Equivalence of $\mathcal{L}_{\Pi^{\sharp}(df)} = 0$ to $[\Pi, \Pi]_{SN} = 0$ for Poisson bivectors?

Question

1. A bivector field $\Pi$ induces a Poisson structure if $[\Pi, \Pi]_{SN} = 0$, where the bracket is the Schouten-Nijenhuis bracket.

2. An $n$-vector field $\Pi$ induces a Nambu-Poisson structure of order $n$ if $\mathcal{L}_{\Pi^{\sharp}(df_1 \wedge \ldots \wedge df_{n-1})} \Pi = 0$ for all smooth functions $f_i$.

3. A Nambu-Poisson structure of order 2 is equivalent to a Poisson structure.

Hence, it should be possible to show the equivalence between $[\Pi, \Pi]_{SN} = 0$ and $\mathcal{L}_{\Pi^{\sharp}(df)} \Pi = 0$ for all $f$.

For $n = 2$ case, we have $\mathcal{L}_{\Pi^{\sharp}(df)} \Pi = \mathcal{L}_{X_f} \Pi = [X_f, \Pi]_{SN}$ where $X_f$ is the Hamiltonian vector field corresponding to $f$.

I do not see how to continue from $[X_f, \Pi]_{SN} = 0$ for all $f$ to $[\Pi, \Pi]_{SN} = 0$. It should be something simple related to some properties of the Schouten-Nijenhuis bracket, but I am stuck.

Answer 1

I guess you have solved the problem after such a long time, but I thought of giving an answer here in case someone else has the same question and comes across this post.

In general, we have that a $k$-vector field $\Lambda\in\Gamma(\wedge^k TM)$ is zero if and only if $i_{df_1}\ldots i_{df_k}\Lambda = 0$ for $f_1,\ldots,f_k\in C^\infty(M)$, where $i_{df_j}$ denotes the interior product with respect to $df_j\in\Omega(M)$. Since the Schouten bracket is of degree $-1$ (i.e. $ \operatorname{deg}([\Lambda_1,\Lambda_2]) = \operatorname{deg}(\Lambda_1) + \operatorname{deg}(\Lambda_2) -1$), we have $$[\pi,\pi]\in \Gamma(\wedge^{3} TM)$$ and, by definition of the Schouten bracket, $$-X_f= -i_{df} \pi =[f,\pi]\in\Gamma (TM).$$ Therefore $[X_f,\pi]\in \Gamma(\wedge^2TM)$. According to the criterion above, we need three functions to check for the validity of the equation $[\pi,\pi]=0$ and two for $[X_h,\pi]=0$.

Using first the Leibniz identity and then the anti-symmetry of the Schouten bracket we compute for $f,g,h\in C^\infty(M)$ $$i_{df} i_{dg} i_{dh}[\pi,\pi]=-i_{df} i_{dg} [h, [\pi,\pi]] = -2 i_{df} i_{dg} [[h,\pi],\pi] = 2 i_{df} i_{dg} [X_h,\pi] .$$ In other words, $[\pi,\pi]=0\Leftrightarrow [X_h,\pi]=0$ for all $h\in C^\infty(M)$.