# What's the general idea behind (rigorously?) proving that a metric space is a manifold?

Perhaps this is a broad question, but I opened Spivak's Differential Geometry and on the first page, it defines a manifold as such:

A manifold is supposed to be "locally" like one of these exemplary metric spaces $\mathbb{R}^n$. To be precise, a manifold is a metric space $M$ with the following property:

If $x\in M$, then there is some neighborhood $U$ of $x$ and some integer $n\geq0$ such that $U$ is homeomorphic to $\mathbb{R}^n$.

I googled around for a couple minutes but I couldn't find any answer directly addressing how to (or how it's impossible to) prove that some metric space is a manifold. That is, according to the last sentence, proving the neighborhood $U$ of each(?) $x\in M$ is homeomorphic to $\mathbb{R}^n$ for some nonnegative integer $n$.

I'm new to differential geometry and topology, so I apologize if this is a meaningless or too-easy-to-be-asking type of question.

• Typically, by exhibiting an explicit atlas (i.e., sufficiently many open sets together with an explicit homeomorphism to an open subset of $\Bbb R^n$) – Hagen von Eitzen Nov 18 '15 at 9:49

TL;DR: It depends.

To show that a given space $M$ is a manifold, one typically exhibits for each $x\in M$ an explicit open neighbourhood $U_x\ni x$ and an explicit homeomorphism $U_x\to \Bbb R^n$. To simplify the task, one need not specifiy different $U_x$ for each point; instead it suffices to cover $M$ with open sets (which are open nighbourhoods for each point they contain). Also, instead of a homeomorphism with all of $\Bbb R^n$ it is sufficient to specify a homeomorphism with an open subset of $\Bbb R^n$ (again, this subset contains an open ball for each of its elements and open balls are in a simple fashion homeomorphic to their containin g$\Bbb R^n$).

Example: The set $S^1=\{\,(x,y)\in\Bbb R^2\mid x^2+y^2=1\,\}$ (with metric/topology induced from $\Bbb R^2$) is a 1-manifold.

Proof: The set $U_N:=S^1\setminus\{(0,1)\}$ is an open subset of $S^1$. On it we can define $f_N\colon U_N\to\Bbb R$, $(x,y)\mapsto \frac{x}{1-y}$ and verify that this is indeed a homeomorphism $U_N\stackrel\approx \longrightarrow\Bbb R$. Similarly, for $U_S:=S^1\setminus\{(0,-1)\}$, we have $f_S\colon U_S\to\Bbb R$, $(x,y)\mapsto \frac{x}{1+y}$. As $U_S\cup U_N=S^1$, we have just shown that $S^1$ is a 1-manifold. $\square$

On the other hand, showing that somehing is not a manifold may be trickier. Details depend on the specific space, of course.

Example. The set $X=\{\,(x,y)\in\Bbb R^2\mid xy=0\,\}$ is not a manifold.

Proof. The intuitive reason is the double point at $(0,0)$. Assume $U\subseteq X$ is an open neighbourhood of $(0,0)$ and $f\colon U\to \Bbb R^n$ a homoemorphism. Then for some $r>0$, $U$ contains the open ball $B_r((0,0))\cap X$. Under $f$, this is homeomorphic to an open subset of $\Bbb R^n$. Now $B_r((0,0))\cap X$ has the interesting property that it is connected, but after removing the single point $(0,0)$ it becomes disconnected (into four connected components). The same should happen with $f(B_r((0,0))\cap X)$ upon removing $f((0,0))$. However, if $n>1$ then removal of a point from a connected open set produces a connected open set (because this is true for the removal of the center from an open ball). And if $n=1$ then the removal of a point from an open set produces two, not four connected components. And if $n=0$, removal of a point from a connected open set produces the empty set. Hence for all $n\in \Bbb N$ we arrive at a contradiction. We conclude that $X$ is not a manifold. $\square$

Caution. The definition for manifold that you quote may not be acceptd by everybody. As quoted, the set $S^1\cup\{(0,0)\}$ would be a manifold - which probably most would disagree with. Also, the space $\Bbb R^\times \Bbb R$ endowed with metric $$d\bigl((x,y),(x',y')\bigr)=\begin{cases}|x-x'|&\text{if }y=y'\\1+|x-x'|&\text{otherwise}\end{cases}$$ is a manifold under the quoted definition, but is not according to more generally accepted definitions. Apart from these two kinds non-manifolds that fit under your definition, there is a third kind, but as you seem to consider metric spaces (as opposed to general topological spaces) this third kind of problem does not apply.

There are also more indirect ways of proving that something is a manifold. For example you have the preimage theorem (not sure how standard that name is):

If $f : M \to N$ is a smooth map between manifold and $y \in N$ is a regular value, then $f^{-1}(y)$ is a submanifold of $M$. (Plus some further results about dimension and tangent spaces.)

Here, $y \in N$ is said to be a regular value of for all $x \in f^{-1}(y)$, the differential $df_x : T_xM \to T_yN$ of $f$ at $x$ is surjective.

So for example if you let $f : \mathbb{R}^2 \to \mathbb{R}$ be given by $f(x,y) = x^2 + y^2$, then $1$ is a regular value: you can easily compute the differential $df_{(x,y)}(u,v) = 2ux + 2vy$, which is surjective if $(x,y) \neq (0,0)$, and of course $f(0,0) \neq 1$. Thus $S^1 = f^{-1}(1) = \{ (x,y) \in \mathbb{R}^2 \mid f(x,y) = x^2 + y^2 = 1 \}$ is a submanifold of $\mathbb{R}^2$. It's a very powerful tool.