Is this the correct understanding of how a geometric surface works?

Question

I'm reading "Elementary Differential Geometry" by Barrett O'Neill. Most of the book is spent looking at surfaces in $\mathbb R^3$, but eventually he introduces the "abstract surface", which I understand to be a surface $M$ which doesn't (necessarily) "live" in $\mathbb R^3$, but still has its points referred to by "abstract patches" $x(u, v)$. I.e., a region of $M$ is covered by the image of some patch $x(u,v)$, but we don't have an explicit form for $x$ like we did when surfaces were embedded in $\mathbb R^3$.

Later, he introduces the concept of a metric tensor $g$, and says:

surface + metric tensor = geometric surface

My understanding is as follows: previously, to do calculations on a surface embedded in $\mathbb R^3$, we might have an explicit equation for a patch $x(u,v)$ and calculate its derivatives $x_u, x_v$, which would let us calculate things like a frame field, unit normal field, curvature, etc. Now, because the geometric surface is an abstract surface that doesn't have some explicit form for $x(u,v)$, we calculate those same things using the metric tensor $g$, but we still don't explicitly have an expression for $x(u,v)$. Now, $g$ is just the tool that tells us what the inner product of $x_u, x_v$ is as a function of $(u,v)$.

Is that correct?

Answer 1

You are correct! From historical point of view, I think you touch upon a great paradigm shift in the development of differential geometry, from the extrinsic viewpoint to intrinsic viewpoint.

The extrinsic viewpoint is what is most familiar to us, since we can naturally investigate the curves and surfaces as they are embedded in $R^3$. Then we parametrize these embedding as $[0,1]\to\mathbb R^3$ or $[0,1]\times [0,1]\to\mathbb R^3$.

However, soon we noticed that the embedding is kind of arbitrary, and it will be affected by rotation and other solid transformations of the surfaces. Moreover, given the same surface, there are numerous ways to parametrize it. Then in some sense the parametrization is only a "scaffolding" for us to study the curves and surfaces, which could be discarded in the end.

This motivates the development of intrinsic characterization of the surfaces: a manifold equipped with a metric. As you said, we developed metric $g$ to quantify the differential distance structure on the surface, which could derive curvature regardless of the embedding.

Answer 2

This is correct in parts. For many things you can do on a surface you don't need to know how it is embedded or if it is embedded at all. Curvature (at least some of it's non-equivalent representations) is an example. For such objects the knowledge on the metric is enough, and you can just do similar stuff if you just take such an 'abstract surface' with a metric tensor.

For example, a unit normal field depends on the embedding. This is not possible to define for an 'abstract surface'.