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Suppose that we have a Lie algebra $\mathfrak{g}$ of a Lie group G, what can I say about the latter if I can find two Lie subalgefbras $\mathfrak{a},\mathfrak{b}\subset\mathfrak{g}$?

I'm asking this question because in my notes that's written that the Lie algebra of the proper orthocronus Lorentz group $\mathcal{L}_{+}^{\uparrow}$ can be decomposed into $$\mathfrak{so}(1,3)=\mathfrak{su}(2)\oplus\mathfrak{su}(2)$$ so that we conclude $$SO(1,3) ≃ SU(2)\otimes SU(2)$$ How do we reach that conclusion?

Filippo
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    What you found in your notes is simply wrong. – Moishe Kohan Oct 11 '23 at 16:39
  • Well, tbh the implication was the opposite with respect to the one I've written above. Is it also wrong? I mean the generators don't commute so I guess is still wrong – Filippo Oct 11 '23 at 16:41
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    Both isomorphisms are wrong. – Moishe Kohan Oct 11 '23 at 16:43
  • Do you know if there is something that we can say more generally? Forgetting about this relation – Filippo Oct 11 '23 at 16:45
  • If you just can find two subalgebras, not much can be said, one needs more info. For instance that the subalgebras commute with each other. – Moishe Kohan Oct 11 '23 at 16:58
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    Note the $\mathfrak{su}(2) \oplus \mathfrak{su}(2) \cong \mathfrak{so}(3,1)$ (incorrect) isomorphism is commonly referred to by physicists but what they really mean is that they both complexify to the same complex Lie algebra $\mathfrak{sl}(2,\mathbb{C}) \oplus \mathfrak{sl}(2,\mathbb{C}) \cong \mathfrak{so}(4,\mathbb{C})$. – Callum Oct 11 '23 at 18:17
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    See here and here for more confused physicists on exactly this issue. – Callum Oct 11 '23 at 18:18
  • Thank you, I still don't understand how can be showed the isomorphism $\mathfrak{su(2)_{\mathbb{C}}} ≃ \mathfrak{sl(2,\mathbb{C})}$, do you know some good reference? – Filippo Oct 12 '23 at 18:45

1 Answers1

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As Moishe says in the comments, both of the stated isomorphisms are incorrect. I will assume that the real question is in the title rather than the body: what can we say if a Lie algebra $\mathfrak{g}$ is the direct sum (of Lie algebras) of two subalgebras $\mathfrak{a}, \mathfrak{b}$?

What we can conclude is that if $G, A, B$ are the unique simply-connected Lie groups with Lie algebras $\mathfrak{g}, \mathfrak{a}, \mathfrak{b}$ then

$$G \cong A \times B.$$

A correct example of this is that $\mathfrak{so}(4)$ decomposes into a direct sum $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$, which gives an exceptional isomorphism

$$Spin(4) \cong SU(2) \times SU(2).$$

However $\mathfrak{so}(1, 3)$ is simple, so it is not the direct sum of two of its subalgebras. It does have an exceptional isomorphism to $\mathfrak{sl}_2(\mathbb{C})$, meaning it can be written as a direct sum of vector spaces $\mathfrak{su}(2) \oplus i \mathfrak{su}(2)$, since $\mathfrak{sl}_2(\mathbb{C})$ is the complexification of $\mathfrak{su}(2)$. However, $i \mathfrak{su}(2)$ is not a subalgebra, so this is not a direct sum of Lie algebras.

Qiaochu Yuan
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