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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.

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"$g$ is invariant" really? Think about that there. Recall that $|g|$ is the determinant of the matrix $(g_{ij})$ as expressed in the coordinate system. How does the matrix change? How then does the determinant change? – Willie Wong Oct 17 '12 at 14:43
In other words, if $u^1 = u^1(v^1, v^2)$, what is $\mathrm{d}u^1$ as a one-form expressed in $v^1, v^2$ coordinates? – Willie Wong Oct 17 '12 at 14:44
I got $\mathrm{d}u^1 \wedge \mathrm{d}u^2=\mathrm{det}(\Phi) \mathrm{d}v^1\wedge\mathrm{d}v^2$. Iknow $g_{ij}=\langle \frac{\partial}{\partial u^i} \frac{\partial}{\partial u^j} \rangle$, also for a surface $f: U \to \mathbb{R}^3$ the components of its first fundamental form $g$ after a change of variables $u=\Phi(v)$ becomes $g'_{ij}(v)=\frac{\partial u^k}{\partial v^i}\frac{\partial u^l}{\partial v^j}g_{kl}(\Phi(v))$. But I don't know how to apply chain rule to the expression $g_{ij}=\langle \frac{\partial}{\partial u^i} \frac{\partial}{\partial u^j} \rangle$ which is not related to any $f$? – user31899 Oct 17 '12 at 22:01

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 $$\mathrm{d}M=\sqrt|g'| \mathrm{d}v^1 \wedge \mathrm{d}v^2=|A|\sqrt|g|\frac{1}{|A|}\mathrm{d}u^1\wedge\mathrm{d}u^2=\sqrt|g|\mathrm{d}u^1\wedge\mathrm{d}u^2$$

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There we go. One small quibble, instead of calling $g'$ the "new Riemannian metric", it should be the presentation of the metric $g$ in $v$ coordinates. (Compare with linear algebra where the same linear transformation $A$ can be expressed as two different matrices $[A]_{\mathcal{B}}$ and $[A]_{\mathcal{C}}$ relative to two different bases $\mathcal{B}$ and $\mathcal{C}$ of the underlying vector space). – Willie Wong Oct 18 '12 at 7:57
Thanks! Yes, it should be called the metric under coordinates $v^1,v^2$ – user31899 Oct 18 '12 at 8:26

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