I am looking at the following exercise:

Show that a curve $\gamma (t) = \sigma (u(t), v(t))$ on a surface patch $\sigma$ is a line of curvature if and only if
$$(EM − FL) \dot u^2 + (EN − GL) \dot u \dot v + (FN − GM ) \dot v^2=0$$


We have that $\gamma$ is a line of curvature if the tangent vector of $\gamma$ is a principal vector of $S$ at all points of $\gamma$, so if $W(\dot\gamma)=-\kappa \dot\gamma$, which is equivalent to $\dot {\textbf{N}}=-\kappa\dot\gamma$.

It holds that $\dot\gamma=\sigma_u\dot u+\sigma_v\dot v$.

We have that $E=\|\sigma_u\|^2, \ F=\sigma_u \cdot \sigma_v , \ G=\|\sigma_v\|^2, \ L=\sigma_{uu}\cdot \textbf{N}, \ M=\sigma_{uv}\cdot \textbf{N}, \ N=\sigma_{vv}\cdot \textbf{N}$.

But how do we get the terms $E, \ F, \ G, \ L, \ M, \ N$ ?




Now an other question of the exericse is the following:

enter image description here

I have done the following:

Suppose that (i) holds, i.e., the second fundamental form of $\sigma$ is proportional to its first fundamental form. Then $$Edu^2+2Fdudv+Gdv^2=\lambda (Ldu^2+2Mdudv+Ndv^2), \ \text{ for some smooth function } \lambda (u,v)$$ so $$E=\lambda L, \ F=\lambda M, \ G=\lambda N$$

Therefore, $$(EM − FL) \dot u^2 + (EN − GL) \dot u \dot v + (FN − GM ) \dot v^2=(\lambda LM − \lambda ML) \dot u^2 + (\lambda LN − \lambda NL) \dot u \dot v + (\lambda MN − \lambda NM ) \dot v^2=0$$ So, we conclude that all parameter curves are lines of curvature.

Is this correct?


Suppose that (ii) holds, i.e., $F=M=0$.

Then, we have that the matrices $\mathcal{F}_I=\begin{pmatrix} E & F \\ F & G \end{pmatrix}=\begin{pmatrix} E & 0 \\ 0 & G \end{pmatrix}$ and $\mathcal{F}_{II}=\begin{pmatrix} L & M \\ M & N \end{pmatrix}=\begin{pmatrix} L & 0 \\ 0 & M \end{pmatrix}$ are diagonal.

Therefore the matrix of the Weingarten map, $\mathcal{F}_I^{-1}\mathcal{F}_{II}$ is diagonal. The principal curvatures are the eigenvalues of this matrix, which are the elements of the diagonal of the Weingarten map. Then the eigenvectors are $(1,0)$ and $(0,1)$. Does this imply that the tangent vector of the curve is a principal vector of the surface?

P.S. Suppose we have $W(t_1) = \kappa_1t_1, \ W(t_2) = \kappa_2t_2$, then $\kappa_1$ and $\kappa_2$ are called the principal curvatures of $S$, and $t_1$ and $t_2$ are called principal vectors corresponding to $\kappa_1$ and $\kappa_2$.

  • 1
    $\begingroup$ use $\sigma_u=\frac{\partial \sigma}{\partial u}$ and $\sigma_v=\frac{\partial \sigma}{\partial v}$ $\endgroup$
    – janmarqz
    Dec 29, 2015 at 17:50
  • $\begingroup$ At which point do we use this? @janmarqz $\endgroup$
    – Mary Star
    Dec 29, 2015 at 17:53
  • $\begingroup$ I edited my initial post... Could you take a look at it? Do you have an idea? @janmarqz $\endgroup$
    – Mary Star
    Jan 17, 2016 at 10:24
  • 1
    $\begingroup$ thanx 4 sharing @Mary Star, i'm going 2 think about it $\endgroup$
    – janmarqz
    Jan 17, 2016 at 14:58

1 Answer 1


I copy and pasted this answer from my own notes (cause I am lazy), the notation is a little different but I hope it is understandable.

$du:dv$ is a principal direction if and only if $$(EM-FL)du^2 +(EN-GL)dudv +(FN-GM)dv^2=0.$$

proof: Let

$\mathbf{e}_1= \mathbf{r}_u du +\mathbf{r}_v dv$ then the vector

$$\mathbf{e}_2= (F\mathbf{r}_u-E \mathbf{r}_v)du +(G\mathbf{r}_u-F \mathbf{r}_v) dv$$ is perpendicular to $\mathbf{e}_1$.

Now $d\mathbf{N}(\mathbf{e}_1) = \mathbf{N}_u du +\mathbf{N}_v dv$ Thus the condition that $\mathbf{e}_1$ be a principal direction is that

$\mathbf{e}_2 \cdot d\mathbf{N}(\mathbf{e}_1) = 0.$ Or $$ ((F\mathbf{r}_u-E \mathbf{r}_v)du +(G\mathbf{r}_u-F \mathbf{r}_v)dv) \cdot (\mathbf{N}_u du +\mathbf{N}_vdv)=0$$ and this works out to be

$$(EM-FL)du^2 +(EN-GL)dudv +(FN-GM)dv^2=0.$$

  • $\begingroup$ Why is the vector $$\mathbf{e}_2= (F\mathbf{r}_u-E \mathbf{r}_v)du +(G\mathbf{r}_u-F \mathbf{r}_v) dv$$ perpendicular to $\mathbf{e}_1$ ? $\endgroup$
    – Mary Star
    Dec 29, 2015 at 18:08
  • 1
    $\begingroup$ For example look at the coefficient of $du^2$ in the dot product: it is $\mathbf{r}_u\cdot (F\mathbf{r}_u-E\mathbf{r}_v)$ now use $E=\mathbf{r}_u\cdot \mathbf{r}_u$ and $F=\mathbf{r}_u\cdot \mathbf{r}_v$. $\endgroup$ Dec 29, 2015 at 18:12
  • $\begingroup$ What do you mean by $dN(e_1)$ ? $\endgroup$
    – Mary Star
    Jan 4, 2016 at 17:40
  • $\begingroup$ I think this is the derivative of the normal vector in direction $e_1$. $\endgroup$ Jan 4, 2016 at 17:41
  • $\begingroup$ Ah ok... Thank you!! :-) $$$$ I edited my initial post... Could you take a look at it? Do you have an idea? $\endgroup$
    – Mary Star
    Jan 17, 2016 at 10:25

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