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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.

user26857
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Dima Moroz
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1 Answers1

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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 $a,b,c$.

Ted
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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 08:09
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    Thanks for the fast replies! – Dima Moroz Jan 26 '13 at 08:34
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    @JyrkiLahtonen: I think you meant $3\times3$ strictly upper triangular matrices. – Marc van Leeuwen Jan 26 '13 at 09:29
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    @Marc: Indeed, sorry about that :-) – Jyrki Lahtonen Jan 26 '13 at 10:13
  • Related: https://math.stackexchange.com/questions/2162019/ – Watson Feb 28 '18 at 16:26
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    Just a note: if the characteristic of the field is not 2, then anticommutativity with associativity already imply the Jacobi identity; if we denote the product by ∗ then the commutator for this product, $[x,y]_∗:=x∗y−y∗x$, satisfies Jacobi identity. But$ [x,y]_∗=2x∗y$ by anticommutativity, so $x∗y=1/2[x,y]_∗$ is a Lie product. Therefore every associative and anticommutative algebra $(A,+,∗)$ over a field of characteristic not 2 satisfies $AAA=0$. – Jose Brox Jun 04 '18 at 21:41