# Does there exist a non-trivial, associative Lie algebra?

In a Linear Algebra text I'm reading the author states that a Lie Algebra $K$ is non-associative unless we have $[x, y] = 0$ for all $x, y \in K$ (i.e, if $[v, [u, w]] = [[v, u], w]$ for all $v, u, w \in K$, then $[x, y] = 0$ for all $x, y \in K$). Is this true? Because I don't see it. By the way, the text is 'The Linear Algebra a Beginning Graduate Student Ought to Know' by Johnathan S. Golan. Thanks.

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It's not true.

Start with the Jacobi identity: $[a,[b,c]] + [b,[c,a]] + [c,[a,b]] = 0.$

If $[,]$ is associative then $[a,[b,c]] = [[a,b],c] = -[c,[a,b]]$, so $[b,[c,a]] = 0$ for all $a,b,c$. In other words, all brackets must lie in the center of the Lie algebra.

Conversely, if we define a bilinear skew-symmetric $[,]$ for which $[b,[c,a]]=0$ holds identically, then we have a Lie algebra because the Jacobi identity is trivially satisfied.

So take a three dimensional vector space with basis $x,y,z$ and define $[x,y]=z$ and $[x,z]=0$ and $[y,z]=0$. This gives a counterexample.

It is true (and we have proved above), that if the Lie bracket is associative then we have $[[a,b],c] = [a,[b,c]] = 0$ for all $a,b,c$.

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+1 was there. Just remarking that the 3-dimensional Lie algebra you use as a counterexample is isomorphic to the Lie algebra of upper triangular 2x2 matrices (bracket= the usual commutator). This may make it easier to believe that it is a Lie algebra. –  Jyrki Lahtonen Jan 26 '13 at 8:09
Thanks for the fast replies! –  Dima Moroz Jan 26 '13 at 8:34
@JyrkiLahtonen: I think you meant $3\times3$ strictly upper triangular matrices. –  Marc van Leeuwen Jan 26 '13 at 9:29
@Marc: Indeed, sorry about that :-) –  Jyrki Lahtonen Jan 26 '13 at 10:13