Why isn't every metric space a manifold?

Question

What I mean with "intuitive": I can handle some formulas, but since I am not a professional mathematician I am not fluent in all the lingo, so I do not know by heart what "second countable" means. If I have to look up all these terms and try to understand them, it takes so much time, that I forget what I was researching in the first place... so basic terminology is appreciated.

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It was previously asked whether every manifold is a metric space, but I have to admit, I did not completely understand the answers. Assuming that a manifold is second-countable, the answer is "yes" (I cannot claim I full understood the property "second countable"). My (non-completely) translation of the answer https://math.stackexchange.com/a/1530066/340174 into an intuitive explanation is

I want to find the distance from $x_0$ to y, both of which are elements of the manifold. Since a manifold is locally Euclidean, I can walk a infinitely small way in an "Euclidean" manner. So, I go a small step from $x_0$ to $x_1$ and I calculate the distance I walked, which is possible, because I can just use the Euclidean distance. I walk from $x_1$ to $x_2$ until I reach y and add up all the distances to the total distance. From all the possible paths I take the one that is the shortest and that is my distance.

First question: It seems intuitively obvious to me that the first three conditions of a metric apply to manifold distances, as I described it above. But how do I know that the triangular condition applies as well to the distance over a manifold? Is there an intuitive explanation in the style I tried above?

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Originally I would have guessed (without too much thinking) that every metric space is a manifold, but not the other way around. Since the second part is wrong, I would guess that now, that the first part is also wrong. (Otherwise there would be no need to differentiate the two, right?) But what is that so? I can come of with a metric space, like one based on the Levenshtein distance, which is not continuous and my usual impression of manifolds is that they are continuous (since they are supposed to be Euclidean locally). However it seem there are also discrete manifolds (which I do not understand either).

Second question: What is an intuitive explanation, why metric spaces are not necessarily manifolds?

Answer 1

To your first question:

Take three points $A, B, C$. Then by construction, $d(A,B)$ is the length of the shortest path from $A$ to $B$. And similarly for $d(B,C)$ and $d(A,C)$.

Now consider the path $\gamma$ that you get by first following the shortest path from $A$ to $C$, and then continuing on the shortest path from $C$ to $B$. Clearly the length of $\gamma$ is the sum of the lengths of the two paths it is composed of, that is, $$\mathrm{length}(\gamma)=d(A,C)+d(C,B). \tag{1}$$ On the other hand, $\gamma$ clearly is a path going from $A$ to $B$, and therefore is at least as long as the shortest path from $A$ to $B$, $$\mathrm{length}(\gamma)\ge d(A,B). \tag{2}$$ Inserting $(1)$ in $(2)$ gives the triangle inequality.

To your second question:

A simple example of a metric space that is not a manifold is $\mathbb Q$. Clearly it is a metric space (with $d(x,y)=\lvert x-y\rvert$), but it is not a manifold because it isn't locally homeomorphic to any $\mathbb R^n$.

Intuitively, in every neighbourhood of some point, there are points missing that would be there in $\mathbb R^n$ (in the case of $\mathbb Q$, that's even true for all points, but if suffices if you can find one such point).

Another way you can fail to be a manifold is when the dimension gets infinite. The set of infinite real sequences with finitely many non-zero entries is such an example. Clearly an infinite-dimensional metric space cannot be locally homeomorphic to a finite-dimensional one, such as $\mathbb R^n$.

Yet another way of failing to be a manifold is if it consists of parts of different dimension. For example, consider an open disc and a straight line not touching that disc, as subsets of $\mathbb R^2$ with the corresponding metric. Both the disc and the straight line separately are manifolds, but together they are not.

Answer 2

The question itself is a bit misleading; a manifold by itself has no metric and a metric space by itself has no manifold structure.

But it is a fact that every manifold can be endowed with a metric so that the metric-topology coincides with the manifold-topology, and not vice versa. But this involves quite a lot of definitions and is probably not the answer you seek.

So to make the story short, just think of curves in a 2-dimensional plane. They are all metric spaces, as you can measure distance in the Euclidean plane. They are manifold if, at every point, the curve looks locally like a line segment -- no ends, no branching, no space-filling curves, no fractals...

Simple counter-examples are letters A, B, E, F, H, K, P... X, Y... (considered as curves in a Euclidean plane). Because those have at least one point where various line segments meet together. However, D and O are clearly manifolds, and C, I, J manifolds with boundary.

Answer 3

Y

The subset of the plane made by three half-open line segments with the closed endpoint of all three identified is a metric space; at the very least it inherits a metric from the plane. It is not a manifold because there is no point on the real line with three distinct tangents (directions in which one can proceed). Here's another:

$\ast$

It is a metric space, with metric inherited from the plane, but it is not a manifold as no point of the line has six closed neighborhoods with only one point in common.

Answer 4

The main point is that a manifold is locally Euclidean: there is a nonnegative integer $n$ (the dimension of the manifold) such that each point of the manifold has a neighbourhood homeomorphic to $\mathbb R^n$. For a simple example of a metric space that is not a manifold, take the closed interval $[0,1]$ (it is a "manifold with boundary", but that is not a manifold).

Answer 5

Look at an infinite metric space that simply assigns distance $1$ to every pair of distinct points. That's a metric space but it doesn't look anything like a manifold.

Or look at the space of bounded real functions on $\mathbb R$ in which the distance between $f$ and $g$ is $\sup_x |f(x)-g(x)|.$

Answer 6

First question (the triangle inequality), as you said in your query:

From all the possible paths I take the one that is the shortest and that is my distance.

So if you have two points A and B and you've found the shortest path between them, then there is no shorter path going via point C. Otherwise that new shorter path would have been the original path you found. It is this feature which gives it the name 'triangle inequality'. The equality holds if the shortest path from A to B already passed through point C because then its not a detour, just a pit stop.

Second question: Think of a cone. There is the obvious metric on the surface of the cone (as it is just the rolled up version of a flat surface) but there is a singular point at the tip of the cone. This stops the surface being a manifold because right at that point the neighbourhood doesn't look like a flat plane.

These kind of structures come up in orbifold theory, and are very useful for understanding symmetries.

Answer 7

In my answer I am assuming that you are familiar with calculus (derivatives, integrals, limits).

1. First of all, a "discrete differential manifold" defined in a linked paper is not a (smooth/differentiable) manifold in the traditional sense.

2. The linked answer skips most of the details. It also deals with smooth manifolds, not topological manifolds. The point of having a smooth manifold $M$ is that (assuming paracompactness!) one can equip the manifold with a Riemannian metric, which is a way to measure length $||v||$ of vectors $v$ in the tangent spaces of $M$.

A smooth structure on $M$ also allows you to define smooth curves $c: [a,b]\to M$ in $M$ and derivatives $c'(t), t\in [a,b]$, where $c'(t)$ is the "velocity vector" of $c$ at the point $c(t)$, i.e. $c'(t)$ belongs to the tangent space $T_{c(t)}M$. Once you have these ingredients, you can use calculus: Define the length of a (piecewise) smooth curve $c(t)$ in $M$ as the integral $$ \ell(c)=\int_a^b ||c'(t)||dt. $$ Assuming that $M$ is connected (any two points $x, y\in M$ can be connected by a piecewise smooth curve) one defines $d(x,y)$ as the infimum of length of curves connecting $x$ to $y$. (If you want to avoid technical details, just think of minimum rather than infimum, but, in general, length-minimizing curves do not exist.)

Now, the triangle inequality with this definition is quite clear: Given three points $x_1, x_2, x_3\in M$ and (almost) length-minimizing curves $c_1: [a,q]\to M, c_2: [q, b]\to M$ connecting $x_1$ to $x_2$ and $x_2$ to $x_3$ respectively, you get: $$ \ell(c_1)+ \ell(c_2)= \int_a^q ||c_1'(t)||dt + \int_q^c ||c_2'(t)||dt. $$ Given this formula, you define a new curve $c$ from $x_1$ to $x_3$ by first following $c_1$ and then following $c_2$. By the properties of integration: $$ \ell(c)= \int_a^b ||c'(t)||dt= \int_a^q ||c_1'(t)||dt + \int_q^c ||c_2'(t)||dt= \ell(c_1)+\ell(c_2). $$ Now, if $c_1, c_2$ were actually length-minimizers, you get that $$ d(x_1,x_3)\le \ell(c)= \ell(c_1)+\ell(c_2)= d(x_1,x_2)+ d(x_2,x_3). $$

This is your triangle inequality. One needs a tiny bit more work in the case when length-minimizers do not exist.

However, how do you know that $d(x,y)\ne 0$ for $x\ne y$? The trouble comes from the fact that there might not be a length-minimizing curve connecting $x$ to $y$.

A proof is not super-hard, but harder than you think.

The actual difficulty is not here, one needs to figure out how the topology of $M$ relates to the metric $d$ defined above. For a manifold to admit a metric, one needs to find a metric defining the same topology as the original topology of the manifold, otherwise, this metric is utterly useless. For a non-mathematician, a topology is a way to make sense of the notion of limits of functions and of sequences. (I.e., given a topology on $M$, we "know" which sequences converge and to which limits in $M$.) This definition is not general enough, but it suffices for the intuition.

One then needs to show that a sequence $p_i\in M$ converges to some $p\in M$ in the topology of $M$ if and only if the sequence of real numbers $d(p_i,p)$ converges to zero. Again the hard part is the one where you assume $$ \lim_{i\to\infty} d(p_i,p)= 0. $$

3. Finally and just for the record (since you did not ask): Every (paracompact) topological manifold $M$ (without any smooth structure) admits a metric. (The linked answer ignores this issue.) Moreover, if one assumes that $M$ is connected then one can find a metric $d$ such that any two points $p, q$ are connected by a length-minimizing rectifiable curve whose length is the distance $d(p,q)$. The proof of the latter is much harder than the "calculus" arguments above.