Abstract Nonsense

Crushing one theorem at a time

Smooth Manifolds (Pt. I)


Point of Post: In this post we define the notion of smooth manifolds, give ample examples, and prove some fundamental results about the construction of smooth manifolds.

\text{ }

Motivation

\text{ }

In previous posts we have defined the notion of a topological manifold as being the end result of a progression through nicer and nicer spaces. In this post we take this progression one step further and describe a class of spaces that, in some sense, are some of the nicest spaces one can reasonably hope to deal with. The spaces come up when we ask ourselves “now that we have spaces that locally look like \mathbb{R}^n, what can we do with them?” Thus, one starts introspecting: what aspect of \mathbb{R}^n, besides being the most natural space for us lowly humans to think of, makes it so special?

\text{ }

This, it turns out, is not such an easy question to ask. I mean, \mathbb{R}^n has a ton of structure: an order structure, a ring structure, a group structure, topological structure, etc. Which do we pick? Well, evidently we have already used up the topological structure of \mathbb{R}^n (we require our spaces to be locally identical in this manner), the order theory clearly isn’t that great. What about the algebra structure–it has some nice algebraic properties? While this may seem initially like a great idea, a second of thought shows two chinks in its armor. First and foremost, algebraic structures are global objects (they are defined everywhere) whereas what we have is that our space locally looks like \mathbb{R}^n. Thus, whatever structure on \mathbb{R}^n we would like to transfer over to our manifolds must be something which is local.

\text{ }

Now, one of the most beautiful, powerful structures we have on \mathbb{R}^n is one that we are so used to by now, that we most likely take for granted–the ability to do calculus. Indeed, while groups, rings, topological spaces or any combination of those are a dime a dozen, spaces that we can do real calculus on are a genuine rarity. Moreover, differential calculus is an entirely local concept–the derivative at a point is defined using entirely local information.

\text{ }

It thus seems advantageous for us to try to transfer over the notion of differential calculus in Euclidean space to our topological manifolds. But, how? The first step will be defining what we mean by a differentiable function fome one manifold to the other. The basic idea is already should be obvious. Suppose that M and N are smooth manifolds of dimension m and n respectively and that f:M\to N is a map. Fix a point p\in M and note that locally around p and f(p) M and N look like \mathbb{R}^m and \mathbb{R}^n respectively. Thus, we can basically pretend that f is a map \mathbb{R}^m\to\mathbb{R}^n which we, of course, know how to verify smoothness.

\text{ }

More rigorously, we can phrase the above paragraph by saying that if f:M\to N and p\in M then we can choose charts (U,\varphi) at p and (V,\psi) at f(p) (with f(U)\subseteq V). We can then consider \psi\circ f\circ\varphi^{-1}:\varphi(U)\to \psi(V). This is exactly the rigorous formulation of thinking about M and N locally as Euclidean and adapting f to this situation. We are then tempted to say that f should be differentiable at p if when we do this, when we locally Euclideanize, that we get a smooth map. There is a slight issue though.

\text{ }

To be one-hundred percent correct, we should really say that the above is a description for what it means for f to be smooth p relative to the charts (U,\varphi) and (V,\psi). Indeed, a topological atlas for a manifold M can contain many overlaping charts, many of which contain the point p. It seems an undesirable property that the differentiability of our function at a point should be chart dependent. So, how can we make it independent? This is exactly where the new criterion on the charts will come into play.

\text{ }

In a sense, the failure of differentiability to be chart dependent is a failure for the charts to “overlap smoothly”. In other words, going from the image of the overlap under one chart to the image of the overlap of the other map should be smooth. Formally, this is described by saying that if two charts (U,\varphi) and (V,\psi) at p then the map \psi\circ\varphi^{-1}:\varphi(U\cap V)\to\psi(U\cap V) should be a smooth map between Euclidean spaces. This is precisely the notion of a smooth manifold–a topological manifold where all of the charts overlap smoothly–allowing us to meaningfully define what a differentiable map between such manifolds means.

\text{ }

Smooth Manifolds

\text{ }

Let M be a topological manifold and let (U,\varphi) and (V,\psi) be two continuous n-dimensional charts on M. We say that these charts are smoothly compatible if either U\cap V=\varnothing or if U\cap V is nonempty and the map \psi\circ\varphi^{-1}:\varphi(U\cap V)\to\psi(U\cap V), a map between open subsets of \mathbb{R}^n, is a diffeomorphism (i.e. a smooth bijective map with smooth inverse). We call the maps \psi\circ\varphi^{-1} an overlap map. We call an atlas \mathscr{A} of M smooth if all of its charts are smoothly compatible.

\text{ }

Note that two prove that a given atlas is smooth it really suffices to prove that all of its overlap are smooth, and not that they are diffeomorphisms. Indeed, if all of the overlap maps \psi\circ\varphi^{-1} are smooth, then its inverse \varphi\circ\psi^{-1} is necessarily smooth, since it is itself an overlap map.

\text{ }

For example, the atlas \{(\mathbb{R}^n,\text{id})\} for \mathbb{R}^n–although this is vacuously true since the only overlap map is the identity map \mathbb{R}^n\to\mathbb{R}^n.

\text{ }

Consider the sphere \mathbb{S}^n with the altas \{(U,\varphi),(V,\psi)\} as described before. We claim that this is a smooth atlas. Indeed, we note that \varphi(U\cap V)=\psi(U\cap V)=\mathbb{R}^n-\{(0,\cdots,0)\} and that \psi\circ\varphi^{-1} is merely the map \displaystyle p\mapsto \frac{p}{|p|^2} and similarly for \varphi\circ\psi^{-1}–from here the fact that this is a smooth atlas follows.

\text{ }

We claim that the standard atlas on \mathbb{RP}^n is a smooth atlas. Indeed, we need to check that \varphi_j\circ\varphi_i^{-1} is smooth for each i,j. We will assume that i>j for the other case is similar. If this is the case then one can easily check that

\text{ }

\displaystyle \varphi_j\circ\varphi_i^{-1}(x_1,\ldots,x_n)=\frac{1}{x_j}\left(x_1,\ldots,\widehat{x_j},\ldots,x_{i-1},1,x_i,\ldots,x_n\right)

\text{ }

which is obviously smooth.

\text{ }

Now, as we have said before, the existence of smooth atlases \mathscr{A} and \mathscr{B} on topological manifolds M and N respectively allows us to define “smooth” maps f:M\to N. Namely, we want to, at each point, “pretend” that our map f is defined on Euclidean spaces which we can do since the manifold looks Euclidean locally around the point and its image. Of course, as we stated before the reason that we require our charts to be smoothly compatible is that it doesn’t matter how we locally look around our point and its image (i.e. it’s independent of chart choice).

\text{ }

\text{ }

References:

[1] Lee, John M. Introduction to Smooth Manifolds. New York: Springer, 2003. Print.

[2] Lee, John M. Introduction to Topological Manifolds. New York: Springer, 2000. Print.

[3] Milnor, John Willard, and David W. . Weaver. Topology from the Differentiable Viewpoint. Charlottesville (Va.): University of Virginia, 1969. Print.

Advertisements

August 30, 2012 - Posted by | Manifold Theory, Topology | , , , , ,

2 Comments »

  1. […] Smooth Manifolds (Pt. II) Point of Post: This is a continuation of this post. […]

    Pingback by Smooth Manifolds (Pt. II) « Abstract Nonsense | August 30, 2012 | Reply

  2. […] anyone who is likely to get a lot out of these posts probably is familiar with the concept of a smooth manifold, which is merely topological spaces with a well-defined notion of how to ‘do calculus’ […]

    Pingback by Riemann Surfaces (Pt. I) « Abstract Nonsense | October 2, 2012 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: