# Riemannian manifolds

1. Mar 10, 2006

### Ratzinger

A manifold is a topological space which locally looks like R^n. Calculus on a manifold is assured by the existence of smooth coordinate system.

A manifold may carry a further structure if it is endowed with a metric tensor.

Why further structure?

If have sphere or a cylinder I can parameterize it with different coordinates ( two patches for sphere), which means different metrics. But does not have any manifold a canonical metric from the start? What is meant with endowing a manifold with a metric?

If I got a cylinder, its metric is euclidean by definition, no need for defining a extra structure on it.

2. Mar 10, 2006

### robphy

Consider a plane.
When doing Euclidean geometry on it, we use a Euclidean metric tensor on it.
When doing Special Relativity (using a Minkowski spacetime diagram), we use a Minkowski metric tensor on it.

As you say, a manifold is a topological space...
But how does one, e.g., measure the separation between points? One needs a rule. A metric provides a rule.

3. Mar 10, 2006

### HallsofIvy

Staff Emeritus
A manifold looks locally like Rn. With a metric tensor, you have global properties as well.

4. Mar 10, 2006

### Hurkyl

Staff Emeritus
No!

Suppose I do nothing but hand you a topological space M which is a one-dimensional compact manifold. I also hand you two points on M. How do you plan on telling me the distance between those two points?

Anothing thing to realize is that the cylinder of radius 1 about the z-axis in Euclidean 3-space is (essentially) the same manifold as the cylinder of radius 17 about the y-axis, and the annulus of radii 3 and 6 about the origin in the Euclidean plane.

Coordinates have nothing to do with metrics. Trying to use the Euclidean metric on the coordinates is a recipe for disaster because, as you observe, all the answers change if you use a different coordinate patch!

But a metric space has a function (the metric) that tells you the distance between two points. No matter how you put coordinates on your manifold, the metric tells you the distance between the two points. (And that distance doesn't have to have anything to do with the Euclidean distance on the coordinates)

5. Mar 10, 2006

### George Jones

Staff Emeritus
There are some nice examples in relativity of spacetimes that share the same underlying manifold, but that are completely different spacetimes because they have different "metrics".

Consider the mandifold M = R^4 for Minkowki spacetime. Removing a point results in the manifold M' = R x S^3. M' is flat spacetime when it inherits its metric from Minkowski spacetime, but M' is the curved spacetime of the Friedmann-Robertson-Walker cosmological spacetime if a different metric is used.

Remove a line from M to get M'' = S^2 x R^2. Again, M'' is flat, or it is the Schwarzschild spacetime of a black hole, depending on the metric used.

Regards,
George

Last edited: Mar 10, 2006
6. Mar 10, 2006

### Ratzinger

Thanks everybody. Metric is tensor function, so it's coordinate-independent.

But here I need more clarification:
If have a coordinate system that parameterized a cylinder, then I still can not measure distances because I don't know the scale of the coordinates?

7. Mar 10, 2006

Staff Emeritus
If you have two points on a manifold, since it's Hausdorff, they can be found in separate neighborhoods that each look like $$R^n$$. But how can you, say, define a derivative, when these neighborhoods and the homeomorphisms to $$R^n$$ are different? To do this you need some way to connect them, and this need is met by the Connection, which is defined on the tangent bundle (the set of copies of $$R^n$$, considered as a vector space, at the different points of the manifold). And so on. See the differential geometry thread.

8. Mar 10, 2006

### Doodle Bob

This question is unanswerable since you don't specify the coordinates. The problem is, in general, there is never any one canonical atlas of coordinates. In fact, I believe it is possible to construct on any manifold more than one incompatible coordinate system.

there is never any one canonical metric on any given manifold, unless you specify more conditions such as curvature conditions. Plus, coordinates and metrics are generally independent of each other, unless you specify that the metric needs to be diffeomorphic, which means that you've settled on a particular atlas and corresponding coordinate system.

That being the case, more often than not, when looking at a particular manifold, the context in which you are viewing it sometimes lends itself to an "obvious" coordinate system and metric. For example, any embedded surface in R^3 has an "obvious" distance measure assoicated to it. But, it's far from canonical.

There are many ways to embed a cylinder in R^3 to generate a completely different metric. For example, coil it up like an infinite thick slinky without any self-intersections. What's cool about this case is that there are infinitely many points with positive Gaussian curvature and infinitely many points with negative Gaussian curvature. This is quite different than the cylinder you described which has everywhere zero Gaussian curvature.

9. Mar 10, 2006

### Ratzinger

I think I got it now.

Even if I have a manifold with coordinates carved on it, it is still up to the defined metric tensor what distance I will measure, because the components of the given metric tensor in that coordinates could be diag(1,1,1,1), diag(1,-1,-1,-1), diag(1/r^2,1,1,1) or something else.

So yes indeed there is an extra structure!

10. Mar 11, 2006

### Doodle Bob

actually, that 2nd one won't give you a distance function, the matrix representing a riemannian metric w/resp. to a coord. system needs to be symmetric and positive definite. Hence, it may not even be a diagonal matrix.

11. Mar 11, 2006

### George Jones

Staff Emeritus
Unfortunately, this is another example of miscommunication caused by the different terminolgies used by mathematicians and physcists. When physicists use the term metric for relativity, they replace positive definiteness by the requirements that the "metric" be non-degenerate, and that the eigenvalues of the (components of the) metric (in any chart) have signs {+, -, -, -}.

For example, {1, -1, -1, -1} is the "metric" for flat Minkowski spacetime, and {1 - 2M/r, -(1 - 2M/r)^(-1), -r^2, -sin^2(theta)r^2} is the "metric" for the spacetime inside and outside a spherically symmetric black hole.

From the "metrics" for relativistic spacetimes, physicists extract a lot of useful information, including times and distances.

I have seen the term semi-Riemannian geometry used for this, but I don't know if this is standard mathematical terminology.

Regards,
George

12. Mar 11, 2006

### Doodle Bob

I understand completely what you're saying here, but I was uncharacteristically careful with my phrasing here. Indeed the second matrix in that list would give a semi-riemannian metric on the tangent space, but not a riemannian metric. Yes, "semi-Riemannian" is the standard term.

Furthermore, I believe that semi-riemannian, non-riemannian metrics will not in general define a coherent distance between any two given points, i.e. satisfying the triangle inequality, etc.: at least if you're going to define the distance as some sort of infimum of all lengths of curves connecting the two given points. If we look at a space-like submanifold, say, we're good to go.

13. Mar 11, 2006

### Hurkyl

Staff Emeritus
Is "pseudoriemannian" synonymous with "semi-riemannian"? (e.g. as ___ metric, or ___ manifold)

14. Mar 11, 2006

### George Jones

Staff Emeritus
Yes, I agree with all this.

Sorry, my eye passed over "function" without really registering it, but I'm not entirely sure what Ratzinger meant by "distance". It might still be possible that the two of you are talking about different things.

In any case, this gives this shows in a quite striking manner that a metric is extra structure. The same underlying differentiable manifold (e.g. R^4) can be either a Riemannian manifold or a semi-Riemannian manifold, depending the metric that is used.

Regards,
George

15. Mar 12, 2006

### Ratzinger

I located two sources of confusion ( at least for mine slow mind).

1. I always thought that the only reason to define vectors on infinitesimal neighbourhoods of points of a manifold is because of curvature. But it’s in the first place because of the fact that manifolds need have no distance relation between points. That’s the initial need for tangent spaces, long before any metric gets involved!

2. I am also beginning to understand that the parameterization of a manifold can happen in plenty of ways. As long as each point is assigned a coordinate and the assignment happens continuous, then any coordinate function goes ( so that’s why mathematicians disrespecting coordinates). Defining a metric then, makes a sphere or a crayon or whatever else out of a quite amorphous manifold. Also the scary word diffeomorphism makes sense now.

And since the thread called Riemannian manifold let me quote this. (Be careful, it could bring tears in your eyes.)

“it’s reading on June 10, 1854 was one of the highlights in the history of mathematics: young, timid, Riemann lecturing to the aged legendary Gauss, who would not live past the next spring, on consequences of ideas the old man must have recognized as his own and which he had long secretly cultivated. W. Weber recounts how perplexed Gauss was, and how with unusual emotion he praised Riemann’s profundity on their way home.”

Last edited: Mar 12, 2006
16. Mar 12, 2006

### Doodle Bob

I believe so.

17. Mar 12, 2006

### Doodle Bob

Well, I'm not quite sure what an infinitesimal neighborhood is, but it is true that there is a lot one can do with a manifold with just the tangent bundle defined, see for example any book on differential topology.

Yes!