I am familiarizing myself with Riemannian manifolds.
Let $M$ be an orientable smooth $n$-manifold with atlas $(U_\alpha, (x_1^\alpha, \dots, x_n^\alpha))$ and let $g$ be a Riemannian metric on $M$. Let $|g_\alpha|$ denote the determinant of the matrix given by $$ (g_\alpha)_{ij} = g\left ( {\partial \over \partial x_i^\alpha} , {\partial \over \partial x_j^\alpha}\right )$$
I want to show that $\sqrt{|g_\alpha|} d x_1^\alpha \wedge \dots \wedge d x_n^\alpha = \sqrt{|g_\beta|} d x_1^\beta \wedge \dots \wedge d x_n^\beta$ on $U_\alpha \cap U_\beta$ but I can't quite do it.
Could someone help me how to show $\sqrt{|g_\alpha|} d x_1^\alpha \wedge \dots \wedge d x_n^\alpha = \sqrt{|g_\beta|} d x_1^\beta \wedge \dots \wedge d x_n^\beta$?
learner - here's one approach.
If $\phi: U \rightarrow V$ is a change of coordinates, say $U$ has coordinates $x_i$ and $V$ has coordinates $y_i$, and the metric is written in $V$ as $(g_{ij})$, and in $U$ as $(h_{ij})$, then the metric transforms as $$(*) \quad \quad (h_{ij}) = d\phi^T (g_{ij}) d\phi$$.
So, the top form on $U$ defined on terms of $(h_{ij})$ is $\lvert \det(h_{ij}) \rvert^{1/2} dx_i$.
On the other hand, the top form on $V$ is similarly $\lvert \det(g_{ij}) \rvert^{1/2} dy_i$, and this pulls back to $\lvert \det(g_{ij}) \rvert^{1/2} \det d\phi dy_i$. Using (*) and asking that your change of coordinates is oriented gives the desired result.