There is this well known theorem (the one in the title), I am reading a proof for it in Spivak's A comprehensive Introduction to Differential Geometry, and I have a question regarding a step in it. First, the definition of orientability used in this book is the following:
A Manifold $M$ is called orientable if there is a collection $\mu=\{\mu_p\}$ on each fibre $\pi^{-1}(p)$ of $TM$, such that for local trivialization $(t,U)$ with $t:\pi^{-1}(U)\rightarrow U\times \mathbb{R}^n$, where $\mathbb{R}^n$ is given a fixed orientation. Then $t$ is either orientation preserving or reversing on all fibres.
The proof of (non vanishing n-form $\Rightarrow $ $M$ is orientable) is pretty straightforward, the proof of the converse is the one I don't really get. We first endow $M$ of cover $\mathcal{O}$ which consist of local coordinate systems $(x,U)$ and a partition of unity $\{\phi_U \}_{U\in \mathcal{O}}$. For each $(x,U)$ we choose an n-form $\omega_U$ on $U$ such that for $v_1,\ldots, v_n\in T_pM$ for $p\in M$ we have $$\omega_U(v_1,\ldots, v_n)>0\Leftrightarrow [v_1,\ldots, v_n]=\mu_p $$.
We next define $$\omega=\sum_{U\in \mathcal{O}}\phi_U \omega_U$$
Then this almost immediately completes the proof since for each $p$, the sum is finite and always non-negative.
Ok, so my question is. Why is $\omega$ smooth? Does the partition of unity is really necessary? Why I mean by the latter is that it seems like we only needed $\mathcal{O}$ to be locally finite, we do not need the functions, we could just replaced them by the constant map $1$.
My idea of the construction of $\omega_U$ is that it should be defined by a given $(x,U)$ from which we generated the cover $\mathcal{O}$ $$\omega_U=dx^1\wedge \ldots \wedge dx^n $$ Here we may assume that $[x^1,\ldots, x^n]=\mu_p$, otherwise just re-arrange them. This is clearly non-zero in $U$. But how is the global definition of $\omega$ smooth?
Best Answer
The reason that $ \omega $ is smooth is that, near any given point, it's defined by a finite sum of smooth terms. An infinite sum of smooth functions is not necessarily smooth; that's why $ \mathcal O $ must be locally finite. And we need a partition of unity built from smooth bump functions, not just something like characteristic functions of a partition of the set of points, so that the $ \phi _ U $ will be smooth. Of course, the $ \omega _ U $ are smooth since they are defined as a smooth function of the local coordinates and their differentials.
If I remember correctly, Spivak has a general discussion of partitions of unity some time before the discussion of orientation; check the index. This should help reinforce the importance of local finiteness and smooth bump functions. You'll want to understand this, because it's a trick that will be used over and over again to prove that something that exists locally can also exist globally.
By the way, you don't just need that each $ \omega _ U $ is nonzero on $ U $ but that $ \sum _ U \phi _ U \omega _ U $ is nonzero at every point. Here it's important that each $ \omega _ U $ has the same orientation. That's why, without the hypothesis that a global orientation exists, you still couldn't prove that a global nonzero $ n $-form exists.