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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?

Qmechanic
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Tim
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2 Answers2

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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.

DanielC
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  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 $o(1,3;\mathbb{R})$.

Qmechanic
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