-1}_{j, x}:F\to F$ which is an element of $G$ for all $x\in U_i\cap U_j$ and $i,j$.

  3. The induced mappings $g_{jk} : U_j\cap U_k \to G$ by $g_{jk}(x) = \tilde{\phi}_{j, x}\circ \tilde{\phi}^{-1}_{i, x}$ are continuous. They are called transition functions and satisfy $g_{jk}g_{ki}=g_{ji}$.

Now, condition one is clear, it says that locally the bundle is a product. Condition $2$ is complicated for me. The reason: I don't understand why there are many parametrizations and why their relation is important. The book says that the transition functions are the all important thing that gives the structure of the bundle, like the twisting of Möebius' band, but I can't see how.

I found an article on that example that said the following:

Say we want to parameterize the Möebius’ band by a point $\theta$ on the circle, and a real number $f$ in the interval $(0,1)$. Concretely, say $\theta= 0$ and $f =\dfrac{3}{4}$. Now, we transport the point around the Möebius' band by increasing $\theta$ and keeping f fixed at $\dfrac{3}{4}$. When $\theta\to 2π$, we should return to the same point $q$ in $E$, since it corresponds to the same $\theta$ and $f$, and hence the same $q$. However, because of the inescapable twist in the band, the point we return to is associated with $\theta= 0$, $f = \dfrac{1}{4}$. Our parameterization for $F$ somehow “flips” when we move one turn around the Möebius' band.

Now, I really can't understand it. How can those functions give two differents parametrizations? I can't get this intuition, on how those functions somehow give us the twist in the band.

How all of that works? Is there any other way to understand this?


Cut the Möbius strip into three parts. Each part has a trivial fibration. Mark an up direction on the fibers of each part. This is completely at will, there is no natural preference of direction. The glueing together to form the Möbius strip can then proceed on the first two overlaps as up-to-up, but on the last it is up-to-down. As group elements of $G=\{-1,1\}$ this is just $(1,1,-1)$. Again, the position of the $-1$ is completely random, and also a glueing together with $(-1,-1,-1)$ is possible.


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