I'm following Spivak's comprehensive guide to differential geometry and I'm getting a bit stuck on a calculation about orientability. The example is showing that the Mobius strip considered as a line bundle over $S^1$ is non-orientable. Intuitively I have no issues, but I would like to construct things a bit more rigorously and am struggling.

Here are some definitions I'm paraphrasing from the book.

We can equip the trivial bundle $\epsilon^n(X)=X\times \mathbb{R}^n$ with the standard orientation on each fibre ${x}\times \mathbb{R}^n$. We write this as $[(x,e_1),...,(x,e_n)]$. Equivalences are fibre-wise isomorphisms, so we can consider the following case. Let $f:\epsilon^n(X)\rightarrow \epsilon^n(X)$ be an equivalence, with $X$ connected. In this case, $f$ either preserves or reverses the orientation on each fibre. To see this, we define the functions $a_{ij}:X\rightarrow \mathbb{R}$ by \begin{align*} f(x,e_i)=\sum_{i=1}^na_{ji}(x)\cdot (x,e_j), \end{align*} which requires that $\text{det}(a_{ij}):X\rightarrow \mathbb{R}$ is continuous and non-zero.

If $\pi:E\rightarrow B$ is a non-trivial $n$-plane bundle, we just need to amend this definition to be local. An orientation of $E$ is defined to be a collection of orientations $\mu_p$ for $\pi^{-1}(p)$ which satisfy the following compatibility condition for any open connected sets $U\subset B$: If $t:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^n$ is an equivalence, and the fibres of $U\times\mathbb{R}^n$ are equipped with the standard orientation, then $t$ is either orientation preserving or reversing on all fibres. Note that if a single equivalence $t$ satisfies this relation, then so must all equivalences $t'$, since $t'\circ t^{-1}:U\times\mathbb{R}^n\rightarrow U\times\mathbb{R}^n$ is also an equivalence. Then as long as we have compatible orientations $\mu_p$ over a collection of sets $U$ that cover $B$, then we obtain an orientation of $E$, $\mu=\{\mu_p\}$.

Here is how I'm trying to do this. What I am imagining is constructing sections on $S^1$; I construct two on each neighborhood, $+1$ and $-1$. I need to do this on a collection of open sets $U$ that cover $S^1$. There is a simple isomorphism $\pi^{-1}(U)\rightarrow U\times \mathbb{R}$ for the problem here, which I can just take to be the identity. Then I want to show that such equivalences cannot define an orientation. Here is my attempt.

We take the bundle \begin{align*} [0,1]\times \mathbb{R} \Big/((0,a)\sim (1,-a)). \end{align*} Consider the following sections on $S^1$: \begin{align*} \sigma^1_\pm:[0,0.4)&\rightarrow \pi^{-1}([0,0.4) )\\ s&\mapsto (s,\pm 1) \\ \sigma^2_\pm:(0.3, 0.7)&\rightarrow \pi^{-1}((0.3,0.7))\\ s&\mapsto (s,\pm 1)\\ \sigma^3_\pm:(0.6,1]&\rightarrow \pi^{-1}((0.6,1] )\\ s&\mapsto (s,\pm 1) \end{align*} I take the equivalences on each neighborhood to be $t_U = \text{Identity}_U$, which either preserves or reverses the orientation on each neighbourhood as required.

Now note the following conditions on each overlapping region (obviously doesn't change choosing a different point):

\begin{align} &t_{[0,0.4)}\circ \sigma^1_\pm(3.5)=(3.5,\pm 1)\\ &t_{(0.3,0.7)}\circ \sigma^2_\pm(3.5)=(3.5,\pm 1)\\ &t_{(0.3,0.7)}\circ \sigma^2_\pm(6.5)=(6.5,\pm 1)\\ &t_{(0.6,1]}\circ \sigma^3_\pm(6.5)=(6.5,\pm 1)\\ \end{align}

So this means that we have $t_U\circ t^{-1}_{U'}(a, 1)=(a, \pm 1)$, and $t_U\circ t^{-1}_{U'}(a, -1)=(a, \pm 1)$, where the sign is changed only if we take different sections (meaning one is $\sigma^i_+$ and the other is $\sigma^j_-$) on the overlapping region $U\cap U'$. But we have the identification $(0,a)\sim (1,-a)$, and so $\sigma^1_+(0)=\sigma^3_-(1)$. But $t_{[0,0.4)} \circ \sigma^1_+(0) = t_{(0.6,1]} \circ \sigma^3_-(1)$, and since these sections are the differently signed ones, this contradicts what we have above, and there is no orientation.

I guess I'm fundamentally unsure that this proves what I want it to. I would appreciate any guidance, but please be as explicit as you can, because it's the construction that I'm really struggling with rather than the concept.

  • $\begingroup$ What is non-rigorous about the reasoning that a real line bundle is orientable iff it is trivial iff it has a nowhere-zero section? The intermediate value theorem shows that the Möbius bundle has no nowhere-zero section. I find your approach hard to follow. Just cover the circle with two open sets, and the intersection will have two components. $\endgroup$ Commented Feb 8 at 17:47
  • $\begingroup$ Yes, as far as I'm aware that reasoning is fine, and intuitively I'm completely happy with that. But what I'm struggling with is turning that logic into a construction, i.e. I'm not quite sure precisely how to construct the objects in your last sentence, and then what calculations to do with these objects. If you could show me what that looks like that would be exactly the answer I'm looking for. And I'd rather do this construction than appeal to theorems just to help my understanding even if it's not necessary. $\endgroup$
    – Bedge
    Commented Feb 8 at 20:30
  • $\begingroup$ My last sentence was a suggestion to simplify your approach. Why is the proof given in my first two sentences not 100% rigorous? What part of that cannot you write down 100% rigorously? $\endgroup$ Commented Feb 8 at 20:51
  • $\begingroup$ I agree your answer is a rigorous proof, I am simply trying to also prove it with a construction of the form I have outlined, to help my understanding. I would like to construct some sections and show that there cannot be an orientation, and this is what I am struggling with. Maybe I'm still not being clear but I'm not disagreeing with your proof, I am just interested in proving this with explicit constructions (i.e. defining sections and showing there are non-orientable equivalences to the local trivializations of each open set), and I am unable to do that myself. $\endgroup$
    – Bedge
    Commented Feb 9 at 10:17
  • $\begingroup$ I do not understand at all what your $3.5$ and $6.5$ are. You must define $\mu_p$ at every $p$ and then check consistency by restricting to an arbitrary connected open set. You're supposedly defining the orientation locally by declaring your sections $\sigma_+$ to define the orientation (i.e., $\sigma_+(x)$ is the positively-oriented basis of $E_x$). Rather than taking a complicated open cover, you know that $\sigma_+(0)=\sigma_-(1)$, and so the orientation defined by $\sigma_+$ in a neighborhood of $x=1$ is wrong. $\endgroup$ Commented Feb 9 at 17:56


You must log in to answer this question.