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I have a Lie algebra comprised of the generators $\{e_1,e_2,e_3,e_4\}$ for which the only non-zero commutators are

$$ [e_4,e_2]=-i e_3 $$ $$ [e_4,e_3]= i e_2 $$

(Excuse the physicist notation, for mathematicians the generators are actually ${ie_1,ie_2,ie_3,ie_4}$). Clearly, this is a solvable Lie Algebra ($[e_3,e_2]=0$).

My question is: Is there a simple way to refer to it without declaring its commutators? Like we can refer to $su(2)$ instead of writing out all of $su(2)$'s generators. I'd like to be able to do that for this algebra. If there's not a way to do this, I'd like to know that instead.

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What is the field? $\mathbb R$, $\mathbb C$ or any? I can't understand completely your notation. –  emiliocba Jan 24 '12 at 19:32
    
Sorry, $\mathbb R$ is implied in physics, so that we get unitary evolution. However, $\mathbb C$ would be just the complexification of the analogous "real" algebra, no? –  thequark Jan 25 '12 at 1:02

2 Answers 2

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Consider using the basis $$\{ f_1, f_2, f_3, f_4 \} = \{ e_1, i e_2 + e_3, e_2 + i e_3, e_4 \}$$ since all basis commutators are zero except for $$[f_4,f_2]=f_2 \quad [f_4,f_3]=-f_3$$

As emiliocba mentioned, the algebra is the direct product of a trivial one-dimensional algebra generated by $e_1=f_1$, and a three dimensional algebra generated by the other basis vectors.

According to wikipedia this three dimensional algebra is the Lie algebra of type Type VI0, the Lie algebra of the group of isometries of Minkowski space.

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Firt at all, your algebra $\mathfrak g$ can be written as $$ \mathfrak g=\mathfrak h\times \mathbb R, $$ where $\mathbb R$ is the generated space by $e_1$ and $\mathfrak h=\langle e_2,e_3,e_4\rangle$.

Now, $\mathfrak h$ is one of the infinite solvable Lie algebras of dimension three. Look at Exercice 2 of Chapter I of Knapp's book. I don't know if these algebras have a name.

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