# Real line bundle is smoothly isomorphic to Möbius bundle

I'm stuck on this question and tried to follow the partial answer of Neal. Erno's answer is fine too but it seems like I need to find the local trivializations of the Mobius bundle, which requires a bit more work. So here is what I got so far:

We have these two local trivializations of $F$:

$\Phi:\pi^{-1}(S^1-\{1\})\to S^1-\{1\}\times\Bbb{R}$ and $\Psi:\pi^{-1}(S^1-\{-1\})\to S^1-\{-1\}\times\Bbb{R}$. Now I need to construct a diffeomorphism between $F$ and $(I\times\Bbb{R})/\sim$. I can do this with the smooth version of gluing lemma but I need to be careful at the intersection and the transition function $\tau$ will take care of it, but how exactly?

• Sir, how could you post the same title again? ;) Dec 21, 2013 at 14:26
• I just added the "is" there ;) Dec 21, 2013 at 14:26
• HAHA, very nice, it was something hot on math.se in this winter! Dec 21, 2013 at 14:28

You've got a map from $[0, 1] \times \mathbb R$ onto the Mobius bundle, so I'm going to give it a name -- $f$ -- and identify each point in the Mobius bundle as $f(x, t)$ for some $x$ and $t$. The only place that's ambiguous is that certain points have two names: $f(0, t)$ is the same point as $f(1, -t)$. But I'll make a rule: I'll always use the name that has a $0$ as its first argument. (IF you like, I've restricted the domain of $f$ to $[0, 1) \times \mathbb R$.) That means that $F$ provides a 1-1 correspondence between its domain and the Mobius bundle. (The correspondence is NOT continuous at the two "ends" of $[0, 1]$, but at least it gives me a unique way to refer to every point of the total space of the bundle.)

I'm going to define a map $H$ from a subset of the domain of $f$ to $V \times \mathbb R$ (in the language of your assignment).

$$H : (0, 1) \times \mathbb R \to V : (x, t) \mapsto (e^{2\pi i x}, t)$$

I'm going to define another map $K$ from a subset of the domain of $f$ to $U \times \mathbb R$:

$$K : [0, \frac{1}{2}) \cup (\frac{1}{2}, 1) \to U \times \mathbb R: (x, t) \mapsto \begin{cases} (e^{2\pi i x}, t) & \text{for x < \frac{1}{2}} \\ (e^{2\pi i x}, -t) & \text{for x > \frac{1}{2}} \end{cases}$$

These two maps together map the total space of the first bundle to the total space of the second in a way that respects the gluing maps on each side.

I believe that this is what you need.

But you may be saying "Wait, that's a map, but how can I know it's a smooth isomorphism of bundles?" OK. Here goes.

First, I'm going to place two coordinate charts on the base space of the Mobius bundle. Let me call that bundle $M$, OK? Once again, I'll be using the map $f$ to identify points in $M$.

The first chart's domain is $$A = (0, 1) .$$ The coordinate map $\phi_A: A \to \mathbb R$ is exactly $x \mapsto x$.

The second chart's domain is $$B = [0, \frac{1}{2}) \cup (\frac{1}{2}, 1)] / \sim$$

And the coordinate map $\phi_B$ is slightly more complicated: $$\phi_B( [x] ) = \begin{cases} x & \text{for 0 < x < \frac{1}{2}} \\ 0 & \text{for [x] = [0] = [1]} \\ x-1 & \text{for \frac{1}{2} < x < 1} \end{cases}$$

The image of $\phi_B$ is the interval $(-\frac{1}{2}, \frac{1}{2})$.

And now I'm going to build coordinate charts on the bundle; I'll call these $\psi_A$ and $\psi_B$, because one of them will trivialize the portion of the bundle over $A$, and the other will trivialize the portion of the bundle over $B$. Once again, I'll use the labelling of points of the bundle provided by $f$. Here goes:

$$\psi_A : \pi^{-1}(A) \to (0, 1) \times \mathbb R : f(x, t) \mapsto (x, t).$$

Clear? $\psi_A$ is just the inverse of the map $f$, with its domain somewhat restricted.

Now for $\psi_B$.

$$\psi_B : \pi^{-1}(B) \to (-\frac{1}{2}, \frac{1}{2}) \times \mathbb R : f(x, t) \mapsto \begin{cases} (x, t) & \text{for x < \frac{1}{2}} \\ (x, -t) & \text{for x > \frac{1}{2}} \end{cases}.$$

Now let $AB$ denote the intersection of $A$ and $B$.

The transition function $\tau_{AB} : \psi_A(AB) \to \psi_B(AB)$ is once again defined by cases. I'll write it down, and then you have to check that it is, in fact, just $\psi_B \circ \psi_A^{-1}$ on the domain $\psi_A(AB)$.

What's $\psi_A(AB)$? It's $((0, \frac{1}{2}) \cup (\frac{1}{2}, 1)) \times \mathbb R$. What's $\psi_B(AB)$? It's $((-frac{1}{2}, 0) \cup (0, \frac{1}{2})) \times \mathbb R$.

Now let's write down the map between them. $$\tau_{AB}(x, t) = \begin{cases} (x, t) & 0 < x < \frac{1}{2} \\ (x, -t) & \frac{1}{2} < x < 1 \end{cases}$$

Whew!

Now you've got a similar structure on the other bundle: charts $U$ and $V$, coordinate functions $\phi_U$ and $\phi_V$. (I'd recommend $\phi_V(e^{2\pi ix}) = x$ for $-\frac{1}{2} < x < \frac{1}{2}$, and something similar for $\phi_U$.) You can also write down the local trivializations $\psi_U$ and $\psi_V$, which are really pretty simple, and the transition function, which will look a lot like the $\tau_{AB}$ that I wrote down.

Then all that's left to do is to show that the function defined by $H$ and $K$ plays nice with these coordinate charts and transition functions (i.e., check the things that Erno suggested in his answer to your previous question).

If you write out $\phi_U, \phi_V,$ and $\tau_{UV}$ here, I'll go ahead and respond with something about checking those conditions on $H$ and $K$.

• where do you use the transition function $\tau$? Dec 21, 2013 at 15:04
• I didn't use it. But that's because I didn't prove that this was a smooth isomorphism -- I left that to you. I'll add some detail to my answer. Dec 21, 2013 at 15:20
• Is $\psi_B$ continuous at $x=0$? Shouldn't local trivializations be continuous? Dec 8, 2019 at 20:09