In special relativity one assume that spacetime can be locally described by 4 coordinates, so it makes sense to model it as a 4-dimensional manifold.

I had the impression that it is assumed that there is an atlas of special charts called inertial reference frames. Their transition maps are restrictions of affine maps on $\mathbb{R}^4 \rightarrow \mathbb{R}^4$. The subgroup of invertable affine maps which are allowed as transition maps is called the kinematical group.

Einstein shows in his paper, I guess, that one can choose the kinematical group to be the Poincare group and for example [Bacry Levy-Leblond] showed that there would only be a few possible choices for these kinematical groups.

In the books, I am aware of, comes now a huge step and one immediately assume one is in Minkowski space.

So my questions are the following: Given a manifold $M^4$ with an atlas such that the transition maps are all restrictions of the affine maps of the Poincare group.

  1. How can one construct a Lorentz metric (a (3,1)-semi-Riemannian metric) on $M$.

  2. Does anyone know a reference for the fact that a manifold $M^4$ with affine transition maps is covered by an affine space (i.e. A simple connected one is topologically $\mathbb{R}^4$)?

One might be tempted to consider the above as something belonging to general relativity, as one uses charts etc. But note affine transition maps are a huge restriction and at the end (maybe after taking a covering map) one talking about Minkowski space anyways.

  • 1
    $\begingroup$ I don't get the last paragraph, i.e. I don't quite see why you would be concerned with patching together subsets of $\mathbb{R}^4$ - in SR, why wouldn't an inertial frame apply to the whole of $\mathbb{R}^4$? In which case the Minkowski metric arises naturally, because that's the structure that the Poincare group is defined to preserve. Maybe I'm missing something but I don't see where this is coming from. $\endgroup$
    – twistor59
    Jul 19, 2013 at 14:50
  • $\begingroup$ If $(\mathbb{R}^4,g)$ is the Minkowski space then consider $\mathbb{Z}^4$ acting on it. It preserves the semimetric and the quotient is a perfectly fine flat spacetime (but compact). Nevertheless you would not be able to describe it with just one chart. $\endgroup$
    – mna
    Jul 19, 2013 at 18:52
  • $\begingroup$ OK I see what you mean. I assume the reasons for preferring "straight" $\mathbb{R}^4$ rather than a compact quotient space like that might be because of a simpler causal structure, i.e avoidance of closed timelike curves? $\endgroup$
    – twistor59
    Jul 19, 2013 at 19:31
  • $\begingroup$ Time orientability is also sometimes imposed as a condition for candidate spacetimes. $\endgroup$
    – twistor59
    Jul 19, 2013 at 19:36

1 Answer 1


Let $\phi: U \rightarrow V\subset \mathbb{R}^4$ be an inertial reference frame and Q the standard Lorentz metric on $\mathbb{R}^4$. The pullback $\phi^* Q$ is independent of the choice of inertial reference frame as the transition maps are from the Poincare group. This way we can define a Lorentz metric on all of $M$.


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