Lie Algebra is Anticommutative

Theorem

Let $L$ be a Lie algebra.

Then $L$ is anticommutative:

$\forall a, b \in L: \sqbrk {a, b} = -\sqbrk {b, a}$

where $\sqbrk {\, \cdot, \cdot \,}$ denotes the bilinear mapping over $L$.

Proof

\(\ds \sqbrk {a + b, a + b}\)

\(=\)

\(\ds \sqbrk {a, a + b} + \sqbrk {b, a + b}\)

Definition of Bilinear Mapping

\(\ds \leadsto \ \ \)

\(\ds \sqbrk {a + b, a + b}\)

\(=\)

\(\ds \sqbrk {a, a} + \sqbrk {a, b} + \sqbrk{b, a} + \sqbrk {b, b}\)

Definition of Bilinear Mapping

\(\ds \leadsto \ \ \)

\(\ds 0\)

\(=\)

\(\ds \sqbrk {a, b} + \sqbrk {b, a}\)

Lie Algebra Axiom $\text L 1$: Alternativity

\(\ds \leadsto \ \ \)

\(\ds \sqbrk {a, b}\)

\(=\)

\(\ds -\sqbrk {b, a}\)

$\blacksquare$

Notes

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The converse, that in a bilinear context anticommutativity implies $[x,x]=0$, is not quite true; however, the only exceptions occur when the vector space underlying the Lie algebra are over a field of characteristic 2.