# How to prove that a flat spacetime admits Minkowski coordinates?

How should I prove the following in general relativity?

1. A flat spacetime can be covered by Minkowski coordinate neighborhoods.

2. A flat spacetime with the trivial topology can be covered by a global Minkowski coordinate chart.

• Would Mathematics be a better home for this question? – Qmechanic Feb 11 '18 at 16:05
• Intuitively, I think both statements should follow fairly straightforwardly from the construction of Riemann normal coordinates. – mmeent Feb 12 '18 at 8:00
• @mmeent: Statement #2 is global rather than local. Riemann normal coordinates are only guaranteed to exist locally. If the topology is nontrivial, Riemann normal coordinates still exist locally, but the global extension of them fails. – Ben Crowell Feb 25 '18 at 19:07
• I don't think that #2 has been studied in the Riemannian case, let alone in the Minkowski case and it seems rather too powerful but I don't have a counter example as the list of flat manifolds seem to be really short... If you assume your manifold to be a symmetric space then #2 holds but it is rather a very restrictive special case. – Gonenc Feb 25 '18 at 21:57
• It seems it was Ben Crowell who introduced it in the thread. Trivial topology is meant in the context of differential geometry and manifolds (as opposed to general topology and topological spaces). – Qmechanic Feb 26 '18 at 19:05

That is a non-trivial theorem of (semi)Riemannian geometry based on Frobenius Theorem: if the Riemann tensor is everywhere zero, then every point belongs to a local chart where the metric has the standard constant diagonal form.

The idea of the proof is the following. One looks for vector felds $X$ such that $$\nabla X =0\:.$$ In coordinates $x^1,\ldots, x^n$, this leads to a first-order equation for the components of $X$.

Next, looking at the found equation and using the condition $Riemann =0$ everywhere written in terms of connection coefficients, Frobenius theorem for first-order PDEs in $\mathbb R^n$ proves that, in a neighborhood of any point $p\in M$, there exist such $X$ satisfying $X(p) = Z_p$ where $Z_p\in T_pM$ is arbitrarily fixed.

So, one can construct $n= \dim(M)$ vector fields $X_{(k)}$, $k=1,\ldots,n$ in a neighborhood $U$ of $p$ such that $\nabla X_{(k)}=0$ and $X_{(k)}(p) = Z_{(k)p}$. Since the scalar product is preserved (from $\nabla X_{(k)}=0$ and the fact that the connection is metric), if $g_p(Z_{(k)p}, Z_{(h)p})= \eta_{kh}$, we have that $g(X_{(k)}, X_{(h)})= \eta_{hk}$ constantly on $U$.

Finally, one has to look for coordinates $y^a= y^a(x^1,\ldots,x^n)$ around $p$ such that $$X_{(k)}= \frac{\partial}{\partial y^k}\:.$$ Writing down this equation in coordinates $x^1,\ldots, x^n$, applying once again Frobenius theorem it is possible to prove that these local coordinates do exist around $p$. This way, shrinking further $U$ around $p$ we end up with a coordinate system $y^1,\ldots,y^n$ covering it where the metric is constant: $$g\left(\frac{\partial}{\partial y^k},\frac{\partial}{\partial y^h} \right)= \eta_{kh}\:.$$

In general, this procedure cannot produce a global chart where the metric is constant. There are trivial counterexamples starting from Minkowski space and assuming some identifications to produce a flat compact torus. This manifold is flat but cannot be covered by a global chart otherwise it would be diffeomorphic to $\mathbb R^n$ which is not compact.

• The question got edited multiple times. I think this answer in its present form does not address statement #2 in the current form of the question. Statement #2 is asking about the case of the trivial topology, and it specifies the trivial topology because we all know that otherwise there are counterexamples, like the torus described in the question. – Ben Crowell Feb 25 '18 at 19:11

I didn't actually realise that I have already solved the 2nd 'problem' for the Riemannian case but the proof most certainly does not carry over to the general Lorentzian. However, I am fairly confident that if you have a Lorentzian manifold $\mathcal M$ that can be written as a 'product of time and space' (more rigorously as a foliation) , $\mathcal M = \mathbb R \times M$ for a Riemanninan manifold $M$ (which is usually the case in physical applications) then the proof that I will give for Riemannian manifolds should carry over to this special case of Lorentizian manifolds:

Our assumptions for the Riemanninan case will be the following:

• $M$ a smooth manifold with vanishing curvature $Riem =0$ i.e. flat and
• there exists a homeomorphism $f:M\to \mathbb R^{\dim M }$ i.e. topologically 'trivial' (so in particular $M$ is simply connected).
• I will further assume that $M$ is complete (i.e. every convergent series has a limit in $M$).

There is a characterisation of locally symmetric spaces via Riemann curvature tensor that is (Cartan-Ambrose-Hicks Theorem):

$M$ is locally symmetric if and only if $\nabla Riem =0$.

In our case this is trivially satisfied. Furthermore, every simply connected complete, locally symmetric space is globally symmetric. Hence, $M$ is a simply connected globally symmetric space.

Luckily there is a classification (Cartan classification) of simply connected globally symmetric spaces, which states:

Let $M$ be a simply connected globally symmetric space. Then $M= M_0\times M_+ \times M_-$, where $M_0$ is an Euclidean factor and $M_\pm$ is a symmetric space of compact ($+$) and non-compact ($-$) type resp. Furthermore, the sectional curvature of $M_\pm$ is larger than (resp. smaller than) or equal to but not identically equal to 0.

Since the curvature is everywhere zero, we must have $M_\pm = \emptyset$ because otherwise the sectional curvature of $M_\pm$ would be identically zero. Hence $M= M_0=\mathbb R^{\dim M}$ $\square$

• This does indeed seem to work for the second question (does trivial topology imply global chart?) in the Riemannian case, but there's absolutely no guarantee the signature doesn't mess it up. You need signatured analogues of the Cartan–Ambrose–Hicks theorem and the classification. The classification especially seems like something that could be highly dependent on signature. – WetSavannaAnimal Mar 1 '18 at 1:40
• @WetSavannaAnimalakaRodVance It is actually clear that this doesn't work for arbitrary Lorentzian manifolds but in physics applications you almost always assume that the Lorentzian manfild $\mathcal M$ is a product $\mathcal M = \mathbb R \times M$ of the time axis and a Riemannian manifold $M$, which is why I thought that it might work. – Gonenc Mar 1 '18 at 10:30
• Maybe: I think that the assumption of that foliation is probably stronger than what either the OP or especially the bounty giver would want to assume; I'm guessing they are hoping for that to be something to be proven from weaker assumptions. But I could be wrong. You might want to add that assumption to your list and your motivation - it's true that one often does see that foliation assumed. – WetSavannaAnimal Mar 1 '18 at 12:15
• @WetSavannaAnimalakaRodVance Tbh if they wanted to have an answer for a general Lorentzian manifold than "physics" stackexchange is the wrong place for that, considering especially that for the "simpler" Riemannian case I needed to use "heavy machinery", none of which is really trivial nor intuitive. – Gonenc Mar 1 '18 at 16:14
• That's a fair comment, and a +1 because it is a worthy proof and the only serious progress here beyond Valter Moretti's answer, although I know that the bounty giver for sure is looking for a proof that the manifold foliates in the way you assume as part of the deal. I believe you should make that assumption explicit at the outset of your answer and also name the Cartan–Ambrose–Hicks theorem and the Cartan classification explicitly to guard against any changes in Wikipedia / link rot. – WetSavannaAnimal Mar 1 '18 at 22:33

I think my answer is consistent with Valter’s, but low-faluting.

Parallel transport preserves inner products of vectors, and the vanishing of the Riemann tensor guarantees the path independence of parallel transport within any local chart. You can start with an orthonormal tetrad and extend the local chart and Minkowski metric from an arbitrary origin. The elements of the metric tensor are defined as pairwise inner products of basis vectors in the tetrad.

To extend the metric globally, exploit the homeomorphism to/from ${{R}^{4}}$ to induce a (possibly curvilinear) coordinate system in which every tuple $(t,x,y,z)$ corresponds to a point of your flat space. A convenient path to reach any point is $(t,x,y,z)=(as,bs,cs,ds)$ with constant coefficients and parameter s running from 0 to 1. This path also happens to be unique, but existence is the key.

Nutshell proof of path-independence: The argument relies on the fact that any path from point A to B is homotopic to any other path in a Euclidean space. The “history” of the continuous deformation sweeps out a two-dimensional surface, which may be broken into plaquettes. The Riemann tensor describes the change in a vector resulting from transport around an infinitesimal plaquette. Integrating the little zeroes gets you a big zero.

Constructing Minkowski cooordinates: Once the orthonormal tetrad of basis vectors or 1-forms has been carried unambiguously to the entire space, we can construct the four coordinate functions by doing path integrals such as $\Delta t=\int{d\mathbf{x}\cdot \mathbf{u}}$, where u is the 1-form that represents the gradient of time, and likewise for x,y,z. The result is path-independent per Stokes’s theorem, because parallel transport guarantees that the covariant derivative and hence the curl of u is zero.

• I should have mentioned one more thing: Path independence implicitly depends on the fact that any path from point A to B is homotopic (continuously deformable) to any other path in Euclidean space. – Bert Barrois Feb 26 '18 at 13:15
• Uniqueness of the path is unimportant (and I regret having emphasized it) when the outcome is path-independent. I'll expand my answer to include a nutshell proof. – Bert Barrois Feb 27 '18 at 13:22
• I understand why the parallel transport of the basis vectors from the origin O to a particular point P is path-independent. What is less obvious to me is that when we integrate along this path to find the coordinates of P, the result of that integration is path-independent. – Ben Crowell Mar 1 '18 at 0:57
• You are not doing it to find the coordinates. You are doing it to extend the tetrad and Minkowski metric to the entire space. Once you have done this, you can choose a Cartesian coordinate system with no sweat. – Bert Barrois Mar 1 '18 at 12:56
• Once you have carried the tetrad unambiguously to the entire space, you could calculate the Minkowski t-coordinate by doing a path integral $\Delta t=\int{d\mathbf{x}\cdot \mathbf{u}}$ of the 1-form u that defines its gradient. Likewise x,y,z. The result is path-independent by Stokes’s theorem, since parallel xport keeps the covariant derivative of u zero. – Bert Barrois Mar 2 '18 at 12:40

It's long to do.

Let's suppose you anyhow know that the spacetime is flat. If so, Christoffel symbols are zero.

According to Einstein's criterion, the covariant derivative of the metric tensor must be zero.

$$g_{\mu\nu;\alpha}=0$$

But the covariant derivative is the normal derivative minus two symbols:

$$g_{\mu\nu;\alpha}\equiv g_{\mu\nu,\alpha} - \Gamma_{\mu\alpha}^{\ \ \rho} g_{\rho\nu} - \Gamma_{\nu\alpha}^{\ \ \rho} g_{\mu\rho}$$

If symbols are zero, then we only have

$$0 = g_{\mu\nu,\alpha} - 0 - 0$$

so the derivative of the metric tensor is 0, and the metric tensor doesn't depend on coordinates, it's constant. If you diagnoalize the tensor and re-scale the axis appropiately, you can have a diagonal matrix $(-1, 1, 1, 1) \equiv \eta_{\mu\nu}$.

• I don't think this is right. You can't infer from flatness that the Christoffel symbols vanish. – Ben Crowell Feb 11 '18 at 15:53
• I was following Einstein's path, on which he shoed that there exists a Reference frame which is locally flat, so that they vanish. Okay, maybe that's not what you are looking for. – FGSUZ Feb 11 '18 at 16:15
• @BenCrowell: vanishing Christoffel symbols certainly imply flatness -- the Riemann tensor is computed from christoffel symbols and their derivatives, after all, but the converse is definitely not true -- you have nonzero christoffel symbols in cylindrical coordinates, after all. – Jerry Schirmer Feb 11 '18 at 16:27
• @JerrySchirmer: Your comment is @ me, but I guess you were really addressing FGSUZ? – Ben Crowell Feb 11 '18 at 17:48