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A smooth manifold $M$ is orientable iff there exists a nowhere-vanishing top form (i.e. volume form). In a coordinate chart $U\subset M$ we can find a volume form over $U$ that corresponds to the standard volume form $\operatorname{vol} = dx^1\wedge\dots\wedge dx^n$ in $\mathbb{R}^n$.

My question is, why can't we just use a partition of unity on the manifold to glue these local volume forms together? I.e. given a partition of unity $\{\rho_i\}$ subordinate to atlas $\{(U_i, \phi_i)\}$, why can't we say that $\sum_i\rho_i\phi_i^*(\operatorname{vol})$ defines a global volume form on $M$? (in the same way that we prove the existence of a Riemannian metric on any manifold: locally define $g_i$ to be, for example, the Euclidean metric, then $g=\sum_i\rho_ig_i$ is a metric on $M$)

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Suppose $p \in U_1\cap U_2$ and consider $\rho_1(p)\phi_1^*(\operatorname{vol})_p + \rho_2(p)\phi_2^*(\operatorname{vol})_p \in \bigwedge^nT_p^*M$. While $\phi_1^*(\operatorname{vol})_p$ and $\phi_2^*(\operatorname{vol})_p$ are non-zero elements of $\bigwedge^nT_p^*M$, they could be negative multiples of each other, in which case $\rho_1(p)\phi_1^*(\operatorname{vol})_p + \rho_2(p)\phi_2^*(\operatorname{vol})_p$ could be zero.

For example, consider the manifold $S^1$. Let $U_1 = S^1\setminus\{1\}$, $\phi_1 : U_1 \to (0, 2\pi)$, $e^{i\theta}\mapsto \theta \bmod{2\pi}$, and $U_2 = S^1\setminus\{-1\}$, $\phi_2 : U_2 \to (-\pi, \pi)$, $e^{i\theta} \mapsto \pi - \theta \bmod{2\pi}$. Then $\phi_1^*(dx) = d\theta$ and $\phi_2^*(dx) = d(\pi - \theta) = -d\theta$ so $\rho_1(p)\phi_1^*(dx)_p + \rho_2(p)\phi_2^*(dx)_p = (\rho_1(p) - \rho_2(p))d\theta$ which is zero whenever $\rho_1(p) = \rho_2(p) = \frac{1}{2}$.

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  • $\begingroup$ So if $M$ is non-orientable, then the standard volume forms in local charts do not string together to give a global $n$-form. Would it be correct to say the converse: $M$ is orientable iff there exist charts $\{(U_i,\phi_i)\}$ covering $M$, compatible with the defining atlas for $M$, such that the "coframe" (is this the right term?) "$dx_1 \wedge...\wedge dx_n$" does in fact define a smooth $n$-form on $M$. This would be equivalent to saying that $\text{det}(D\phi_{ij})=1$ for all transition functions $\phi_{ij}$. Are these statements correct? $\endgroup$
    – Tobyhas
    Commented Nov 22, 2016 at 23:19
  • $\begingroup$ By "$dx_1 \wedge...\wedge dx_n$" I mean a function $M \to \bigwedge^nT^*M$ which at each point $p\in M$ gives the basis for $\bigwedge^nT_p^*M$ corresponding to the coordinates $\{x_i\}$ of $\phi_i(U_i) \subset \mathbb{R}^n$ $\endgroup$
    – Tobyhas
    Commented Nov 22, 2016 at 23:20
  • $\begingroup$ @Alex I'm not sure I understand what you are getting at, but it is true that all transitions functions have a positive Jacobian determinant if and only if the manifold is orientable. $\endgroup$
    – Nick Alger
    Commented Nov 24, 2016 at 5:57
  • $\begingroup$ @NickAlger Basically I was wondering if your statement is equivalent to the seemingly stronger statement that there's a representative in the equivalence class of atlases defining the manifold, which contains charts in which any volume form have local expression $\omega_0=dx^1\wedge...\wedge dx^n$. It is true that a volume form admits a local expression $h\omega_0$ in an oriented chart, where $h$ is a positive function, so I wondered if we can impose a smooth change of coordinates $\psi$ on each oriented chart such that it just becomes $\omega_0$. (In these new charts then... $\endgroup$
    – Tobyhas
    Commented Nov 24, 2016 at 21:27
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    $\begingroup$ I think the proposed procedure works if the atlas belongs to the orientation of the manifold. In the example provided by Michael Albanese, my quick observation shows that the transition map has derivative -1. This answers why a combination of volume forms could vanish. $\endgroup$
    – QA Ngô
    Commented May 27, 2019 at 17:16
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A convex combination of volume forms could turn out to be zero, violating the nowhere-vanishing condition. For example, $$\frac{1}{2}dx \wedge dy + \frac{1}{2}(-dx) \wedge dy = 0$$

This is actually exactly what will happen if you try to do the proposed procedure on a mobius strip with standard parameterization.

In contrast, at each point a metric corresponds to a positive definite matrix (or operator more generally), and convex combinations of positive definite matrices are themselves positive definite.

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