Why can one discriminate between the trivial $S^1$ line bundle and the Möbius strip by knowing the fibre transformation group?

Both $S^1\times\mathbb R$ and the Möbius strip can be regarded as line bundles over $S^1$.

I have read that one can reconstruct a fibre bundle by knowing its base space, its fibre and the bundle group. In the standard definition of both $S^1$ line bundles (with two charts on $S^1$), both have the same base space $S^1$, the same fibre $\mathbb R$. Their difference is in the bundle group, which is trivial in the case of the trivial bundle, and $\mathbb Z_2$ in the case of the Möbius strip, because at one "coordinate patch", we have to switch the 1-dimensional coordinate $x^1\mapsto-x^1$.

What if we do this on both coordinate patches? If the $\mathbb Z_2$ transformation $x^1\mapsto-x^1$ is done at both patches, then we have a line bundle over $S^1$ that looks like this: I hope the image is clear. It should be a "double Möbius strip", meaning instead of just one twist in the strip there are two twists. The second twist reverses the action of the first one, so the resulting line bundle is the trivial bundle $S^1\times\mathbb R$.

This line bundle has the same parameters as the Möbius strip (same base manifold, same fibre, same group), but it's the trivial one.

What did I understand wrong? What's the precise mechanism of "re-constructing" a bundle?

To construct a vector bundle $p:E \to M$ with fibre $V$, it's not enough to specify an abstract "bundle group" $G$; instead you need something like a locally finite open covering $\{U_{\alpha}\}$ of the base space, a subgroup $G$ of the general linear group $GL(V)$, and a collection of $G$-valued clutching functions $g_{\alpha\beta}:U_{\alpha} \cap U_{\beta} \to G$ satistfying the cocycle condition \begin{align*} g_{\alpha\alpha} &= \text{Id} && \text{in $U_{\alpha}$,} \\ g_{\beta\alpha} &= g_{\alpha\beta}^{-1} && \text{in $U_{\alpha} \cap U_{\beta}$,} \\ g_{\beta\gamma}\, g_{\alpha\beta} &= g_{\alpha\gamma} && \text{in $U_{\alpha} \cap U_{\beta} \cap U_{\gamma}$.} \end{align*} The total space of $E$ is the quotient of the disjoint union $\coprod_{\alpha} U_{\alpha} \times V$ by the equivalence relation $$(x, v) \in U_{\alpha} \times V \sim \bigl(x, g_{\alpha\beta}(x)v\bigr) \in U_{\beta} \times V.$$

Let $U_{1}$ and $U_{2}$ denote the sets in your covering of the circle, and let $U^{-}$ and $U^{+}$ denote the connected components of $U_{1} \cap U_{2}$.

For the "untwisted" bundle, the clutching function is $g_{12} = 1$.

For the Möbius strip, $g_{12}(x) = \pm x$ if $x \in U^{\pm}$.

For the twice-twisted bundle, $g_{12}(x) = -1$. Here, re-trivializing over $U_{2}$ (say) "converts" the clutching function to $g_{12} = 1$. In other words, the twice-twisted bundle is trivial because its clutching function has continuous, non-vanishing extensions to the trivializing neighborhoods $U_{1}$ and $U_{2}$. That the clutching function "doesn't take values in the trivial group" is, in this sense, beside the point.

• +1 for the answer, and (if I could) +1 for the last remark about the global extension of the clutching function. Is it correct that if the clutching functions can be extended globally, then the bundle is trivial? Since one can trivialize the bundle by "globally inverting" the group action?
– Bass
Dec 25 '15 at 18:04
• I was a bit careless; the right way (well, the cohomological way) to think of line bundles is as Čech cocycles. In this setting, trivial bundles are coboundaries: What matters is not that the clutching functions themselves have non-vanishing extensions, but that there exist non-vanishing functions $f_{\alpha}$ on $U_{\alpha}$ such that $g_{\alpha\beta} = f_{\beta}/f_{\alpha}$. The functions $\{f_{\alpha}\}$ then glue together, giving a global section. :) Dec 25 '15 at 22:04