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I am aware that there are 6 independent infinitesimal Lorentz transformations that can be separated into 3 rotations and 3 boosts. Is it possible for a quantum field theory to be invariant under the boosts but not invariant under the rotations?

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    $\begingroup$ No, the Lie algebra cannot be separated like that. The commutator of two boosts is a rotation. $\endgroup$ – ClassicStyle Jan 6 '17 at 2:29
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    $\begingroup$ So I know the commutator of two boosts is a rotation and I suspected that that was the key to this but I can't see exactly why that rules it out $\endgroup$ – Paul Malinowski Jan 6 '17 at 2:31
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Every quantum field theory has a symmetry group under which its Lagrangian is invariant. Like every group, it must be closed. The boosts are not closed under composition, so they cannot form a symmetry group by themselves.

Any rotation can be achieved by a composition of four boosts, so if each boost leaves the Lagrangian invariant then the resulting rotation will as well.

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    $\begingroup$ To add further to this answer and the one below it should be noted that there exist quantum field theories that are invariant under SO(3) but not under boosts (typically condensed matter systems). $\endgroup$ – I.E.P. Jan 6 '17 at 6:22
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    $\begingroup$ @I.E.P. Good point - the three spatial rotations in a given frame are closed under composition, so SO(3) rotational symmetry is a perfectly good symmetry group for a nonrelativistic theory. $\endgroup$ – tparker Jan 6 '17 at 7:12
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To add to tparker's answer that the boosts by themselves are not closed under composition, you can either prove this by direct computation - take the product of two boost matrices and you can show that the polar decomposition of the product is generally a boost composed with a nontrivial rotation - or by far the easiest way to prove this is by the Lie correspondence - see Rossmann, "Lie Theory- an Introduction through Linear Groups", section 2.5. Here we learn that the connected analytic subgroups of any connected Lie group correspond bijectively to the Lie subalgebras of the group's Lie algebra. So we only need to witness that the Lie bracket of two infinitesimal boosts is an infinitesimal rotation to prove that the smallest group containing the boosts must needfully include rotations as well.

Indeed, this non closure gives rise to the phenomenon of Thomas Precession / Wigner Rotation.

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  • $\begingroup$ Lie algebras actually only correspond bijectively to simply connected Lie groups, not to all connected Lie groups. $SU(2)$ and $SO(3)$ are not isomorphic, but their Lie algebras are. They're only locally isomorphic but differ in their global topological structure, and they share the same universal cover $SU(2)$, which is simply connected by definition. $\endgroup$ – tparker Jan 6 '17 at 20:36
  • $\begingroup$ @tparker No, that's a different, categorical statement - that you have to consider simply connected ones to make the a functor "Lie" (mapping groups to algebras as categories) bijective. The Lie correspondence is saying something different: given one particular connected Lie group, its connected Lie subgroups correspond one-to-one with the Lie subalgebras of that particular group's algebra. (Replace "Lie subgroup" with "analytic subgroup" if you're one of the many people who insist on reserving the word "Lie subgroups" exclusively for embedded Lie subgroups :) ) $\endgroup$ – WetSavannaAnimal Jan 6 '17 at 23:28

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