Lorentz boost transformations form a group?

Question

In the QFT book of Ryder, he states that Lorentz boost transformations do NOT form a group. This is due to the boost generators $\textbf{K}$, i.e. they do not form a closed algebra under commutation. Mathematically:

\begin{equation} [ K^{i}, K^{j} ]=-i{\epsilon^{ijk}}J^{k}.\tag{1} \end{equation} This makes sense to me since boosts cause the Lorentz group (group?) to be non-compact ( you can keep boosting the system till you reach $c$). Is that what he means?

I think the Lorentz transformations compose the restricted Lorentz group (?) — Daddy Kropotkin, Jun 05 '21 at 13:04
Combining two Lorentz boosts results to a rotation followed by a Lorentz boost or by a Lorentz boost followed by a rotation, any order you like more. See the answers here Combining two Lorentz boosts — Frobenius, Jun 05 '21 at 13:29
Proper homogeneous Lorentz transformations form a 6-parametric group. What are the proper homogeneous Lorentz transformations and why they form a group see here Show that any proper homogeneous Lorentz transformation may be expressed as the product of a boost times a rotation — Frobenius, Jun 05 '21 at 13:35
possible duplicate: https://physics.stackexchange.com/q/525974/226902 — Quillo, Oct 07 '22 at 11:01

score 5 · Answer 1 · answered Jun 05 '21 at 13:23

All this means is that the pure boosts do not form a subgroup of the Lorentz group. That commutator tells you that it is possible to do a series of boosts which result, overall, in a spatial rotation.

The boosts plus the spatial rotations on the other hand do of course form a subgroup, specifically the restricted Lorentz group, usually denoted $SO^+(1,3)$.

It is a bit unhelpful to say "the Lorentz transformations do not form a group", since we usually think of "the Lorentz transformations" as simply being the elements of $SO^+(1,3)$.

Lorentz boost transformations form a group?

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