Abstract Nonsense

Crushing one theorem at a time

Topological Manifolds (Pt. III)

Point of Post: This is a continuation of this post.

\text{ }

Some Topological Properties of Topological Manifolds

\text{ }

What we’d now like to do is discuss some of the nice topological properties that topological manifolds have. The first of the them is the fact that they have a countable cover to topological Euclidean balls (i.e. spaces homeomorphic to a ball B_\delta(0)\subseteq\mathbb{R}^n for some n). The basic idea is simple, at each point p we know that there is a neighborhood which is homeomomorphic to an open subset of \mathbb{R}^n. Now, this open subset of Euclidean space contains a Euclidean ball around the image of p–the preimage of this ball will be a topological Euclidean ball containing p. Now, we have an open cover of our manifold by topological Euclidean balls–the argument is then over once we recall that every second countable space is Lindelof and so we may find a countable subcover. Formally:

\text{ }

Theorem: Let M be a topological manifold. Then, M admits a countable atlas for topological Euclidean balls.

\text{ }

This is a good time to point out a non-example of a topological manifold. Now, while you may be trying to think of some crazy, pathological beast of a space, there is in fact a very nice non-manifold living (embedded) in your computer screen. Indeed, consider the topological space represented by the plus sign: +. More rigorously, we can consider the infinite analogue gotten by considering the union of the two lines \mathbb{R}\times\{0\} and \{0\}\times\mathbb{R} sitting inside \mathbb{R}^2 as a subspace. Why is + not a topological manifold? Well, suppose for a second that it was, then by the above theorem there exists a neighborhood U of the point p where the two lines meet (the origin in the more rigorous version) which is homeomorphic to B_\delta(0)\subseteq\mathbb{R}^n for some (clearly nonzero) n. That said, if U\approx B_\delta(0) then U-\{p\} is homeomorphic to B_\delta(0) minus a point. That said, U-\{p\} has four connected components whereas B_\delta(0) minus a point has at most two connected components (two in dimension n=1 and zero in higher dimensions)–this is a contradiction.

\text{ }

With this theorem we can prove many, very interesting, facts about the topology of a topological manifold. For example, we immediately get the following theorems:

\text{ }

Theorem: Every topological manifold is locally simply connected. 

\text{ }

Theorem: Every topological manifold is locally compact. In fact, it has a countable basis of precompact Euclidean topological balls.

\text{ }

Now, perhaps a more interesting (because it is not as obvious) fact is the following:

\text{ }

Theorem: Let M be a topological manifold. Then, M is connected if and only if it’s path connected.

\text{ }

This follows immediately from the following lemma:

\text{ }

Lemma: Let X be a locally path connected connected space, then X is path connected.

Proof: Let x_0\in X and let \Omega be the set of all points in X for which there is a path from x_0 to that point. Since \Omega\ne\varnothing (x_0\in \Omega) it suffices to prove that it is both open and closed (and the result will then follow from X‘s connectedness). So, let x\in\Omega. By assumption there exists some neighborhood U of it such that U is path connected, we claim that U\subseteq\Omega. To see this, let x'\in U. Concatenating the paths from x to x', and from x to x_0 we can then construct a path from x' to x_0 and so x'\in \Omega. It follows that \Omega is open.

Now, to show that \Omega is closed we show it is invariant under closure, and since \Omega\subseteq\overline{\Omega} it suffices to show the reverse inclusion. So, let x\in\overline{\Omega}. Then, by X‘s local connectedness there exists a neighborhood U of x which is path connected. But, since x\in\overline{\Omega} there must be some point x'\in\Omega\cap U. So, using the exact same technique described in theorem (*) again we may connect the path from x to x' and  the path from x' to x_0 to get a path from x to x_0. It follows that x\in\Omega.

Thus, \Omega is open and closed and since it’s non-empty it must be that \Omega=X. But, that means that every point of X may be connected to x_0 by a path. This clearly implies the theorem. \blacksquare

\text{ }

Last but not least we prove the, very interesting, fact that the fundamental group of a topological manifold is necessarily countable. This, once again, is due to the fact that our manifold has a countable cover by topological Euclidean balls. Note, since connectedness is equivalent to path connectedness, the independence of base point for fundamental groups comes to connected manifolds. Our proof heavily follows [1].

\text{ }

Theorem: Let M be a connected topological manifold, then \pi_1(M) is countable.

Proof: We know that M has a cover ty topological Euclidean balls \left\{(B_n,\varphi)n)\right\}_{n\in\mathbb{N}}. Let B_{n,m}=B_n\cap B_m. Since our space is second countable we know that B_{n,m} has countably many path components. For each n\geqslant m let X_{n,m} be a collection of representatives from each path component of B_{n,m}. Let then, X=\bigcup_{n\geqslant m} X_{n,m}–evidently X is countable. If x,x'\in X and x,x'\in B_n then let p_{x,x'}^n to be a path x\to x' in B_n. Fix some x_0\in X, we will show that \pi_1(M,x_0)\cong\pi_1(M) is countable. Let G be the set of all finite products of paths of the form p_{x,x'}^n as mentioned above. Call a loop \alpha at x_0 special if \alpha\in G.

\text{ }

Evidently the set of special loops, being a subset of G (which is clearly countable) is countable. If we can show that every element of \pi_1(M,x_0) is homotopic to a special loop, then we’ll be done. To this end, let \alpha:[0,1]\to M be a loop anchored at x_0. Since [0,1] is compact, there exists a finite cover of the set \{\alpha^{-1}(B_n)\}_{n\in\mathbb{N}}. Thus, there exists finitely many points 0=a_1<\cdots<a_k=1 such that \alpha^{-1}(B_n)\subseteq[a_i,a_{i+1}] for some n. Let \alpha_i be the map one obtains by restricting \alpha to [a_{i},a_{i+1}] and reparamaterizing so that it is defined on [0,1] and let B_{n_i} be such that \alpha_i([0,1])\subseteq B_{n_i}. By definition, for each i we have that \alpha(a_i)\in B_{i,i+1} and that there exists some point x_i in the same component of B_{i,i+1}–we can take x_1=x_k=x_0. Let \beta_i be a path x_i\to \alpha(a_i) where we take \alpha_0=\alpha_k to be the constant loop at x_0. Note then that

\text{ }

\begin{aligned}\text{ }[\alpha] &= [\alpha_1]\cdots[\alpha_k]\\ &= ([\beta_0]\cdot[\alpha_1]\cdot[\beta_1]^{-1})\cdot([\beta_1]\cdot[\alpha_2][\beta_2]^{-1})\cdots([\beta_{k-1}]\cdot[\alpha_k]\cdot[\beta_k])\end{aligned}

\text{ }

Now, each [\beta_i][\alpha_{i+1}][\beta_{i+1}]^{-1} is a path x_i\to x_{i+1} and thus (since B_{n_i} is simply connected) is homotopic to p_{x_i,x_{i+1}}^{n_i}. Thus, we see that [\alpha] is a special loop. Since [\alpha] was arbitrary the conclusion follows. \blacksquare

\text{ }

\text{ }


[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.


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

No comments yet.

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: