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.


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 $a,b,c$.

  • 1
    $\begingroup$ +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. $\endgroup$ – Jyrki Lahtonen Jan 26 '13 at 8:09
  • $\begingroup$ Thanks for the fast replies! $\endgroup$ – Dima Moroz Jan 26 '13 at 8:34
  • $\begingroup$ @JyrkiLahtonen: I think you meant $3\times3$ strictly upper triangular matrices. $\endgroup$ – Marc van Leeuwen Jan 26 '13 at 9:29
  • $\begingroup$ @Marc: Indeed, sorry about that :-) $\endgroup$ – Jyrki Lahtonen Jan 26 '13 at 10:13
  • $\begingroup$ Related: math.stackexchange.com/questions/2162019 $\endgroup$ – Watson Feb 28 '18 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.