Why do manifolds require a Riemannian metric?

[Alesak]

When reading other threads, following question crept into my mind:

When given a manifold, why shouldn't I give it distance function by giving it a simple metric function, that is MxM→ℝ with the usual axioms? I could happily measure distances in coordinate-independent way for ever after...

Why do I need to use the horrid construction of symmetric positive-definite covariant tensor field of second degree?

Also, what stops me from defining Riemannian metric on topological manifold? In definition of metric space we need no smoothness either.

 

[Bacle2]

I think one of the issues is that you want a metric that gives rise to the topology in the manifold, and this would not happen with just any metric.

Re the Riemannian metric , the argument I know defines a Riemannian metric on the manifold by pulling back the inner-product from R^n.

We then need the inverse of the pullback to be a diffeomorphism to preserve the positive-definiteness of the inner-product (we then use

second countability+ paracompactness to define partitions of unity, which patch all the local inner-product pullbacks together into

a single, globally-defined one).

 

[micromass]

When reading other threads, following question crept into my mind:

When given a manifold, why shouldn't I give it distance function by giving it a simple metric function, that is MxM→ℝ with the usual axioms? I could happily measure distances in coordinate-independent way for ever after...

Why do I need to use the horrid construction of symmetric positive-definite covariant tensor field of second degree?

Also, what stops me from defining Riemannian metric on topological manifold? In definition of metric space we need no smoothness either.

Note that what you saying is possible. All (reasonable) topological manifolds are indeed metrizable. Very example, a compact manifold M is always embeddable in \mathbb{R}^n. This metric will agree with the topology.

There are two reasons why this metric is uninteresting:
1) Consider the circle S^1. This can be embedded in \mathbb{R}^2. So we can restrict the metric in \mathbb{R}^2 to S^1. However, what will be the distance between (0,1) and (0,-1)?? Well, the distance between these points is the usual distance in \mathbb{R}^2 and it will be 2. But this is not good: this distance is measured by drawing a straight line through the points and to get the length. But if you were to live on S^1, then you actually need to stay on the circle to get from (0,1) and (0,-1). You cannot go in a straight line, you will need to go in a curved line. In that case, the distance you really want is \pi. So you will need to have an actual distance function that actuall "follows the paths on the manifold". This is significantly harder to do and will require the Riemannian metric.

2) The metric is not intrinsic. The circle S^1 can be embedded in \mathbb{R}^2 the usual way, but it is also diffeomorphic to 2S^1 (just double everything). It is clear that the distance function in the first case will be different from the distance function in the second case.
This is no problem with S^1 since it has a "canonical" embedding. But what about less canonical examples such as the torus or the projective space?

 

[arkajad]

You may like to check: Smooth quasimetric spaces at

http://en.wikipedia.org/wiki/Finsler_manifold

This looks similar to what you are looking for. Riemannian geometry is just a particular case. Pseudo-Riemannian is somewhat more complicated.

 

[Bacle2]

Arkajad, all:

Do you agree with my explanation? I'd like to think it is "spiritually correct", but

I'm not 100% . Basically, I think a Riemannian manifold can be made into a length

space :

http://en.wikipedia.org/wiki/Length-metric_space

(see 3rd example)

But, as stated in the same article, the intrinsic metric topology does not necessarily

agree with the original metric, so there seems to be something I am missing.

Any Ideas?

 

[arkajad]

I think that what is missing in the Wikipedia article is the assumption of smootheness.

It also says: "Every Riemannian manifold can be turned into a path metric space by defining the distance of two points as the infimum of the lengths of continuously differentiable curves connecting the two points. (The Riemannian structure allows one to define the length of such curves.) Analogously, other manifolds in which a length is defined included Finsler manifolds and sub-Riemannian manifolds."

Of course you are free to look for all kinds of generalizations of Riemann or Finsler geometry. In physics, as fars as I know, only Riemann and Finsler geometries (with their indefinite generalizations) have been considered till now. Of course there are theories based on connections instead on metrics - but that is another story.

 

[lavinia]

I think one of the issues is that you want a metric that gives rise to the topology in the manifold, and this would not happen with just any metric.

Re the Riemannian metric , the argument I know defines a Riemannian metric on the manifold by pulling back the inner-product from R^n.

We then need the inverse of the pullback to be a diffeomorphism to preserve the positive-definiteness of the inner-product (we then use

second countability+ paracompactness to define partitions of unity, which patch all the local inner-product pullbacks together into

a single, globally-defined one).

- Any metric on a manifold, by definition, gives the same topology. If not, it would not be a metric on the manifold.

- A Riemannian metric requires a smoothly varying inner product on the tangent spaces. This does not have to come from from the pull pack of the standard inner product on Euclidean space.

For 2 dimensional Riemannian manifolds the best you can do is find a coordinate system where the inner product is conformal to the standard inner product on Euclidean space. These coordinates are called isothermal coordinates and were discovered by Gauss.

For instance, isothermal coordinates on the Riemann sphere for the metric of constant Gauss curvature are

ds = |dz|/(1 + |z|^2)
For higher dimensional manifolds I do not believe that there are isothermal coordinates in general although I am not sure, maybe for complex manifolds.

- Not all topological manifolds can be given a smoothness structure. For them, there are no Riemannian metrics. However any metric on the manifold would give it the same topology.

- The theorem that any smooth manifold has a Riemannian metric, as you say, pieces together local Riemannian metrics with partitions of unity. However, these local metrics do not have to be pull backs to the standard metric on Euclidean space.

 

[arkajad]

Perhaps at this point it is worthwhile to recall that in exceptional cases topological space can be endowed with several inequivalent smooth manifold structures:

http://www.math.harvard.edu/~tiozzo/public_html/exotic.pdf

 

[Bacle2]

lavinia:
I don't understand what you mean when you say that any metric on a manifold gives rise to the same topology:

R^n with the discrete metric does not have the same topology as R^n with the Euclidean topology,for one. In

one case you get an n-manifold, and the other you get a 0-dimensional manifold. Of course, if you start with

a given choice of topology, and this topology is metric--as manifolds are metrizable--your choice of metric is

restricted; this is tautological. But even then, there are actually infinitely-many choices of metrics, since

all you need is that the different metrics induce the same topology. And there are infinitely-many metrics that

induce the same topology. Morover, a manifold (4-manifold that I know off , e.g.) can have uncountably-many

incompatible structures; each structure is associated to a unique topology, which I believe is not even comparable

to the topology of other structures.

 

[Bacle2]

The difficult technical question --at least to me-- is : once you're given the topology and you know that it is metrizable, how do you construct the/an actual metric function
d(x,y) that gives rise to the topology? I think the Urysohn metrization theorem may be a recipe.

 

[lavinia]

lavinia:
I don't understand what you mean when you say that any metric on a manifold gives rise to the same topology:

R^n with the discrete metric does not have the same topology as R^n with the Euclidean topology,for one. In

one case you get an n-manifold, and the other you get a 0-dimensional manifold. Of course, if you start with

a given choice of topology, and this topology is metric--as manifolds are metrizable--your choice of metric is

restricted; this is tautological. But even then, there are actually infinitely-many choices of metrics, since

all you need is that the different metrics induce the same topology. And there are infinitely-many metrics that

induce the same topology. Morover, a manifold (4-manifold that I know off , e.g.) can have uncountably-many

incompatible structures; each structure is associated to a unique topology, which I believe is not even comparable

to the topology of other structures.

R^n with the discrete metric is a different space. It is no longer R^N but just a pile of points,

 

[homeomorphic]

When given a manifold, why shouldn't I give it distance function by giving it a simple metric function, that is MxM→ℝ with the usual axioms? I could happily measure distances in coordinate-independent way for ever after...

You can try to do it that way. I think there are geometers who like that stuff. CAT0 spaces, length spaces. I'm not an expert on it, so I'm not sure how far you can really go with it.

Why do I need to use the horrid construction of symmetric positive-definite covariant tensor field of second degree?

Seems like you are making it sound uglier than it is. It's just a dot-product, basically. Nothing that bad, really.

Also, what stops me from defining Riemannian metric on topological manifold? In definition of metric space we need no smoothness either.

A topological manifold doesn't even have a tangent space at each point in an obvious way, so the idea of Riemannian metric doesn't even apply. But, yes, you could make it a metric space.

The difficult technical question --at least to me-- is : once you're given the topology and you know that it is metrizable, how do you construct the/an actual metric function
d(x,y) that gives rise to the topology? I think the Urysohn metrization theorem may be a recipe.

It is difficult, but it's covered in Munkres, for example, in chapter 6. Urysohn metrization isn't the most powerful metrization theorem.

 

[Bacle2]

R^n with the discrete metric is a different space. It is no longer R^N but just a pile of points,

Sorry, don't mean to be argumentative --after all, all I want is a pepsi, just a pepsi :) -- but R^N with the discrete is still the same R^N, collection-of-points-wise. I guess to really dot the t's, cross the i;s, one should say: given a topological space (X,T,S) that can be made into a manifold, i.e., having differential structure S , there is only one metric d:XxX-->R (up to equivalence, e.g., the metric in R^2 with basis given by open squares, and the metric with basis the open balls) that gives rise to the topology T . No doubt on that. But maybe you'll say potatoe, or tomatoe. I guess it's hard to use shorthand when peoplehave different perspectives/assumptions.

Homeomorphic: yes, I have seen the lemma, I have also seen Nagata-Smirnov (not in as much detail as I would have liked), but I have had trouble reconstructing the original metric

 

[homeomorphic]

Homeomorphic: yes, I have seen the lemma, I have also seen Nagata-Smirnov (not in as much detail as I would have liked), but I have had trouble reconstructing the original metric

Yes, I'm not sure how practical a construction those theorems give if you wanted to produce the actual metric. They just tell you that there is one. I can't imagine that using them to construct actual metrics would be a very productive way of producing a metric.

 

[lavinia]

Sorry, don't mean to be argumentative --after all, all I want is a pepsi, just a pepsi :) -- but R^N with the discrete is still the same R^N, collection-of-points-wise. I guess to really dot the t's, cross the i;s, one should say: given a topological space (X,T,S) that can be made into a manifold, i.e., having differential structure S , there is only one metric d:XxX-->R (up to equivalence, e.g., the metric in R^2 with basis given by open squares, and the metric with basis the open balls) that gives rise to the topology T . No doubt on that. But maybe you'll say potatoe, or tomatoe. I guess it's hard to use shorthand when peoplehave different perspectives/assumptions.

Homeomorphic: yes, I have seen the lemma, I have also seen Nagata-Smirnov (not in as much detail as I would have liked), but I have had trouble reconstructing the original metric

yes you are right. A collection of points is the same no matter what the metric. I was just saying that any metric on the Manifold must give the same topology. If it is not a manifold then you are right. I think I started a silly semantic argument. Apologies. Pizza would be nice also.

 

[julian]

When reading other threads, following question crept into my mind:

When given a manifold, why shouldn't I give it distance function by giving it a simple metric function, that is MxM→ℝ with the usual axioms? I could happily measure distances in coordinate-independent way for ever after...

Why do I need to use the horrid construction of symmetric positive-definite covariant tensor field of second degree?

Also, what stops me from defining Riemannian metric on topological manifold? In definition of metric space we need no smoothness either.

From a pysicists perspective with reference to general relativity...start with Einstein'shole argument - say you have two reference systems, say labelled by x and y, the field equations should take the same form in both systems, that is you have exactly the same diff equation to solve in both reference systems. So once you find a solution in the x-coordinates, simply write down the same function but replace x by y. This is also a solution but as it belongs to a different coordinate system but takes the same functional form it induces a different spacetime geometry! Now say the two reference systems coincided at first then were allowed to differ after say t = 0 - they would have the same initial conditions but would impose different spacetime geometries! They must then be gauge related...the fact that they assume all the same values,they assume them at different point suggests that that gauge transformation is simultaneously dragging the fields around - a diffeomorthism (not a coordinate transformation as the metrics within this equivalence class impose different geometries! Nearly 90 years on since the birth of GR and most pysicists still think diffeormphism invariance is a coordinate transformation!)...Anyway, in GR a solution (metric) is only unique up to spacetime diffeomorphism. In Hawking and Ellis on page 227, where they consider the Cauchy problem, they talk about introducing a fixed 'background' metric and use it to impose four gauge conditions to obtain a unique solution for the mtric components. Also in the chapter on differential geometry they talk about manifolds admitting positive defininative metrics and Lorentz metrics.

 

[julian]

Possibly relavent is in Hawking and Ellis, just a physics nerd, they say "In fact, any non-compact mamifold admits a line element field, while a compact manifold does so if and only if its Euler number is zero (e.g. the torus does, but the shpere does not, admit a line elemnet)"...

 

[Alesak]

Thanks everyone a lot, I need to go through this thread once again I understand this more.

Possibly relavent is in Hawking and Ellis, just a physics nerd, they say "In fact, any non-compact mamifold admits a line element field, while a compact manifold does so if and only if its Euler number is zero (e.g. the torus does, but the shpere does not, admit a line elemnet)"...

Is Hawking and Ellis good book? I've been eyeing it for some time, since I want to learn GR. After I learn some stuff about Lie groups and Riemannian manifolds, I plan to read Dirac and Wald.

Also, do you about two volume text by Penrose? It seems rather good too.