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Show that, if $\sigma (u, v)$ is a surface patch, the set of linear combinations of $\sigma_u$ and $\sigma_v$ is unchanged when $\sigma$ is reparametrized.

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I have done the following:

Let $\tilde{\sigma}(\tilde{u}, \tilde{v})$ be a reparametrization of $\sigma$, i.e., $\sigma (u, v)=\tilde{\sigma}(\tilde{u}(u,v), \tilde{v}(u,v))$.

Then we have $$\sigma_u=\frac{\partial{\overline{u}}}{\partial{u}}\tilde{\sigma}_{\tilde{u}}+\frac{\partial{\overline{v}}}{\partial{u}}\tilde{\sigma}_{\tilde{v}} \\ \sigma_v=\frac{\partial{\overline{u}}}{\partial{v}}\tilde{\sigma}_{\tilde{u}}+\frac{\partial{\overline{v}}}{\partial{v}}\tilde{\sigma}_{\tilde{v}}$$

Is this correct?

How could we continue?

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2 Answers 2

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You note that the Jacobian matrix of a change of variables is an isomorphism( a change of variables is a diffeomorphism). In particular the tangent plane spanned by $\sigma_u , \sigma_v$ is equal to the tangent plane spanned by $\sigma_\tilde{u} , \sigma_\tilde{v}$ .

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  • $\begingroup$ Do you mean to show that $\sigma_u \times \sigma_v =\overline{\sigma}_{\overline{u}}\times \overline{\sigma}_{\overline{v}}$ ? $\endgroup$
    – Mary Star
    Dec 18, 2015 at 18:00
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You need to show that $\left \{ \vec r_u(u,v), \vec r_v(u,v)\right \}$ and $\left \{ \vec r_\tilde u(\tilde u,\tilde v), \vec r_\tilde v(\tilde u,\tilde v)\right \}$ span the same vector space. The easiest way to do this is to compute the Jacobian $\left | \frac{\partial (u,v)}{\partial (\tilde u,\tilde v)} \right |$. The definition of reparameterization requires that this determinant $\neq 0$ which in turn says that the two pairs of basis vectors span the same space.

If you are being asked to do the calculation from scratch, then proceed as follows:

$\vec r(\tilde u,\tilde v)=x(u(\tilde u,\tilde v),v(\tilde u,\tilde v))\vec i+y(u(\tilde u,\tilde v),v(\tilde u,\tilde v))\vec j$

$\vec r_\tilde u=\left ( \frac{\partial x}{\partial u}\frac{\partial u}{\partial \tilde u}+\frac{\partial x}{\partial v}\frac{\partial v}{\partial \tilde u} \right )\vec i+\left ( \frac{\partial y}{\partial u}\frac{\partial u}{\partial \tilde u}+\frac{\partial y}{\partial v}\frac{\partial v}{\partial \tilde u} \right )\vec j$

$\vec r_\tilde v=\left ( \frac{\partial x}{\partial u}\frac{\partial u}{\partial \tilde v}+\frac{\partial x}{\partial v}\frac{\partial v}{\partial \tilde v} \right )\vec i+\left ( \frac{\partial y}{\partial u}\frac{\partial u}{\partial \tilde v}+\frac{\partial y}{\partial v}\frac{\partial v}{\partial \tilde v} \right )\vec j$

which is

$\vec r_\tilde u=\frac{\partial u}{\partial \tilde u}\vec r_u+\frac{\partial v}{\partial \tilde u}\vec r_v$

and

$\vec r_\tilde v=\frac{\partial u}{\partial \tilde v}\vec r_u+\frac{\partial v}{\partial \tilde v}\vec r_v$

and it is precisely because the Jacobian is non-zero that we can conclude that

$\left \{ \vec r_\tilde u, \vec r_\tilde v \right \}$ are independent and therefore span the same space as $\left \{ \vec r_u, \vec r_v \right \}$.

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  • $\begingroup$ So do we have to show that the following is $\neq 0$ ? $$\begin{vmatrix} \frac{\partial{\overline{u}}}{\partial{u}}\tilde{\sigma}_{\tilde{u}}+\frac{\partial{\overline{v}}}{\partial{u}}\tilde{\sigma}_{\tilde{v}} & \frac{\partial{\overline{u}}}{\partial{v}}\tilde{\sigma}_{\tilde{u}}+\frac{\partial{\overline{v}}}{\partial{v}}\tilde{\sigma}_{\tilde{v}} \\ \tilde{\sigma}_{\tilde{u}} & \tilde{\sigma}_{\tilde{v}} \end{vmatrix}$$ $\endgroup$
    – Mary Star
    Dec 18, 2015 at 18:14
  • $\begingroup$ $$=\left (\frac{\partial{\overline{u}}}{\partial{u}}\tilde{\sigma}_{\tilde{u}}+\frac{\partial {\overline{v}}}{\partial{u}}\tilde{\sigma}_{\tilde{v}} \right )\tilde{\sigma}_{\tilde{v}}-\left (\frac{\partial{\overline{u}}}{\partial{v}}\tilde{\sigma}_{\tilde{u}}+\frac{\partial{\overline{v}}}{\partial{v}}\tilde{\sigma}_{\tilde{v}} \right )\tilde{\sigma}_{\tilde{u}} $$ $\endgroup$
    – Mary Star
    Dec 18, 2015 at 18:14
  • $\begingroup$ please see my edit for details. $\endgroup$
    – Matematleta
    Dec 18, 2015 at 18:35
  • $\begingroup$ Why is the Jacobian non-zero? $\endgroup$
    – Mary Star
    Dec 20, 2015 at 1:08
  • $\begingroup$ The jacobian is assumed to be nonzero. This is a part of the def of parameterization $\endgroup$
    – Matematleta
    Dec 20, 2015 at 3:19

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