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It is a common argument in the theory of kinematic groups (the groups of motions for a spacetime) that the subgroups generated by boosts must be non-compact[1][2][3]. This is true of all commonly used kinematic groups such as the Poincaré group, Galilean group, De Sitter group, Anti De Sitter group, and also the less commonly used examples.

Why is this a requirement of kinematic group? The only idea that I have been able to come up with so far is that compactness may spoil the causal structure of a spacetime otherwise.

Consider this argument : Take some spacetime vector space $V$, with one time dimension and the rest spatial. Now take the oriented projective space $OV \cong V\setminus \{0\} / \mathbb{R}+$, where we contract every type of vector onto the $n-1$ sphere. Vectors of various types, future timelike, past timelike, null, and spacelike if they exist, all form separate regions on that sphere. In the Poincaré case, it's two disks on that sphere for past and future vectors, in the Galilean case the sphere is split in two regions by a circle of null vectors, etc.

The stabilizer subgroup of the kinematic group $\mathrm{Stab}(K)$ spans the entire region for each : the orbit space of any vector of any type by $\mathrm{Stab}(K)$ is the entire region. However, if the stabilizer subgroup is entirely compact (rotations are usually assumed compact as well), then the orbit generated will be compact as well, by continuity of the action map $g \to g \cdot x$. Since the projective space $OV$ is compact and connected, there exists only two possible compact open sets on it, which are the whole space itself or the empty set. Therefore if our stabilizer subgroup is compact, there is no clear differentiation of vector types as in the other cases.

Does this sound like a proper argument for it, and also, what would be a good argument for the case where only one of the subgroups generated by boosts was compact?

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If a set is compact, any sequence of points of this set has a subsequence converging to a point of the set, whereas, if we consider a boost $a$ corresponding to a nonzero speed, then the sequence $a^n$ will not have a subsequence converging to a point of the set of boosts: for large $n$, the speed can be arbitrarily close to $c$, but there are no boosts corresponding to speed $c$.

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  • $\begingroup$ Doesn't this implicitly assume that the topology on the set of boosts is the same as that of their relative velocities $\vec{v}$? Can we necessarily assume that's the case? If I instead said that "the space of all boosts has the topology induced by the image of the vector $(1,0,0,0)$ in $\mathbb{R}^4$" then the sequence you've described would be pretty clearly non-convergent under this topology. $\endgroup$ Jul 3, 2023 at 14:44
  • $\begingroup$ Also, this argument doesn't seem to apply to the Galilean group, which was another group the OP asked about. $\endgroup$ Jul 3, 2023 at 14:45
  • $\begingroup$ @MichaelSeifert : "Doesn't this implicitly assume that the topology on the set of boosts is the same as that of their relative velocities $v$?" I am not sure. My understanding is the choice of topology on a Lie group can be pretty limited (doi.org/10.1016/j.aim.2011.07.019), but I am not sure (either way) any topology on the set of boosts that is "natural" in some sense for the Lorentz group is necessarily equivalent to the topology you mention. $\endgroup$
    – akhmeteli
    Jul 3, 2023 at 18:16
  • $\begingroup$ @MichaelSeifert : "Also, this argument doesn't seem to apply to the Galilean group, which was another group the OP asked about." No, it does not apply directly, but it seems to apply if "$c$" is replaced with "infinity". $\endgroup$
    – akhmeteli
    Jul 3, 2023 at 18:18
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The very (extremely) low tech way of thinking is as follows. A compact group has finite volume. This means if you apply a transformation enough times, you should come back to where you started. This is the case with $SU(2)$ or any other rotation group: for some angle of rotation you have to come back where you started. However, this is not the same with boosts. You can perform infinitely many boosts in the $x$ axis but you will never end up with your original frame. Therefore, the boost part of this group must have infinite volume. This means it is not compact.

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  • $\begingroup$ This definitely needs to be reworded, particularly This means if you apply a transformation enough times, you should come back to where you started. This sounds like you’re saying every element of a compact group has finite order, which is clearly wrong. Even in 2D if you take a rotation by an angle $2\pi\alpha$ with $\alpha$ irrational you get a counterexample. I know you said low-tech way of thinking but what does finiteness of volume of the group have to do with having finite order? Also, last sentence should be “not compact”. $\endgroup$
    – peek-a-boo
    Jul 3, 2023 at 14:30
  • $\begingroup$ Thank you for your comment! I apologise for not being clear. What I meant is, in a 1-parameter subgroup, there will be a finite t for which you get the identity again. $\endgroup$
    – emir sezik
    Jul 3, 2023 at 14:43

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