# Confusing about coordinate curves and quadrilateral formed?

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.

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.

• 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? Feb 23, 2020 at 23:38

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 $$\int_{u_{1}}^{u_{2}}$$.