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> A Lorentzian manifold is not necessarily locally hyperbolic and certainly the existence of light-cones do not follow from its hyperbolic nature. A Lorentzian

by bollu 6y ago

> A Lorentzian manifold is not necessarily locally hyperbolic and certainly the existence of light-cones do not follow from its hyperbolic nature. A Lorentzian manifold may be flat or curved, if it's flat we usually call it also a Minkowski space. Curved spaces come in many shapes of forms depending on the curvature which is a complicated geometric object. One particular case of curved space is a hyperbolic space which is one with constant negative curvature. A hyperbolic (constant negative curvature) Lorentzian (space+time) manifold is also called an Anti-de Sitter (AdS) space and plays a very important role in physics (see AdS/CFT correspondence). I am confused. Let me write down the definition I know, and let's error-correct to a point of agreement? - A Lorentzian manifold is a pseudo-Riemannian manifold with a metric tensor that can be positive or negative definite, but not zero. - A neighbourhood of a Lorentzian manifold is flat if one can assign a coordinate system where we have the minowski metric ds^2 = dx1^2 dx2^2 + ... + dx(n-1)^2 - dxn^2. So it's hyperbolic space of dimension n, Hn. This space has constant negative curvature; You can check by computing the curvature tensor. - Thanks for that particular model of Schwarzschild's solution! I wasn't aware of it. - Do you have more examples of a similar flavour? I don't care if the solution obeys the einstein field equations. I'm mostly just interested in examples of Lorentzian manifolds with curvature. - Thanks for the advice regarding visualizing higher dimensions. I am quite comfortable with the algebra. However, it is often the case that people have great insight into the structure of hyperbolic space, which I am trying to build. Consider Thurston and his famous insight into the structure of 3-manifolds and hyperbolic space.

pa7x1 6y ago
> I am confused. Let me write down the definition I know, and let's error-correct to a point of agreement? Fixing the definitions is a good idea because they may vary slightly in the literature. > - A Lorentzian manifold is a pseudo-Riemannian manifold with a metric tensor that can be positive or negative definite, but not zero. A bit unconventional definition as it places some of the characteristics of a pseudo-Riemannian manifold into Lorentzian territory. Usually the Lorentzian manifold restricts the signature of the metric to (1, n-1) or (n-1, 1). That is to say, restricts the pseudo-Riemannian manifold to one with only 1 time dimension. > - A neighbourhood of a Lorentzian manifold is flat if one can assign a coordinate system where we have the minowski metric ds^2 = dx1^2 dx2^2 + ... + dx(n-1)^2 - dxn^2. So it's hyperbolic space of dimension n, Hn. This space has constant negative curvature; You can check by computing the curvature tensor. If the space is flat (Riemann curvature tensor = 0) how can it have constant negative curvature? Hope you see with this question that something is off. Here is a more detailed explanation: There is a tight link between Hyperbolic space and the Lorentz transformations. The proper (i.e. no space inversion) ortochronous (i.e. no time inversion) Lorentz group of transformations SO^+(1,n) is an isometry group of H^n. To be noted here the dimensionality difference, this means that a hyperbolic space H^n can be embedded in R^(1,n) and that the action of Lorentz transformations leaves H^n invariant. You can see a neat visualization in the wiki: https://en.wikipedia.org/wiki/Lorentz_group#orthochronous https://en.wikipedia.org/wiki/Lorentz_group#orthochronous, the Lorentz transformations leave invariant the hyperboloids shown in the picture. The action of a Lorentz transformation in SO^+(1,n) on any vector in one those hyperboloids H^n produces another vector in the same hyperboloid. But H^n is not Minkowski space, it's just one "shell" embedded in a Minkowski space of one additional dimension that is left invariant by SO^+(1,n). These shells have constant negative curvature but the space they are embedded into (Minkowski) is flat. This is equivalent to the relation between Euclidean space and spheres. The group of rotations in n-dimensional Euclidean space E^n leaves invariant the spheres S^(n-1). The spheres have positive constant curvature and are embedded in a higher dimensional flat space. > - Thanks for that particular model of Schwarzschild's solution! I wasn't aware of it. > - Do you have more examples of a similar flavour? I don't care if the solution obeys the einstein field equations. I'm mostly just interested in examples of Lorentzian manifolds with curvature. Not that I can think of, to visualize a curved space you always have to embed it in a higher dimensional flat space. To "see" one curved dimension, you need (at least) 2 flat ones in which to embed it. This is what Flamm's paraboloid does, it shows the curved radial dimension of Schwarzschild solution embedded in 2D so we can see how distances change along the way. The problem is that we run very quickly out of "real estate" to show the curvature. If you are a highly visual person I could recommend you to get a copy of Gravitation by Misner, Thorne, Wheeler. It is a masterpiece and comes with huge amounts of visual intuition.
- bollu 6y ago
  > If the space is flat (Riemann curvature tensor = 0) how can it have constant negative curvature? You are wrong. Locally our space is _hyperbolic_. So "flat" means "flat space-time" means "constant negative curvature", not "zero curvature". This is unlike the Riemannian case where "flat = zero curvature". You seem to be confusing the Riemannian and Lorentzian cases?
  - codethief 6y ago
    > "flat" means "flat space-time" means "constant negative curvature", not "zero curvature" So Minkowski space is not flat according to your definition? This way of defining flatness seems quite unusual to me both from a math and a physics perspective. Could you provide a source? > Locally our space is _hyperbolic_. How would this even be possible? A hyperbolic manifold is a Riemannian manifold by definition[0] but here we're dealing with a Lorentzian one! [0] https://en.wikipedia.org/wiki/Hyperbolic_manifold https://en.wikipedia.org/wiki/Hyperbolic_manifold
    - bollu 6y ago
      So I think I understand the confusion here. - In Riemannian manifolds, we take R^n as the building block - Every hyperbolic space can be locally represented by R^n. If you do this, the hyperbolic manifold will have negative curvature. Hence, hyperbolic space is a Riemannian manifold. - On the other hand, one can take Minkowski space as a given, and then build spaces where we have space that is locally hyperbolic. So we are trying to complete the analogy: Locally R^n :: Riemannian Locally Minkowski [signature (n-1,1) :: ??? The answer to `???` is _Lorentzian/Psuedo Riemannian_. You can check the wiki link for more(https://en.wikipedia.org/wiki/Pseudo-Riemannian_manifold#Properties_of_pseudo-Riemannian_manifolds https://en.wikipedia.org/wiki/Pseudo-Riemannian_manifold#Pro...), to quote: > Just as Euclidean space can be thought of as the model Riemannian manifold, Minkowski space with the flat Minkowski metric is the model Lorentzian manifold. Notice the use of the word flat here: Just as flatness in the Riemannian case means "globally like R^n", flatness in the psuedo-Riemannian case means "globally like Minkowski". You can also notice that the tangent space structure of the psuedo-Riemannian manifold is different from the Riemannian case: > Every tangent space of a pseudo-Riemannian manifold is a pseudo-Euclidean vector space. [Once again from the wiki page] As for _why_ we do this, my understanding is that we want to generalize SR(special relativity). For SR, we use a single space of signature (3,1). For GR, we want to 'put together' many such spaces, which are then distorted by gravity (this is curvature of space-time). Hence, we take the local space to be (3, 1), and the global space connects versions of (3, 1). Does this make sense? Please tell me if this is incoherent; This is my current understanding of the state of things, I could well be wrong.

by bollu 6y ago

pa7x1 6y ago

> I am confused. Let me write down the definition I know, and let's error-correct to a point of agreement? Fixing the definitions is a good idea because they may vary slightly in the literature. > - A Lorentzian manifold is a pseudo-Riemannian manifold with a metric tensor that can be positive or negative definite, but not zero. A bit unconventional definition as it places some of the characteristics of a pseudo-Riemannian manifold into Lorentzian territory. Usually the Lorentzian manifold restricts the signature of the metric to (1, n-1) or (n-1, 1). That is to say, restricts the pseudo-Riemannian manifold to one with only 1 time dimension. > - A neighbourhood of a Lorentzian manifold is flat if one can assign a coordinate system where we have the minowski metric ds^2 = dx1^2 dx2^2 + ... + dx(n-1)^2 - dxn^2. So it's hyperbolic space of dimension n, Hn. This space has constant negative curvature; You can check by computing the curvature tensor. If the space is flat (Riemann curvature tensor = 0) how can it have constant negative curvature? Hope you see with this question that something is off. Here is a more detailed explanation: There is a tight link between Hyperbolic space and the Lorentz transformations. The proper (i.e. no space inversion) ortochronous (i.e. no time inversion) Lorentz group of transformations SO^+(1,n) is an isometry group of H^n. To be noted here the dimensionality difference, this means that a hyperbolic space H^n can be embedded in R^(1,n) and that the action of Lorentz transformations leaves H^n invariant. You can see a neat visualization in the wiki: https://en.wikipedia.org/wiki/Lorentz_group#orthochronous https://en.wikipedia.org/wiki/Lorentz_group#orthochronous, the Lorentz transformations leave invariant the hyperboloids shown in the picture. The action of a Lorentz transformation in SO^+(1,n) on any vector in one those hyperboloids H^n produces another vector in the same hyperboloid. But H^n is not Minkowski space, it's just one "shell" embedded in a Minkowski space of one additional dimension that is left invariant by SO^+(1,n). These shells have constant negative curvature but the space they are embedded into (Minkowski) is flat. This is equivalent to the relation between Euclidean space and spheres. The group of rotations in n-dimensional Euclidean space E^n leaves invariant the spheres S^(n-1). The spheres have positive constant curvature and are embedded in a higher dimensional flat space. > - Thanks for that particular model of Schwarzschild's solution! I wasn't aware of it. > - Do you have more examples of a similar flavour? I don't care if the solution obeys the einstein field equations. I'm mostly just interested in examples of Lorentzian manifolds with curvature. Not that I can think of, to visualize a curved space you always have to embed it in a higher dimensional flat space. To "see" one curved dimension, you need (at least) 2 flat ones in which to embed it. This is what Flamm's paraboloid does, it shows the curved radial dimension of Schwarzschild solution embedded in 2D so we can see how distances change along the way. The problem is that we run very quickly out of "real estate" to show the curvature. If you are a highly visual person I could recommend you to get a copy of Gravitation by Misner, Thorne, Wheeler. It is a masterpiece and comes with huge amounts of visual intuition.

bollu 6y ago
> If the space is flat (Riemann curvature tensor = 0) how can it have constant negative curvature? You are wrong. Locally our space is _hyperbolic_. So "flat" means "flat space-time" means "constant negative curvature", not "zero curvature". This is unlike the Riemannian case where "flat = zero curvature". You seem to be confusing the Riemannian and Lorentzian cases?
- codethief 6y ago
  > "flat" means "flat space-time" means "constant negative curvature", not "zero curvature" So Minkowski space is not flat according to your definition? This way of defining flatness seems quite unusual to me both from a math and a physics perspective. Could you provide a source? > Locally our space is _hyperbolic_. How would this even be possible? A hyperbolic manifold is a Riemannian manifold by definition[0] but here we're dealing with a Lorentzian one! [0] https://en.wikipedia.org/wiki/Hyperbolic_manifold https://en.wikipedia.org/wiki/Hyperbolic_manifold
  - bollu 6y ago
    So I think I understand the confusion here. - In Riemannian manifolds, we take R^n as the building block - Every hyperbolic space can be locally represented by R^n. If you do this, the hyperbolic manifold will have negative curvature. Hence, hyperbolic space is a Riemannian manifold. - On the other hand, one can take Minkowski space as a given, and then build spaces where we have space that is locally hyperbolic. So we are trying to complete the analogy: Locally R^n :: Riemannian Locally Minkowski [signature (n-1,1) :: ??? The answer to `???` is _Lorentzian/Psuedo Riemannian_. You can check the wiki link for more(https://en.wikipedia.org/wiki/Pseudo-Riemannian_manifold#Properties_of_pseudo-Riemannian_manifolds https://en.wikipedia.org/wiki/Pseudo-Riemannian_manifold#Pro...), to quote: > Just as Euclidean space can be thought of as the model Riemannian manifold, Minkowski space with the flat Minkowski metric is the model Lorentzian manifold. Notice the use of the word flat here: Just as flatness in the Riemannian case means "globally like R^n", flatness in the psuedo-Riemannian case means "globally like Minkowski". You can also notice that the tangent space structure of the psuedo-Riemannian manifold is different from the Riemannian case: > Every tangent space of a pseudo-Riemannian manifold is a pseudo-Euclidean vector space. [Once again from the wiki page] As for _why_ we do this, my understanding is that we want to generalize SR(special relativity). For SR, we use a single space of signature (3,1). For GR, we want to 'put together' many such spaces, which are then distorted by gravity (this is curvature of space-time). Hence, we take the local space to be (3, 1), and the global space connects versions of (3, 1). Does this make sense? Please tell me if this is incoherent; This is my current understanding of the state of things, I could well be wrong.