question regarding Jacobi's identity [duplicate]

Let ${\frak g}$ be a Lie algebra. Then $$ [[[X,Y],Z],W] +[[[Y,X],W],Z]+[[[Z,W],X],Y]+[[[W,Z],Y],X]=0 $$for all $X,Y,Z,W \in {\frak g}$.

I started using the Jacobi identity four times to get: \begin{align} [[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y] &= 0 \\ [[Y,X],W]+[[X,W],Y]+[[W,Y],X] &= 0 \\ [[Z,W],X]+[[W,X],Z]+[[X,Z],W] &= 0 \\ [[W,Z],Y]+[[Z,Y],W]+[[Y,W],Z] &= 0 \end{align}

Then I apply $[\cdot, W]$ to the first equality, $[\cdot, Z]$ to the second, $[\cdot, Y]$ to the third and $[\cdot, X]$ to the last one to get

\begin{align} \color{blue}{[[[X,Y],Z],W]}+[[[Y,Z],X],W]+[[[Z,X],Y],W] &= 0 \\ \color{blue}{[[[Y,X],W],Z]}+[[[X,W],Y],Z]+[[[W,Y],X],Z] &= 0 \\ \color{blue}{[[[Z,W],X],Y]}+[[[W,X],Z],Y]+[[[X,Z],W],Y] &= 0 \\ \color{blue}{[[[W,Z],Y],X]}+[[[Z,Y],W],X]+[[[Y,W],Z],X] &= 0 \end{align}

Summing, we get what we want (in blue), plus some crap consisting of eight terms, equaling zero. I don't know what to do to see that the crap is zero. Help?

Solution 1:

I want to add an additional reference, which also contains a proof, but also shows how to generalise such Jacobi-like identities. It is the paper Jacobi - type identities in algebras and superalgebras. The identity for $n=4$ is given in $(5.3)$, on page $9$.

Solution 2:

You can generate additional identities using the Jacobi identity for $T = [X,Y], Z$, and $W$ and for $S = [Z,W], X$, and $Y$: $$ \begin{align*} 0 & = [[T,Z],W] + [[W,T],Z] + [[Z,W],T] \\ & = [[[X,Y],Z],W] + [[W,[X,Y]],Z] + [[Z,W],[X,Y]]\\ 0 & = [[S,X],Y] + [[Y,S],X] + [[X,Y],S] \\ & = [[[Z,W],X],Y] + [[Y,[Z,W]],X] + [[X,Y],[Z,W]] \end{align*} $$ Using anticommutativity twice on the second terms lets us rewrite these in the following form: $$ [[[X,Y],Z],W] + [[[Y,X],W],Z] + [[Z,W],[X,Y]]=0\\ [[[Z,W],X],Y] + [[[W,Z],Y],X] + [[X,Y],[W,Z]]=0 $$ Adding these together gives you the desired relation (again using anticommutativity).

(You can use this method to show that the eight remaining terms you mention add to zero as well.)