# How to compute the metric without an embedding

1. Oct 16, 2009

### gorkiana

Hello

I know that the metric is an intrinsic property of a manifold, hence it should be possible to compute it without any use of an embedding in a higher dimensional space.

That is, one can easily compute the metric on a surface of a 2-sphere just by computing the inner products of the basis tangent vectors as they appear in R^3, with the trivial Euclidean inner product.

My question is, can one compute the metric without recurring to the embedding?

-artur palha

2. Oct 16, 2009

### Hurkyl

Staff Emeritus
I'm going to be rather literal for a moment: this is not true. Manifolds don't come equipped with metrics. They don't even have the infrastructure needed to define tangent vectors!

A manifold with a differentiable structure is a differentiable manifold.
A differentiable manifold with metric tensor is called a Riemannian manifold.
Only when you have a Riemannian manifold can you say it has a metric tensor as an intrinsic property.

A Riemannian manifold is essentially, by definition, a differentiable manifold together with a choice of a specific metric tensor. There are lots of ways to specify the metric tensor -- one of which is an embedding as you described. Another way is to specify its components relative to a choice of basis. There are surely others.

3. Oct 16, 2009

### wofsy

Metrics can be described without embeddings. I suggest you look up the Poincare disc as a simple example. But Hurky is right - metrics are not intrinsic on manifolds. A given differentiable manifold has infinitely many different metrics.

4. Oct 16, 2009

### zhentil

I agree with the guys before me, but I'll add that you might be putting the cart before the horse. One defines the metric, and then computing it depends on how one defines it. In the special case you described (where one has a fixed embedding), computing the metric is identical to what you've described.

Think of it this way: a metric is a way of assigning angles and distances that follows certain rules. If I follow those rules, I can define it however I want. Say I want to put a metric on the plane where vectors in the x-direction are 10 times "longer" (in the traditional sense) than those in the y-direction. What you're asking is a way to compute lengths and angles in my made-up metric without me telling you what it is.

To get an idea of how these things are somewhat elastic, think of this example: a bug is walking on a circular plate (maybe the unit disk in the plane), with the plate room temperature at the center and temperature inversely proportional to the square of the distance to the edge. Say he wants to get from the point (1/2, 0) to the point (0, 1/2) and minimize the amount of scorching his feet take. What would his path look like? It certainly wouldn't be a straight line (in our sense of the word).

5. Oct 17, 2009

### firearrow

Ok so everyone has pointed out that the manifold doesn't come with a unique metric.

Just for fun though, you could take an n-dimensional smooth manifold, which can be embedded in R^2n by the Whitney Embedding Theorem, and write the formula for the metric that you get from the construction used in the proof of the theorem. Then once you have a formula, try and remove any reference to the higher space.

I guess you would have something rather canonical. Perhaps there is an intrinsic geometric flow that would take any metric to this one.

6. Oct 17, 2009

### Office_Shredder

Staff Emeritus
This assumes you have some 'canonical' embedding to begin with. Different embeddings will give different metrics

7. Oct 17, 2009

### zhentil

To quote another theorem (this one by Nash): given any metric, there is an isometric embedding into Euclidean space. That's about as anti-canonical as it gets, I fear.

8. Oct 17, 2009

### firearrow

Yes good point. I suppose the class of metrics obtained from a Whitney-type imbedding followed by inducing the metric has been studied but I don't know any results. It would be interesting to know.

9. Oct 17, 2009

### wofsy

Whitney embeddings are not unique and also give infinitely many metrics. As a simple proof just follow one embedding by any diffeomorphism of R^2n into itself. If this diffeomorphism is not an isometry in a neighborhood of the embedded manifold then you get a new metric.

However the restriction to R^2n does restrict the possible metrics. For example any surface embedded in R^3 must have a point of positive Gauss curvature ( the proof is easy). On the other hand there is an embedding of the torus in R^4 where the Gauss curvature is everywhere zero ( also easy to prove).

Generally speaking one may need a high embedding dimension to realize a metric.

10. Oct 17, 2009

### firearrow

Thanks wofsy.
Do you happen to know if the class of metrics that are induced from some Whitney embedding are more useful/ more nicely behaved than the full set of metrics? Is it a preserved set under any of the standard flows?
Just idly wondering really, don't mean to pester...

11. Oct 17, 2009

### wofsy

I don't know the answer. The only systematic application of the Whitney embedding theorem that I know is Thom's cobordism theory for smooth manifolds. This theory is topological and metric independent.

Here is an example of the kind of thing embedding dimension can tell you. Suppose that the manifold can be embedded as a hypersurface in R^n+1. Then it must be orientable. Also it's Euler characteristic must be even. So in particular there are no orientable 2 dimensional surfaces of odd Euler charateristic since they can all be embedded in R^3. Also the manifold must be a boundary of an n+1 manifold.

12. Oct 18, 2009

### gorkiana

First of all thank you for all the replies

A metric is intrinsically connected to a specific basis for used for the tangent spaces. In this way one can have as many metrics as basis for the tangent spaces.

But all this metrics are connected to each other by a pull back no?

I say this because, if i pick two vectors, a and b, of the tangent space of a Riemannian manifold M at a point, P, and compute their inner product it will always be the same, regardless of the metric I use. If I change the metric I am also changing the basis vectors hence the representation (components) of a and b will be different.

So, metric is not intrinsic, to the geometry in the sense that it can have different appearances, but it encodes the same geometrical properties of the manifold M, right?

Again, my question is: besides embedding the manifold M in a higher dimensional space, what other ways are there of computing a metric?

13. Oct 18, 2009

### wofsy

What you are saying here is not right.

The inner product of two vectors definitely depends upon the metric. Changing the metric does not change any bases at all. It changes the inner product of the vectors.

14. Oct 19, 2009

### gorkiana

I do not agree with what you say, at least in what you say as a reply to what I said.

If I have a certain Riemannian manifold I can always find an embedding for it in a higher dimensional space, by Nash's theorem, right? This embedding is just a "visualization" of the Riemannian manifold, the manifold in itself is not changed. It's geometric properties are unchanged. So any angles or lenghts on the manifold do not change with the embedding, they are intrinsic to the manifold.

With doing this I can easily compute the metric. Hence, I can compute any inner product of two vectors. And the norm of a vector. If I change the embedding, the parameterization, of the manifold, I get a different metric. But I also get a different set of basis vectors. Again if I take two pairs of vectors their inner product is the same, regardless of the metric used. As long as the Riemannian manifold is the same. If I change the basis vectors, the representation of any vector changes, the same happens to the metric.

This is clearly stated in Riemannian geometry and geometric analysis, by Jost. See section 1.4 - Riemannian metrics. Equation 1.4.3 give the pullback of the metric. Also you can see in 1.4.2 that the inner product does not change with the transformed metric. This metric transforms as do the basis vectors. Which makes sense, since the metric is simply the collection of all inner products between the basis vectors used in the tangent space.

15. Oct 19, 2009

### zhentil

If you have a fixed Riemannian manifold and embed it isometrically, the metric doesn't change (this is correct). If you change the metric, it's a different Riemannian manifold. So what Wofsy is saying is completely correct: the tangent bundle is independent of any metric structure. You can put a metric on it and assign lengths and angles, but that is in no way intrinsic to the underlying smooth manifold.

16. Oct 19, 2009

### zhentil

Unless you're saying that the metric is intrinsic to the Riemannian manifold, but that's by definition.

17. Oct 19, 2009

### gorkiana

I agree with you, the metric is specific to the Riemannian manifold, as you said, it is so by definition. The change of metric, and this was an abuse of language from me, I meant to be the appearance of the metric in different coordinate systems. So, as you say this change is not considered to be a metric change, right?

I don't understand what you mean by: "the tangent bundle is independent of any metric structure"

What you say is that the tangent bundle is simply a collection of linear vector spaces, not linear vector spaces equipped with an inner product. So, as was said before, you can take the manifold and generate infinitely many metrics, one of the ways being an embedding in a higher dimensional Euclidean space.

One can also take a Riemannian manifold an isometrically embedded it on an Euclidean space, then the metric does not change.

Thanks.

18. Oct 19, 2009

### wofsy

If the metric is unchanged the Riemannian manifold is the same - up to an isometry. If you change coordinates, the representations of the vectors will change but the vectors will not. The vectors are the same. They are just written in different coordinates.