Problem with definition of regular surface in classical differential geometry

I am reading Do Carmo's differential geometry book and the definition of a regular surface in the second chapter is given to be this: 1) Why we need to find a neighbourhood of point $p$? Is it because we can't always define a map $X$ that will work for the whole surface and we are trying to find local maps for every point. And later in chapter 3, when author talks about this parametrization/map for a surface, he says surface parametrized at point $p$, what does "at point $p$" means, are we talking about local parametrizations?

2) $X$ is differentiable to infinite order....is this really necessary? What if map $X$ is differentiable to a large but finite order?

3) X is continuous by condition 1. But I don't understand the use of continuity as described in the book. In book it is given that we need condition 2 for one to oneness of the map so that we can have single tangent plane at each point (basically to avoid self-intersections). So why don't just make condition 2 to be $X$ being one to one function?

4) I couldn't understand why we need condition 3 very clearly. Could you please give a geometric intuition for this?

1) Because you usually can't map the whole surface to a planar set. Like the sphere, for example: every world map has to cut it somewhere.

2) Not really necessary. Much of the theory works for $C^k$ instead of $C^\infty$ as long as $k$ is large enough. But some constructions become more difficult and statements cumbersome, because one has to keep track of how many derivatives you took so far, and how many you can still take. So there's a nonzero cost for unclear benefit (especially at the textbook level).

3) Continuous and one-to-one is not the same as being a homeomorphism. (There are some situations when they are, but that is something one has to think through.) Stating the definition as it is allows us to say: locally, as far as topology is concerned, the surface is exactly like a plane. That's a nice thing to have. All local topological properties of the plane are immediately available to us, because they are preserved by homeomorphisms.

4) This is easier to illustrate with an example of a curve. The curve $y= x^{2/3}$ is visibly non-smooth - it has a cusp at $(0,0)$. But it admits a parametrization $(x,y)=(t^3,t^2)$ which is infinitely differentiable. There seems to be a disconnect between the geometric roughness of the curve and the smoothness of its parametrization. Imposing regularity ensures that we don't have this problem; by some form of the implicit function theorem, a surface with a smooth regular parametrization is indeed a smooth-looking geometric object.

A related surface example: $z=x^{2/3}$ with parametrization $(u,v)\mapsto (u^3,v,u^2)$. The differential map has rank $1$ at some points, and this is what allows the cusp to form.

Ultimately, the proof of the pudding is in the eating. Complicated as it is, this definition succeeds at matching a formal mathematical concept to an informal geometric notion of smooth surface. This is something you get convinced of when you actually use it.

• Could you please elaborate point 4. – Andrew Miller Oct 14 '13 at 10:19
• Also why we go looking for neighbourhood $V$ in $R^3$? Why not just search for the neighbourhood on the surface $S$? – Andrew Miller Oct 14 '13 at 10:42

1): Yes. Arbitrary surfaces may not be globally parametrized by an open subset in $\mathbb{R}^2$. This is a preliminary definition of the more general concept of a manifold, where you think of a manifold of dimension $n$ as gluing together pieces of open subsets of $\mathbb{R}^n$, for which this is a special case.

2):You could relax the definition and not require that the partial derivatives be infinitely differentiable, but this is a simplifying assumption that allows one to not care about about how many times you differentiate these functions in proofs. For example, you could define say a `$k$-regular surface' by replacing the sentence "partial derivatives of all orders" by "partial derivatives up to order $k$", but this is a much more cumbersome assumption to deal with.

3):Even if $x$ is one-to-one its inverse may not be continuous as well.

4):Basically we want paths in $U$ to be in one to one correspondence with paths in $S$. That is, if a particle is traveling through $U$ at a certain velocity at the point $P$ the image of the path in $S$ should have a well defined velocity at $x(P)$.

• 1, 2 and 3 are alright but could you please elaborate how condition 3 implies (4). – Andrew Miller Oct 14 '13 at 10:04
• I understand $dx_q$ represents the velocity map $(dv)$, but why it has to be one to one. – Andrew Miller Oct 14 '13 at 19:14
• what happens to the velocity map at singular points? – Andrew Miller Oct 16 '13 at 12:29