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

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- May 11th 2013, 11:26 AMheejenna55line 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 AMxxp9Re: line of curvature is developable
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. - May 12th 2013, 06:36 PMhughc1992Re: line of curvature is developable
is B the binormal vector?

- May 13th 2013, 05:13 AMxxp9Re: 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 AMxxp9Re: line of curvature is developable
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. - May 13th 2013, 12:47 PMxxp9Re: line of curvature is developable
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.