MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In most physics situations one gets the metric as a positive definite symmetric matrix in some chosen local coordinate system.

Now if on the same space one has two such metrics given as matrices then how does one check whether they are genuinely different metrics or just Riemannian Isometries of each other (and hence some coordinate change can take one to the other).

If in the same coordinate system the two matrices are different then is it proof enough that they are not isometries of each other? (doing this test over say a set of local coordinate patches which cover the manifold)

Asked otherwise, given two ``different" Riemannian Manifolds how does one prove the non-existence of a Riemannian isometry between them?

There have been two similar discussions on mathoverflow at this and this one. And this article was linked from the later.

In one of the above discussions Kruperberg had alluded to a test for local isometry by checking if the Riemann-Christoffel curvatures are the same locally. If the base manifold of the two riemannian manifolds is the same then one can choose a common coordinate system in which to express both the given metrics and then given softwares like mathtensor by Mathew Headrick this is probably not a very hard test to do.

So can one simply patch up such a test through out a manifold to check if two given riemannian manifolds are globally isometric?

How does this compare to checking if the metrics are the same or not in a set of common coordinate patches covering the manifold?

I somehow couldn't figure out whether my query above is already getting answered by the above two discussions.

share|cite|improve this question
It seems that you are just curious. In principle such problem can appear, but I did not ever see it to appear naturally. – Anton Petrunin Jan 27 '10 at 19:32
No. This situation is pretty common in physics when 2 different ways of calculation give very different looking metrics but actually they are just coordinate transforms of the other. This discussion seems to provide insights about why testing such equivalence in general can be a hard problem. Simplest example where this arises is when one sees that the Rindler metric just describes a certain wedge in the minkowski space although the 2 metrics a priori are not obvious to be the same – Anirbit Jan 28 '10 at 4:02
up vote 2 down vote accepted

This goes by the name of the "equivalence problem" in riemannian geometry and it is an important problem in the classification effort of solutions to gravity field equations. Malcolm MacCallum has many results in this area. For example this paper from 1985 might be a place to start reading about it.

share|cite|improve this answer
I started tracking SHEEP and ended up on this helpful wikipedia page: – Vít Tuček Jun 26 '13 at 20:58

Here's an irresponsible under-referenced response.

On the one hand, this is hard! Indeed, I've read somewhere --- not handy --- that this question lurks among quantum gravity's many difficulties. For instance, in general, given combinatorial descriptions of two topological manifolds, it's undecidable whether they are homeomorphic.

On the other hand, it's not too hard, in that given two metrics on contractible patches you can study how their curvature tensors act on orthonormal frames; and thus for a point in each you can ask if the curvatures are equivalent, which boils down to a not-too-large linear algebra problem (that much is essentially the answer to your first linked question) and if they are you can study whether such an equivalence extends to a neighborhood of those points, e.g. comparing the exponential maps (and the second linked question is a special case of this).

I suspect you could even do such a comparison in a Computable Analysis setting, assuming some kind of oracle access to the two metrics and their curvatures, to decide whether two metrics were Gromov-Hausdorf closer than $\varepsilon$, or Gromov-Hausdorf/Sobolev close as spaces-with-2nd-order-data --- but I digress.

share|cite|improve this answer

It all depends on what you mean by "the same space". If you have two possibly different Riemannian metrics on the same underlying space (i.e. smooth manifold) and you want to know if the two metrics are the same (i.e., the distance between any two points in space given by each metric agree), then indeed all you need to do is check to see whether the symmetric matrices you get from local co-ordinates agree at each point in space.

If on the other hand, you have a metric on one space and another metric on a second space and you want to know whether there is an isometry between the two spaces (note that the two spaces could actually be the same space but you're looking for an isometry that is not necessarily the identity map), then problem is considerably more difficult (unless one of the spaces is special, like flat or constant curvature) and I defer to the answer by José.

share|cite|improve this answer
Thanks for the clarification. Could also comment on whether patching up the test of local isometry that Kruperberg alluded to would result in a test of global isometry? – Anirbit Jan 27 '10 at 18:47
The only thing that could go wrong globally with Kuperberg's local isometry test is that the manifolds are (globally) topologically different. But then one manifold would have to be a covering of the other (like the natural map from R^2 to the flat torus). – Deane Yang Jan 27 '10 at 21:19

(I wanted to make a comment but it is bit too long)

The answer depends very much on what you are able to calculate. So try to understand first what is calculable in your case...

For example: Assume you can calculate all eigenvalues of Laplacian for both manifolds (assume they are compact) --- if the two obtained sequences are different then manifolds are also nonisometric. If they all the same you might have different manifolds but it is very unlikely (it means that usually, one can hear shape of drum).

On the other hand --- I do not think you are able to perform these calculations.

share|cite|improve this answer
Thanks for your comment. When you point out the inability to perform these calculations, you mean that it is the general difficulty of being able to analytically solve the complicated differential equation that finding eigen-values of laplacians involve? Or are you alluding to some other difficulty? – Anirbit Jan 29 '10 at 5:28
It is question for you, nt for me --- if you are able to perform any kind of computations then there is no problem... – Anton Petrunin Jan 29 '10 at 18:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.