# Abstract Nonsense

## Smooth Manifolds (Pt. III)

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

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August 30, 2012

## Topological Manifolds (pt. I)

Point of Post: In this post we define the notion of a topological manifolds, discuss some of their importance, and prove some elementary topological facts about them.

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Motivation

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Anyone who has taken a general topology course before can tell you the following simple fact: topological spaces can get messy. It is not hard to create topological spaces which completely break our intuition for what a topological space should be! It’s definitively possible that a topological space doesn’t locally look like connected space (i.e. isn’t locally connected), doesn’t locally look like a compact space (i.e. isn’t locally compact), and (in a more dramatic way) it doesn’t even have to be able to distinguish points (i.e. doesn’t have to be Kolomogorov)!

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August 30, 2012

## Surfaces (Pt. III)

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

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October 9, 2011

## Local Properties of Manifolds

In this post we will discuss some of the local properties manifolds attain from $n$ dimensional Euclidean space. It should be intuitive that all of the local attributes should transfer since manifolds “locally look like Euclidean space”. We formalize them in what follows.

We start with a theorem which when put together with the definition of manifolds will give us an amazing revelation.

Theorem: Let $\mathfrak{M}$ be a $n$-manifold. Then, $\mathfrak{M}$ is locally compact.

Proof: Let $x\in\mathfrak{M}$ be arbitrary. As was proven earlier every manifold has an open base of Euclidean balls, and so let $U$ be the guaranteed Euclidean ball containing $x$. So, by assumption $U\overset{\varphi}{\approx} B$ for some open ball $B\subseteq\mathbb{R}^n$. So, by the regularity of $\mathbb{R}^n$ there exists some open ball $B'\subseteq B$ such that $\overline{B'}\subseteq B$. So, $\varphi^{-1}(B')=U'$ is a neighborhood of $x$ in $U$. Also,

$\overline{U'}=\overline{\varphi^{-1}\left(B'\right)}=\varphi^{-1}\left(\overline{B'}\right)\subseteq\varphi^{-1}\left(B\right)=U$

And so, in particular

$\varphi\mid_{\overline{U'}}:\overline{U'}\to\varphi\left(\overline{U'}\right)=\overline{B'}$

is a homeomorphism and so $\overline{U'}$ is compact in $U$. Thus, $U'$ is a precompact neighborhood of $x$ in $U$. But, since compactness in a subspace implies compactness in the ambient space and $U$ is open we see that $U'$ is in fact a precompact neighborhood of $x$ in $\mathfrak{M}$. The conclusion follows. $\blacksquare$

The following corollary, to me, was astounding albeit intuitively clear.

Corollary: Since every locally compact Hausdorff space is regular and every second countable regular space is normal we see in particular that every manifold is second countable and normal and thus metrizable by Urysohn’s embedding theorem.

Also, it follows that every manifold which is not compact has an Alexandroff compactification.

We next discuss some issues related to local connectedness/path connectedness.

Theorem: Let $\mathfrak{M}$ be a $n$-manifold. Then, $\mathfrak{M}$ is locally path connected and consequently locally connected.

Proof: Let $x\in \mathfrak{M}$ be arbitrary and let $U$ be the guaranteed Euclidean ball around it. By assumption $U\approx B$ for some open ball $B\in\mathbb{R}^n$. But, as was proven earlier every open ball in a normed vector space is convex and thus path connected. Thus, $U$ is path connected and the conclusion follows. $\blacksquare$

This was just a quick post to prove some results which are true, but need to be proven at least once.

April 30, 2010

## 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.

April 1, 2010

## Locally Euclidean Spaces

We wish to begin discussing the concept of manifolds (the motivation and definition will be given in a subsequent post). We have  covered to a degree much more than is necessary all the concepts need to define a manifold except one. Namely, the concept of a space being “locally Euclidean”. Intuitively a locally Euclidean space of dimension $n$ is one that “looks like” $\mathbb{R}^n$ on a small enough scale. Consider our planet, it is locally Euclidean of dimension $n$ for although it is embedded in $\mathbb{R}^n$ to us it looks like $\mathbb{R}^2$ thus giving the age-old misconception that the world is flat. It is clear (intuitively) then that the unit sphere $\mathbb{S}^2$ is locally euclidean of dimension $n$ although it is not homeomorphic to any Euclidean space (since it is compact and any Euclidean space is not). Another concrete example is $(-\infty,0)\cup(0,\infty)$ with the usual topology. Clearly, given any point in this space we can find a neighborhood contained in either of these components which then clearly looks like $\mathbb{R}$. Once again this space is not homeomorphic to $\mathbb{R}$ since it is not connected.

Enough with the motivation, let’s move onto the formal definitions.

Locally Euclidean of Dimension $n$: Let $X$ be a topological space such that for every $x\in X$ there exists some neighborhood $U$ of $X$ which may embedded as an open set in $\mathbb{R}^n$. In, other words there exists some homeomorphism $\varphi:U\to V$ where $V\subseteq\mathbb{R}^n$ is open.

Chart: If $U\subseteq X$ is open and $\varphi:U\to\varphi(U)$ is a homeomorphism where $\varphi(U)\subseteq\mathbb{R}^n$ is open we call the ordered pair $(U,\varphi)$ a chart on $X$ and if $x\in U$ we call it a chart at $x$.

Remark: It is clear from the above that a space is locally Euclidean if and only if there is chart at every point of $X$.

We first show that $V$ being an open set may be replaced with either an open ball or $\mathbb{R}^n$ itself.

Theorem: Let $X$ be locally Euclidean, then given any $x\in X$ there exists some neighborhood $U$ of $x$ such that $U\approx B_\varepsilon(\varphi(x))$ for some $\varepsilon>0$.

Proof: Let $x\in X$ be arbitrary. Since $X$ is locally Euclidean there exists neighborhood $V$ of $x$ such that there exists some homeomorphism $\varphi:V\to O$ where $O\subseteq\mathbb{R}^n$ is open. But, since $O$ is open and $\varphi(x)\in O$ there exists some $B_\varepsilon(\varphi(x))\subseteq O$. It follows that if $U=\varphi^{-1}(B_{\varepsilon}(\varphi(x)))$ that $\varphi\mid U:U\to B_{\varepsilon}(\varphi(x))$ is a homeomorphism. The conclusion follows. $\blacksquare$.

Corollary: Since every open ball in $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$ the above is still valid if $B_{\varepsilon}(\varphi(x))$ is replaced with $\mathbb{R}^n$.

It turns out that the property of being locally Euclidean is inherited by open subspaces as is shown in the following:

Theorem: Let $X$ be a locally Euclidean space of dimension $n$ and $\mathcal{S}$ an open subspace of $X$, then $\mathcal{S}$ is locally Euclidean of dimension $n$.

Proof: Let $x\in\mathcal{S}$ be arbitrary. Then, by $X$‘s assumed local Euclidean property there exists some neighborhood $U$ of $x$ such that there exists some homeomorphism $\varphi:U\to O$. Clearly then $U\cap\mathcal{S}$ is an open neighborhood of $x$ in $\mathcal{S}$ and $\varphi\mid U\cap\mathcal{S}:U\cap\mathcal{S}\to\varphi\left(U\cap\mathcal{S}\right)$ is the desired homeomorphism. $\blacksquare$.

It is of course natural to ask under what conditions does a map preserve the property of being locally Euclidean. It should come as a surprise to no one that being locally Euclidean is invariant under homeomorphism. But, as with most topological properties there is a weaker condition for this to be true. But before we may give it in it’s fullest generality we need a new definition.

Local Homeomorphism: If $X,Y$ are topological spaces and $\varphi:X\to Y$ a continuous map such that for every $x\in X$ there exists some neighborhood $U$ of $x$ such that $\varphi\mid U:U\to\varphi(U)$ is a homeomorphism and $\varphi(U)$ is open in $Y$. If so, we say that $X$ and $Y$ are locally homeomorphic (symbolized by $X\overset{\text{loc.}}{\approx}Y$ and $\varphi$ is called a local homeomorphism.

Theorem: Let $\varphi:X\to Y$ be a local homeomorphism, then $\varphi$ is open.

Proof: Let $O\subseteq X$ be open and let $\varphi(x)\in\varphi(O)$. By assumption there exists some neighborhood $U$ of $x$ such that $\varphi(U)$ is open in $Y$ and $\varphi\mid U:U\to\varphi(U)$ is a homeomorphism. Clearly then we have that $O\cap U$ is an open subspace of $U$ and thus by assumption $\varphi\mid_U\left(O\cap U\right)$ is an open subset of $\varphi(U)$. But, since $\varphi(U)$ is a subspace we know that $\varphi(O\cap U)=\varphi(U)\cap V$ for some open set $V$ in $Y$. It follows that $\varphi\mid_U(O\cap U)$ is open in $Y$. But, this means that $\varphi\mid_U(O\cap U)$ is a neighborhood of $x$ contained in $\varphi(U)$. The conclusion follows. $\blacksquare$

Theorem: Let $X$ be a locally Euclidean space of dimension $n$. Then, if $\varphi:X\to Y$ is a surjective local homeomorphism, then $Y$ is a locally Euclidean space of dimension $n$.

Proof: Let $\varphi(x)\in Y$ be arbitrary. By assumption there exists some $U$ of $x$ such that $\varphi\mid_U :U\to\varphi(U)$ is a homeomorphism and $\varphi(U)$ is open in $Y$. But, by assumption there also exists some neighborhood $V$ of $x$, some open set $O$ in $\mathbb{R}^n$, and some $\psi:V\to O$ which is a homeomorphism. Clearly then $V\cap U$ is an open subspace of $U$ and thus using previous methods we see that $\varphi\mid_U(V\cap U)$ is a neighborhood of $x$ in $Y$. But, clearly since $U\cap V$ is open we also have that $\psi\mid_{U\cap V}:U\cap V\to\psi\left(U\cap V\right)$ is a homeomorphism and $\psi(U\cap V)\subseteq\mathbb{R}^n$ is open (this is clear it is an open subset of the open subspace $O$). It follows that $\varphi\mid U(V\cap U)$ is a neighborhood of $\varphi(x)$ and $\psi\mid_{U\cap V}\circ\left(\varphi\mid_{U\cap V}\right)^{-1}:\varphi(U\cap V)\to\psi(U\cap V)$ is a homeomorphism with an open subset of $\mathbb{R}^n$. The conclusion follows. $\blacksquare$.

Remark: A similar idea will appear later when we discuss overlap maps. Also, the idea of local homeomorphisms is much richer than what we have shown here. Maybe one day we will explore the concept in its entirety.

The next theorem should come as a surprise to no one.

Theorem: Let $X_1,\cdots,X_m$ be a finite number of topological spaces such that $X_k$ is locally Euclidean of dimension $n_k$ for each $k=1,\cdots,m$. Then, $\displaystyle \prod_{j=1}^{m}X_j$ under the product topology  is locally Euclidean of dimension $\displaystyle \sum_{j=1}^{n}n_j$.

Proof: Let $\bold{x}\in X$ be arbitrary. By assumption there exists a neighborhood $U_k$ of $\pi_k(\bold{x})$ and a homeomorphism $\varphi_k:U_k\to O_k$ where $O_k\subseteq \mathbb{R}^n_k$ is open. Clearly then $U_1\times\cdots\times U_m$ is a neighborhood of $\bold{x}$ and $O_1\times\cdots\times O_m$ and open subset of $\mathbb{R}^{n_1+\cdots+n_m}$. It follows from an old theorem that $\varphi_1\times\cdots\times\varphi_m:U_1\times\cdots\times U_m\to O_1\times\cdots\times O_m$ given by

$(x_1,\cdots,x_m)\to \{\varphi_1(x_1)\}\times\cdots\times\{\varphi_m(x_m)\}$

or equivalently (if it’s hard to picture)

$\left(\pi_1(\varphi_1(x_1))),\cdots,\pi_{n_1}(\varphi_1(x_1))),\cdots,\pi_1(\varphi_m(x_m))),\cdots,\pi_{n_m}(\varphi_m(x_m)))\right)$

is a homeomorphism (since it’s the product of homeomorphisms). The conclusion follows $\blacksquare$

We now discuss particularly nice theorem regarding a base that can be imposed on $X$. But, first a definition.

Euclidean Ball (E.B.): Let $X$ be a locally Euclidean space of dimension $n$ and let $U\subseteq X$ be open and such that $U\approx B$ where $B\subseteq\mathbb{R}^n$ is an open ball. Then, we call $U$ a Euclidean ball (E.B).

Our next theorem in effect says that the Euclidean balls are sufficient to describe $X$‘s topology.

Theorem: Let $X$ be a locally Euclidean space of dimension $n$ then $X$ has an open base consisting entirely of E.B.s

Proof: Let $x\in X$, we have by assumption there exists some neighborhood $U_x$ of $x$ such that $U_x\approx B_x$ for some open ball $B_x\subseteq\mathbb{R}^n$, and let $\varphi_x$ be the associated homeomorphism. Define then $\Omega_x=\left\{B_{\varepsilon}(\varphi_x(x)):B_{\varepsilon}(\varphi(x))\subseteq B_x\right\}$ and $\Lambda_x=\left\{\varphi_x^{-1}(\omega):\omega\in\Omega_x\right\}$. It is clear that each element of $\Lambda_x$ is an open subspace of $U_x$ but since $U_x$ is open we have that it is an open subspace of $X$. Furthermore, it is not to hard to see that given some $\varphi_x^{-1}(B_{\varepsilon}(\varphi_x(x)))$ that $\varphi\mid_{\varphi_x^{-1}(B_{\varepsilon}(x)))}:\varphi^{-1}(B_{\varepsilon}(\varphi(x)))\to B_{\varepsilon}(\varphi(x))$ is a homeomorphism. Thus, each element of $\Lambda_x$ is an E.B. Finally, let $\displaystyle \Lambda=\bigcup_{x\in X}\Lambda_x$. It is clear that $\Lambda$ is a collection of E.B.s in $X$ it remains to show that is a base.

To see this let $y\in X$ be arbitrary and $N$ any neighborhood of it. Letting the notation be as in the previous paragraph we have that $E_y\cap N$ is an open subspace of $E_y$ and so $\varphi_y(E_y\cap N)$ is an open subspace of $B_y$ and so there exists some $B_{\varepsilon}(\varphi_y(y))$ such that $B_{\varepsilon}(\varphi_y(y))\subseteq B_y$. It follows that $\varphi_y^{-1}(B_{\varepsilon}(\varphi_y(y)))\subseteq E_y\cap N\subseteq N$ and $\varphi_y^{-1}(B_{\varepsilon}(\varphi_y(y)))\in\Lambda$. The conclusion follows. $\blacksquare$

It is not surprising that most of the “local properties” of $\mathbb{R}^n$ are inherited by locally Euclidean spaces. While we will discuss most of them in subsequent posts we prove one here.

Theorem: Let $X$ be locally Euclidean of dimension $n$. Then, $X$ is first countable.

Proof: Let $x\in X$ be arbitrary and let $U_x,\varphi_x,B_x$ be as in the last theorem. Then, just as before there exists some $B_{\varepsilon}(\varphi_x(x))$ such that $B_{\varepsilon}(\varphi_x(x))$. By the Archimedean principle there exists some $N\in\mathbb{N}$ such that $\displaystyle \frac{1}{N}<\varepsilon$. Define

$\mathcal{N}=\left\{\varphi_x^{-1}\left(B_{\frac{1}{n}}(\varphi_x(x))\right):n\in\mathbb{N}-\{1,\cdots,N\}\right\}$

Clearly $\text{card }\mathcal{N}=\aleph_0$ so it remains to show that it is an open base at $x$. To do this let $N$ be any neighborhood of $x$. Clearly then $N\cap U_x$ is a neighborhood of $x$ contained in $U_x$. Thus, we have that $\varphi_x(U_x\cap N)$ is open in $B_x$ and so there exists some $\delta>0$ such that $B_{\delta}(\varphi_x(x))\subseteq \varphi(E_x\cap N)$. Thus, appealing to the Archimedean principle again we may find some $M\in\mathbb{N}-\{1,\cdots,N\}$ such that $\displaystyle \frac{1}{M}<\delta$. Clearly then $\varphi_x^{-1}\left(B_{\frac{1}{M}}(\varphi_x(x))\right)\subseteq E_x\cap N\subseteq N$ and since it’s also in $\mathcal{N}$ the conclusion follows. $\blacksquare$

This ends our discussion for now. Next time we will talk about manifolds.

March 31, 2010