I am looking at the following exercise:

Show that the following are equivalent conditions on a surface patch $\sigma (u, v)$ with first fundamental form $Edu^2 + 2F dudv + Gdv^2$ :

  1. $E_v = G_u = 0$.
  2. $σ_{uv}$ is parallel to the standard unit normal $N$.
  3. The opposite sides of any quadrilateral formed by parameter curves of $\sigma$ have the same length.


I have done the folowing:

$(1)\Rightarrow (2)$:

We have that $E_v=0 \Rightarrow \sigma_{uv}\cdot \sigma_u=0$ and $G_u=0\Rightarrow \sigma_{uv}\cdot \sigma_v=0$.

So, $\sigma_{uv}$ is perpendicular to $\sigma_u$ and $\sigma_v$. Since the unit normal, $N$, is perpendicular to $\sigma_u$ and $\sigma_v$, we have that $\sigma_{uv}$ is paralle to $N$.


$(2) \Rightarrow (1)$:

We have that $\sigma_{uv}$ is parallel ot the unit normal $N$. Since $\sigma_u$ and $\sigma_v$ are tangent to the surface, so perpendicular to $N$, we have that they are perpendicular also to $\sigma_{uv}$. That means that $\sigma_{uv}\cdot \sigma_u=0$ and $\sigma_{uv}\cdot \sigma_v=0$.

Since $E_v=2\sigma_{uv}\cdot \sigma_v$ we have that $E_v=0$ and since $G_u=2\sigma_{uv}\cdot \sigma_u$ we have that $G_u=0$.


Is everything correct so far?

How can we show that $(1)-(3)$ are equivalent? Could you give me some hints?



$\quad$ If $\gamma$ is a curve lying in the image of a surface patch $\boldsymbol{\sigma}$, we have $$\gamma(t)=\boldsymbol{\sigma}(u(t),v(t))$$ for some smoth functions $u(t)$ and $v(t)$. Then, denoting $d/dt$ by a dot, we have $\dot\gamma=\dot{u}\boldsymbol{\sigma}_u+\dot{v}\boldsymbol{\sigma}_v$ by the chain rule, so $$\langle\dot\gamma,\dot\gamma\rangle=E\dot u^2+2F\dot u\dot v+G\dot v^2,$$ and the length of $\gamma$ is given by $$\int(E\dot u^2+2F\dot u\dot v+G\dot v^2)^{1/2}dt.\tag{6.1}$$




$(3) \Rightarrow (1)$:

From the IVT we have $$\left (\sqrt{E(u,\tilde{v})}\right )_v=\frac{\sqrt{E(u,v^\ast )}-\sqrt{E(u,0)}}{v^\ast}, \ \ \tilde{v}\in (0,v^\ast)$$

So $$\frac{E_v(u,v)}{2\sqrt{E(u,v)}} \big |_{v=\tilde{v}}=\frac{\sqrt{E(u,v^\ast )}-\sqrt{E(u,0)}}{v^\ast}$$

Taking the derivative we have $$\int_0^{\epsilon}\frac{E_v(u,\tilde{v})}{2\sqrt{E(u,\tilde{v})}}du=\int_0^{\epsilon}\frac{\sqrt{E(u,v^\ast )}-\sqrt{E(u,0)}}{v^\ast}du=\frac{\int_0^{\epsilon}\sqrt{E(u,v^\ast )}du-\int_0^{\epsilon}\sqrt{E(u,0)}du}{v^\ast}=0$$

So $$\int_0^{\epsilon}\frac{E_v(u,\tilde{v})}{2\sqrt{E(u,\tilde{v})}} du=0 \Rightarrow E_v(u,\tilde{v})=0$$

Is this correct?

Or is the last implication wrong?



Could you give me also a hint how we could show the following?

enter image description here


How can we find such a reparametrization?

  • 2
    $\begingroup$ Have you tried to calculate lengths of parameter curves from (3) ? $\endgroup$
    – Evgeny
    Jan 4, 2016 at 12:19
  • $\begingroup$ How can we calculate these lengths? @Evgeny $\endgroup$
    – Mary Star
    Jan 4, 2016 at 12:45
  • 2
    $\begingroup$ We can calculate them using first fundamental form. If you parameterized part of surface with $(u, v)$ coordinates, then opposite sides of small quadrilateral are parameterized, for example, by $(u, 0)$ and $(u, v^\ast)$ for $u \in \lbrack 0, \varepsilon \rbrack$ and $v^\ast$ small. If I am not mistaken expressions for length would be like $\int_{0}^{\varepsilon} \sqrt{E(u, 0)}\, du$ and $\int_{0}^{\varepsilon} \sqrt{E(u, v^\ast)}\, du$ (but I'm not 100% sure). $\endgroup$
    – Evgeny
    Jan 4, 2016 at 12:57
  • $\begingroup$ Ok... Is this formula related to the one of the picture that I added in my initial post? @Evgeny $\endgroup$
    – Mary Star
    Jan 4, 2016 at 15:06
  • 1
    $\begingroup$ Yep, exactly. One final step left: to prove $(3) \Rightarrow (1)$. $\endgroup$
    – Evgeny
    Jan 4, 2016 at 15:17

2 Answers 2


Let's take the $u-$parametric curve. The length is $$\int_a^b \sqrt{E(u,v)} \mathrm du$$

You need to show that this does not depend on $v$. Since $E_v = 0$ $E$ doesn't depend on $v$... Hope that you can continue from here. For the $v-$parametric lines the length is $$\int_a^b \sqrt{G(u,v)} \mathrm dv$$ and you repeat the same argument as above. This proves $(1) \implies (3)$.

  • $\begingroup$ Ok... Is the formula of the length related to the one of the picture that I added in my initial post? Or how do we get this formula? $\endgroup$
    – Mary Star
    Jan 4, 2016 at 15:08
  • $\begingroup$ yes that is the same formula. For the $u-$parametric lines, $u$ is a parameter and $v$ is constant, so plug into the formula from the book $\dot u = 1$ and $\dot v = 0$ $\endgroup$
    – user26977
    Jan 4, 2016 at 15:11
  • $\begingroup$ At the formula we have $dt$, how do we get $du$ ? $\endgroup$
    – Mary Star
    Jan 4, 2016 at 16:09
  • $\begingroup$ This was a sort of an abuse of notation. Or you can remember that $\dot{u}\, dt $ is the same as $du$ if $u(t)$ is a nice transformation. $\endgroup$
    – Evgeny
    Jan 4, 2016 at 16:40
  • $\begingroup$ So the formula is $$\int \sqrt{Edu^2+2Fdudv+Gdv^2}$$ Which are the limits of the interval? Since $v=v_0$ is a constant we have that $dv=0$, so the formula is $$\int \sqrt{Edu^2}=\int \sqrt{E}du$$ right? @Evgeny $\endgroup$
    – Mary Star
    Jan 4, 2016 at 17:34

First fundamental form

$$ ds^2 = E(u) du^2 + 2 F (u,v) du dv + G(v) dv^2 \tag{1} $$

and as you simplified

$$ \Delta s = \int_a ^b \sqrt{E(u)} du = \int_c ^d \sqrt{G(v)} dv \tag {2} $$

It is a curvilinear "parallelogram " becoming a "rhombus", if $ a=c, b=d.$ The sum of four internal angles here is less than $2 \pi$

Taking cross product between $ \sigma_u, \sigma_v $ having variable direction $ \theta$ between them, since we are allowed to use cos rule and other trig relations to differential lengths also like the following..


$$ ds_1^2 = du^2 + 2 \cos \theta\, du dv + dv^2 \tag{3} $$

Using the Christoffel coefficients ( several ones vanish) calculate Gauss curvature as

$$ K (u,v) = - \frac{\partial ^2 \theta}{\partial u \partial v } /\sin \theta \tag {4} $$

which is a negative constant. This results in the Chebychev fishnet. In another form it is also stated as the Sine-Gordon Equation

$$ \frac{\partial ^2 \theta}{\partial u \partial v } - K (u,v) \sin \theta =0 \tag{5} $$

The Gauss curvature can be either constant when K is negative or positive constant, or even variable as in case of a dance artist fishnet leg stockings.

  • $\begingroup$ I haven't really understood how you got the relation $(3)$, $$ds_1^2 = du^2 + 2 \cos \theta\, du dv + dv^2$$ Could you explain it to me? $\endgroup$
    – Mary Star
    Jan 14, 2016 at 21:19
  • $\begingroup$ There is a scaling. Apply Cosine rule for a differential triangle SAS . Sides are $du,dv$ and and angle in between them in a parallelogram $ \theta \, $ $\endgroup$
    – Narasimham
    Jan 14, 2016 at 21:40
  • $\begingroup$ What do you mean by "differential triangle SAS" ? $\endgroup$
    – Mary Star
    Jan 14, 2016 at 22:00
  • $\begingroup$ As in the parallogram diagonal construction, etc.. $\endgroup$
    – Narasimham
    Jan 14, 2016 at 22:52
  • $\begingroup$ Do you mean to apply the law of cosine? Do we not get $$ds^2=du^2+dv^2-2dudv\cos\theta$$ How do we get the plus sign instead of the minus? $\endgroup$
    – Mary Star
    Jan 17, 2016 at 23:31

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