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