# Curved spaces that locally aren't flat

Gold Member

## Main Question or Discussion Point

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.

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jbriggs444
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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?

Dale
martinbn
What do you mean by locally flat?

Gold Member
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
2019 Award
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.

MathematicalPhysicist, Pencilvester and George Jones
Dale
Mentor
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
Staff Emeritus
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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
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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.

Gold Member
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
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2019 Award
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).

Gold Member
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
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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
Gold Member
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
Staff Emeritus
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
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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
Mentor
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

Last edited:
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
Dale
Mentor
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!

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
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Do metric coefficients have to be smooth functions?
I believe that is part of the definition of a manifold with metric.

Ibix
Gold Member
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
Staff Emeritus
Gold Member
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