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The sphere has a parameterization map for a surface patch $\phi(u,v)=(u,v,\sqrt{1-u^2-v^2})$.

It has another parametrization map for a surface patch $\beta (x,y)=(\sin x \cos y,\sin x \sin y,\cos x)$.

For the first the first fundamental form comes here. $E= \frac{1-v^2}{1-v^2-u^2}$ , $F=\frac{uv}{1-v^2-u^2}$ and $G=\frac{1-u^2}{1-v^2-u^2}$.

If we calculate the first fundamental form then second case we have. $ E= 1 , F=0 , G= \sin^2 x$.

My question is how can we relate them in the common domain.

abac
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    Use the chain rule. (It's unfortunate that you've chosen the same letters for both parametrisations.) – Zhen Lin Dec 12 '14 at 17:22
  • I couldn't get your point. How should I use chain rule and why? Can you explain please. Thanks for your time. – abac Dec 12 '14 at 17:31
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    The point is that, for coordinates $(u, v)$ and $(u', v')$ and respective f.f.f. $(E, F, G)$ and $(E', F', G')$, we get $E , \mathrm{d}u^2 + 2 F , \mathrm{d}u , \mathrm{d}v + G , \mathrm{d}v^2 = E' , \mathrm{d}u'^2 + 2 F' , \mathrm{d}u' , \mathrm{d}v' + G' , \mathrm{d}v'^2$. – Zhen Lin Dec 12 '14 at 17:33
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    @abac, because once you understand how bases and components of vectors change under change of coordinates, then how do rank two tensors change is easy to establish. – janmarqz Dec 12 '14 at 20:12

1 Answers1

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For each point in their common domain, you can find a neighbourhood so that $\phi_u,\phi_v$ and $\beta_u,\beta_v$ form a basis for $T_pS$. In what you wrote above, let $E,F,G$ be the coefficients of the first fundamental form with respect to $\phi$ and $\widetilde E,\widetilde F,\widetilde G$ be the coefficients with respect to $\beta$. In these respective bases, the bilinear form $\langle \ ,\ \rangle$ corresponding to the first fundamental form (quadratic form) has matrix representation given by these coefficients.

These matrices are both matrix representations of the same bilinear form (i.e. the metric), but in different bases. So to relate these two matrices, you just need to find the change of basis matrix between $\phi_u,\phi_v$ and $\beta_u,\beta_v$. As suggested in the comments, this is where the chain rule will come in.

JonHerman
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