Let $(M,g)$ be an oriented Riemannian surface. Then globally $(M,g)$ has a canonical area-$2$ form $\mathrm{d}M$ defined by $$\mathrm{d}M=\sqrt{|g|} \mathrm{d}u^1 \wedge \mathrm{d}u^2$$ with respect to a positively oriented chart $(u_{\alpha}, M_{\alpha})$ where $|g|=\mathrm{det}(g_{ij})$ is the determinant of the Riemannian metric in the coordinate frame for $u_{\alpha}$.
Let $u^{i}=\Phi^{i}(v^1,v^2)$ be a change of variables (so $\Phi: V \to U$ is the diffeomorphism of the coordinate change). Calculate the effect on $\sqrt{|g|}$ and $\mathrm{d}u^1 \wedge \mathrm{d}u^2$ to prove $\mathrm{d}M$ is independent of the choice of positively oriented coordinates.
Remark: I know $g$ is invariant under an orientation preserving change of variables ($\mathrm{det}(\Phi)>0$), but how to compute explicitly the effect of $\Phi$ on $\mathrm{d}u^1 \wedge \mathrm{d}u^2$? I want to use this as an example to learn the exterior calculus.
Best Answer
Followed by Willie's hint. Write $u^1=u^1(v^1,v^2)$ and $u^2=u^2(v^1,v^2)$ to get $$\mathrm{d}u^1 \wedge \mathrm{d}u^2=\mathrm{det}(A)\mathrm{d}v^1 \wedge \mathrm{d}v^2$$ where $A:=\begin{vmatrix}\frac{\partial u^1}{\partial v^1}&\frac{\partial u^1}{\partial v^2}\\ \frac{\partial u^2}{\partial v^1}&\frac{\partial u^2}{\partial v^2}\end{vmatrix}$ is the Jaobian for $\Phi$.
Denote the metric $g$ under coordinates $v^1,v^2$ by $g'$ and apply chain rule to get $$g'_{ij}=g_{kl}\frac{\partial u^k}{\partial v^i}\frac{\partial u^l}{\partial v^j}.$$ Which is equivalent to $(g'_{ij})=(A)(g_{ij})(A)^t$, take determinant on both sides yield $|g'|=|A|^2|g|$.
Now $$\mathrm{d}M=\sqrt|g'| \mathrm{d}v^1 \wedge \mathrm{d}v^2=|A|\sqrt|g|\frac{1}{|A|}\mathrm{d}u^1\wedge\mathrm{d}u^2=\sqrt|g|\mathrm{d}u^1\wedge\mathrm{d}u^2$$