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Can we define a group operation of pure Lorentz Boosts?

Consider the Lorentz group. By the Wigner rotation, the composition of two pure boosts results in a non-pure boost. However, fortunately a Lorentz group element can always decomposited into a composition of a pure boost and a rotation by polar decomposition theorem, i.e. for any pure boosts $B_1$ and $B_2$ there exists unique pure boost $B_3$ and unique rotation $R$ such that $B_2 \circ B_1 = R \circ B_3$.

I am wondering if we can define a group operation of pure boosts. My thought is that the group operation is defined by $$ B_2 B_1 = B_3 \text{ where } B_2 \circ B_1 = R \circ B_3 \text{ by polar decomposition.} $$ It is a kind of looking like a quotient space of Lorentz group modulo the rotation group, e.g. the oriented hyperbolic space: $SO^+(1, n)/SO(n)$.

Po C.
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    Related: https://physics.stackexchange.com/q/161313/2451 , https://physics.stackexchange.com/q/395036/2451 , https://physics.stackexchange.com/q/525974/2451 and links therein. – Qmechanic Feb 23 '23 at 01:03
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    The multiplication rule is likely not associative. – Prahar Feb 23 '23 at 01:21
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    Collinear boosts form a group. – Ghoster Feb 23 '23 at 04:25
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    Since this binary operation clearly has totality, an identity & inverses, if as @Prahar suspects it's not associative it only defines a loop, not a group. – J.G. Feb 23 '23 at 22:04
  • A coset space is not a group, however: your construction is reminiscent of the axial generators of current algebra which definitely do not close to a group, but specify projective coordinates for the analogous coset space. – Cosmas Zachos Feb 23 '23 at 23:44
  • This is a strictly math question, suitable for MSE. The subgroup H of rotations specifies a right coset Hg of the Lorentz group consisting of the left action of rotations on the group, $H\times G \to G$. So the coset space G/H is the set of such orbits and is not a group for the set of boosts. Asking it there might yield pithy answers.... – Cosmas Zachos Feb 24 '23 at 15:08
  • If you are inclined to compute an explicit (counter)example, you could use this shortcut. – Cosmas Zachos Feb 24 '23 at 21:03

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