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Let $\sigma: U\subset\Bbb{R^2}\to V\subset S$ a parameterization of a surface $S$, and let $g:\tilde{U}\to U$ be a diffeomorphism between open sets of $\Bbb{R^2}$. I need to obtain a formula for the coefficients of the first fundamental form associated to $\sigma\circ g$ in terms of the coefficients of the first fundamental form of $\sigma$.

Since $g$ is a diffeomorphism, then $g^{-1}:U\to\tilde{U}$ exists and is differentiable.

If $g^{-1}(u,v)=(\tilde{u},\tilde{v})$, so $g(\tilde{u},\tilde{v})=(u,v)$, then $\sigma(u,v)=[\sigma\circ(g\circ g^{-1})](u,v)=(\sigma\circ g)(\tilde{u},\tilde{v})$ but I'm stuck at this point.

Can I say after that $\frac{\partial\sigma}{\partial u}=\frac{\partial(\sigma\circ g)}{\partial\tilde{u}}$?

I think this is wrong and I need to consider $g^{-1}$ but I'm not sure.

Armando Rosas
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1 Answers1

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It is worthwhile to consider this picture

enter image description here

where we can plainly see the maps involved.

We have $\partial_1=J\sigma e_1$ and $\partial_2=J\sigma e_2$ as the generators of the tangent space and $\sigma\circ g$, which is another parametrization, serves to define another tangent base $$\tilde{\partial_1}=J(\sigma\circ g)e_1\quad,\quad \tilde{\partial_1}=J(\sigma\circ g)e_2$$ But by the Chain Rule $J(\sigma\circ g)=J\sigma\cdot Jg$ then $$J(\sigma\circ g)e_i=(J\sigma\cdot Jg )e_i$$ which evaluated at some point $a$ in the domain of $g$ would give $$J(\sigma\circ g)|_ae_i=(J\sigma|_{g(a)}\cdot Jg|_a)e_i$$

If we choose the coordinate functions $\tilde{v},\tilde{w}$ in the domain of $g$ and $v,w$ in the domain of $\sigma$, then $$Jg= \left[\begin{array}{cc} \dfrac{\partial v}{\partial\tilde v}&\dfrac{\partial v}{\partial\tilde w}\\ \\ \dfrac{\partial w}{\partial\tilde v}&\dfrac{\partial w}{\partial\tilde w} \end{array}\right]$$ So $$Jge_1= \dfrac{\partial v}{\partial\tilde v}e_1+ \dfrac{\partial w}{\partial\tilde v}e_2 \quad ,\quad Jge_2= \dfrac{\partial v}{\partial\tilde w}e_1+ \dfrac{\partial w}{\partial\tilde w}e_2$$ which plugged above \begin{eqnarray*} \tilde{\partial_1}&=& J\sigma\left(\dfrac{\partial v}{\partial\tilde v}e_1+ \dfrac{\partial w}{\partial\tilde v}e_2\right)\\ &=&\dfrac{\partial v}{\partial\tilde v}J\sigma e_1+ \dfrac{\partial w}{\partial\tilde v}J\sigma e_2\\ \\ &=& \dfrac{\partial v}{\partial\tilde v}\partial_1+ \dfrac{\partial w}{\partial\tilde v}\partial_2 \end{eqnarray*} and similarly $$\tilde{\partial_2}= \dfrac{\partial v}{\partial\tilde w}\partial_1+ \dfrac{\partial w}{\partial\tilde w}\partial_2$$

So the first fundamental form will be $$\tilde{g}_{11}= \left(\dfrac{\partial v}{\partial\tilde v}\partial_1+ \dfrac{\partial w}{\partial\tilde v}\partial_2\right) \bullet \left(\dfrac{\partial v}{\partial\tilde v}\partial_1+ \dfrac{\partial w}{\partial\tilde v}\partial_2\right)$$ i.e. $$\tilde{g}_{11}= \left(\dfrac{\partial v}{\partial\tilde v}\right)^2g_{11}+ 2\dfrac{\partial v}{\partial\tilde v} \dfrac{\partial w}{\partial\tilde v}g_{12}+ \left(\dfrac{\partial w}{\partial\tilde v}\right)^2g_{22}, $$ similarly with $$\tilde{g}_{12}= \dfrac{\partial v}{\partial\tilde v}\dfrac{\partial v}{\partial\tilde w} g_{11}+ 2\dfrac{\partial v}{\partial\tilde v}\dfrac{\partial w}{\partial\tilde w}g_{12}+ \dfrac{\partial w}{\partial\tilde v}\dfrac{\partial w}{\partial\tilde w} g_{22}. $$ $$\tilde{g}_{22}= \left(\dfrac{\partial v}{\partial\tilde w}\right)^2g_{11}+ 2\dfrac{\partial v}{\partial\tilde w}\dfrac{\partial w}{\partial\tilde w}g_{12}+ \left(\dfrac{\partial w}{\partial\tilde w}\right)^2g_{22}. $$

janmarqz
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