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Below is a problem which states a fact about "Tchebyshef net". I don't understand meaning of bolded part.

The coordinate curves of a parametrization $x(u, v)$ constitute a Tchebyshef net if the lengths of the opposite sides of any quadrilateral formed by them are equal. Show that a necessary and sufficient condition for this is $$\frac{\partial E}{\partial v} =\frac{\partial G}{\partial u}=0.$$ Reference: Differential Geometry of Curves and Surfaces [Manfredo P.do carmo] Page 100 Problem 7.

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2 Answers 2

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The definition is equivalent to

  1. two curves $x(u_1,v)$, $x(u_2,v)$ determine segments of equal lengths on all curves $x(u,\mathrm{const})$.

  2. two curves $x(u,v_1)$, $x(u,v_2)$ determine segments of equal lengths on all curves $x(\mathrm{const},v)$.

For statement 1, it means that, for any constant $v$, the curve $\alpha:(u_1,u_2)\to \mathbb{R}^3$ given by $\alpha(u)=x(u,v)$ has the same length on the surface. That is$$\int_{u_1}^{u_2} \sqrt{E(u,v)} \, du =\mathrm{const} \quad \forall \, \mathrm{const} \, v $$ Differentiate above equation with respect to $v$, we get $$\int_{u_1}^{u_2} \partial_v\sqrt{E(u,v)} \, du =0 \quad \forall \, \mathrm{const} \, v $$

Hence we get $\frac{\partial E}{\partial v}=0$. Similarly, we can get the second condition.

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  • $\begingroup$ Sorry for the late comment, but this means being a Tchebyshef net is a property of the parametrization, not of the actual coordinate curves, right? $\endgroup$ Feb 23, 2020 at 23:38
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The curves $x(u_{1},v)$ and $x(u_{2},v)$ determine segments of equal length along the curves $x(u,v_{0})$.

The curves $x(u,v_{0})$ and $x(u,v_{1})$ determine segments of equal length along the curves $x(u_{0},v)$.

$\Longrightarrow)$ The curve $\alpha:(u_{1},u_{2})\rightarrow S$, where $\alpha(u)=x(u,v)$, satisfies

\begin{eqnarray*} \text{constant} & = & \int_{u_{1}}^{u_{2}}\left|\alpha^{\prime}(u)\right|du\\ & = & \int_{u_{1}}^{u_{2}}\left|x_{u}(u,v)\right|du\\ & = & \int_{u_{1}}^{u_{2}}\sqrt{E(u,v)}du. \end{eqnarray*}

Since taking the derivative of this function with respect to $v$ results in $0$, we have that $f(u_{2}):=\int_{u_{1}}^{u_{2}}\sqrt{E(u,v)}du$ does not depend on $v$. Therefore, its derivative with respect to $u_{2}$ does not depend on $v$, which means $\sqrt{E(u,v)}$ does not depend on $v$, and consequently, $E_{v}=0.$

$\Longleftarrow)$ \begin{eqnarray*} 0 & = & E_{v}\\ & = & \frac{E_{v}}{2\sqrt{E}}\\ & = & \partial_{v}\sqrt{E}\\ & = & \int_{u_{1}}^{u_{2}}\partial_{v}\sqrt{E}du\\ & = & \partial_{v}\int_{u_{1}}^{u_{2}}\sqrt{E}du\\ & = & \partial_{v}\int_{u_{1}}^{u_{2}}\left|x_{u}(u,v)\right|du\\ & = & \partial_{v}\int_{u_{1}}^{u_{2}}\left|\alpha^{\prime}(u)\right|du.\\ \text{constant} & = & \int_{u_{1}}^{u_{2}}\left|\alpha^{\prime}(u)\right|du. \end{eqnarray*}

Since $E$ is differentiable, we were able to interchange $\partial_{v}$ with $\int_{u_{1}}^{u_{2}}$.

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