# Which are the underlying Lie group and algebra related to the translation invariance in field theories?

I'm new to Physics SE. I've seen a lot of interesting questions and answers, and thought it will be very useful to participate a little.

I'm currently stuck in a, probably, very simple matter, regarding the nature of the linear momentum $$P_{\alpha}$$ in field theory. I know it is commonly known as the "infinitesimal generator of translations", pointing an obvious relation with the Lie algebra generators. But which are the Lie group, and correspondingly Lie algebra, associated with $$P_{\alpha}$$?

I thought of something like the following: if $$M$$ is the Minkowski spacetime, let $$G = (M,+)$$ be the Lie group, therefore $$G$$ acts on $$M$$. So if $$x^{\alpha}\in M$$ and $$g\in G$$, then:

$$g(x^{\alpha}) = x^{\alpha} + \delta^{\alpha}$$,

but I've been struggling to mathematically write the relation between the algebra generators, the exponential map and the momentum operator $$P^{\alpha}$$.

Does anyone knows how to point me in the right direction? Thank you all!

• Translations, rotations, and boosts, oh my. You may be looking at a subgroup of the Poincare group – ggcg Jan 6 at 16:26
• Thanks for your answer! Yes, I agree with you on that, but how can I mathematically write, or how to algebraically deduce, the relation between $P_{\alpha}$ and the Lie algebra? Other than the statement "Oh, it is in the exponential map (like $e^{-i\delta^{\alpha}P_{\alpha}}$) so it must be the generator!" – Condereal Jan 6 at 16:32
• I see. Also, are you looking at local in invariance in the tangent space or large scale invariance in M, there it would be part of the diffeomorphism group. – ggcg Jan 6 at 16:59

## 1 Answer

The group is $$\mathbb{R}^4$$ (tuples of 4 real numbers), group product is vector space addition, group inverse is the opposite vector, and group identity is the zero vector. Easy to see that the group is Abelian (commutative).

Corresponding Lie algebra is $$4 \mathfrak{u}_1$$, which also consists of tuples of 4 numbers, with the Lie bracket identically zero for any two elements.

This is reflected in the relation $$[ P_{\mu}, P_{\nu} ] = 0.$$

It follows that 4-momenta components are simultaneously diagonalizable, which enables us to speak about the spectrum of the quantum field theory, and also that the momentum conservation law holds (as the components of 4-momenta commute with the Hamiltonian $$P_0$$).

$$\mathbb{R}^4$$ is the maximally noncompact form of $$4 \mathfrak{u}_1$$, the other forms having one or more of their dimensions compactified on a circle. Of course in reality space is infinitely extended... probably. Or not.

As mentioned in the comments, this is only part of the Poincare group/algebra, which also contains rotations and boosts (Lorentz transformations of the reference frame).

• Nice, thank you for your answer. I sounds logic using $\mathbb{R}^4$ instead of $M$. Do you know if there is a way to explicitly show that $P_{\alpha}$ is related to the tangent space of $\mathbb{R}^4$ in the origin? That would be the generator definition as far as I understand. – Condereal Jan 6 at 16:56
• @Condereal translations are only symmetries of the flat Minkowski space. They are global symmetries, not related to local topological structures such as the tangent bundle. If you want a generalization for general relativistic space time, look up Killing vector fields and their Lie algebra (the closest thing you can get to global Poincare in GR). However, QFTs with these symmetries don’t work. You are well aware that QFT and GR are incompatible, right? :) – Prof. Legolasov Jan 6 at 17:00
• Yes yes, of course, sorry if I'm probably messing myself while trying to explain. What I'm trying to write down is the path that leads from the unitary transformation $U = exp(-i\delta^{\alpha}P_{\alpha})$ as typically deduced in field theories, to the idea that $P_{\alpha}$ is the generator of a Lie algebra. I do not see a trivial relation between a group that acts more or less as I explained in my question, and $P_{\alpha}$ being the generator. – Condereal Jan 6 at 17:07
• @Condereal ah well that is true for any Lie group/algebra. Just differentiate the exponential wrt the coordinate on the group $\delta$ at the group identity $\delta = 0$. Or are you asking why we conclude that $P$ are 4-momenta? In that case the answer is — Noether theorem. – Prof. Legolasov Jan 6 at 17:15
• Thank you for your help! So, $U(\delta^{\alpha})\in G$, and $\left.\frac{dU}{d\delta}\right|_{\delta=0}=-iP_{\alpha}$ is enough as a statement that "$P_{\alpha}$ is the Lie algebra generator because it is evaluated in the group identity $\delta^{\alpha}=0$? If that's the case I think I've understood. – Condereal Jan 6 at 17:23