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
Suppose is the curvature line, denote the tangent vector of , 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
For the surface , the two coordinate tangent vectors are:
and
And since
From the above 3 equations we get (this can also be obtained by the Frenet Serret formulas)
So and
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.
For the converse, since we have
, and , we can compute the first fundamental form
let and so that
So
, ,
According to Gaussian curvature
Use the formula .
We get
=
Only when b = 0 the last term vanishes and then we get , according to the post line of curvature question T is a principal vector.
Somebody asked me via private messages that "why the second fundamental form is ?", and let me explain here:
The vector is a unit principal vector, the unit vector 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.