# The Differential Geometry of a 2-D Surface

I'm currently self-studying the differential geometry of embedded surfaces. My question is, how am I to chose the appropriate coordinates and derive the covariant basis for the surface I'm interested in? The book I'm studying out of derives the equations for the surface normal derivative, curvature tensor, mean curvature, etc for the simple sphere of radius R. This is a straight forward and easy to deal with case, however, I'm have trouble generalizing the methods used to arbitrary surfaces.

Consider the surface $z= \sin(x) + \sin(y)$. By simply looking at the graph Mathematica generates, I get the impression that I should choose coordinates that lie on the surface the same way the $x$ and $y$ axes lie in the $xy$-plane. This is where I run into a snag.

In the book, when the formula $\overrightarrow{S} _{\alpha} = \frac{\partial \overrightarrow{R}}{\partial S^{\alpha} }$ was used to derive the surface covariant basis, it was done in a purely geometric way and for something as simple as a sphere. Now, I can parametrize the surface with respect to $x$ and $y$. And from here generate the position vector which lies purely on the surface

$$\overrightarrow{R} =x \hat{i} + y \hat{j} + (\sin(x)+\sin(y)) \hat{k}$$

Here, using $x$ and $y$ as the surface coordinates (which I'm not certain even makes sense). This allows me to generate a surface covariant basis (in terms of the ambient basis), a metric tensor, and a covariant metric tensor. However, if I want to calculate the contravariant basis vectors (in order to use in covariant derivatives, to get the curvature tensor, etc) all of the sudden I have to deal with the fact that my covariant basis is in terms of the ambient basis and prohibits me from doing the necessary contraction with the contravariant metric tensor.

Essentially, I'm looking for someone to set me straight and lay some guidelines on the general procedure that one would go through when trying to set the framework to analyze such a surface as the one mentioned.

How do I conveniently choose the surface coordinates and from there, derive the covariant basis?

• Are you familiar with the notion of a manifold? – Sempliner Aug 21 '15 at 6:09
• @Sempliner I heuristically understand what a manifold is. However, I don't think I know enough topology to grasp a lot of the technicalities. – Seenathin Aug 21 '15 at 15:07

Using $x$ and $y$ as parameters makes perfect sense. This means you are using the diffeomorphism $\Phi(x,y) = (x,y,\sin x+\sin y)$ to cover the whole surface with a coordinate patch. Pushing forward the standard vector fields $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ from $\mathbb{R}^2$ to the surface gives a basis of the tangent space, $$\frac{\partial }{\partial x}+\cos x \frac{\partial }{\partial z }\quad \text{ and }\quad \frac{\partial }{\partial y}+\cos y \frac{\partial }{\partial z }$$ I don't think you should worry about using the ambient space here: after all, your manifolds gets its metric from the ambient space. For example, the norm of $\frac{\partial }{\partial x}+\cos x \frac{\partial }{\partial z }$ is $\sqrt{1+\cos^2x}$.
In addition to pushing forward vector fields, you can pull back the differential forms $dx$ and $dy$ on $\mathbb{R}^2$ under the projection map $\Phi^{-1}(x,y,z)=(x,y)$. This gives the forms that can still be denoted $dx$ and $dy$, forming a basis of the cotangent space of the surface.