Definition: The big bracket is the graded Lie algebra structure on algebraic functions on $V\oplus V^*$ defined as follows: $\bullet$ For $u,v\in B^{-1}\oplus B^{-1}=k\oplus V\oplus V^*$: $$[u,v]=\begin{cases} \langle u,v\rangle & \text{if}\ u\in V\oplus V^*\ \text{and}\ v\in V\oplus V^*;\\ 0 & \text{if}\ u\in k\ \text{or}\ v\in k . \end{cases}$$ $\bullet$ The bracket on other terms is defined by linearity and the graded Leibniz rule: for $u\in B^k$, $v\in B^l$, $w\in B^m$, $$[u,v\wedge w]=[u,v]\wedge w+(-1)^{kl}v\wedge[u,w].$$ $$B=\bigoplus_{j\geq-2}B^j, \text{where}\ B^j=\bigoplus_{p+q=j}\left(\bigwedge^{p+1}V^*\otimes\bigwedge^{q+1}V\right), j\geq-1, \text{and}\ B^{-2}=\mathbb{K}.$$
The above definition is copied from https://arxiv.org/pdf/math/0601301.pdf page 4. And the Proposition 7 in the same paper states that for a vector space $V$ and an element $l\in V^*\wedge V^*\otimes V$, $[l,l]=0$ will give a Lie algebra structure on $V$ by defining the Lie bracket as $\{x,y\}=[[l,x],y]$. The proof is very simple, just say that $[l,l]=0$ is exactly the Jacobi identity. But when I write out as $l=f\wedge g\otimes v$ and try to compute directly, I still cannot figure out the equivalence of the condition $[l,l]=0$ and the Jacobi identity. This is how I compute: $$\begin{aligned} &[l,l]\\ =&[f\wedge g\otimes v,f\wedge g\otimes v]\\ =&[f\wedge g\otimes v,f]\wedge g\otimes v-f\wedge[f\wedge g\otimes v,g\otimes v]\\ =&[f,f\wedge g\otimes v]\wedge g\otimes v+f\wedge[g\otimes v,f\wedge g\otimes v]\\ =&([f,f]\wedge g\otimes v-f\wedge[f,g\otimes v])\wedge g\otimes v+f\wedge([g\otimes v,f]\wedge g\otimes v+f\wedge[g\otimes v,g\otimes v])\\ =&2f\wedge[g\otimes v,f]\wedge g\otimes v. \end{aligned}$$ I searched it on google, and find no proof to it. I guess it should be very simple. Maybe there is a stupid mistake in my computation. Or maybe I totally misunderstood the whole concept. Can anyone help me with it? Thanks a lot in advance.