3
$\begingroup$

Suppose a Riemannian manifold is such that the metric tensor in a coordinate system is given by

$$g_{ij}=\delta_{ij}$$ so that $$ds^2=g_{ij}dx^i dx^j=(dx^1)^2+(dx^2)^2+(dx^3)^2$$

(i) Is this space called (a) flat or locally flat, (b) Euclidean or locally Euclidean?

(ii) Can we think of a simple space which is locally flat but not locally Euclidean or vice-versa?

(iii) What is the difference between a flat space and an Euclidean space?

(iv) What is a space called, if it has a metric $g_{ij}=C_i\delta_{ij}$ (where $C_i$ are some constants independent of the coordinates)?

Not-too-technical answer will be helpful because my understanding of differential geometry and manifolds is limited.

$\endgroup$

migrated from physics.stackexchange.com Oct 16 '16 at 19:33

This question came from our site for active researchers, academics and students of physics.

  • 2
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Oct 15 '16 at 7:35
  • $\begingroup$ Yes, this is a maths question, and they're also more tolerant of homework than we are. $\endgroup$ – John Rennie Oct 15 '16 at 7:40
  • 6
    $\begingroup$ Is it really? I thought these questions are very basic for someone trying to understand general relativity and could be relevant questions in a classroom. Moreover, I don't see how is it a homework problem. $\endgroup$ – SRS Oct 15 '16 at 7:45
  • 1
    $\begingroup$ @SRS Here on Physics.SE, we use homework or "homework-like" to mean exactly "questions which could be relevant in a classroom." $\endgroup$ – rob Oct 15 '16 at 14:31
9
$\begingroup$

I think this question deserves an answer here in physics SE, since these notions are very common in Relativity.

For a manifold $M$ equipped with a metric (either Euclidean or Lorentzian) $g$, locally flat means that every point $p\in M$ admits an opens set $U \ni p$ endowed with coordinates $$\psi : U \ni q \mapsto (x^1(q),\ldots, x^n(q)) \in \psi(U) \subset \mathbb R^n$$ such that the metric in these coordinates takes the constant canonical form $$g_{ab}(q) = \delta_{ab}\quad \mbox{Riemannian case}\:,$$ or $$g_{ab}(q) = \eta_{ab}\quad \mbox{Lorentzian case}\:.$$ The point is that there is no guarantee that there is such a $U$ which covers the whole manifold. For this reason we use the adverb locally in front of flat. An example is a two dimensional cylinder in $\mathbb R^3$ equipped with the metric induced by the standard metric on $\mathbb R^3$.

For a manifold $M$ equipped with a metric (either Euclidean or Lorentzian) $g$, globally flat means that there is a global coordinate system $$\psi : M \ni q \mapsto (x^1(q),\ldots, x^n(q)) \in \psi(M) \subset \mathbb R^n$$ such that the metric in these coordinates takes the constant canonical form $$g_{ab}(q) = \delta_{ab}\quad \mbox{Euclidean case}\:,$$ or $$g_{ab}(q) = \eta_{ab}\quad \mbox{Lorentzian case}\:.$$ In this case $(M,g)$ is isometrically identified with an open portion (possibly the whole space) of $\mathbb R^n$ equipped with the standard (Euclidean or Lorentzian) metric.

In view of the definitions above, the answer to your question (i) depends on the extension of your coordinate system. If it covers the whole manifold we have a globally flat manifold. If it is not the case, we can only say that a region of the manifold, the one covered by the coordinates, is flat.

Flat space and Euclidean space have more or less the same meaning. They usually indicate Minkowski spacetime (i.e. a spacetime isometrically isomorphic to $\mathbb R^n$ equipped with the standard Minkowsky metric $\eta_{ab}$) or $\mathbb R^n$ equipped with the standard Euclidean metric $\delta_{ab}$. However are a bit vague terms whose meaning (if global or local) should be understood from the context.

Regarding your last question. If all $C_i$ are positive, the metric can be recast into the standard form $\delta_{ij}$ simply rescaling (changing) the coordinates with a factor $1/\sqrt{C_i}$. If some constant is negative, with a similar procedure the metric can be reduced into a constant form $h_i \delta_{ij}$ where $h_i = \pm 1$, which is Lorentzian only if all $h_i$ are $+1$ but one (or all $h_i$ are $-1$ but one depending on the adopted convention). It is impossible that a $C_i$ vanishes because it would imply $det [g_{ij}]=0$ which is not admitted by definition of (non-degenerate) metric.

ADDENDUM. An apparently related question concerns the so called locally flat coordinates around a given point. This is an independent issue. These coordinates are also called Riemannian normal coordinates or Gaussian normal coordinates. Are coordinates $x^1,\ldots, x^n$ defined in a neighborhood of a point $p\in M$ such that exactly at $p$ the components of the metric have their canonical form (e.g. $g_{ab}(p)=\delta_{ab}$ or $g_{ab}(p)=\eta_{ab}$ depending on the nature of the metric) and $\frac{\partial g_{ab}}{\partial x^c}|_p =0$. It is possible to prove that every manifold equipped with a smooth metric admits such a coordinate system for every given point $p\in M$.

It is also possible to extend the definition referring to a geodesic $\gamma\subset M$ instead of a point $p$: Exactly on $\gamma$ the components of the metric have their canonical form (e.g. $g_{ab}(\gamma)=\delta_{ab}$ or $g_{ab}(\gamma)=\eta_{ab}$ depending on the nature of the metric) and $\frac{\partial g_{ab}}{\partial x^c}|_\gamma =0$. This form gives rise to a very precise mathematical description of Einstein's equivalence principle in GR when $\gamma$ is a timelike geodesic.

$\endgroup$
  • 1
    $\begingroup$ I think your definition of "locally flat" is the mathematical one, while physicists often mean that they can choose coordinates such that at a point they can choose the metric to be Euclidean/Lorentzian, i.e. their "locally flat" is the mathematician's "existence of Riemannian/Lorentzian normal coordinates", cf. e.g. this question for at least one such usage (Of course, "locally flat" is a silly name for this since the curvature won't vanish at the point, but that's how I think it is used). $\endgroup$ – ACuriousMind Oct 15 '16 at 9:07
  • 1
    $\begingroup$ All manifolds are locally flat at every point with your definition since Riemannian normal coordinates exist in a neighborhood of every point! I use the above terminology in my courses of physics, I think is appropriate. You are instead mentioning (perhaps) the terminology, in my opinion quite confused, adopted in Landau-Liftsits' textbook. $\endgroup$ – V. Moretti Oct 15 '16 at 9:08
  • 1
    $\begingroup$ Yes, that's another reason that (L-L's) is a silly definition :) I just thought it might be worth to mention it because I'm not confident that physics texts always make it clear which notion of locally flat they are talking about. $\endgroup$ – ACuriousMind Oct 15 '16 at 9:11
  • 1
    $\begingroup$ As far as I remember there is also a logical loop in a statement (in LL book) just due to the confused terminlogy $\endgroup$ – V. Moretti Oct 15 '16 at 9:12
  • $\begingroup$ On the other hand I like that book for some very interesting discussions about synchronization procedures and the spatial metric when adopting different synchronization procedures... $\endgroup$ – V. Moretti Oct 15 '16 at 9:16
-1
$\begingroup$

Once you have $ds^2=(dx^1)^2+(dx^2)^2+(dx^3)^2$, Pythagorean theorem holds locally, you get triangles with sum of angles = $\pi$ and you are locally homeomorphic to an Euclidean space. This may not hold globally, as for example you could very well be on a cylinder, which is not an Euclidian space. The same holds if $g_{ij}=C_i\delta_{ij}$ for positive constants $C_i$, as you can interpret it as an anisotropic scaling. I am not aware about a fundamental difference between locally flat and locally Euclidean, apart local flatness referring to embedded manifolds whereas locally Euclidean does not involve embedding.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.