# For which of the pairs of surfaces shown below there is a local isometry?

I just started to study exponential maps in differential geometry. I'm using the book Differential Geometry of Curves and Surfaces, by Manfredo P. Carmo. In section $4.6$ I'm having trouble at the exercise $5$. The problem is that I don't know how I'm supposed to use the exponential map to solve this. I need to someone solve only one of the items, then I can try for myself the other ones.

For which of the pairs of surfaces shown below there is a local isometry?

a. Revolution torus and a cone.

b. Cone and sphere.

c. Cone and cylinder.

## 2 Answers

The question is about curvature and if the surface can isometrically transformed to the plane. The cone has zero curvature so it is only locally isometric with another surface of zero curvature. The cone and also be unfolded ? or opened up to lie on the plane. Eg, you can make a cone from a piece of paper, but not a torus or a sphere. I think the exercise is in reference to Minding's theorem.

• The cone has zero gaussian curvature, but not zero mean curvature right? – Ellya May 31 '14 at 19:18
• That is right, for sure the Gaussian is zero, if you think about mean, one principle direction is along a generator, this is a straight line so zero principle curvature here, the other normal direction is orthogonal this is along a circle so a positive curvature. – Rene Schipperus May 31 '14 at 20:11

By Gauss's Theorema Egregium, if two surfaces are locally isometric, then they must have the same Gaussian curvature at corresponding points. The converse holds with appropriate hypotheses for surfaces; as Rene indicates, Minding's Theorem deals with the case of constant curvature. But you don't need anything fancy to do this exercise. Most of it comes from the first sentence I wrote. :)