Easy way to identify manifolds with different co-ordinates?

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Summary:
Can we use the Riemann curvature tensor to identify different manifolds?
Hi. I have a question.
Two manifolds can be equivalent but have different co-ordinates and a correspondingly different metric defined on them.
Is there an easy way to identify when two such manifolds are equivalent but just have different co-ordinates?

What do I mean by "equivalent"? The manifolds are diffeomorphic so that the metrics are just the usual tensor transformations of each other due to a change in co-ordinates and the manifolds themselves are smoothly and bijectively mapped to/from each other by the co-ordinate transformation. However, I'll take whatever type of identification and sense of equivalence may be possible.

It doesn't seem to be as simple as just using the Riemann curvature tensor to identify different spaces but this is the kind of thing I was hoping for.

Here's an example:
The Riemann curvature vanishes
<--> There exists co-ordinates such that the metric has constant components.
This is great and it's almost enough to identify and categorise the space but not quite enough (for example the metric could have any signature).
 

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  • #2
Dale
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The usual way is to find the transformation equations for going from one set of coordinates to the other.
 
  • #3
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Thanks (but obviously that doesn't sound like an easy or consistent method).

If we can't get a necessary and sufficient condition from the Riemann curvature tensor, can we get something like half of this? A sufficient OR necessary condition?

If the Riemann curvature tensor of one space has identical components to that of another space EXCEPT for one component. Is that sufficient to show the spaces can NOT be equivalent?

(Genuinely, thanks for your time and there is no need to spend more time here).
 
  • #4
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It seems like one could construct a simple counter example. Take a flat Minkowski space. It has zero curvature. Now impose periodic boundary conditions, make a donut space. These are topologically distinct. So curvature alone is insufficient.
 
  • #5
Ibix
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If the Riemann curvature tensor of one space has identical components to that of another space EXCEPT for one component. Is that sufficient to show the spaces can NOT be equivalent?
The problem is that if you take a manifold and define some coordinates ##x^\mu## on it then you can calculate the Riemann tensor in terms of those coordinates and perhaps even something like the Kretschmann scalar, ##K=R_{abcd}R^{abcd}##, and you have ##K(x^\mu)##. Then you take some other coordinates ##y^\nu## and follow the same process and you get ##K(y^\nu)##. Unless the scalar happens to have a fairly simple dependency on the coordinates it doesn't help you because you can't recognise ##K(x^\mu)## as a transform of ##K(x^\nu)##.

For example, you can define ##(t,x,y,z)## coordinates on Schwarzschild spacetime (they aren't Cartesian, but perhaps Cartesian inspired). The Kretschmann scalar is $$K\propto\frac 1{z^6+3y^2z^4+3x^2z^4+3y^4z^2+6x^2y^2z^2+3x^4z^2+y^6+3x^2y^4+3x^4y^2+x^6}$$If you've got sharp eyes you might recognise that as $$K\propto\frac 1{(x^2+y^2+z^2)^3}=\frac 1{r^6}$$and hence that this might be Schwarzschild spacetime, but that depends on exactly what @Dale said - sorting out an algebraic mess and guessing a coordinate transform.

Characterising a spacetime will certainly help you. Are there event horizons? Singularities? CTCs? Symmetries? And looking at the Riemann and its scalars might well help with that. But I don't think there's any foolproof mechanical way of comparing two manifolds, no.
 
  • #6
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Thanks to Dale, Ibix and Paul.
This is much as I thought.
 
  • #7
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What do I mean by "equivalent"? The manifolds are diffeomorphic so that the metrics are just the usual tensor transformations of each other due to a change in co-ordinates and the manifolds themselves are smoothly and bijectively mapped to/from each other by the co-ordinate transformation. However, I'll take whatever type of identification and sense of equivalence may be possible.
So by equivalent you mean isometric? Riemann curvature vanishing is equivalent to the manifold being isometric to a flat space IIRC
 
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  • #8
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Hi Killtech and thank you for your time.
I think I already have all the answers I needed (they just aren't the answers I wanted to hear :frown: but that is the way it goes sometimes). I'll update the thread here using your comments to build around but it's likely to be the last post unless someone feels I have badly misunderstood what they were saying or (being optimistic ) someone turns up and does know of a technique.

So by equivalent you mean isometric?
Yes and no. I want an isomorphism of some kind. An isometry is one of the strongest forms of isomorphism and is not required. The original post used the phrase "diffeomorphism": I would consider two manifolds with metric tensor to be "equivalent" if there is a (sensible) co-ordinate transformation between the two.

Riemann curvature vanishing is equivalent to the manifold being isometric to a flat space IIRC
Yes. There are some additional requirements but we can assume we have them. Technically, ..[this statement].. should be restricted to a region of the manifold that is simply connected... - page 124, Spacetime and Geometry, Sean Carroll.

If we knew the Riemann Curvature tensor of two manifolds both vanished then this goes a long way toward identifying the two manifolds as being equivalent. There are some fine details, such as the metric signature remaining arbitrary but it is exactly the sort of thing I was looking for. The purpose of my original post was more about the hope that there would be some similar identification possible when the curvature tensors of two manifolds are compared but they aren't both zero everywhere. Or to phrase this another way: Can we generalise the result to identify when manifolds are equivalent based on comparison of their Riemann Curvature tensors?

Conclusion -
this is what I've understood about the situation:
The replies received so far indicate that no one is aware of an established practical procedure or fool-proof method. It would be more conventional to try and seek a co-ordinate transformation between the two manifolds directly. There are obviosuly a few cases when we can be certain the two manifolds are not equivalent: For example, if one has a vanishing Riemann curvature tensor (so it is "flat") but the other does not. However, the general concensus seems to be that the Riemann Curvature tensor is not likely to be the primary quantity or tool that we can use to establish the equivalence of two manifolds.
 
  • #9
PeterDonis
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I want an isomorphism of some kind. An isometry is one of the strongest forms of isomorphism and is not required.
If you want anything weaker than isometry, then you are contradicting yourself:

The original post used the phrase "diffeomorphism": I would consider two manifolds with metric tensor to be "equivalent" if there is a (sensible) co-ordinate transformation between the two.
And that requires them to be isometric, since the only way to even define a coordinate transformation between two manifolds with metric is to use the invariant geometric properties of the manifold with metric to define which points are "the same" in the two manifolds, which is required to define a coordinate transformation (since any such transformation depends on identifying which 4-tuples of coordinates in the two coordinate systems refer to "the same" points in the two manifolds). In other words, the concept of coordinate transformation between two manifolds with metric is only meaningful if the two manifolds are isometric.
 
  • #10
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Hi Killtech and thank you for your time.
[...]
Yes and no. I want an isomorphism of some kind. An isometry is one of the strongest forms of isomorphism and is not required. The original post used the phrase "diffeomorphism": I would consider two manifolds with metric tensor to be "equivalent" if there is a (sensible) co-ordinate transformation between the two.
Hmm, if the least thing you have is diffeomorphism between two manifolds they are at least topologically equivalent. If you are looking for something stronger then that, yet weaker then isometry there are a few relation in between. For example conformality is a quite a lot weaker then isometry. This leaves at least some some quantities invariant. But if you want to explore relations like this then you should look deeper into differential geometry rather then physics because such type of classifications are of interest there - especially since you keep your question so general that nothing here refers to physics specifically.
 
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  • #11
PeterDonis
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if the least thing you have is diffeomorphism between two manifolds they are at least topologically equivalent
Yes, but that's not sufficient for a coordinate transformation to exist, since if the manifolds are only topologically equivalent, there is no way to identify "the same" points in the two manifolds.

If you are looking for something stronger then that, yet weaker then isometry there are a few relation in between.
Yes, but none of them are sufficient for a coordinate transformation to exist, for the same reason as above.
 
  • #12
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Hi and thanks @PeterDonis and @Killtech

I think, if we started with a diffeomorphism between manifolds and the requirement that the metric on one manifold is just the tensor transformation of the metric on the other (much as stated in the OP) then it does turn out that the diffeomorphism will actually be an isometry using these two metrics. I just didn't notice that an isometry was forced. I didn't require an isometry but I would get one anyway.
 
  • #13
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Yes, but that's not sufficient for a coordinate transformation to exist, since if the manifolds are only topologically equivalent, there is no way to identify "the same" points in the two manifolds.


Yes, but none of them are sufficient for a coordinate transformation to exist, for the same reason as above.
In diff geo terminology every smooth manifold has coordinate charts by definition. But a smooth manifold all by itself doesn't have a metric. So on that level you cannot tell if coordinates are curvilinear or not yet - so you cannot formally name them cartesian or spherical because they look quite the same at this point. In fact such a coordinate chart ist nothing else then just a diffeomorphisms to ##R^n##. So you can use the word of transformation in the sense of chaining up diffeomorphisms

I am not sure what he means by coordinate transformation precisely and do not intend to do the guesswork for him so i leave it to him to figure that out.
 
  • #14
PeterDonis
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I think, if we started with a diffeomorphism between manifolds and the requirement that the metric on one manifold is just the tensor transformation of the metric on the other (much as stated in the OP) then it does turn out that the diffeomorphism will actually be an isometry using these two metrics.
Yes, that is correct.
 
  • #15
PeterDonis
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In diff geo terminology every smooth manifold has coordinate charts by definition.
Sure, but that doesn't mean you can define a coordinate transformation between charts on two different manifolds. The conditions for that are much stronger.

I am not sure what he means by coordinate transformation precisely
Then please do not clutter this thread with irrelevant statements that then need to be addressed.
 
  • #16
PeterDonis
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for example the metric could have any signature
If you're talking about relativity, which you are in this forum, then you are talking about manifolds with the Lorentz signature. If you want to talk about differential geometry more generally, including manifolds of any signature, please post in the math forum, not this one.
 
  • #17
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If you're able to access a copy of "General Relativity: An introduction to the theory of the gravitational field" by Hans Stephani (2nd edition!), he has a chapter on invariant characterizations of exact solutions. Eventually you learn about the Weyl tensor, and Petrov classifications. This may be what you're looking for.
 
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