# Why is there no natural metric on manifolds?

One of the things that always bothered me after learning introductory differential geometry (as a physics student) and then delving deeper into this field on my own is that, the usual construction of differential manifolds are such that $M$ is an $n$-dimensional manifold if it is 1) a Hausdorff-space, 2) it is locally homeomorphic to $\mathbb{R}^{n}$, 3) the transistion functions between charts are invertable and continuous/differentiable/analytic etc. depending on manifold type.

Now as far as I understand, norm, metric, etc. are topological properties and homeomorphisms are maps that preserve topological structure. Yet, metric is an extra structure on a manifold, not part of a manifold's given properties. But if manifolds are locally homeomorphic to a set that has a very strong topological structure, including a metric ($\mathbb{R}^{n}$) then why is that structure not preserved locally?

Secondly, which is probably related to this, if the transition functions are differentiable, as it is demanded for differential manifolds, does that mean that the coordinate maps themselves are also differentiable? Because then they are basically diffeomorphisms from $M$ to $\mathbb{R}^{n}$. And if it is so, then given the natural metric tensor $g$ on $\mathbb{R}^{n}$, why can't that metric be induced on the manifold via the pullback?

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Different charts will induce different metrics on the same manifold. –  Dustan Levenstein Aug 25 '14 at 14:05
It should be mentioned that (for an appropriate meaning of the words, which is a rather subtle point) in dimension 2 and 3, (most) manifolds do have certain metrics which are in a sense natural. This is classic for surfaces and Thurston's geometrization program (whose proof was completed by Perelman) for dimension 3. –  Mariano Suárez-Alvarez Aug 25 '14 at 16:33

First, I think you have some misconceptions about what it means to be a "topological property." A topological property is a property that is preserved by homeomorphisms. Norms and metrics are definitely not topological properties. For example, the unit ball in $\mathbb R^n$ is homeomorphic (in fact, diffeomorphic) to $\mathbb R^n$ itself, but the two spaces have very different metric properties. (One is bounded and the other is not, for example.) You can certainly use a metric to induce a topology; but many different metrics will induce the same topology.

Second, if $M$ is a smooth manifold and $\phi$ is a coordinate map from an open set $U\subseteq M$ to an open set $\widehat U\subseteq\mathbb R^n$, then yes, $\phi$ is a diffeomorphism from $U$ to $\widehat U$.

Third, given a coordinate map $\phi$ as above, you can certainly use the pullback by $\phi$ to induce a metric on $U$. But note that this metric will only be defined on the subset $U$, typically not on all of $M$, and different coordinate charts will induce different metrics. That's why a metric is an extra piece of data that has to be chosen; it's not determined intrinsically by the smooth manifold structure.

EDIT: Your comment suggests that you're thinking of "Riemannian" as a property that a smooth manifold might or might not have. It's not; instead, it's an additional layer of structure that we can add to a smooth manifold if we wish.

In fact, there are four different layers of structure here. Let's take $\mathbb S^2 = \{(x,y,z)\in\mathbb R^3: x^2 + y^2 + z^2 = 1\}$ as an example. It has the following four structures:

1. It's a set: By definition, $\mathbb S^2$ is just a certain set of ordered triples of real numbers.
2. We can make it into a topological space: By specifying a topology on $\mathbb S^2$ (which is just a certain collection of subsets satisfying certain conditions), we make it into a topological space. Usually, when talking about a subset of a Euclidean space, we give it the subspace topology: that is, we declare a subset of $\mathbb S^2$ to be open iff it is the intersection of $\mathbb S^2$ with an open subset of $\mathbb R^3$. Note that this topology was originally defined in terms of a metric (the Euclidean distance function), but we're not going to use that metric for anything other than deciding which sets are open. Once we've decided on a topology, we can note that this topology happens to have the properties that make it a topological manifold: it's Hausdorff, second-countable, and locally Euclidean.
3. Then we can make it into a smooth manifold: Note that "smoothness" is not a topological property. If it were, we'd certainly want to say that $\mathbb S^2$ is smooth; but $\mathbb S^2$ is homeomorphic to the surface of a cube, which we would probably not want to say is smooth. Instead, a "smooth structure" is an additional piece of structure that we have to put on a manifold: it's a choice of a particular collection of coordinate charts that overlap smoothly. Fortunately, because of the way $\mathbb S^2$ sits in $\mathbb R^3$, there's a natural choice of smooth structure on it, and that's generally the one we use (often without mentioning that there's a choice).
4. Then we can make it into a Riemannian manifold: "Riemannian" is not a property that a smooth manifold might or might not have. Instead, a Riemannian metric is a choice of inner product on each tangent space, in such a way that it varies smoothly from point to point. For $\mathbb S^2$, a very natural choice is to declare the inner product on each tangent space to be the restriction of the Euclidean dot product. But we don't have to choose this one. For example, let $M$ be a smooth hot-dog-shaped surface in $\mathbb R^3$, endowed with the Riemannian metric obtained by restricting the Euclidean dot product. There's a diffeomorphism $F\colon \mathbb S^2\to M$, and we could also choose to give $\mathbb S^2$ the metric obtained by pulling back $M$'s metric via $F$. How you define the metric depends on what your purposes are.

One reason it's easy to get confused about these layers of structure is that when there are obvious "natural" choices such as the ones I described above, we often don't even mention that a choice is being made. For example, if an author writes "Let $\mathbb S^2$ be the unit sphere in $\mathbb R^3$," you have to decide from the context whether she's thinking of it as a set, or a topological space with the subspace topology, or a smooth manifold with the induced smooth structure, or as a Riemannian manifold with the induced Riemannian structure.

You can find lots more detail about these things in Chapter 1 of my book Introduction to Smooth Manifolds.

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Thank you. I see what you all say but I am not sure of the implications. Allow me to indulge. Let us take something like $S^{3}$. I assume that this manifold is well defined and is Riemannian. Does this mean that $S^{3}$ is defined through an embedding into $\mathbb{R}^{4}$ and the metric is pulled from that embedding, and THEN I remove the manifold from the context of embedding while keeping said metric as "an extra structure"? –  Uldreth Aug 25 '14 at 15:05
Yes: Let's restrict to $S^2$, note that if you embedded this as something more pear-shaped , or like a peanut with two large bulges connected by a slimmer bridge, then the pullback metrics would be very different (for example, geodesics would have completely different structure: On the bridge of the peanut there could be a closed geodesic such that for any two points on it, the shortest path between them has to lie on that circle, whereas for any geodesic on standard $S^2$ there are antipodes having multiple shortest paths, not all on the geodesic) –  JHance Aug 25 '14 at 15:25
Yes, I see that the embedding map defines the metric through the pull back, my question, how shall I say, is about how riemannian manifolds "get" their metrics. In my example, I WANT $S^3$ to be a 3-dim analogue to a sphere surface so in this case I DEFINE my manifold via the embedding ($S^3=\{x:||x-O||=a}$, where $x\in\mathbb{R}^{4}$), pull the metric back that way and THEN I say I want to study my manifold intrinsically, so I "ignore" the embedding but I keep the metric I got this way. Am I correct in that? –  Uldreth Aug 25 '14 at 15:32
@Uldreth: See my additions above. –  Jack Lee Aug 25 '14 at 16:03
Yes, it's still $\mathbb S^3$ without its metric. In fact, technically speaking, $\mathbb S^3$ is only a set. The topological space is an ordered pair $(\mathbb S^3,\mathscr T)$, where $\mathscr T$ is a topology; the smooth manifold is an ordered triple $(\mathbb S^3,\mathscr T,\mathscr A)$, where $\mathscr A$ is a smooth structure; and the Riemannian manifold is an ordered quadruple $(\mathbb S^3,\mathscr T,\mathscr A,g)$, where $g$ is a Riemannian metric. To keep the notation from getting too cumbersome, we usually omit all those additional structures when they're understood from context. –  Jack Lee Aug 25 '14 at 17:40

"Why is there no natural metric on manifolds?"

Think of three physical variables, like $p$, $V$, $T$, dependent on each other via some natural law $$W(p, V, T)=0\ ,\tag{1}$$ where $W$ is an explicit expression in three variables (say, van der Waals' law) chosen by a professor of theoretical physics. The equation $(1)$ defines a two-dimensional manifold $\Omega\subset{\mathbb R}^3$, maybe with singularities. The theory of differentiable manifolds then can describe how the physical variables $p$, $T$, $V$, and their mutual partial derivatives are dependent on each other, etc., and it can even describe the nature of $\Omega$'s singularities in terms of algebraic properties of the expression $W$.

But any physicist would laugh at the idea of a "natural metric" on this $(p,V,T)$-manifold $\Omega$. Note that the actual values of the variables $p,\ V,\ T$ are heavily dependent on the chosen units, and it is by no means God-given whether $1^\circ$ difference in temperature is of the same order of magnitude as a volume increase of $1\>{\rm cm}^3$. But any metric on $\Omega$ at once creates such an identification of scales, which is utterly arbitrary.

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This is a great physics answer. –  Sean D Aug 25 '14 at 21:48
One might argue that it's already wrong to call the domain of $W$ simply $\mathbb{R}^3$. I would rather say that it's some product space $\mathbb{R}_p\times\mathbb{R}_V\times\mathbb{R}_T$, where each component is linearly isomorphic to $\mathbb{R}$, but not the same space. –  leftaroundabout Jan 30 at 10:51

Metric is not a topological structure. You can only say a metric defines a metric topology, or a topological manifold is metrizable (the metrization theorems).

1) 2) 3) do not ensure a differentiable manifold structure, you also need the second countability axiom.

You will benefit from local Euclidean property the following:

It says nothing about the metric being unique. In fact, a common way to construct a global Riemannian metric from metrics on local coordinate charts is to use the partition of unity, whose existence is ensured by the manifold being Hausdorff and paracompact. It seems to me the non-uniqueness of Riemannian metric arises as a result of non-uniqueness of POU.

Yes, coordinate mappings are differentiable by definition. However, the diffeomorphism is obviously not global, otherwise you have every $n$-dimension manifold is homeomorphic (topologically equivalent) to $\mathbb R^n$ (topologists will starve).

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One can also put different Riemannian metrics on a coordinate chart. There's obviously a canonical one, but it's not the only one, so non-uniqueness of POU isn't necessarily the only cause. –  JHance Aug 25 '14 at 14:53
@JHance You are right. Thanks for the comment. –  Troy Woo Aug 25 '14 at 14:55

Yes, in general, there is no "god-given" (canonical, natural,...) Riemannian metric on a smooth manifold. However, sometimes, there is one! Much of the research in the modern differential geometry is about finding such canonical metrics. For instance, this is how Poincare conjecture was solved in dimension 3 by Perelman. See my answer here for details and references.

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