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.

  • 3
    $\begingroup$ You may first expand all exponentials to infinitesimal order in the parameters ε and a... $\endgroup$ Commented Mar 30, 2023 at 22:13
  • $\begingroup$ 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 $\epsilon_{\mu,\nu}$ $\endgroup$ Commented Mar 31, 2023 at 9:18
  • $\begingroup$ @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? $\endgroup$
    – user132849
    Commented Mar 31, 2023 at 17:41
  • $\begingroup$ @ThomasTappeiner Presumably we just have that $\epsilon^\mu_\nu$ = $\epsilon_{\alpha \nu} g^{\mu \alpha}$, right? $\endgroup$
    – user132849
    Commented Mar 31, 2023 at 17:44
  • 1
    $\begingroup$ 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... $\endgroup$ Commented Mar 31, 2023 at 19:01

1 Answer 1


This is an important step, but most books leave it as an exercise to the index-shuffling-gifted student, out of boredom. Below, feel free to relabel summed-over dummy indices to get to the point.

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

Can you take it from there for non-vanishing a?

  • 1
    $\begingroup$ Yeah this makes sense. I should be able to get it for the non vanishing $a$. Thanks a lot for your time! $\endgroup$
    – user132849
    Commented Apr 3, 2023 at 16:15
  • 1
    $\begingroup$ I was able to get it. Thanks again! $\endgroup$
    – user132849
    Commented Apr 3, 2023 at 19:30

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