general relativity - 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.

   
   

Answer

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.

ADDENDUM

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.