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The area form of a Riemannian surface

  1. Oct 17, 2012 #1
    Let $$(M,g)$$ be an oriented Remannian 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 Remannian 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.
     
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  3. Oct 17, 2012 #2

    quasar987

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    The intelligible way to introduce the canonical volume form dM on a riemannian manifold is by declaring it to be the top form that takes the value 1 on any positively oriented g-orthonormal basis. This is obviously well-defined simply because the transformation matrix btw any 2 such basis is an element of SO(n) which has determinant 1. Then, we may ask what is the local expression of this form in terms of local coordinates u.

    For this, pick an arbitrary positively oriented g-orthonormal basis v. Then
    [tex]dM=\det(A)du^1\wedge ... \wedge du^n[/tex]
    where A= "transformation matrix from basis ∂/∂u to basis v" = (g(∂/∂ui,vj)).

    On the other hand, go ahead and compute gij=g(∂/∂ui,∂/∂uj) to be A².

    Thus conclude that wrt local coordinates, dM has the form from your definition.
     
  4. Oct 17, 2012 #3
    I got $$\mathrm{d}u^1 \wedge \mathrm{d}u^2=\mathrm{det}(\Phi) \mathrm{d}v^1\wedge\mathrm{d}v^2$$. I also know $$g_{ij}=\langle \frac{\partial}{\partial u^i} \frac{\partial}{\partial u^j} \rangle$$, moreover, 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 $$f$$?
     
  5. Oct 18, 2012 #4
    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 new Remannian metric 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 $$\mathrm{d}M=\sqrt|g'| \mathrm{d}v^1 \wedge \mathrm{d}v^2=|A|\sqrt|g|\frac{1}{|A|}\mathrm{d}u^1\wedge\mathrm{d}u^2=\sqrt|g|\mathrm{d}u^1\wedge\mathrm{d}u^2$$
     
    Last edited: Oct 18, 2012
  6. Oct 18, 2012 #5

    quasar987

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    Well done!
     
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