# Poincaré group representation generator commutators

I am currently trying to understand how to derive the commutation relations for the generators of the Poincaré group. What I am reading instructs to start with:

$$U( \Lambda, a) = e^{\frac{i}{2} \epsilon_{\mu \nu} M^{\mu \nu}} e^{i a_{\mu} P^{\mu}}$$

And use the representation relations:

$$U(\Lambda , a) U(\Lambda', a') U^{-1}(\Lambda, a) = U(\Lambda \Lambda' \Lambda^{-1} , a + \Lambda a' - \Lambda \Lambda' \Lambda^{-1} a)$$

Along with an infinitesimal Lorentz transformation such that: $$\Lambda^\mu_\nu = \delta^\mu_\nu + \epsilon^\mu_\nu$$. I have no issue with determining the commutator $$[ P^\mu, P^\nu]$$ since it can be determined by letting $$\Lambda = 1$$. The issue I have is determining the commutators:

$$i \big[ M^{\mu \nu}, M^{\rho \sigma} \big] = g^{\mu \sigma} M^{\nu \rho} + g^{\nu \rho} M^{\mu \sigma} - g^{\mu \rho} M^{\nu \sigma} - g^{\nu \sigma} M^{\mu \rho}$$

$$i\big[P^{\mu} , M^{\rho \sigma} \big] = g^{\mu \rho} P^\sigma - g^{\mu \sigma} P^\rho$$

My confusion is this: When I derive the commutator for the $$P^\mu$$, I know exactly how to relate the infinitesimal parameter $$a$$ to the exponential $$e^{i a_{\mu} P^{\mu}}$$ since $$e^{i a_{\mu} P^{\mu}} = 1 + i a_\mu P^\mu$$. Then, setting $$\Lambda = 1$$ then makes it easy to determine the commutators. However, it is a little less clear to me how to relate the infinitesimal parameters in the infinitesimal Lorentz transformation: $$\Lambda^\mu_\nu = \delta^\mu_\nu + \epsilon^\mu_\nu$$ to the exponential $$e^{\frac{i}{2} \epsilon_{\mu \nu} M^{\mu \nu}}$$ and to the operator $$U(\Lambda \Lambda' \Lambda^{-1} , a + \Lambda a' - \Lambda \Lambda' \Lambda^{-1} a)$$. If there is just one $$\Lambda$$, then it is clear that: $$\Lambda^\mu_\nu = \delta^\mu_\nu + \epsilon^\mu_\nu$$, with $$\epsilon^\mu_\nu$$ infinitesimal. This gives: $$e^{i \epsilon_{\mu \nu} M^{\mu_\nu}} = 1 + \epsilon_{\mu \nu} M^{\mu \nu}$$ Then the expression $$U(\Lambda , a) U(\Lambda', a') U^{-1}(\Lambda, a)$$ can be determined rather easily. But for $$U(\Lambda \Lambda' \Lambda^{-1} , a + \Lambda a' - \Lambda \Lambda' \Lambda^{-1} a)$$, I am unsure as to whether I need to just keep up to $$O(\epsilon)$$ in the argument of the operator and then expand, or if I should keep up to $$O(\epsilon \epsilon')$$. If anyone could help that would be great.

Edited for clarity. Thanks to Cosmas Zachos for a clear answer.

• You may first expand all exponentials to infinitesimal order in the parameters ε and a... Commented Mar 30, 2023 at 22:13
• For the relation between the infinitesimal trafo, and the exponential it is helpful to consider the explicit form of $M$ in the fundamental rep. to relate $\epsilon^\mu_\nu$ and $\epsilon_{\mu,\nu}$ Commented Mar 31, 2023 at 9:18
• @CosmasZachos I am aware of how to do that, but the issue is with expanding the $U(\Lambda \Lambda' \Lambda^{-1}, ...)$ in this fashion. Would each $\epsilon$ for each transformation be attached to the $M^{\mu \nu}$ in this case? Commented Mar 31, 2023 at 17:41
• @ThomasTappeiner Presumably we just have that $\epsilon^\mu_\nu$ = $\epsilon_{\alpha \nu} g^{\mu \alpha}$, right? Commented Mar 31, 2023 at 17:44
• Your pathological expression $\Lambda^\mu_\nu = 1 + \epsilon^\mu_\nu$ is meaningless nonsense. You actually mean $\Lambda^\mu_\nu = \delta^\mu_\nu + \epsilon^\mu_\nu$. Now you know how to compute $\Lambda \Lambda'\Lambda ^{-1}$, etc. Show your work, so the reader knows what you are asking... Commented Mar 31, 2023 at 19:01

I'll start you with the case of vanishing translations, a=0, $$U( \Lambda, 0) = e^{\frac{i}{2} \epsilon_{\mu \nu} M^{\mu \nu}} \approx {\mathbb I} + \frac{i}{2} \epsilon_{\mu \nu} M^{\mu \nu} + O(\epsilon^2),\leadsto \\ U( \Lambda, 0)^{-1} = e^{-\frac{i}{2} \epsilon_{\mu \nu} M^{\mu \nu}} \approx {\mathbb I} - \frac{i}{2} \epsilon_{\mu \nu} M^{\mu \nu} + O(\epsilon^2),$$ so that $$U(\Lambda , 0)~ U(\Lambda', 0)~ U^{-1}(\Lambda, 0)\\ = \left ({\mathbb I} + \frac{i}{2} \epsilon_{\mu \nu} M^{\mu \nu} + O(\epsilon^2)\right ) \left({\mathbb I} + \frac{i}{2} \epsilon'_{\rho \sigma} M^{\rho \sigma} + O(\epsilon^2)\right ) \left({\mathbb I} - \frac{i}{2} \epsilon_{\kappa \lambda} M^{\kappa \lambda} + O(\epsilon^2)\right )\\ = {\mathbb I} + \frac{i}{2} \epsilon'_{\rho \sigma} M^{\rho \sigma} -\frac{1}{4} \epsilon_{\mu \nu} \epsilon'_{\rho \sigma} [M^{\mu \nu}, M^{\rho \sigma}]+ O(\epsilon^2) + O(\epsilon'^2). \tag{1}$$
Now, $$\Lambda^\mu_{~~\nu}\Lambda'^\nu_{~~~\kappa}(\Lambda^{-1})^\kappa_{~~\rho}\\ = \delta^\mu_\rho+\epsilon'^\mu_{~~~~\rho}+ \epsilon^\mu_{~~\kappa}\epsilon'^\kappa_{~~\rho}-\epsilon'^\mu_{~~~~\kappa}\epsilon^\kappa_{~~\rho} + O(\epsilon^2)+O(\epsilon'^2),$$ hence, $$U(\Lambda \Lambda' \Lambda^{-1},0) \\ = {\mathbb I} + \frac{i}{2}( \epsilon'_{\mu \nu} + \epsilon_{\mu\kappa}\epsilon'^\kappa_{~~\nu}-\epsilon'_{\mu\kappa}\epsilon^\kappa_{~~\nu} ) M^{\mu \nu} + O(\epsilon^2)+O(\epsilon'^2) .\tag{2}$$
Equating (1) to (2), after an orgy of relabeling and exploitation of the antisymmetry of the indices of the generators M yields your target Lie algebra, $$i \big[ M^{\mu \nu}, M^{\rho \sigma} \big] = g^{\mu \sigma} M^{\nu \rho} + g^{\nu \rho} M^{\mu \sigma} - g^{\mu \rho} M^{\nu \sigma} - g^{\nu \sigma} M^{\mu \rho} .$$
• Yeah this makes sense. I should be able to get it for the non vanishing $a$. Thanks a lot for your time! Commented Apr 3, 2023 at 16:15