A Curved spaces that locally aren't flat

[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

 

[lavinia]

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.

That is true but if a metric is everywhere locally flat - which is what I was responding to - then the curvature tensor must be identically zero.

On a manifold with boundary, the boundary manifold inherits a connection from the connection on the entire manifold. This is not the same connection. For instance on a solid ball in Euclidean space with the standard connection, the Riemann curvature tensor is identically zero. The boundary sphere has curvature that it derives from a different connection.

 

[martinbn]

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

No, because she said around ANY point. In your example outside the shell the condition is not satisfied.

 

[Dale]

No, because she said around ANY point. In your example outside the shell the condition is not satisfied.

She said

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

For a spherical shell of mass the curvature tensor is zero around any point on the interior, but it is not zero around any point outside the shell, therefore it is not zero everywhere as claimed.

 

[lavinia]

She said For a spherical shell of mass the curvature tensor is zero around any point on the interior, but it is not zero around any point outside the shell, therefore it is not zero everywhere as claimed.

If is is not zero outside the shell then it is not locally zero around the shell.

 

[martinbn]

She said For a spherical shell of mass the curvature tensor is zero around any point on the interior, but it is not zero around any point outside the shell, therefore it is not zero everywhere as claimed.

Exactly. You are saying that around any point outside it isn't, only around point inside. She said around any point on the manifold.

 

[Dale]

It is is not zero outside the shell then it is not locally zero around the shell.

Correct. And this is different from the claim above. Above you said if it was zero around any point then it is zero everywhere.

An interior point is one of “any” points. So the claim is that the curvature is zero “everywhere”. That includes points in the exterior region where the curvature is non zero.

 

[martinbn]

Correct. And this is different from the claim above. Above you said if it was zero around any point then it is zero everywhere.

An interior point is one of “any” points. So the claim is that the curvature is zero “everywhere”. That includes points in the exterior region where the curvature is non zero.

But any point means an arbitrary point i.e. every point.

 

[Dale]

But any point means an arbitrary point i.e. every point.

”Any” is not the same as “every”. “Any” means pick one element of the set, and “every” means pick all of the elements of the set.

“Any” is a single arbitrary point. “Every” is not arbitrary, it is all of the points, the entire set.

The claim is wrong.

 

[martinbn]

”Any” is not the same as “every”. “Any” means pick one element of the set, and “every” means pick all of the elements of the set.

“Any” is a single arbitrary point. “Every” is not arbitrary, it is all of the points, the entire set.

The claim is wrong.

Well, when one says that something is true for any element of a set it means that it is true for all the elements, not just one single one.

How is this wrong!

 

[weirdoguy]

How is this wrong!

Especially that it is a standard phrasing in maths literature, both in english and polish and probably in other languages as well. When we try to prove some statement we say "let's take any point belonging to a set", or "let's take every point"?

 

[Dale]

Not according to my understanding which is captured here: https://math.stackexchange.com/a/430787

Note that the claim in question is part of an if-then so it follows the second pattern.

 

[lavinia]

I think the pedagogic point has been lost here. In normal coordinates the metric is not only diagonal at the central point but it is diagonal up to first order in a neighborhood of the central point. The second order term involves the curvature tensor so infinitesimally the metric looks flat. However the curvature is not necessarily zero at any point in the neighborhood around the central point. Normal coordinates would seem in some sense - don't know the physics - to correspond to free fall since they are generated by the flow of geodesics emanating from the central point. The physics then would say that the connection would not only have to be metric compatible but also torsion free. So an example of a curved space which is not locally flat in this hypothetical sense of the words would be a connection that is metric compatible but not torsion free.

 

[Dale]

I have no disagreement with that.

 

[martinbn]

Not according to my understanding which is captured here: https://math.stackexchange.com/a/430787

Note that the claim in question is part of an if-then so it follows the second pattern.

I disagree with your reading of that answer. Here is an example. Let $f: X\rightarrow Y$ be a map. If for any $y\in Y$ there exists an $x\in X$ such that $f(x)=y$, then $f$ is surjective. How do you understand that? That it refers to on single $y$ or all elements of $Y$?

 

[Dale]

I disagree with your reading of that answer. Here is an example. Let $f: X\rightarrow Y$ be a map. If for any $y\in Y$ there exists an $x\in X$ such that $f(x)=y$, then $f$ is surjective. How do you understand that? That it refers to on single $y$ or all elements of $Y$?

I understand that as referring to a single $y$. That is clearly what is described in the SE link.

Note that the Wikipedia definition of Surjective says “if for every ...” instead of “if for any...”

https://en.m.wikipedia.org/wiki/Surjective_function

 

[martinbn]

I understand that as referring to a single $y$. That is clearly what is described in the SE link.

Note that the Wikipedia definition of Surjective says “if for every ...” instead of “if for any...”

https://en.m.wikipedia.org/wiki/Surjective_function

Well, the same wiki article has this

Any function induces a surjection by restricting its codomain to the image of its domain. Every surjective function has a right inverse, and every function with a right inverse is necessarily a surjection. The composition of surjective functions is always surjective. Any function can be decomposed into a surjection and an injection.

Does it refer to a single function or all functions?

 

[martinbn]

I understand that as referring to a single $y$. That is clearly what is described in the SE link.

The SE link says that it depends on the context and gives an example. It doesn't say that for "if.., then.." statements it always means one single and not every.

 

[Dale]

Well, the same wiki article has this

Does it refer to a single function or all functions?

That refers to all functions.

 

[martinbn]

That refers to all functions.

Why do you have a problem with the other cases?

 

[lavinia]

It seems that a foundational idea that leads to General Relativity is that in a small enough free falling coordinate system the deviation from flat i.e. the Minkowski metric is too small to be measured. And if one tries more sensitive measuring instruments, one can take a smaller coordinate frame where the instruments can no longer detect the deviations. This would seem to be what is meant by locally flat. Mathematically, this can be interpreted as trajectories of free fall are geodesics with respect to a Levi-Civita connection on a Lorentz metric. Then the deviation from Minkowski in normal coordinates will be of second order. But this is no way means that the curvature tensor is locally zero.

 

[Dale]

Why do you have a problem with the other cases?

I don’t have a problem with it. English has lots of weird rules. I just accept them.

If you want to understand “if for any” as referring to the entire set then you can think of it as saying test the entire set and if at least one element meets the condition then ... It does not mean to test the entire set and if all elements meet the condition then ... That is why the claim fails.

This is the difference between testing the entire set and combining the results with “or” (if for any) vs with “and” (if for every).