It can be shown easily, by introducing new generators from the usual ones that we can think of the Lie algebra of the Lorentz group as being built up by two copies of the $SU(2)$ Lie algebra:

$$ \mathfrak{so}(3,1) \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2) $$

The Poincare group is a semidirect product of the translations and the Lorentz group.

Is there a similar relation for the Lie algebra of the Poincare group?

  • 1
    $\begingroup$ Careful, it is not true that $\mathfrak{so}(3,1) = \mathfrak{su}(2)\oplus\mathfrak{su}(2)$, see this answer by Qmechanic. $\endgroup$
    – ACuriousMind
    Apr 19 '15 at 15:12
  • $\begingroup$ Since the Poincare group contains the Lorentz group, it inherits the nice group theoretic property you reference here. But that is all, there is no corresponding factorization for the translations. $\endgroup$ Apr 20 '15 at 2:07

Well, as it is noted in the first comment, it is not true that the Lorentz group algebra is isomorphic to the vector space sum of two su(2) algebras, but the complexification of the Lorentz algebra is isomorphic to the vector space sum of two copies of the sl(2) algebra seen as a vector space over $\mathbb C$. Mathematically:

$$ \mathfrak{lor}^{\mathbb C} (1,3) \simeq \mathfrak{sl}_\mathbb{C} (2,\mathbb C) \oplus \mathfrak{sl}_\mathbb{C} (2,\mathbb C) $$

Moving to the Poincaré group, there is no corresponding relation, because the Poincaré algebra, unlike the Lorentz one, is not semisimple, it has a non-trivial abelian subalgebra, namely, the algebra of 4-translations. So any Cartan or Iwasawa decomposition of the algebra does not exist. One can still have a Levi decomposition as in the answer by Qmechanic below.

  • $\begingroup$ I would also like to mention that $\mathfrak{so}(1,3)\sim\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ is actually one of the well-known accidental isomorphisms. One does not expect such a decomposition for, say, the Lorentz Group in higher dimensions or other relevant symmetry algebras (Galilei, etc.). $\endgroup$ Nov 1 '17 at 22:19
  1. The restricted Lorentz group $$SO^+(1,3;\mathbb{R})~\cong~ SL(2,\mathbb{C})/\mathbb{Z}_2 \tag{1} $$ has Lie algebra $$so(1,3;\mathbb{R})~\cong~sl(2,\mathbb{C}),\tag{2}$$ not $su(2)\oplus su(2)$, as OP writes in the title (v3). This is e.g. explained in this Phys.SE post.

  2. The complexified proper Lorentz group $$SO(1,3;\mathbb{C})~\cong~ [SL(2,\mathbb{C})\times SL(2,\mathbb{C})]/\mathbb{Z}_2 \tag{3} $$ has Lie algebra $$so(1,3; \mathbb{C})~\cong~sl(2,\mathbb{C})\oplus sl(2,\mathbb{C}).\tag{4}$$

  3. The Poincare group is a semidirect product $$ O(1,3;\mathbb{R}) \ltimes \mathbb{R}^{1,3} \tag{5}$$ of the Lorentz group $O(1,3;\mathbb{R})$ and the abelian normal subgroup $(\mathbb{R}^{1,3},+)$ of translations. The Poincare algebra is a vector space sum of two Lie subalgebras $$ o(1,3;\mathbb{R}) \oplus \mathbb{R}^{1,3}.\tag{6}$$ It is not a direct sum of two Lie algebras, i.e. the Lie bracket between the two summands is not zero. In fact, $\mathbb{R}^{1,3}$ furnishes a 4-dimensional irreducible representation of $o(1,3;\mathbb{R})$.

  • $\begingroup$ I would like to add that this point about complexification first and the subtleties of going back to the real form from the complexification is very often overlooked. For instance the adjoint representation is irreducible over the reals, but reducible over the complex. $\endgroup$ Nov 5 '17 at 15:57

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