# Abstract Nonsense

## Topological Manifolds (Pt. II)

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

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In the above example, with $M=\mathbb{R}^n$ and $U$ just any open subset of $\mathbb{R}^n$ we see that the inclusion $\{(U,\iota)\}$ (where $\iota$ is the inclusion) constitutes an $n$-dimensional continuous atlas for $U$. Thus, open subsets of Euclidean space are themselves topological manifolds.

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We claim that the circle $\mathbb{S}^1$ is a topological manifold of dimension $1$. Indeed, let $(U,\varphi)$ be the chart described previously and let $(V,\psi)$ be the chart defined by $V=\mathbb{S}^1-\{(0,-1)\}$ and $\psi(x,y)=\frac{x}{1+y}$–one can check, exactly as in the case of $\varphi$, that this is chart. Clearly then $U\cup V=\mathbb{S}^1$ and thus $\{(U,\varphi),(V,\psi)\}$ constitutes a $1$-dimensional atlas for $\mathbb{S}^1$.

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In fact, the $n$-sphere $\mathbb{S}^n=\{(x_1,\cdots,x_{n+1})\in\mathbb{R}^{n+1}:x_1^2+\cdots+x_{n+1}^2=1\}$ is an $n$-dimensional topological manifold. We basically do the same procedure as the sphere. Namely, we pick two points, and successively exclude those points from the sphere, from where we “unfold” the remains to get all of $n$-space. Indeed, define $N=(0,\ldots,0,1)$ and $S=(0,\ldots,0,-1)$. Consider then the sets $U=\mathbb{S}^n-\{N\}$ and $V=\mathbb{S}^n-\{S\}$ and the maps

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$\displaystyle \varphi(x_1,\cdots,x_{n+1})=\frac{1}{1-x_{n+1}}(x_2,\cdots,x_{n+1})$

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and

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$\displaystyle \psi(x_1,\cdots,x_{n+1})=\frac{1}{1+x_{n+1}}(x_1,\cdots,x_n)$

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One can easily verify (using similar logic to the one dimensional case) that $\{(U,\varphi),(V,\psi)\}$ is an atlas for $\mathbb{S}^n$.

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We now discuss one last example of a topological manifold. We define the real projective space $\mathbb{RP}^n$ to be the quotient space $\mathbb{R}^{n+1}/\sim$ where $v\sim w$ if $v=\alpha w$ for some $\alpha\in\mathbb{R}$. In other words, we can think about $\mathbb{RP}^n$ as being the space of one-dimensional subspaces of $\mathbb{R}^{n+1}$. It is easy to show that the map $\pi:\mathbb{R}^{n+1}\to\mathbb{RP}^n$ given by sending $x$ to it’s span, is a quotient map. We leave it to the reader to prove that this is second countable and Hausdorff (or see any standard texts on manifolds). To see that $\mathbb{RP}^n$ is locally Euclidean of dimension $n$. Denote the elements of $\mathbb{RP}^n$ as $[x_1:\cdots:x_{n+1}]$ (the colons are supposed to make one think of ratios–the thing constant on subspaces). Let $U^i=\left\{(x_1:\cdots:x_{n+1}):x_i\ne0\right\}$ and let $V^i=\pi(U^i)$. This is open because it’s preimage, $U^i$, is open. Moreover, $\{U_1,\cdots,U_{n+1}\}$ form an open cover of $\mathbb{RP}^n$. Noting then that the map $\varphi_i:V^i\to\mathbb{R}^n$ given by

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$\displaystyle [x_1,\cdots,x_{n+1}]\mapsto \frac{1}{x_i}(x_1,\cdots,\widehat{x_i},\cdots,x_{n+1})$

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is a well-defined homeomorphism.

Ways to Manufactures Topological Manifolds

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There is many a-way to manufacture topological manifolds from old ones, and even sometimes from thin air! Probably the nicest is the following:

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Theorem: Let $U\subseteq\mathbb{R}^n$ be open and let $f:U\to\mathbb{R}^m$ be a continuous map. Then, if $\Gamma_f$ denotes the graph $\{(x,f(x)):x\in U\}$ of $f$ one has that $\Gamma_f$ is a topological manifold of dimension $n$.

Proof: Define $\varphi:\Gamma_f\to U$ by $\varphi(x,f(x))=x$. Clearly this map is continuous, and is bijective. It’s inverse is $U\to\Gamma_f$ given by $x\mapsto (x,f(x))$ and since each coordinate function is continuous so is the total map. Thus, $\{(U,\varphi)\}$ is an atlas for $\Gamma_f$. $\blacksquare$

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While there is a more snappy result to the above theorem–namely that $\Gamma_f\approx U$ (for us, $\approx$ will mean homeomorphic)–it is much nicer to have written down an explicit atlas.

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This tells us that things like parabolas and sine curves are topological manifolds.

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The next obvious way to get manifolds is to take their product and sum.

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Let $M$ and $N$ be topological manifolds of dimensions $m$ and $n$ respectively. Suppose then that we have atlases $\{(U_\alpha,\varphi_\alpha)\}$ and $\{(V_\beta,\psi_\beta)\}$ for $M$ and $N$. Define, for each $\alpha,\beta$, the map $\varphi_\alpha\times\psi_\beta:U_\alpha\times V_\beta\to \mathbb{R}^m\times\mathbb{R}^n\approx\mathbb{R}^{m+n}$ as follows: let $(x,y)\in U_\alpha\times V_\beta)$, then $(\varphi_\alpha\times\psi_\beta)(x,y)=(\varphi_\alpha(x),\psi_\beta(y))$. By definition of the product of topological spaces the map $\varphi_\alpha\times\psi_\alpha$ is a homeomorphism $U_\alpha\times V_\beta\to \varphi_\alpha(U_\alpha)\times\psi_\beta(V_\beta)$. Since $\varphi_\alpha(U_\alpha),\psi_\beta(V_\beta)$ are open in $\mathbb{R}^m$ and $\mathbb{R}^n$ respectively we have that $\varphi_\alpha(U_\alpha)\times\psi_\beta(V_\beta)$. Moreover, it’s clear that $\{U_\alpha\times V_\beta\}$ constitutes a cover for $M\times N$. Now, since Hausdorffness and second countability are preserved under finite products we may summarize all of this with the following:

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Theorem: Let $M$ and $N$ be an $m$-dimensional and $n$-dimensional topological manifold respectively. Then, $M\times N$ is a topological manifold of dimension $m+n$. In particular, if $\{(U_\alpha,\varphi_\alpha)\}$ and $\{(V_\beta,\psi_\beta)\}$ are atlases for $M$ and $N$ respectively then $\{(U_\alpha\times V_\beta,\varphi_\alpha\times\psi_\beta)\}$ is an atlas for $M\times N$.

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Of course, the theorem above extends to the product of finitely many manifolds So for example the $n$-torus $\mathbb{T}^n=(\mathbb{S}^1)^n$ is a topological manifold of dimension $n$. Of course, the $2$-torus is homeomorphic  to the surface-of-a-doughnut that we usually call a torus.

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The second way we can construct new manifolds out of old ones is via the disjoint union of topological spaces. Indeed, suppose that we have two $n$-dimensional topological manifolds $M$ and $N$ with atlases $\{(U_\alpha,\varphi_\alpha)\}$ and $\{(V_\beta,\psi_\beta)\}$ respectively, and let us consider their disjoint union $M\sqcup N$. For each chart $(U,\varphi)$ on $M$ define $\varphi':U\times\{1\}\to\varphi(U)$ by $(x,1)\mapsto \varphi(x)$. Similarly, define $\psi'$ for each chart $(V,\psi)$ on $N$. It’s fairly easy to see that $(U\times\{1\},\varphi')$ and $(V\times\{2\},\psi')$ are charts on $M\sqcup N$ and their domains cover $M\sqcup N$. Thus, modulo some technical details (which I have previously worked out) we have the following theorem:

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Theorem: Let $M$ and $N$ be two topological manifolds of dimension $n$, then $M\sqcup N$ is a topological manifold of dimension $n$. In particular, if $\{(U_\alpha,\varphi_\alpha)\}$ and $\{(V_\beta,\psi_\beta)\}$ are atlases for $M$ and $N$ respectively then $\{(U_\alpha\times\{1\},\varphi')\}\cup\{(V_\beta\times\{2\},\psi_\beta')\}$ is an atlas for $M\sqcup N$.

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This tells us that things like the disjoint union of two circles $\mathbb{S}^1\sqcup\mathbb{S}^1$ is a one dimensional topological manifold.

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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.

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August 30, 2012 -

## 2 Comments »

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

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2. […] that and are smooth manifolds of the same dimension with atlases and respectively. We already know that with is a topological atlas for . That said, all of the charts in this atlas are clearly […]

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