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I've seen it claimed in physics classes that there is an isomorphism of Lie algebras $$\mathfrak{so}\left(1,3\right)\cong \mathfrak{su}\left(2\right)\oplus\mathfrak{su}\left(2\right)$$ However, the argument used relied on taking complex linear combinations of a basis for $\mathfrak{so}\left(1,3\right)$ so that the resulting basis contained two copies of $\mathfrak{su}\left(2\right)$.

Namely, there is a basis for $\mathfrak{so}\left(1,3\right)$ given by $\left\{J_1,J_2,J_3,K_1,K_2,K_3\right\}$ with $$\left[J_i,J_j\right]=\sum _{k=1}^3\varepsilon _{ijk}J_k$$ $$\left[K_i,K_j\right]=\sum _{k=1}^3\varepsilon _{ijk}J_k$$ $$\left[J_i,K_j\right]=\sum _{k=1}^3\varepsilon _{ijk}K_k$$ The new elements $J_i^{\pm}=J_i\pm iK_i$ then have the commutation relations of two copies of $\mathfrak{su}\left(2\right)$.

The problem I have with this derivation is that $\mathfrak{so}\left(1,3\right)$ and $\mathfrak{su}\left(2\right)$ are real Lie algebras. So, taking complex linear combinations doesn't really make sense.

Is the stated result true?

Joshua Tilley
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  • Nope, this isomorphism is false. The Lie algebra $so(1,3)$ is noncompact (it has indefinite Killing form) while $su(2)\oplus su(2)$ is compact. Another way to see that such an isomorphism does not exist is to observe that $so(1,3)$ is simple while $su(2)\oplus su(2)$ is not. – Moishe Kohan Oct 06 '19 at 13:02
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    Possible duplicate of Relationship of $\mathfrak{so}(1,3)$ to $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$. – Dietrich Burde Oct 06 '19 at 13:03
  • Looks unlikely to be. Your RHS is the Lie algebra of a compact Lie group, your LHS is the Lie algebra of a non-compact Lie group. – Angina Seng Oct 06 '19 at 13:03
  • @Lord Shark the Unknown Which theorem are you using to go from topological properties of the group to properties of its Lie algebra? – Joshua Tilley Oct 06 '19 at 13:04
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    @JoshuaTilley: Every finite-dimensional real Lie algebra is isomorphic to the Lie algebra of a unique (up to a continuous isomorphism) simply-connected real Lie group. – Moishe Kohan Oct 06 '19 at 13:19
  • @MoisheKohan, what then is the signature of the Killing form in the case of $\mathfrak{so}\left(1,3\right)$? – Joshua Tilley Oct 06 '19 at 18:29
  • @LordSharktheUnknown, Can you tell me which are the simply connected Lie groups in either case to show the compactness vs non-compactness? – Joshua Tilley Oct 06 '19 at 18:31
  • @JoshuaTilley: The signature is $(3,3)$: The maximal compact subalgebra in $o(3,1)$ is $o(3)$ contributes a maximal negative subspace of the Killing form. – Moishe Kohan Oct 06 '19 at 18:38
  • @MoisheKohan, Thanks for the info. Any chance you could post a more step by step explanation if you have time? It is not clear to me that $\mathfrak{o}\left(n\right)$ has a +ve definite Killing form for instance, I think the rest I can then get. – Joshua Tilley Oct 06 '19 at 18:40
  • @JoshuaTilley: https://math.stackexchange.com/questions/37323/proof-that-lie-group-with-finite-centre-is-compact-if-and-only-if-its-killing-fo?rq=1 – Moishe Kohan Oct 06 '19 at 18:42
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    It's the complexifications of those Lie algebras which are isomorphic, and apparently many sources are too sloppy to write this, and so every couple of weeks somebody asks about this here. Besides @DietrichBurde's link, cf. for example https://math.stackexchange.com/q/639749/96384 and https://math.stackexchange.com/q/1109369/96384. – Torsten Schoeneberg Oct 06 '19 at 21:01

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$\mathfrak{so}(1, 3)$, while a real Lie algebra, happens to be isomorphic as a real Lie algebra to the complex Lie algebra $\mathfrak{sl}_2(\mathbb{C})$. This Lie algebra is the complexification of the real Lie algebra $\mathfrak{su}(2)$, so we have a direct sum decomposition

$$\mathfrak{sl}_2(\mathbb{C}) \cong \mathfrak{su}(2) \oplus i \mathfrak{su}(2)$$

but this is not a Lie algebra direct sum; the factor $i \mathfrak{su}(2)$ isn't closed under Lie bracket. It is just a way of expressing that $\mathfrak{sl}_2(\mathbb{C})$ is the complexification of $\mathfrak{su}(2)$.

Qiaochu Yuan
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