# Zero gaussian curvature implies developability proof

1. May 9, 2012

### Pinko

1. The problem statement, all variables and given/known data

Hey guys, I have a small question on a proof in Struik's Differential Geometry book. It concerns the proof that all surfaces with K = 0 (gaussian curvature zero) are developable, i.e. all surfaces with K = 0 are ruled and the tangent planes along the rulings are parallel.

2. Relevant equations

Suppose there is a surface with surfaace coordinates {u,v} and K = 0 everywhere. Since K = 0 the coefficients of the first fundamental tensor I must be I_uu = I_vv = 1, I_uv = I_vu = 0. The basis vectors g are thus orthonormal everywhere on the surface. The normal vector to the surface is given by N = gu × gv

3. The attempt at a solution

I've already derived the part which proves that:
K = 0 --> N . (N_u × N_v) = 0 (N_u means dN/du)
For this to be true, either N_u or N_v must be zero, or N . N_u and N . N_v must be zero. If the latter is true, the length || N_u × N_v || must be zero, which means that there exists another coordinate system for which N_u or N_v is zero. However, how do I prove that if N_u or N_v are zero vectors, that this implies that the curvature in that direction is zero along a straight line? The orthonormality of the basis vectors might support the claim, however I don't think this proves the matter...

Many thanks!

Last edited: May 9, 2012