# Partition of unity and volume form on a manifold

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 $M$)

## 2 Answers

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 $\rho_1(p) = \rho_2(p) = \frac{1}{2}$.

• 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? – AMFS Nov 22 '16 at 23:19
• 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 $\phi_i(U_i) \subset \mathbb{R}^n$ – AMFS Nov 22 '16 at 23:20
• @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. – Nick Alger Nov 24 '16 at 5:57
• @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... – AMFS Nov 24 '16 at 21:27
• ...all the Jacobian determinants would not only be positive but would in fact just be 1). By how differential forms transform under a change of coordinates, this means that we need to find each $\psi$ such that it satisfies: for each $y$ in the new coordinates, det$(D\psi^{-1}|_y)h(\psi^{-1}(y))=1$. I suspect this can always be done... – AMFS Nov 24 '16 at 21:28

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.