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I am a bit struggling with one of the many definitions of orientability. In what follows $M$ will always denote a smooth, connected manifold, $T_{m}M$ will be the tangent space at $m$, $\mathcal{B}_{m}$ will be the set of ordered bases of $T_{m}M$, i.e. $$\mathcal{B}_{m}:=\{(v_{1},\ldots,v_{n})\in (T_{m}M)^{n};\operatorname{span}_{\mathbb{R}}\{v_{1},\ldots,v_{n}\}=T_{m}M\}$$$\sim_{m}$ will denote the equivalence relation on $\mathcal{B}_{m}$ obtained by $v\sim_{m}w:\Leftrightarrow \det(A(v,w))>0$, where $v=(v_{1},\ldots,v_{n})$ and $w=(w_{1},\ldots,w_{n})$ are elements in $\mathcal{B}_{m}$ and $A(v,w)$ is the unique element in $\operatorname{GL}(T_{m}M)$ satisfying $A(v,w)v_{i}=w_{i}$ for all $1\leq i\leq n$. Finally $\Lambda_{n}(T_{m}M)$ denotes the space of alternating $n$-forms on $T_{m}M$ and $\Lambda_{n}^{\ast}(M):=\bigcup_{m\in M}\{m\}\times\Lambda_{n}(T_{m}M)$. I have topologized $\Lambda_{n}^{\ast}(M)$ by requiring that if $(U,\rho)$ is a chart in the maximal atlas on $M$, then the map $\tilde{\rho}:\tilde{U}\to\mathbb{R}^{n+1}$ is smooth, where: $$\tilde{U}:=\bigcup_{m\in U}\Lambda_{n}(T_{m}M) \text{ and }\tilde{\rho}(m,\omega):=\left(\rho(m),\omega\left(\frac{\partial}{\partial x_{1}}\bigg|_{m},\ldots,\frac{\partial}{\partial x_{n}}\bigg|_{m}\right)\right)$$ Indeed the image of $\tilde{\rho}$ is open in $\mathbb{R}^{n+1}$ and so on. The main goal is to put more sense into the following statement:


M is orientable iff $\Lambda_{n}^{\ast}(M)\setminus N$ has two connected components, where $N$ is the range of the map $\omega:M\to\Lambda_{n}^{*}(M): \omega(m):=(m,0)$.

I am struggling with the following statement:


$\Lambda_{n}^{\ast}(M)\setminus N$ has at most two components.

That is it has either one or two. I approached it by comparing different notions of orientability. So far I have managed to prove the following statement:


Let $M$ be a smooth manifold and $\dim M=n$. Then the following are equivalent:

  1. There exists a smooth, no-where vanishing differential $n$-form on $M$
  2. There exists an atlas $\mathcal{A}$ on $M$ such that for all $(U,\rho),(V,\psi)\in\mathcal{A}$ holds $\det(\operatorname{d}_{m}(\rho\circ\psi^{-1}))>0$ for all $m\in U\cap V$.
  3. There exists a map $\epsilon:M\to\bigcup_{m\in M}\mathcal{B}_{m}/\sim_{m}$ and for all $m\in M$ there exists $U\subseteq M$ an open neighbourhood of $m$ and smooth vector fields $X_{1},\ldots,X_{n}$ on $U$ such that for all $p\in U$ holds $(X_{1}(p),\ldots,X_{n}(p))\in\epsilon(p)$.
  4. $\Lambda_{n}^{\ast}(M)\setminus N$ has two components.

This does not tell me why we have either one or two components (and definitely not more). I have no idea what to do (I even tried a direct proof).

Edit: It seems to me to be most promising to prove that non-3 implies that $\Lambda_{n}^{\ast}(M)$ has only one component, i.e. I would try to show that if for every $\epsilon:M\to\bigcup_{m\in M}\mathcal{B}_{m}/\sim_{m}$ there exists $m\in M$ such that for every open neighbourhood $m\in U\subseteq M$ and all smooth vector fields $X_{1},\ldots,X_{n}$ on $U$ satisfying $(X_{1}(p),\ldots,X_{n}(p))\in\mathcal{B}_{p}$ for all $p\in U$ we can find $p,q\in U$ such that $(X_{1}(p),\ldots,X_{n}(p))\in\epsilon(p)$ and $(X_{1}(q),\ldots,X_{n}(q))\not\in\epsilon(q)$, then $\Lambda_{n}^{\ast}(M)\setminus N$ has one component. What do you think of this? Is this feasible?

Thank you so much


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up vote 2 down vote accepted

A sketch of a possible proof:

let $\pi: \Lambda_n^*(M) \to M$ be the natural projection map.

  1. Let $C$ be a connected component of $\Lambda_n^*(M)$. Show that $\pi(C) = M$.
  2. Let $U$ be a nonempty connected trivializing open subset of $M$, then $\pi^{-1}(U) - U = U \times \mathbb{R}_{> 0} \cup U \times \mathbb{R}_{< 0}$. (1) shows that $C$ contains at least one of these two components.

a. If for some $U$, $C$ contains both components, prove that $C = \Lambda_n^*(M)$.

b. If for all $U$, $C$ contains only one component, let $D$ be the union of all the other components. Show that $D$ is connected and $\Lambda_n^*(M) = C \cup D$.

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Thank you. I will try it out within the next few days. Have you done this already, i.e. are you sure that this will work or do you need any feedback? – M. Luethi Jan 21 '13 at 18:27
@M.Luethi, I'm pretty certain that this works. – user27126 Jan 21 '13 at 18:31
A quick question: do you mean by $\pi^{-1}(U)-U$ the preimage without the part intersecting the range of the zero-section, i.e. in fact $\pi^{-1}(U)-U\equiv\pi^{-1}(U)-N$? – M. Luethi Jan 21 '13 at 18:52
@M.Luethi, yes. – user27126 Jan 21 '13 at 18:57

You can easily show that $\dim\Lambda_n^*(V) = 1$ (as a vector bundle), so you only need to show, that for every section $\omega\neq 0$ and $\lambda > 0$, $\lambda\omega\in\Lambda_n^*(V)\setminus\{0\}$. That way, for all $\omega\neq 0$ you'd get path-connectedness to all other $\omega'$ on the same “side” of $0$ (always remembering that you are working on the real axis here), so you have at most 2 connected components.

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What do you mean by for all $\omega \neq 0$ you would get path-connectedness? – user27126 Jan 21 '13 at 9:26
I've edited it, is it clearer now? – filmor Jan 21 '13 at 10:12
I still don't quite understand it. 1. You seem to be connecting sections. Can you make it a bit more precise how this connects to the connectedness of the line bundle - zero section? – user27126 Jan 21 '13 at 10:19
2. Non-vanishing global sections may not exist. How do you deal with the zeros of sections? – user27126 Jan 21 '13 at 10:21
@filmor: Thank you. The first part ("you can easily show ...") is clear. So let's stick with the rest of your suggestion. Am I mistaken by interpreting it as "if there exists a non-vanishing section, then $\Lambda_{n}^{\ast}(M)\setminus N$ has two components"? This is what I have done in mentioned theorem. What I need is somehow that if I can NOT find such a section, then $\Lambda_{n}^{\ast}(M)\setminus N$ has ONE component. Please let me know what's wrong about my "understanding". – M. Luethi Jan 21 '13 at 10:24

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