Abstract Nonsense

Crushing one theorem at a time

Topological Manifolds


We now begin our talk on topological manifolds (to be defined shortly). From there we will begin to study some of the effects that prior subjects have on such manifolds. For example, what are necessary and sufficient conditions for a topological manifold (now just called manifolds) to be compact? Connected? Metrizable? We will also introduce some concepts previously left out for no particular reason in that it didn’t fit. For example, we will discuss the concepts of path connectedness and quotient spaces. Both are very relevant in the realm of point-set topology but the full realization of their power is when they are coupled with the idea of manifolds. So, let us begin:

Topological Manifold of Dimension n (n-manifold): Let \mathfrak{M} be a topological space that is : locally Euclidean of dimension n, second countable, and Hausdorff. Then, we call \mathfrak{M} a topological manifold of dimension n (an n-manifold).

We first prove a small theorem and then give some concrete examples of manifolds.

Theorem: Let \mathfrak{M} be an n-manifold and \mathfrak{N} an open subspace of \mathfrak{M}. Then, \mathfrak{N} is an n-manifold.

Proof: Since any subspace of a Hausdorff space is Hausdorff and  any subspace of a second countable space second countable it remains to show that an open subspace of a locally Euclidean space of dimension n is locally Euclidean of dimension n.

To do this let (U_x,\varphi_x) be the guaranteed chart at each x\in \mathfrak{N} and let (V_x,\psi_x) be such that V_x=U_x\cap \mathfrak{N} and \psi_x=\varphi_x\mid_{V_x}\to\varphi(V_x). This is clearly a chart at x. Since x was arbitrary the conclusion follows. \blacksquare.

We now give some examples

Example: Let \mathbb{S}^n=\left\{\bold{x}\in\mathbb{R}^{n+1}:\|\bold{x}\|=1\right\} . It is clear that \mathbb{S}^n being the subspace of \mathbb{R}^n is both second countable and Hausdorff. It remains to show that it is locally Euclidean. We only outline this procedure since the full construction can be found in any textbook on geometric topology. It is called stereographic projection. We define two maps \varphi_1,\varphi_2 on subspace of \mathbb{S}^n. So, let N=(\underbrace{0,\cdots,0}_{n\text{ times}},1) and S=(\underbrace{0,\cdots,0}_{n\text{ times}},-1) and define \varphi_1:\mathbb{S}^n-\{N\}\to\mathbb{R}^n by \displaystyle (x_1,\cdots,x_{n+1})\mapsto\frac{(x_1,\cdots,x_n)}{1-x_{n+1}} and \varphi_2:\mathbb{S}^n-\{S\}\to\mathbb{R}^n given by \displaystyle (x_1,\cdots,x_{n+1})\mapsto\frac{(x_1,\cdots,x_n)}{1+x_{n+1}}. In the three-dimensional case this can be thought of as removing the north or south pole and smoothing out what’s rest onto the plane. It turns out that both these maps are homeomorphisms. And so, let \bold{x}\in\mathbb{S}^n if \bold{x} if \bold{x}\ne N take the chart at \bold{x} to be \left(H^{-},\psi_1\right) where H^{-}=\left\{(x_1,\cdots,x_{n+1}):x_{n+1}<0\right\} and \psi_1=\varphi_1\mid_{H^{-}}:H^{-}\to\varphi(H^{-}). If \bold{x}=N take the cart to be (H^{+},\psi_2) where H^{+}=\left\{(x_1,\cdots,x_{n+1}):x_{n+1}>0\right\} and \psi_2=\varphi_2\mid_{H^{+}}:H^{+}\to\varphi(H^+).

It makes sense though. That although \mathbb{S}^2 is three dimensional since cutting out a smaller portion and examining it reveals that you are really looking at a portion of a plane (a two dimensional object) but just bent slightly.

Clearly any open subspace of Euclidean space is itself a manifold. So are tori (they are 2-manifolds). Also, though I will not prove it so is the real projective spaces \mathbb{RP}^n

Our next definition is a specific kind of n-manifold.

n-manifold with boundary: Let \mathfrak{M} be a second countable Hausdorff space such that for every x\in \mathfrak{M} there exists some neighborhood U_x and some open O_x\subseteq\mathbb{H}^n=\left\{(x_1,\cdots,x_n):x_n\geqslant0\right\} such that U_x\approx O_x. Then \mathfrak{M} is called an n-manifold with boundary. We define \partial\mathfrak{M}=\left\{x\in\mathfrak{M}:x\in\varphi^{-1}\left(\partial\mathbb{H}^n\right),\text{ }(U_x,\varphi)\text{ is some chart at }x\right\}. We define \mathfrak{M}^{\circ} similarly.

It will take some serious work (probably well-beyond what we’ll get into in the near future) to show that the boundary and interior of a manifold with boundary are disjoint.

For the sake of notation convenience we define \mathbb{R}^0 to be the two point discrete space \{0,1\}.  Our next theorem completely characterizes 0-manifolds.

Theorem: Let \mathfrak{M} be a 0-manifold. Then, \mathfrak{M} is a countable discrete space.

Proof: We claim that \mathfrak{M} is discrete space. To see this let x\in \mathfrak{M} be arbitrary. By assumption there exists some neighborhood U_x such that U_x\approx O_x where O_x=\{0\},\{1\},\{0,1\}. If it’s either of the first two we are done because then x\in U_x and \text{card }U_x=1 which would imply that U_x=\{x\}. Otherwise, U_x\approx\{0,1\} and so U_x=\{x,y\} for some y\in Y. We may assume WLOG that \varphi(x)=0 and since \{0\} is open and \varphi continuous we have that \varphi^{-1}(\{0\})=\{x\} is an open subspace of U_x and since U_x is open it follows that \{x\} is open in X. The conclusion follows.

To see that it’s countable notice that \displaystyle \left\{\{x\}\right\}_{x\in\mathfrak{M}} is an open cover for \mathfrak{M} with no proper subcover. But, by Lindeolf’s theorem it must have a countable subcover. Thus, it follows from previous comment that \left\{\{x\}\right\}_{x\in\mathfrak{M}} must be a countable subcover. The conclusion follows. \blacksquare

That is all for right now, there will be plenty more to come.

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April 1, 2010 - Posted by | Manifold Theory | , , , ,

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