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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

by 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.

codethief 6y ago
> In Riemannian manifolds, we take ℝ^n as the building block We take ℝ^n as the building block for any smooth manifold, not just Riemannian ones. Note that there is a difference between ℝ^n and Euclidean space. ℝ^n is a smooth manifold without a metric structure, so it is not a Riemannian manifold automatically. In contrast, Euclidean space is ℝ^n with a metric structure, namely the one given by the standard Euclidean metric δ. So Euclidean space is a Riemannian manifold and it is often denoted by the tuple (ℝ^n, δ). Finally, if we denote by η the (n-1, 1) Minkowski metric on ℝ^n, then (ℝ^n, η) is just Minkowski space (in particular, it is a Lorentzian / pseudo-Riemannian manifold). > Just as flatness in the Riemannian case means "globally like ℝ^n" This is not correct. There are many flat Riemannian manifolds that are not globally isometric (or even diffeomorphic) to Euclidean space (ℝ^n, δ). Take e.g. the torus T^n = S¹ × … × S¹ of any dimension. > On the other hand, one can take Minkowski space as a given, and then build spaces where we have space that is locally hyperbolic Again, I think you're confusing the terms "Lorentzian" and "hyperbolic" here. As I said (and you then yourself said), a hyperbolic manifold is a Riemannian manifold, not a Lorentzian one. A Lorentzian manifold is locally Lorentzian, not locally hyperbolic. You might want to read pa7x1's excellent explanation again. Also note that the quotes you gave from Wikipedia don't contradict what I (or pa7x1) said in the slightest. > 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). This is certainly one of way of thinking about it, yes: Locally, we see Minkowski space but globally the geometry (and topology) might be more complicated and this is what GR allows us to implement through a general Lorentzian manifold.
- bollu 6y ago
  Thank you, It does appear I am hopefully confused between hyperbolic and Lorentzian :) Thanks for patiently showing me where I am going wrong. For my selfish benefit, could you please collect the definitions of (i) Hyperbolic space (ii) Minkowski space (iii) Lorentzian manifold (iv) Pseudo Riemannian manifold?
  - codethief 6y ago
    I'm glad I could help! > could you please collect the definitions of (i) Hyperbolic space (ii) Minkowski space (iii) Lorentzian manifold (iv) Pseudo Riemannian manifold? Sure: A Riemannian manifold is a smooth manifold M together with a positive-definite metric g on M. A hyperbolic manifold is a Riemannian manifold M of constant negative sectional curvature. It is often required in addition that M be a complete manifold. A pseudo-Riemannian manifold is a smooth manifold M together with a non-degenerate metric g on M (of arbitrary signature). A Lorentzian manifold of dimension n is a pseudo-Riemannian manifold (of dimension n) whose metric has signature (1, n-1) or, equivalently, (n-1, 1). (n+1)-dimensional Minkowski space is the Lorentzian manifold (ℝ^(n+1), η), where η = - (dx⁰)² + (dx¹)² + … + (dx^n)² with respect to the standard coordinates (x⁰, x¹, …, x^n) on ℝ^(n+1).
    - bollu 6y ago
      And what is the precise relationship between hyperbolic space and Minkowski space? am I right in understanding that one can _embed_ a model of hyperbolic space (the hyperboloid moidel, obeying the equation `-x0^2 + x1^2 + ... xn^2 = 0` as a subspace of the Minkowski space? But Minkowski space is far larger than hyperbolic space?
      - codethief 6y ago
        pa7x1 already answered that question: https://news.ycombinator.com/item?id=23549361 https://news.ycombinator.com/item?id=23549361

by bollu 6y ago

codethief 6y ago

> In Riemannian manifolds, we take ℝ^n as the building block We take ℝ^n as the building block for any smooth manifold, not just Riemannian ones. Note that there is a difference between ℝ^n and Euclidean space. ℝ^n is a smooth manifold without a metric structure, so it is not a Riemannian manifold automatically. In contrast, Euclidean space is ℝ^n with a metric structure, namely the one given by the standard Euclidean metric δ. So Euclidean space is a Riemannian manifold and it is often denoted by the tuple (ℝ^n, δ). Finally, if we denote by η the (n-1, 1) Minkowski metric on ℝ^n, then (ℝ^n, η) is just Minkowski space (in particular, it is a Lorentzian / pseudo-Riemannian manifold). > Just as flatness in the Riemannian case means "globally like ℝ^n" This is not correct. There are many flat Riemannian manifolds that are not globally isometric (or even diffeomorphic) to Euclidean space (ℝ^n, δ). Take e.g. the torus T^n = S¹ × … × S¹ of any dimension. > On the other hand, one can take Minkowski space as a given, and then build spaces where we have space that is locally hyperbolic Again, I think you're confusing the terms "Lorentzian" and "hyperbolic" here. As I said (and you then yourself said), a hyperbolic manifold is a Riemannian manifold, not a Lorentzian one. A Lorentzian manifold is locally Lorentzian, not locally hyperbolic. You might want to read pa7x1's excellent explanation again. Also note that the quotes you gave from Wikipedia don't contradict what I (or pa7x1) said in the slightest. > 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). This is certainly one of way of thinking about it, yes: Locally, we see Minkowski space but globally the geometry (and topology) might be more complicated and this is what GR allows us to implement through a general Lorentzian manifold.

bollu 6y ago
Thank you, It does appear I am hopefully confused between hyperbolic and Lorentzian :) Thanks for patiently showing me where I am going wrong. For my selfish benefit, could you please collect the definitions of (i) Hyperbolic space (ii) Minkowski space (iii) Lorentzian manifold (iv) Pseudo Riemannian manifold?
- codethief 6y ago
  I'm glad I could help! > could you please collect the definitions of (i) Hyperbolic space (ii) Minkowski space (iii) Lorentzian manifold (iv) Pseudo Riemannian manifold? Sure: A Riemannian manifold is a smooth manifold M together with a positive-definite metric g on M. A hyperbolic manifold is a Riemannian manifold M of constant negative sectional curvature. It is often required in addition that M be a complete manifold. A pseudo-Riemannian manifold is a smooth manifold M together with a non-degenerate metric g on M (of arbitrary signature). A Lorentzian manifold of dimension n is a pseudo-Riemannian manifold (of dimension n) whose metric has signature (1, n-1) or, equivalently, (n-1, 1). (n+1)-dimensional Minkowski space is the Lorentzian manifold (ℝ^(n+1), η), where η = - (dx⁰)² + (dx¹)² + … + (dx^n)² with respect to the standard coordinates (x⁰, x¹, …, x^n) on ℝ^(n+1).
  - bollu 6y ago
    And what is the precise relationship between hyperbolic space and Minkowski space? am I right in understanding that one can _embed_ a model of hyperbolic space (the hyperboloid moidel, obeying the equation `-x0^2 + x1^2 + ... xn^2 = 0` as a subspace of the Minkowski space? But Minkowski space is far larger than hyperbolic space?
    - codethief 6y ago
      pa7x1 already answered that question: https://news.ycombinator.com/item?id=23549361 https://news.ycombinator.com/item?id=23549361