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I have been given an anticommutative $\mathbb{k}$-algebra $L$ with basis $\{a,b,c,d,e\}$ . I need to verify that $L$ is a Lie algebra, i.e the Jacobi identity $=0$ for any three elements $\in L$.

My question is, is there a quicker way to show the Jacobi identity is $0$ for any three elements other than showing $j(a,b,c), j(a,b,d), j(a,b,e) \dots = 0$ for all 10 combinations of three basis elements?

user3450230
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1 Answers1

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No, there is no quicker way. You have to check all these combinations for the Jacobi identity. Of course, $j(a,b,a)=[a,[b,a]]+[b,[a,a]]+[a,[a,b]]=0$ is satisfied automatically, if two elements are equal. You can take the classification of $5$-dimensional Lie algebras, to have some examples.

Dietrich Burde
  • 130,978