Let $(\Lambda,a)$ be a Poincaré coordinate transformation. Let $U$ be an unitary representation of the Poincaré group on some vector space.
Is it always possible to express $U(\Lambda,a)$ in the following way?

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

If no, under which condition is it possible? If the exponential espression is valid, how can be the coefficients $\epsilon_{\mu}$ and $\omega_{\mu \nu}$ retrieved from $(\Lambda,a)$?

Note: The exponential expression should hold for pure translations or spatial rotations, but what about the "mixed" cases or the case of boosts?


1 Answer 1


This is a Lie group, and its connected component around the identity results by exponentiating the Lie algebra, whose most general element you already wrote down.

Expanding the generic group element around the identity, you find, as is standard in Lie theory, $$G(\epsilon, \omega)=e^{i\epsilon_{\mu}P^{\mu}-i\omega_{\mu \nu}M^{\mu\nu}/2}= 1\!\!1 + i\epsilon_{\mu}P^{\mu}-i\omega_{\mu \nu}M^{\mu\nu}/2 + O(\epsilon^2,\omega^2,\epsilon \omega). $$

Normally, however, one writes the full Poincare transformation, always possible, (by theorem, as your text might review), as a product of a translation T and a proper Lorentz transformation Λ, $$ U(a,\Lambda)=T(a) \Lambda =e^{ia_{\mu}P^{\mu}} e^{-i {\tilde \omega}_{\mu \nu}M^{\mu\nu}/2}, $$ so that $$ U( a',\Lambda')U(a,\Lambda)= U(\Lambda' a+ a',\Lambda ' \Lambda). $$ This is your expression U (left-hand side). Notationally, $e^{ia P} f(x)=f(x+a)$ and $\tilde \omega$ is the "angle" of the "generalized noncompact rotation" matrix $\Lambda= 1\!\!1 -i\tilde\omega\cdot M/2+ O(\tilde\omega^2)$.

You may, always, as per Lie's theorem, compose the two exponential factors into a single exponential that looks like the above G, but with its parameters ε,ω as messy functions of those of each factor exponential in your U, as you saw from the composition law. (You may compute the first few orders for fun. Do you see that $\tilde\omega$ will never be modified, so $\omega=\tilde \omega$? Do you see the first subleading term in the translation parameter, $\tilde \omega \cdot a$?)

  • $\begingroup$ Thank you! Is the "connected component" of the Lie group (for which the exponential expression is valid) the entire subgroup of $U(\Lambda,a)$ s.t. $\Lambda$ is a proper orthochronous Lorentz transformation? Second sub-question: could you please suggest me a simple textbook to understand these topics better? (I'm interested just in applying these arguments in quantum field theory) $\endgroup$ Commented May 5, 2020 at 21:16
  • 1
    $\begingroup$ Yes, the orthochronous part, the one that connects to the identity. Students seem to like Wu-ki Tung's book Group theory in physics, but many QFT books have decent reviews, as well... There is a virtually closed form of the mutated translation parameter, reachable from the "special cases" of the WP article, involving coth ... but do you really need it? The conventional factor form is friendlier. $\endgroup$ Commented May 5, 2020 at 21:23
  • $\begingroup$ ...but of course, this question addresses the issue.... $\endgroup$ Commented May 6, 2020 at 14:49

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