# Why does one study the representations of Lorentz group instead of studying only the representations of Poincare group?

Why does one separately study the representations of the Lorentz group and the Poincaré group, instead of directly and only studying the representations of the Poincaré group? After all, the Lagrangians of field theory should not only be Lorentz invariant but should also be Poincaré invariant.

It is not so much true that we separately study the representations of the Poincaré and the Lorentz group as that simply the two coincide in some cases:

1. In the case where we are interested in the finite-dimensional representation on the classical target space of the fields, the translation part of the semi-direct product $\mathrm{P}(3,1) = \mathrm{O}(3,1)\ltimes\mathbb{R}^4$ acts trivially on the fields ($A_\mu(x)\mapsto A_\mu(x-a)$) and we are merely interested in the representation $\rho : \mathrm{SO}(3,1)\to\mathrm{GL}(V)$ of the Lorentz part as $A_\mu(x)\mapsto \rho^\nu_\mu(\Lambda)A_\nu(\Lambda^{-1}x)$. That is, in this case there is no distinction between representations of the Poincaré and the Lorentz groups.

In particular, this means that "invariance under the Poincaré group" for a field theory Lagrangian is no stronger a demand than "invariance under the Lorentz group".

In essence, what I'm saying here is that when you have a function of spacetime (a field), the behaviour under translation is already fixed by the natural way such a function transforms under translation, no matter where it takes values - translating a vector-valued function does nothing to the vector values, it just shifts them around. However, a Lorentz transformation also rotates the vectors at every point. So it is non-trivial to determine how exactly the vectors transform, but the part of the Poincaré group that's just translations acts on all those function in the same trivial way - it doesn't do anything to the values, it just shifts the point at which the values are taken.

2. In the quantum mechanical case where we are interested in projective representations, we are seeking unitary irreducible representations of the universal cover $\bar{P}(3,1) = \mathrm{SL}(2,\mathbb{C})\ltimes\mathbb{R}^4$ of the identity component of the Poincaré group on the infinite-dimensional space of states, not the finite-dimensional target space of fields. Now, the theory of induced representations developed by Mackey says that for $G\ltimes H$ with Abelian $H$ we must find the orbits of the natural action of $G$ on $H$. Then, every irreducible representation of $G\ltimes H$ is obtained by choosing one orbit, choosing a unitary irreducible representation of the stabilizer subgroup of that orbit and a unitary character $\chi:\mathbb{R}^4\to\mathbb{C}$.

Now, if one translates this into "physics language", one fixes $\chi = 1$ and recognizes that the "stabilizers of the orbits" are nothing but Wigner's little groups as they are employed in Wigner's classification.

In this case, the translation part acts non-trivially because what it acts on are not functions of spacetimes, but abstract quantum states that have no clear "dependence" on position, so we can't just shift their argument as for the fields. Here, the crucial requirement is that the representation of the Poincaré group be unitary/projective, and this leads to non-trivial transformation behaviour under the translations.

There is usually a lot of confusion about which of the two cases we are in in a given situation in quantum field theory. That is because QFT inherently mixes both situations by the Wightman axiom that demands that the (Lorentz) representation $\rho_\text{fin}$ on the classical target space of the fields is compatible with the representation $\mathrm{U}$ on the quantum mechanical Hilbert space of states such that $$\rho_\text{fin}(\Lambda^{-1})\phi(\Lambda x + a) = \mathrm{U}(\Lambda,a)\phi(x)\mathrm{U}(\Lambda,a)^\dagger$$ for every element $(\Lambda,a)\in\bar{\mathrm{P}}(3,1)$. What this axiom is saying is that QFT is only consistent if transforming a field when you consider it as a classical field is the same as transforming it when considering it as an operator on the quantum space of states.

• @ ACuriousMind-Under homogeneous Lorentz transformation a vector field $A_\mu(x)$ transform as: $A_\mu(x)=\Lambda^\nu_\mu A_\nu(\Lambda^{-1}x)$. I couldn't understand what is $\rho^\mu_\nu$ and $u(\Lambda)$ in your transformation rule as given in point 1. Can you give the transformation rule under an arbitrary Poincare transformation (i.e., a transformation rule for $A_\mu$ under a simultaneous homogeneous and inhomogeneous Lorentz transformations)? – SRS Nov 18 '16 at 9:11
• I think, under a Poincare transformation, a vector field should behave as $A_\mu(x)=\Lambda^\nu_\mu A_\nu(\Lambda^{-1}(x-a))$. Am I correct? – SRS Nov 18 '16 at 9:50
• @SRS There was a typo in the Lorentz transformation, thanks. The transformation rule under an arbitrary Poincaré transformation in 1. is simply obtained by combining what I wrote for the translation part and the Lorentz part, and gives the l.h.s. of the Wightman axiom I wrote at the end. – ACuriousMind Nov 18 '16 at 14:42
• Goodness, that got technical fast. Is it possible to summarize for more of a dumb-theorist audience? Or is it just intrinsically technical? – Emilio Pisanty Nov 18 '16 at 15:04
• @EmilioPisanty I tried to explain some parts in a less technical way, but especially the way we determine the allowed representations on the space of states (Wigner's classification) is intrinsically technical, I'm afraid. – ACuriousMind Nov 18 '16 at 15:17