# Is zero curvature space flat?

Jianbing_Shao
TL;DR Summary
zero curvature
In relativity, a flat space is always regraded to be endowed with an invariant metric field ##g_{\mu\nu}(x)= \eta_{\mu\nu}##, So in a flat space the corresponding connection ##\Gamma_{\mu\nu}^\rho(x)=0## It means that if we parallel transport a vector ##v^\nu(x_0)## in the space. Then it obviously will obey rule ##\partial_\mu v^\nu(x)=0 ##. So if we parallel transport a vector ##v^\nu(x_0)## along an arbitrary loop ##C##, then we will get an invariant vector field ## v^\nu(x)= v^\nu(x_0)##.

From the definition of the curvature of a space, we know that if the curvature ##R_{\mu\nu\rho}^\lambda(x)=0##, then when we parallel transport a vector ##v^\nu(x_0)## from point ##x_0## along an arbitrary loop ##C## in the space, then when it returns back to point ##x_0##. We will find that ##\delta v^\nu(x_0)=0##. Here when vector moves along the loop, there is no need to restrict the vector must be invariant when it moves. In fact, if there exists a vector field ## v^\mu(x) =G^\mu_\nu(x)v^\nu(x_0), G^\mu_\nu(x_0)=I##, then it obviously fulfill the relation:
$$\partial_\mu v^\nu(x)= \partial_\mu G^\nu_\alpha(x) (G^{-1})^\alpha_{\rho } v^\rho(x)$$.
So if we define ##\Gamma_{\mu\rho}^\nu(x)= -\partial_\mu G^\nu_\alpha(x) (G^{-1})^\alpha_{\rho }(x) ##. So the vector field confined on the loop ## v^\mu(x) ## can be regarded to be generated by parallel transporting a vector ##v^\mu(x_0)## along loop ##C## in the space with a connection ##\Gamma_{\mu\rho}^\nu(x)##. Directly from the definition of curvature we can conclude that the curvature ##R^\rho_{\mu\nu\lambda}(x)=0##. And after a simple calculation we can find the conclusion is right. So we can draw such a conclusion the zero curvature space is not flat, but a flat space is obviously a zero curvature space. And the geodesics in the two different kinds of space will be different. How do we distinguish between the two in general relativity.

Staff Emeritus
Homework Helper
Gold Member
Summary:: zero curvature

In relativity, a flat space is always regraded to be endowed with an invariant metric field gμν(x)=ημνgμν(x)=ημνg_{\mu\nu}(x)= \eta_{\mu\nu},
This is not correct. While you can find coordinates where ##g_{\mu\nu} = \eta_{\mu\nu}##, this will not be true in all coordinate systems. Consider polar coordinates on regular 2D Euclidean space, where the non-zero components of the metric are ##g_{rr} = 1## and ##g_{\phi\phi} = r^2##.

• bhobba and dextercioby
Jianbing_Shao
This is not correct. While you can find coordinates where ##g_{\mu\nu} = \eta_{\mu\nu}##, this will not be true in all coordinate systems. Consider polar coordinates on regular 2D Euclidean space, where the non-zero components of the metric are ##g_{rr} = 1## and ##g_{\phi\phi} = r^2##.
I am sorry, I really can not understand what do you mean. Do you mean that the definition of a flat space is not ##g_{\mu\nu}(x)= \eta_{\mu\nu}##? then can you give a more precise one?
Logically I think the definition of flatness of a space can use metric and when ##g_{\mu\nu}(x)= \eta_{\mu\nu}## the space can be regarded as flat. but you say it only valid in some coordinate system, Do you mean that the metric is not an effective one to describe the flatness of a space? then I wonder which way you will use to define the flatness of a space.
Further more, it is not wrong to say that if the metric ##g_{\mu\nu}(x)= \eta_{\mu\nu}##, then the space is flat. and you talk about other coordinate system. and perhaps what do you mean is some kinds of coordinate transformation do not change the flatness of a space. but the metric ## \eta_{\mu\nu}## will change under such coordinate transformation. and usually we can get to an arbitrary metric field from ## \eta_{\mu\nu}## through coordinate transformation. then it means only some kinds of coordinate transformation will keep the flatness of a flat space, others will not. then what kind of coordinate transformation will keep the flatness of a flat space?

Jianbing_Shao
This is not correct. While you can find coordinates where ##g_{\mu\nu} = \eta_{\mu\nu}##, this will not be true in all coordinate systems. Consider polar coordinates on regular 2D Euclidean space, where the non-zero components of the metric are ##g_{rr} = 1## and ##g_{\phi\phi} = r^2##.
Or perhaps you can directly tell me that zero curvature space is flat?

Staff Emeritus
Homework Helper
Gold Member
then can you give a more precise one?
”A space is flat if the Riemann curvature tensor is zero.”

Do you mean that the metric is not an effective one to describe the flatness of a space?
Of course it is, the curvature can be computed from the metric. (Assuming you even have a metric and that you have chosen the Levi-Civita connection - but that is assumed in relativity.)

You just cannot assume that the metric takes the form ##\eta## in all coordinate systems because the components of any tensor are generally going to change under coordinate transformations. However, if your space is flat it is possible to find such coordinates - it just might not be the case in your original coordinates.

but the metric ημνημν \eta_{\mu\nu} will change under such coordinate transformation. and usually we can get to an arbitrary metric field from ημνημν \eta_{\mu\nu} through coordinate transformation.
No, you cannot get an arbitrary functional form of the metric. However, you can get a form that is not ##\eta##.

then it means only some kinds of coordinate transformation will keep the flatness of a flat space, others will not. then what kind of coordinate transformation will keep the flatness of a flat space?
No. The curvature tensor of a flat space is a tensor. If it is zero in one coordinate system it will be zero in all coordinate systems and therefore flatness is a property of the space itself, not of particular coordinates.

Or perhaps you can directly tell me that zero curvature space is flat?
This is the definition of a flat space.

• PeroK
Homework Helper
Gold Member
2022 Award
Or perhaps you can directly tell me that zero curvature space is flat?

Flat spacetime can be described by any number of different coordinate systems: spherical, hyperbolic, general curvilinear systems etc. Note that all these systems of coordinates can be used to describe 3D Euclidean space, for example. Using an exotic coordinate system does not mean your underlying space or spacetime changes.

The simplest example is using polar coordinates in a Euclidean plane.

By contrast, if you define your space as the surface of a sphere, then that is curved. That space is different from a Euclidean plane. Regardless of what coordinate systems you use.

• dextercioby
Jianbing_Shao
No, you cannot get an arbitrary functional form of the metric. However, you can get a form that is not ηη\eta.
Then what kind of metric we can not get through coordinate transformation?

Homework Helper
Gold Member
2022 Award
Then what kind of metric we can not get through coordinate transformation?
One that represents a different curvature of spacetime.

Jianbing_Shao
Flat spacetime can be described by any number of different coordinate systems: spherical, hyperbolic, general curvilinear systems etc. Note that all these systems of coordinates can be used to describe 3D Euclidean space, for example. Using an exotic coordinate system does not mean your underlying space or spacetime changes.

The simplest example is using polar coordinates in a Euclidean plane.

By contrast, if you define your space as the surface of a sphere, then that is curved. That space is different from a Euclidean plane. Regardless of what coordinate systems you use.
I know to a space with a zero connection, then under an arbitrary coordinate transformation, the connection will be such a form: ##\Gamma_{\mu\rho}^\nu(x)= \partial_\mu J^\nu_\alpha(x) (J^{-1})^\alpha_{\rho }(x) ##, obviously the curvature of the connection is zero. so the coordinate transformation do not change the zero curvature of the space, then if the zero curvature space means the space is flat. Then we will draw such a conclusion: For an arbitrary metric field we can get through coordinate transformation from ##\eta_{\mu\nu}##. the corresponding Rieman curvature tensor is zero. then what kind of metric we can not get through coordinate transformation? and the reason we can not get it through coordinate transformation is just the corresponding Rieman curvature tensor is not zero.

Homework Helper
Gold Member
2022 Award
I know to a space with a zero connection, then under an arbitrary coordinate transformation, the connection will be such a form: ##\Gamma_{\mu\rho}^\nu(x)= \partial_\mu J^\nu_\alpha(x) (J^{-1})^\alpha_{\rho }(x) ##, obviously the curvature of the connection is zero. so the coordinate transformation do not change the zero curvature of the space, then if the zero curvature space means the space is flat. Then we will draw such a conclusion: For an arbitrary metric field we can get through coordinate transformation from ##\eta_{\mu\nu}##. the corresponding Rieman curvature tensor is zero. then what kind of metric we can not get through coordinate transformation? and the reason we can not get it through coordinate transformation is just the corresponding Rieman curvature tensor is zero.

You cannot change the geometry by a coordinate transformation. Coordinate transformations give you different ways to describe the same geometry: a flat 2D Euclidean plane; 4D flat spacetime; the black hole geometry.

In each of these examples, you can (and do) use different coordinate systems to study different problems. For the black hole geometry we have Schwarzschild, Eddington-Finkelstein, Kruskal-Szekeres etc. All of these coordinates are used to describe a simple black hole. You cannot do a coordinate transformation and get the geometry of a rotating black hole; or globally flat spacetime; or anything else.

That's the physical/geometric argument.

Spacetime is flat if and only if the Riemann curvature tensor is zero.

More generally, mathematically, certain curvature invariants do not change under coordinate transfromations. That's what can be used to classify spacetimes according to their curvature.

In practice, it's not easy to see whan two metrics are equivalent in terms of there being a coordinate transformation between them.

Jianbing_Shao
Spacetime is flat if and only if the Riemann curvature tensor is zero
In general relativity we always calculate curvature from metric, and obviously zero curvature will corresponding to a series of metrics, and coordinate trasformation can connect these metrics to each other and transform one metric to another.
Here lies a question. because metric and curvature are all tensors, so under coordinate transformation they will obey the same rule, simply say matrix multiplication, but because if curvature is 0, then arbitrary matrix multiply 0 is 0, so coordinate transformation will not change zero curvature, but metric is not 0, so under coordinate transformation we will get plenty of metrics.
For example, if we multiply an arbitrary real number times 0, we only can get 0,but if we multiply an arbitrary real number times 1, then we can get almost all the real number, so 0 is very different from other real number.
So I want to know what kind of metric can not transform to ##\eta_{\mu\nu}## under coordinate transformation.
In practice, it's not easy to see whan two metrics are equivalent in terms of there being a coordinate transformation between them.
Using the law the tensors obeys under coordinate transformation, we can easily transform a metric to another, and the two metrics obviously are equivalent

Homework Helper
Gold Member
2022 Award
In general relativity we always calculate curvature from metric, and obviously zero curvature will corresponding to a series of metrics, and coordinate trasformation can connect these metrics to each other and transform one metric to another.
Here lies a question. because metric and curvature are all tensors, so under coordinate transformation they will obey the same rule, simply say matrix multiplication, but because if curvature is 0, then arbitrary matrix multiply 0 is 0, so coordinate transformation will not change zero curvature, but metric is not 0, so under coordinate transformation we will get plenty of metrics.
For example, if we multiply an arbitrary real number times 0, we only can get 0,but if we multiply an arbitrary real number times 1, then we can get almost all the real number, so 0 is very different from other real number.
So I want to know what kind of metric can not transform to ##\eta_{\mu\nu}## under coordinate transformation.

You are over-thinking this. A metric can transform (globally) to ##\eta_{\mu\nu}## iff the Riemann tensor is globally zero.

That is the mathematical answer. That doesn't mean you can just look at a metric and see immediately that it represents flat spacetime. You must either calculate the Riemann tensor, or find the coordinate transformation.

Using the law the tensors obeys under coordinate transformation, we can easily transform a metric to another, and the two metrics obviously are equivalent

If you are simply given two metrics, it may not be easy to find the coordinate transformation between them. Take this metric for example:
$$ds^2 = -X^2dT^2 + dX^2$$
Is that flat spacetime? If so, find the coordinate transformation that transforms it to:
$$ds^2 = -dt^2 + dx^2$$

Staff Emeritus
Homework Helper
Gold Member
You are over-thinking this. A metric can transform (globally) to ##\eta_{\mu\nu}## iff the Riemann tensor is globally zero.
There are flat spaces that do not allow global coordinate systems where this is true.

• dextercioby
Mentor
There are flat spaces that do not allow global coordinate systems where this is true.

Staff Emeritus
Homework Helper
Gold Member
The circle ##\mathbb S^1##. It is flat (being one-dimensional) but does not allow a global coordinate system.

Mentor
It is flat (being one-dimensional)

Hm, ok, I agree this is a valid example but it's kind of disappointing since, as you note, it's automatically flat by being 1-D. Are there any examples in 2 or more dimensions?

Staff Emeritus
Homework Helper
Gold Member
Hm, ok, I agree this is a valid example but it's kind of disappointing since, as you note, it's automatically flat by being 1-D. Are there any examples in 2 or more dimensions?
The (flat) torus.

I also suspect the cylinder although it does allow global coordinates. The typical ##\phi##-##z## coordiates where the metric takes the ##\delta## form are not global and I suspect no such global coordinates exist.

Staff Emeritus
Homework Helper
Gold Member
Also, the cone with the apex removed should be an example. It has the same topology as the cylinder, but parallel transport around the apex does not return the same vector due to the global geometry. If there was a global coordinate system it would because it would just be ##\mathbb R^2## with a closed set removed.

Jianbing_Shao
You are over-thinking this. A metric can transform (globally) to ημνημν\eta_{\mu\nu} iff the Riemann tensor is globally zero.

That is the mathematical answer. That doesn't mean you can just look at a metric and see immediately that it represents flat spacetime. You must either calculate the Riemann tensor, or find the coordinate transformation.
I agree that if Riemann tensor is globally zero, then we can find a transformation which can transform a metric globally to ##\eta_{\mu\nu}##, actually as a symmetric nonsingular matrix, we always can find a transformation which can transform a metric globally to ##\eta_{\mu\nu}##.

I wonder the problem is if such a transformation can be induced from a coordinate transformation?

In fact if there exist a coordinate transformation ##x’(x)##, then each component ## {x’}^\mu(x)## is a scalar field. Then vector field ##\partial_\nu {x’}^\mu ## should be a gradient vector field, and we know the curl of ##\partial_\nu {x’}^\mu ## should be zero. But to an arbitrary vector field, There not always exists such a scalar field, so similarly to an arbitrary transformation ##J^\mu_\nu(x)##, it perhaps can not induced from a coordinate transformation then ##J^\mu_\nu(x)=\frac{\partial {x'}^\mu}{\partial x^\nu}##, and ##J^\mu_\nu(x)## can transform a globally orthogonal normalized basis into a new basis field, and this new basis field is not coordinate basis.

So I am not sure if you think that if the metrics generated from coordinate basis, then the corresponding curvature tensor is 0, and if the metrics generated from non-coordinate basis, then the corresponding curvature tensor is not 0?

To me the problem is because an arbitrary transformation ##J^\mu_\nu(x)## do not change the zero curvature of the space, and according it also change a metric of a flat space to another metric of a flat space? Then can we find a transformation can transform a metric with non-zero curvature to ##\eta_{\mu\nu}##, if not, why?

Mentor
I wonder the problem is if such a transformation can be induced from a coordinate transformation?

Such a transformation is a coordinate transformation. The components of the metric, considered as a matrix, depend on your choice of coordinates.

Jianbing_Shao
There are flat spaces that do not allow global coordinate systems where this is true.
Because Rieman curvatrue ##R^\rho_{\mu\nu\lambda}(x)## is a tensor, so if ##R^\rho_{\mu\nu\lambda}(x)=0##, an arbitrary transformation ##\frac{\partial {x'}^\mu}{\partial x^\nu}## can keep ##R'^\rho_{\mu\nu\lambda}(x)=0##, but the choose of ##\frac{\partial {x'}^\mu}{\partial x^\nu}## is arbitrary, It is not always can be induced from a coordinate transformation.

Mentor
change a metric of a flat space to another metric of a flat space?

There is only one "flat space", with one metric. The transformations you are talking about don't "change the metric". They only change the components of the metric, since, as I have just noted, those components depend on your choice of coordinates.

I think you are confusing the metric itself, which is a coordinate-independent geometric object, with its components in a particular coordinate chart.

• PeroK
Mentor
I am not sure if you think that if the metrics generated from coordinate basis, then the corresponding curvature tensor is 0, and if the metrics generated from non-coordinate basis, then the corresponding curvature tensor is not 0?

There is no such thing as "metrics generated from coordinate basis" vs. "metrics generated from a non-coordinate basis". The metric is a coordinate-independent geometric object. A choice of basis is separate from the metric itself and changing your basis doesn't change the metric.

Staff Emeritus
Homework Helper
Gold Member
Because Rieman curvatrue ##R^\rho_{\mu\nu\lambda}(x)## is a tensor, so if ##R^\rho_{\mu\nu\lambda}(x)=0##, an arbitrary transformation ##\frac{\partial {x'}^\mu}{\partial x^\nu}## can keep ##R'^\rho_{\mu\nu\lambda}(x)=0##, but the choose of ##\frac{\partial {x'}^\mu}{\partial x^\nu}## is arbitrary, It is not always can be induced from a coordinate transformation.
You have some serious misunderstandings about what a metric is. Please see Peter’s posts.

Jianbing_Shao
Such a transformation is a coordinate transformation. The components of the metric, considered as a matrix, depend on your choice of coordinates.
Of course a vector field or a tensor field is independent of the choice of coordinate transformation, but only the components changes under coordinate transformation, So when I say metric changes under coordinate transformation, I obviously indicate the change of the components of the tensor field .

So let's return to the intial problem, in a cartisen coordinate, the components of metric is ##\eta_{\mu\nu}## and the corresponding connection is 0, so to a vector ##v(x_0)## at point ##x_0##,when it parallel transport along a loop in the flat space, vector field ##v(x)## is invariant, and under coordinate transformation. the invariance will keep.

But if we only demand curvature is 0, then to a vector ##v(x_0)## at point ##x_0##,when it parallel transport along a loop in the zero curvature space, it only require that when the vector return to point ##x_0##, ##\delta v(x)=0##, when it move along the loop, it can changes,

So there are significant differences between the two situations, obviously the description is independent of coordinate transformation. So I think zero curvature is not equivallent to flat.

Staff Emeritus
Homework Helper
Gold Member
So there are significant differences between the two situations, obviously the description is independent of coordinate transformation. So I think zero curvature is not equivallent to flat.
No there is not. A flat space has curvature zero everywhere, not just at ##x_0##. You also cannot think of vector components being constant meaning that vectors are ”the same” for the same reason as tensor components.

2022 Award
But if we only demand curvature is 0, then to a vector ##v(x_0)## at point ##x_0##,when it parallel transport along a loop in the zero curvature space, it only require that when the vector return to point ##x_0##, ##\delta v(x)=0##, when it move along the loop, it can changes,
What do you mean by "it can change" as it goes around the loop? To what are you comparing it?

Jianbing_Shao
You have some serious misunderstandings about what a metric is. Please see Peter’s posts.
Perhaps I can not express my opnion clearly, what I stressed is just that: to a vector field ##v(x)##, and we can express ##v(x)## using basis ##e^\mu(x)##.then
$$v(x)=v_\mu(x)e^\mu(x)$$
also we can express ##v(x)## using basis ##e'^\mu(x)##, and
$$v(x)=v_\mu(x)e^\mu(x)=v'_\mu(x)e'^\mu(x)$$
then because all the basis ##e^\mu(x)## are coordinate basis, so sometimes when we express ##v(x)## from one basis to another, then the changes of components can not be equivalantly described as change under coordinate transformation

Staff Emeritus
Homework Helper
Gold Member
No there is not. A flat space has curvature zero everywhere, not just at ##x_0##. You also cannot think of vector components being constant meaning that vectors are ”the same” for the same reason as tensor components.
To expand slightly on that. A space (with trivial first homotopy group) being flat means that the parallel transport of a vector ##v## at ##x_0## to any other point ##x## does not depend on the path and the parallel transport of ##v## from ##x_0## to ##x## is unique. It is only in this respect that the vector field ##v## is "constant".

Gold Member
2022 Award
Hm, ok, I agree this is a valid example but it's kind of disappointing since, as you note, it's automatically flat by being 1-D. Are there any examples in 2 or more dimensions?
In 2D just extend @Orodruin 's example of the 1D circle to a 2D cylinder surface Staff Emeritus
Homework Helper
Gold Member
Perhaps I can not express my opnion clearly, what I stressed is just that: to a vector field ##v(x)##, and we can express ##v(x)## using basis ##e^\mu(x)##.then
$$v(x)=v_\mu(x)e^\mu(x)$$
also we can express ##v(x)## using basis ##e'^\mu(x)##, and
$$v(x)=v_\mu(x)e^\mu(x)=v'_\mu(x)e'^\mu(x)$$
then because all the basis ##e^\mu(x)## are coordinate basis, so sometimes when we express ##v(x)## from one basis to another, then the changes of components can not be equivalantly described as change under coordinate transformation
I am sorry, it is impossible to deduce what you are trying to say here. If anything, this post made your argument murkier, not clearer.

Staff Emeritus
Homework Helper
Gold Member
In 2D just extend @Orodruin 's example of the 1D circle to a 2D cylinder surface Actually, as I hinted to in a later post (#17), the cylinder is not as easy to get as the circle as the cylinder does admit a global coordinate system (being homeomorphic to ##\mathbb R^2 \setminus \{0\}##). The problem becomes showing that there is no way to arrange such a global coordinate system such that the metric becomes diagonal with the diagonal entries equal to one.

Edit: Just to be a little more specific. The global coordinate system on the cylinder ##x^2+y^2=r^2_0## as a submanifold of ##\mathbb R^3## using coordinates ##\xi## and ##\eta## on ##\mathbb R^2 \setminus \{0\}## can be constructed as
$$x = \frac{r_0 \xi}{\sqrt{\xi^2 + \eta^2}}, \quad y = \frac{r_0 \eta}{\sqrt{\xi^2 + \eta^2}}, \quad z = \frac{r_0}{2} \ln\left(\frac{\xi^2 + \eta^2}{r_0^2}\right).$$
Of course, in these coordinates, the induced metric from the embedding in ##\mathbb R^3## does not take the ##\delta## form, but it shows that a global coordinate system exists, making the argument a bit muddier than for the circle.

Last edited:
• vanhees71
Jianbing_Shao
What do you mean by "it can change" as it goes around the loop? To what are you comparing it?
I know where I was wrong, thanks ! everyone

Staff Emeritus