Curved spaces that locally aren't flat

[MathematicalPhysicist]

What sort of spaces are there that locally aren't necessarily flat?

Do they have applications in physics?

I agree this question seems naive, cause it seems we cannot say anything meaningful thing in such spaces, physically speaking.

But I sense that for QG you would need to generalize the notion of curved spacetime.

 

[jbriggs444]

What sort of spaces are there that locally aren't necessarily flat?

For instance, the space of the surface of a piece of paper rolled into a cone shape? Flat except for a point with undefined curvature at the tip?

 

[martinbn]

What do you mean by locally flat?

 

[MathematicalPhysicist]

What do you mean by locally flat?

Well the definition in Schutz's book which I am reading states that locally flat refers to the fact that in specific local coordinates which define the manifold, Riemann curvature tensor is zero.

 

[PAllen]

Well the definition in Schutz's book which I am reading states that locally flat refers to the fact that in specific local coordinates which define the manifold, Riemann curvature tensor is zero.

I think you must be misunderstanding. You can have coordinates that make the metric Minkowski at a point, and also for which the metric compatible connection vanishes. However, if the curvature tensor vanishes in any coordinates at a point, it vanishes for all coordinates at that point. This is never true in most GR manifolds, e.g. it is nowhere true in the Kruskal manifold, and is also nowhere true in FLRW manifolds except the degenerate Milne case, in which case it is true everywhere.

Locally flat normally means you can define a tangent space, and this implies what I said about the metric and connection. All of this is part of any definition of either a Riemannian or pseudo Riemannian manifolds.

 

[Dale]

For instance, the space of the surface of a piece of paper rolled into a cone shape? Flat except for a point with undefined curvature at the tip?

This is exactly the example I was going to give.

 

[George Jones]

You can have coordinates that make the metric Minkowski at a point, and also for which the metric compatible connection vanishes. However, if the curvature tensor vanishes in any coordinates at a point, it vanishes for all coordinates at that point. This is never true in most GR manifolds, e.g. it is nowhere true in the Kruskal manifold, and is also nowhere true in FLRW manifolds except the degenerate Milne case, in which case it is true everywhere.

Locally flat normally means you can define a tangent space, and this implies what I said about the metric and connection. All of this is part of any definition of either a Riemannian or pseudo Riemannian manifolds.

Schutz basically states all of this, so I agree,

I think you must be misunderstanding.

 

[PeterDonis]

the definition in Schutz's book which I am reading states that locally flat refers to the fact that in specific local coordinates which define the manifold, Riemann curvature tensor is zero.

It might help to give an exact page/section/paragraph reference, since at least two posters suspect you are misunderstanding something.

 

[MathematicalPhysicist]

Yes, I got confused between a flat manifold and locally flat manifold.

SO which cases are there besides the cone that aren't locally flat?

I think you must be misunderstanding. You can have coordinates that make the metric Minkowski at a point, and also for which the metric compatible connection vanishes. However, if the curvature tensor vanishes in any coordinates at a point, it vanishes for all coordinates at that point. This is never true in most GR manifolds, e.g. it is nowhere true in the Kruskal manifold, and is also nowhere true in FLRW manifolds except the degenerate Milne case, in which case it is true everywhere.

Locally flat normally means you can define a tangent space, and this implies what I said about the metric and connection. All of this is part of any definition of either a Riemannian or pseudo Riemannian manifolds.

 

[PeterDonis]

Flat except for a point with undefined curvature at the tip?

Strictly speaking, I think the "point" at the tip of the cone is not part of the manifold. As a manifold with metric, I think the cone is flat; it's a flat manifold with topology ##S^1 \times R## (or ##R^2## with a point removed).

which cases are there besides the cone that aren't locally flat?

In order for "locally flat" to have a meaning, I think there must be a metric on the manifold. But if there is a metric on the manifold, I think it has to be locally flat; I don't think a metric that isn't locally flat is even possible. Or, in more technical language, I think any manifold with metric has to be Riemannian or pseudo-Riemannian (depending on whether its metric is positive definite or not).

 

[MathematicalPhysicist]

Strictly speaking, I think the "point" at the tip of the cone is not part of the manifold. As a manifold with metric, I think the cone is flat; it's a flat manifold with topology ##S^1 \times R## (or ##R^2## with a point removed).
In order for "locally flat" to have a meaning, I think there must be a metric on the manifold. But if there is a metric on the manifold, I think it has to be locally flat; I don't think a metric that isn't locally flat is even possible. Or, in more technical language, I think any manifold with metric has to be Riemannian or pseudo-Riemannian (depending on whether its metric is positive definite or not).

So Riemannian or pseudo Riemannian manifolds are necessarily locally flat, and the metric makes it such.

SO only spaces without a metric defined on them can be candidates of spaces that aren't locally flat.
Do you have any examples of such spaces?

 

[PeterDonis]

SO only spaces without a metric defined on them can be candidates of spaces that aren't locally flat.

No. Without a metric "locally flat" makes no sense. You need a metric for "flat" or "locally flat" to even have a meaning.

Do you have any examples of such spaces?

There can't be any. See above.

 

[MathematicalPhysicist]

Thanks for clearing this matter to me @PeterDonis

 

[Ibix]

Do metric coefficients have to be smooth functions? If not, it seems like discontinuous functions would give you undefined derivatives at the discontinuity.

 

[stevendaryl]

I'm trying to understand what the question might mean. All manifolds are "locally flat" in the sense that curvature tends to become less important as the size of the region goes to zero. Within a city of radius 10 kilometers, the maps of the city are not going to indicate that the Earth is curved. But it's not that the curvature tensor is any smaller, it's just less important in getting around using maps.

The other interpretation of "locally flat" might mean manifolds where the curvature tensor is actually zero at every point. A cylinder is a 2-dimensional example. Just looking locally, there is no indication that it is curved. The "curvature" is only reflected in the connectivity---the fact that traveling far enough in any direction will get you back to where you started. That's intuitively "curved", but the mathematical definition of "curvature" doesn't cover it, by definition. Mathematically, when people talk about curvature, they usually mean Riemann curvature, which is zero for a cylinder, so technically, it isn't curved in the mathematical sense.

I think that the intuitive notion of a space being flat would mean: Riemannian curvature is zero AND the space is simply connected (no way to go "around" the universe ).

 

[martinbn]

I looked at Schutz, he uses locally flat in the sense that coordinates can be chosen so that at a point the metric is Mikowski and the connection coefficients are zero at the point. Of course that is true for all manifolds (pseudo-Riemannian) so the original question has the answer: none, no such manifolds.

 

[Dale]

So Riemannian or pseudo Riemannian manifolds are necessarily locally flat,

Yes, so the whole cone, including the point, is an example of a topological space that is not a Riemannian manifold precisely because it is not locally flat at the point. By removing the point you get a manifold.

 

[Dale]

No. Without a metric "locally flat" makes no sense. You need a metric for "flat" or "locally flat" to even have a meaning.

Are you sure about this? I am pretty sure that you can define a connection without defining a metric, and I thought that you could define curvature in terms of the connection.

Edit: the proofs I was thinking of all seem to start with a Riemannian manifold and measure the area enclosed by a path or the radius. So I could be wrong

 

[martinbn]

Are you sure about this? I am pretty sure that you can define a connection without defining a metric, and I thought that you could define curvature in terms of the connection.

Edit: the proofs I was thinking of all seem to start with a Riemannian manifold and measure the area enclosed by a path or the radius. So I could be wrong

You are right, you don't need a metric to have curvature, a connection is enough. But "local flatness", surprisingly to me, doesn't refer to curvature, but to a property of the metric.

 

[Dale]

You are right, you don't need a metric to have curvature, a connection is enough. But "local flatness", surprisingly to me, doesn't refer to curvature, but to a property of the metric.

Thanks, that helps!

 

[romsofia]

A more intuitive approach to this question is to consider what GR needs to recover. If we had a curved space that wasn't locally flat, could we still recover SR and eventually Newton?

If the answer is no, then GR "fails". By having this requirement of spaces being locally flat, we will always recover the above in some region (which is needed).

 

[PeterDonis]

Do metric coefficients have to be smooth functions?

I believe that is part of the definition of a manifold with metric.

 

[MathematicalPhysicist]

A more intuitive approach to this question is to consider what GR needs to recover. If we had a curved space that wasn't locally flat, could we still recover SR and eventually Newton?

If the answer is no, then GR "fails". By having this requirement of spaces being locally flat, we will always recover the above in some region (which is needed).

I was more into, are there physical phenomena in reality that cannot be described by local flat spaces?

I know already that GR is actually SR locally, but what happens when locally the space isn't flat, and cannot be described by the metric?
Are there any such instances?

Perhaps one need to learn more of modern geometry and see perhaps there's something there.

 

[George Jones]

I know already that GR is actually SR locally, but what happens when locally the space isn't flat, and cannot be described by the metric?
Are there any such instances?

Normal classical spacetimes are modeled by a (pseudo-)Riemannian differentiable manifolds that have smooth metrics. It is, however, sometimes convenient to join two (or more) such spacetimes together. In neigbourhoods that contain part of "the join", the metric has to be continuous, but not necessarily differentiable; think of the absolute value function, which is continuous, has a step function as first (distributional) derivative, and has a Dirac delta function as second derivative (and thus curvature tensor). The idealized physical meaning of this is that there is a (hyper)surface layer of mass whose density is given by a delta function, so the stress-energy tensor involves delta functions. The LHS of the Einstein equation involves (contractions of) curvature delta functions, and the RHS of the Einstein equation is given by the stress-energy tensor, so this can all be made to hang together.

Just as idealized situations with surface charge layers and distributional (delta function) volume charge densities, and with electric field discontinuities are useful in undergrad electromagnetism, mass hypersurface layers with metric component discontinuities and distributional stress-energy tensors are useful in general relativity, e.g., for domain walls. This is called the the thin shell/junction condition formalism.

 

[PAllen]

I believe that is part of the definition of a manifold with metric.

For special purposes, physicists sometimes cheat and use metric that is only continuous, not smooth on some boundary. This allows quick and dirty gluing of e,g. an arbitrary material ball solution to an external vacuum. Synge covers this, and the cautions needed from connection and curvature being undefined on the boundary (while e.g proper time along a timelike path across the boundary is perfectly well defined).

 

[PeterDonis]

It is, however, sometimes convenient to join two (or more) such spacetimes together. In neigbourhoods that contain part of "the join", the metric has to be continuous, but not necessarily differentiable

For special purposes, physicists sometimes cheat and use metric that is only continuous, not smooth on some boundary. This allows quick and dirty gluing of e,g. an arbitrary material ball solution to an external vacuum.

Yes, you're right, these are examples where the metric does not have to be smooth, just continuous. (I believe there are also junction conditions on some of the first derivatives of the metric in these cases.)

 

[George Jones]

The junction conditions also are given in section 3.7 of Poisson's notes,

http://www.physics.uoguelph.ca/poisson/research/agr.pdf

Note in particular equation (3.7.9) and the explanation given below the equation.

These notes evolved into the excellent book, "A Relativist's Toolkit: The Mathematics of black hole Mechanics".

 

[lavinia]

I think you must be misunderstanding. You can have coordinates that make the metric Minkowski at a point, and also for which the metric compatible connection vanishes. However, if the curvature tensor vanishes in any coordinates at a point, it vanishes for all coordinates at that point. This is never true in most GR manifolds, e.g. it is nowhere true in the Kruskal manifold, and is also nowhere true in FLRW manifolds except the degenerate Milne case, in which case it is true everywhere.

Locally flat normally means you can define a tangent space, and this implies what I said about the metric and connection. All of this is part of any definition of either a Riemannian or pseudo Riemannian manifolds.

In normal coordinates the metric tensor is diagonal at the central point and the Christoffel symbols vanish at that point. This requires the connection to be torsion free as well as metric compatible. In a neighborhood of the central point the metric tensor is diagonal up to first order. The second order term is proportional to the curvature and there is an error term of order O##(|x^3|)##. Not sure if this is what is meant by locally flat.

While I don't know the Physics, it would seem that since a freely falling coordinate patch is infinitesimally Minkowski, the connection would have to be torsion free. One seems to be driven by the Physics to a Levi-Civita connection.

In any case, if the connection is not torsion free then one might say that it is not locally flat.

 

[lavinia]

Well the definition in Schutz's book which I am reading states that locally flat refers to the fact that in specific local coordinates which define the manifold, Riemann curvature tensor is zero.

If around any point on the manifold there is a coordinate system in which the curvature tensor is zero, then it is zero everywhere. So a manifold with non-zero curvature tensor is not locally flat by this definition. Locally flat by this definition means globally flat. Minkowski space is an example but there are many others, e.g. a torus can be give a flat Lorentz metric.

Take a look at post #28

 

[Dale]

If around any point on the manifold there is a coordinate system in which the curvature tensor is zero, then it is zero everywhere.

This is not correct. Consider a spherical shell. In the interior the curvature is zero at every point, but outside of the shell the curvature is non-zero.

 

[lavinia]

This is not correct. Consider a spherical shell. In the interior the curvature is zero at every point, but outside of the shell the curvature is non-zero.

No. The curvature tensor is zero. The inherited geometry of the embedded sphere has non zero curvature. But the curvature of the ambient space even in neighborhoods of points on the sphere is exactly zero.

Whether a tensor is zero is independent of the coordinate system - since it is a tensor.

 

[Ibix]

No. The curvature tensor is zero. The inherited geometry of the embedded sphere has non zero curvature. But the curvature of the ambient space in neighborhoods of points on the sphere is exactly zero.

I think @Dale is imagining a Dyson sphere, while you seem to be discussing S² embedded in R³, or something of that nature.

 

[Dale]

Whether a tensor is zero is independent of the coordinate system - since it is a tensor.

That is true but irrelevant. The curvature is invariant but it need not be the same everywhere in the manifold. The curvature of the exterior Schwarzschild is not zero.

 

[lavinia]

I think @Dale is imagining a Dyson sphere, while you seem to be discussing S² embedded in R³, or something of that nature.

That doesn't change the point.

 

[Dale]

I think @Dale is imagining a Dyson sphere, while you seem to be discussing S² embedded in R³, or something of that nature.

I am imagining a spherical shell of mass. The exterior metric is Schwarzschild and the interior metric is Minkowski. Minkowski is flat and Schwarzschild is not. This disproves @lavinia’s incorrect assertion