Is the Lie algebra with generators $a$, $b$, $c$ and commutators $$ [a,b]=2c, \quad [c,b]=2a, \quad [c,a]=2b $$ isomorphic to something well-known?
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2This is almost a presentation of $\mathfrak{so}(3)$ except that the middle relation should be $[b, c] = 2a$. – Qiaochu Yuan Jun 15 '17 at 22:34
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1@QiaochuYuan $2+3=4$ except that $3$ should be replaced by $2$. – YCor Jun 16 '17 at 08:05
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@YCor: well, I was implicitly asking if the OP intended the other presentation. – Qiaochu Yuan Jun 16 '17 at 23:13
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It's isomorphic to $\mathfrak{sl}_2$ (over any field of characteristic $\neq 2$).
Indeed, write $x=(c-a)/2$, $y=(c+a)/2$, $h=b$. Then $[x,y]=h$, $[h,x]=2x$, $[h,y]=-2y$, which is the standard Lie algebra law for $\mathfrak{sl}_2$.
(Of course in characteristic 2, it's abelian.)
YCor
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Except that I get $[x,y]=-h$: am I making a mistake or is there a way to also fix this? – jj_p Jun 16 '17 at 16:34
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Denote by $L$ the above Lie algebra over a field $K$. Since $\dim L=3$ and obviously $[L,L]=L$ we already know that, for $K=\mathbb{C}$, $L\cong \mathfrak{sl}_2(K)$. For $K=\mathbb{R}$ one could compute the Killing form to see that it is not negative definite. Thus again $L\cong \mathfrak{sl}_2(K)$. But clearly, Yves answer is the best one, as we see the answer for all fields with $2\neq 0$ directly.
Dietrich Burde
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2What I actually did was to also first observe that the Lie algebra is simple, and then to check if we have some nonzero element $h$ such that $\mathrm{ad}(x)$ is diagonalizable over the ground field (since this is enough to be isomorphic to $\mathfrak{sl}_2$), and in the original basis $(a,c,b)$ indeed $\mathrm{ad}(b)=\begin{pmatrix}0 & -2 & 0\ -2 & 0 & 0\ 0 & 0 & 0\end{pmatrix}$, which is diagonalizable, and I just changed the other basis element to make it diagonal. – YCor Jun 16 '17 at 10:19