# Two metric tensors describing same geometry

• I
Consider two coordinate systems on a sphere. The metric tensors of the two coordinate systems are given. Now how can I check that both coordinate systems describe the same geometry (in this case spherical geometry)?
(I used spherical geometry as an example. I would like to know the process in general.)

Orodruin
Staff Emeritus
Homework Helper
Gold Member
The metric tensors of the two coordinate systems are given.
You mean that the components of the metric tensors are given? The metric tensor itself is not coordinate system dependent.

In either case, the way of checking whether the metric tensors are the same is to make sure that they act in the same way on all combinations of vectors. If you want to stay with components, you have to make sure that the components satisfy the transformation rules for coordinate transformations on the sphere.

FactChecker
Consider two coordinate systems on a sphere. The metric tensors of the two coordinate systems are given. Now how can I check that both coordinate systems describe the same geometry (in this case spherical geometry)?
(I used spherical geometry as an example. I would like to know the process in general.)
Construct the curvature tensor from the first and second metric. Find the transformation from the first to second metric. Show that this transformation changes the first curvature tensor into the second.

Is this method correct?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
No. You have multiplied with ##g^{lm}## but you already have ##l## as a summation variable in the right-hand side of your expression. Then you are changing which is the ##l## that actually belongs to the sum and which is a free index.

Sorry for the mistake.
If you want to stay with components, you have to make sure that the components satisfy the transformation rules for coordinate transformations on the sphere.
Is it necessary to know the transformation matrix? The metric tensor defines the geometry. Isn't it sufficient just to know the components of the metric tensor?
Moreover, if I know the components of metric tensor in the two different coordinate systems, I can actually find the transformation matrix by solving this equation:
$$g_{ij}=\frac{\partial x'^k}{\partial x^i}\frac{\partial x'^l}{\partial x^j}g'_{kl}$$
##n^2## equations, ##n^2## unknowns, thus solvable.

pervect
Staff Emeritus
Consider two coordinate systems on a sphere. The metric tensors of the two coordinate systems are given. Now how can I check that both coordinate systems describe the same geometry (in this case spherical geometry)?
(I used spherical geometry as an example. I would like to know the process in general.)
Any method I can think of involves finding the transformation equations between the two coordinate systems. Though it seems like there should be a better way. In special cases there certainly may be, but I'm not sure if there is in general a better way or not.

Once you have those, there are several ways to proceed, but your idea of showing that said transformation equations do in fact transform the first metric into the second via the tensor transformation rules seems to be the most direct, if carried out correctly.

I'd start out by writing out the transformation equations. Primed/unprimed notation gets a bit confusing, I think, it would probably be better to use different symbols for the two coordinate systems.

So you might use ##x^1, x^2, x^3## and ##\chi^1, \chi^2, \chi^3## for the two sets of coordinates, then in the 3 dimensional case you'll have three functions for the transformation, for instance you might write ##\chi^1## as a function of x^1, x^2, x^3. Or you could write x^1 as a function of ##\chi^1, \chi^2, \chi^3##. Similarly for the other components.

We will get an overdetermined set of equations. Making use of the fact that the metric tensor is symmetric, in 2 dimensions we'll have 3 constraint equations in two unknowns, in 3 dimensions 6 constraints in 3 unknowns, in 4 dimensions 10 constraints in 4 unknowns.

We will get an overdetermined set of equations. Making use of the fact that the metric tensor is symmetric, in 2 dimensions we'll have 3 constraint equations in two unknowns, in 3 dimensions 6 constraints in 3 unknowns, in 4 dimensions 10 constraints in 4 unknowns.
Aren't there ##N^2## unknowns as the transformation matrix ##\frac{\partial \chi^i}{\partial x^j}## has ##N^2## components in ##N##-dimension?

pervect
Staff Emeritus
Aren't there ##N^2## unknowns as the transformation matrix ##\frac{\partial \chi^i}{\partial x^j}## has ##N^2## components in ##N##-dimension?
When you note that the metric tensor is symmetric, because ##g_{ij} = g_{ji}## there are fewer independent equations. For instance, ##g_{12}## is trivially equal to t ##g_{21}##. So there are in general N diagonal terms, which generate unique equations, and N^2 - N off-diagonal terms, which result in only half the number of unique equations, i.e. (N^2-N)/2 due to symmetry. So the total number of unique equations is ##N + (N^2-N)/2 = (N+N^2)/2##

When you note that the metric tensor is symmetric, because ##g_{ij} = g_{ji}## there are fewer independent equations. For instance, ##g_{12}## is trivially equal to t ##g_{21}##. So there are in general N diagonal terms, which generate unique equations, and N^2 - N off-diagonal terms, which result in only half the number of unique equations, i.e. (N^2-N)/2 due to symmetry. So the total number of unique equations is ##N + (N^2-N)/2 = (N+N^2)/2##
So, there are ##\frac{N+N^2}{2}## equations and ##N^2## unknowns (the ##N^2## components of ##\frac{\partial \chi^i}{\partial x^j}##). As ##N^2>\frac{N+N^2}{2}##, there will be infinite number of solutions.

martinbn
So, there are ##\frac{N+N^2}{2}## equations and ##N^2## unknowns (the ##N^2## components of ##\frac{\partial \chi^i}{\partial x^j}##). As ##N^2>\frac{N+N^2}{2}##, there will be infinite number of solutions.
I don't think it as simple as that. Here is one equation ##x^2+y^2=0## in two unknowns, which doesn't have an infinite number of solutions even though ##2>1##.

I don't think it as simple as that. Here is one equation ##x^2+y^2=0## in two unknowns, which doesn't have an infinite number of solutions even though ##2>1##.
Complex numbers are not allowed? Why?

martinbn
Complex numbers are not allowed? Why?
Because ##g_{ij}## are real.

Because ##g_{ij}## are real.
The components of transformation matrix can be complex number, can't they?
Here, I am looking for the transformation matrix between two coordinate systems while the components of metric tensor are given for the two coordinate systems.

PAllen
There are a whole family of invariants (scalars) that can be constructed from the curvature tensor and the metric. Certainly a necessary (not sufficient) condition for same geometry is that all of these are identical. This test does not involve solving for any coordinate transforms. These can be computed directly in each case and must come out the same however different the coordinate components are.

Having passed this test, you might then solve for a coordinate transform. This is very hard, in general, and should not be possible if the metrics describe different geometry. In practice, it could be exceedingly difficult to distinguish between solution difficulty and non-existence of solution.

[ edit: oops, this only works in trivial cases, where the invariant is constant. Otherwise you need the transform to know which coordinate values correspond, to compare the invariants. ]

Last edited:
If complex numbers are allowed, then there always exists a transformation matrix between two metric tensors. So I guess complex numbers are not allowed.
Could you explain me why?

PAllen
If complex numbers are allowed, then there always exists a transformation matrix between two metric tensors. So I guess complex numbers are not allowed.
Could you explain me why?
A coordinate transform is supposed to be a set of 4 (for D=4) real (differentiable) functions of 4 real variables. The transform matrix is defined to be a matrix of partial derivatives of these. The latter must obviously be real under these definitions.

Last edited:
PAllen