# line of curvature is developable

• May 11th 2013, 11:26 AM
heejenna55
line of curvature is developable
Let β be a curve on M. Show that β is a line of curvature on M iff the surface defined by y(u,v)= β(u)+vU(u) is developable. Here U(u) denotes the unit normal to M along β
• May 12th 2013, 07:40 AM
xxp9
Re: line of curvature is developable
Suppose $\beta$ is the curvature line, denote $T = \beta_u$ the tangent vector of $\beta$, and let B be the unit tangent vector of M
that is orthogonal to T. Thus T, B is the two principal directions forming a basis of the tangent plane, and denote K, k the corresponding principal curvatures.
In this basis the second fundamental form of M is a diagonal matrix $\mathrm{I\!I}=diag( K, k)$

For the surface $\gamma(u,v)= \beta(u)+v \cdot U(u)$, the two coordinate tangent vectors are:
$\gamma_u = \beta_u + v \cdot U_u$ and $\gamma_v = U$

And since $U_u \cdot U = \frac{1}{2}(U\cdot U)_u=0$
$U_u\cdot T=(U\cdot T)_u - U\cdot T_u = - U\cdot T_u=-\beta_{uu} \cdot U = -\mathrm{I\!I}_{11}=-K$
$U_u\cdot B=(U\cdot B)_u - U\cdot B_u = - U\cdot B_u = -\mathrm{I\!I}_{12} = 0$
From the above 3 equations we get $U_u=-K T$ (this can also be obtained by the Frenet Serret formulas)

So $\gamma_u = \beta_u + v \cdot U_u = (1-vK)T$ and $\gamma_v = U$
Then the normal vector of this surface is orthogonal to both T and U thus is constant along the v-lines on the surface, which implies the differential of the Gauss map is degenerate along U, thus having a 0 Gaussian curvature and hence developable.

I'll try to show the converse in a new post.
• May 12th 2013, 06:36 PM
hughc1992
Re: line of curvature is developable
is B the binormal vector?
• May 13th 2013, 05:13 AM
xxp9
Re: line of curvature is developable
probably yes in this case but not for general cases. Only when the normal vector of the curve coincides with the normal vector of the surface, B will be the binormal vector.
• May 13th 2013, 06:59 AM
xxp9
Re: line of curvature is developable
For the converse, since we have
$\gamma_u = T + v U_u$, and $\gamma_v = U$, we can compute the first fundamental form
let $k=\langle T, U_u \rangle$ and $b=\langle U_u, B \rangle$ so that $U_u = kT+bB$
So $\gamma_u = T + v (kT+bB) = (1+vk)T+bB$
$E = (1 + vk)^2 + b^2$, $F = \langle T, U \rangle + \langle v U_u, U \rangle = v\langle U_u, U \rangle = \frac{v}{2} \langle U, U \rangle_u = 0$,
$G = \langle U, U \rangle = 1$
According to Gaussian curvature
Use the formula $K = -\frac{1}{2\sqrt{EG}}\left(\frac{\partial}{\partial u}\frac{G_u}{\sqrt{EG}} + \frac{\partial}{\partial v}\frac{E_v}{\sqrt{EG}}\right)$.
We get $K = -\frac{1}{2\sqrt{E}}\frac{\partial}{\partial v}\frac{E_v}{\sqrt{E}}$
= $-\frac{1}{\sqrt{E}}\frac{\partial}{\partial v}\frac{(1+vk)k}{\sqrt{(1 + vk)^2 + b^2}}$
Only when b = 0 the last term vanishes and then we get $U_u = kT$, according to the post line of curvature question T is a principal vector.
• May 13th 2013, 12:47 PM
xxp9
Re: line of curvature is developable
Somebody asked me via private messages that "why the second fundamental form is $\mathrm{I\!I}=diag(K,k)$?", and let me explain here:
The vector $T=\beta_u$ is a unit principal vector, the unit vector $B=U \times T$ is a tangent vector to M and orthogonal to T. So (T, B) form a orthonormal basis of the tangent plane of M. Since T is a principal direction B is the other principal direction, that is, they're the eigen-vectors of the second fundamental form. Then the result follows from the fact that a matrix is diagonal when expressed with respect to its eigen-vector basis, with eign-values as the diagonal values.