Two definitions of the super Jacobi identity

Question

In this paper, page 149, the super Jacobi identity is given by \begin{align} J(x, y,z) := (-1)^{|x||z|}[[x, y],z] +(-1)^{|z||y|}[[z,x], y]+(-1)^{|y||x|}[[y,z],x] = 0. \end{align} But in this article, the super Jacobi identity is given by \begin{align} [x,[y,z]]=[[x,y],z]+(-1)^{|x| |y|}[y,[x,z]]. \end{align} Are these two definitions equivalent? Thank you very much.

Answer 1

By super skew-symmetry $[x, y] = - (-1)^{|x||y|}[y, x]$. Also note that $\left|[x, y]\right| = |x| + |y| \bmod 2$, in particular, $(-1)^{|[x, y]|} = (-1)^{|x|+|y|}$. So we see that

\begin{align*} [x, [y, z]] &= -(-1)^{|x||[y, z]|}[[y, z], x]\\ &= -(-1)^{|x||y| + |x||z|}[[y, z], x]\\ &= -(-1)^{|x||z|}\left(-(-1)^{|x||z|}[[x, y],z] -(-1)^{|z||y|}[[z,x], y]\right)\\ &= [[x, y], z] + (-1)^{|x||z| + |y||z|}[[z, x], y]\\\ &= [[x, y], z] - (-1)^{|x||z| + |y||z|}(-1)^{|[z, x]||y|}[y, [z, x]]\\ &= [[x, y], z] - (-1)^{|x||z| + |y||z|}(-1)^{|x||y| + |y||z|}[y, [z, x]]\\ &= [[x, y], z] - (-1)^{|x||y| + |x||z|}[y, [z, x]]\\ &= [[x, y], z] - (-1)^{|x||y| + |x||z|}[y, -(-1)^{|x||z|}[x, z]]\\ &= [[x, y], z] + (-1)^{|x||y|}[y, [x, z]]. \end{align*}

Answer 2

I believe these should be the same. The reason being, $J(x,y,z)$ reduces to the ordinary Jacobi identity in all cases except one: when any two of the elements in $\{x,y,z\}$ are Fermi and the third one is Bose, in which case one of the three (usual) Jacobi terms has its sign flipped.