Schouten-Nijenhuis bracket

Let $M$ be a smooth manifold of dimension $M$. Recall
$$\begin{align*} \Om^p(M) &= \Ga(\La^p T^*M)\text{ differential $p$-forms, }p\geq ... ...{\bullet}(M) &= \bigoplus_{p}\gX^p(M) \text{ graded vector spaces.} \end{align*}$$
External product gives both spaces a structure of graded, associative algebra, $\bZ_2$-commutative (supercommutative) i.e.

$$P\land Q=(-1)^{\deg Q \deg P}Q\land P.$$

The natural duality pairing between $T_xM$ and $T_x^*M$ extends to a natural pairing between $\Om^{\bullet}(M)$ and $\gX^{\bullet}(M)$ as follows
$$\begin{align*} \bracket{\al}{X}_x &:= \bracket{\underbrace{\al(x)}_{\in T_x^* M}... ...=X_1\wyw X_p,\\ & \al_i\in\Om^1(M),\; X_j\in \gX^1(M). \end{cases}} \end{align*}$$


Inner product for $P\in\gX^{\bullet}(M)$, $\om\in\Om^{\bullet}(M)$:

$$\bracket{i_P\om}{Q}=\bracket{\om}{P\land Q}\quad \forall\; Q\in\gX^{\bullet}(M).$$

It is the left transpose of external product.

$\gX^1(M)$ is a Lie algebra with bracket $[X, Y]=XY-YX$ of vector fields. Let $\gerg$ be a Lie algebra over $\bK$.

Remark: It is not correct to say that $\La^{\bullet }\gerg$ is a graded Lie algebra. In fact in a graded Lie algebra the 0-component should be a Lie subalgebra, therefore the 0-component should be $\gerg$. $\La^{\bullet + 1}\gerg$ is a graded Lie algebra.









Let $(\gA, \land, [-,-])$ be a Gerstenhaber algebra. An operator $D\: \gA^{\bullet}\to \gA^{\bullet -1}$ is said to generate the Gerstenhaber algebra if for all $a\in\gA^i$, $b\in \gA$

$$[a, b]=(-1)^i(D(a\land b)- Da\land b- (-1)^ia\land Db).$$

If $D^2=0$ we say that our Gerstenhaber algebra is exact or Batalin-Vilkovisky algebra.

We will show that the Gerstenhaber algebra of differential forms on a Poisson manifold is a Batalin-Vilkovisky algebra. Its generating operator will be called Poisson (or canonical or Brylinski) differential.