How Is Intrinsic Curvature Measured in Higher Dimensions Like 4D Spacetime?

[geordief]

The intrinsic curvature of a 2d sphere is measured (as far as I can tell) by the means of its tangent planes that lie in the 3d dimensions it is embedded in.(I think parallel transport relies on these tangent planes)

So "intrinsic seems to still rely on the embedding dimensions as far as I can see.

Can we extrapolate to 4d spacetime in order to measure its curvature?
Does it also have tangent planes and are they aldo embedded in a higher dimension?

 

[A.T.]

The intrinsic curvature of a 2d sphere is measured (as far as I can tell) by the means of its tangent planes that lie in the 3d dimensions it is embedded in.(I think parallel transport relies on these tangent planes)

So "intrinsic seems to still rely on the embedding dimensions as far as I can see.

You can define intrinsic curvature based on purely intrinsic measurements. So it doesn't rely on the embedding dimensions.

 

[geordief]

@A.T.

Without recourse to parallel transport?

 

[Nugatory]

by the means of its tangent planes that lie in the 3d dimensions it is embedded in.

The tangent space at a point is not a plane in the 3D embedding space (although that is a natural way of visualizing it, continuing our general tendency to think of the 2D surface of a sphere as embedded in the 3D euclidean space containing the sphere) and can be defined without reference to that embedding space.

 

[geordief]

The tangent space at a point is not a plane in the 3D embedding space (although that is a natural way of visualizing it, continuing our general tendency to think of the 2D surface of a sphere as embedded in the 3D euclidean space containing the sphere) and can be defined without reference to that embedding space.

So I take your (and,presumably eveyone else's) word that the tangent planes of a sphere (and presumably of any "uneven sphere" ) can be formulated solely from measurements on the sphere.

That being so what relation do those tangent planes have to the sphere itself?
They are not "on" the sphere(except at one point) Are they a property of the sphere?

And is the sphere a purely mathematical object that is used to approximate what we see as our external and physical reality? (so that I am reading too much into it and seeing the sphere as a physical ball when it is not)

 

[PeterDonis]

its tangent planes that lie in the 3d dimensions it is embedded in.

No, they aren't. Tangent spaces do not rely on any embedding. People often visualize tangent spaces using such an embedding, but that's a crutch that we humans use, not part of the actual math.

 

[PeterDonis]

what relation do those tangent planes have to the sphere itself?

They are mathematical abstractions that rely on the fact that, locally, the sphere looks flat (the effects of its curvature get smaller and smaller as you look at a smaller and smaller piece of them).

 

[PeterDonis]

is the sphere a purely mathematical object that is used to approximate what we see as our external and physical reality?

The mathematical sphere is a model. It's not anything in reality.

We often use models to make predictions about reality, but that doesn't mean the models are reality. For example, if the surface of a real, physical ball has the right properties, we can use the abstract, 2-sphere model to make accurate predictions about measurements on the surface of the real, physical ball. But the mathematical 2-sphere is still just a model; it's not the ball itself.

 

[Nugatory]

That being so what relation do those tangent planes have to the sphere itself?
They are not "on" the sphere(except at one point) Are they a property of the sphere?

They aren't "on" the sphere anywhere. There is a mapping from each point on the surface of the sphere to a two-dimensional vector space specific to that point, which we call the "tangent space" at that point. But the elements of that vector space are neither vectors in the embedding three-dimensional space nor vectors on the two dimensional surface of the sphere.

The wikipedia article on tangent spaces has a subsection entitled "tangent vectors as directional derivatives" which will get you started on uderstanding the relationship between the surface of the sphere and the elements of tthe tangent space.

 

[FactChecker]

For one thing, if the sum of the angles of a triangle are not 180 deg., then there is curvature.

 

[A.T.]

That being so what relation do those tangent planes have to the sphere itself?

From:
https://en.wikipedia.org/wiki/Tangent_space

In the context of physics, the tangent space to a manifold at a point can be viewed as the space of possible velocities for a particle moving on the manifold.

They are not "on" the sphere (except at one point) Are they a property of the sphere?

Go one dimension down. You often see diagrams like below, that show the velocity vector v at point P along a curved path. While those diagrams are intuitive, you have to keep in mind that position and velocity vectors are part of two different vector spaces. Also note that P and v can both can be represented intrinsically to the path by single number (1D vector) each, instead of 2D vectors of the embedding space.

The same applies one dimension up to those diagrams of tangent planes at curved surfaces.

 

[PAllen]

Note, there are other ways to define intrinsic curvature constants. For example, consider the area of a small circle on the surface of a sphere. Measure its area and take the difference with the plane area formula for the same radius as measured on the sphere. Divide by the plane area to get a proportional difference. Divide by the plane area again to get the proportional difference per area. Take the limit as the circle radius goes to zero. This gives the sectional curvature at the point.

This can be generalized and related to the curvature tensor. Look up treatments of sectional curvature.

 

[Dale]

@A.T.

Without recourse to parallel transport?

Why would you do it without recourse to parallel transport?

 

[Dale]

In the context of physics, the tangent space to a manifold at a point can be viewed as the space of possible velocities for a particle moving on the manifold

I hadn’t seen that, but it is a good quote. It also helps that it is clear that physics depends not only on positions but also on velocities/momenta. So it is natural that our mathematical model will contain both positions and velocities at a pretty basic level in the model.

 

[Nugatory]

Why would you do it without recourse to parallel transport?

When @geordief wrote that he was still under the misunderstanding that “I think parallel transport relies on these tangent planes [that is, planes in the 3D space]” from the original post.

 

[Orodruin]

Why would you do it without recourse to parallel transport?

Not only why, but ”you cannot”. It is impossible to do it without parallel transport - or at least without parallel transport being defined - as curvature is a property of the affine connection that defines what parallel means.

OP’s misunderstanding that parallel transport requires embedding is of course the culprit here.

 

[Ibix]

Not only why, but ”you cannot”. It is impossible to do it without parallel transport - or at least without parallel transport being defined

What about @PAllen's area method? I can define a ball as the set of points for which I can find a route to a chosen point with length less than R. Then I can compare the area/volume/whatever to the n-dimensional Euclidean equivalent.

I'm clearly using the metric but I don't immediately see where the connection comes in. Unless it's in the integral for the area...?

 

[Orodruin]

What about @PAllen's area method? I can define a ball as the set of points for which I can find a route to a chosen point with length less than R. Then I can compare the area/volume/whatever to the n-dimensional Euclidean equivalent.

I'm clearly using the metric but I don't immediately see where the connection comes in. Unless it's in the integral for the area...?

I believe that is intrinsically applying the Levi-Civita connection as the geodesics of the Levi-Civita connection are the paths that extremise path lengths.

More generally, for an arbitrary connection you could define the surface of the ball as the set of points reached from the center after moving a distance R along a geodesic (and <R for the interior).

The method will also not always give a non-zero result if there is curvature as it doesn’t capture the full structure of the curvature tensor.

 

[PAllen]

I believe that is intrinsically applying the Levi-Civita connection as the geodesics of the Levi-Civita connection are the paths that extremise path lengths.

The sectional curvature as described can be directly related to the Riemann tensor as written in terms of the metric. BUT, for connection different from the Levi-Civita connection, this would not be the 'right' curvature tensor. So the reliance is there, but rather hidden.

More generally, for an arbitrary connection you could define the surface of the ball as the set of points reached from the center after moving a distance R along a geodesic (and <R for the interior).

The method will also not always give a non-zero result if there is curvature as it doesn’t capture the full structure of the curvature tensor.

If you consider the sectional curvature on an appropriate number of independent planes through a point (depending on the dimensionality of the manifold), complete curvature information can be recovered (again, with curvature being the one corresponding to the Levi-Civita connection).

 

[Orodruin]

If you consider the sectional curvature on an appropriate number of independent planes through a point

Planes, I agree, but not with a single ball, which was my understanding from the discussion.