P. Aluffi's "Algebra: Chapter ${\it 0}$", (part of) exercise VI.$1.4.$
Let $V$ be a vector space over a field $k$. A Lie bracket on $V$ is an operation $[\cdot,\cdot]\colon V\times V\to V$ such that
- $(\forall u,v,w\in V),(\forall a,b\in k),~[au+bv,w]=a[u,w]+b[v,w],~[w,au+bv]=a[w,u]+b[w,v]$
- $(\forall v\in V),~[v,v]=0$
- and $(\forall u,v,w\in V),~[[u,v],w]+[[v,w],u]+[[w,u],v]=0$
(This axiom is called the Jacobi identity). A vector space endowed with a Lie bracket is called a Lie algebra. $[\dots]$ Prove the following
- In a Lie algebra $V$, $[u,v]=-[v,u]$ for all $u,v\in V$
As it appears Aluffi's definition is a bit unusual. After looking around I have seen Lie bracket commonly defined as antisymmetric. From here it follows easily that $[v,v]=0$ for all $v\in V$. Indeed, by the antisymmetry of the Lie bracket take $u=v$ and then $[v,v]=-[v,v]\implies 2[v,v]=0$. Thus $[v,v]=0$ for all $v\in V$ as desired.
Anyway, lets consider the listed axioms and try to derive antisymmetry. First off, note that $[v,0]=[0,v]=0$ for all $v\in V$ as $[v,0]=[v,0v]=0[v,v]=0$, by the first axiom, and similarly for $[0,v]$. Now, take the Jacobi identity with $v=w$ to obtain \begin{align*} [[v,w],w]+[[w,w],v]+[[w,v],w]&=0\\ [[v,w]+[w,v],w]+[0,v]&=0\\ [[v,w]+[w,v],w]&=0 \end{align*} Now, if $w=0$ this holds trivially, so take $w\ne0$. Then either $[v,w]+[w,v]=0$ or $[v,w]+[w,v]=w$. But I was not able to derive a contradiction from the assumption that $[v,w]+[w,v]=w$ which would yield antisymmetry right away.
Is it possible to finisht this attempt? If not, how would one approach proving the antisymmtery of the Lie bracket from the given axioms?
Thanks in advance!
By the axioms we have $$ 0=[u+v,u+v]=[u,u]+[u,v]+[v,u]+[v,v]=[u,v]+[v,u] $$ so that $[u,v]=-[v,u]$. Note that we do not need to use the Jacobi identity.