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For the polynomial ring $A=k[X_1,\dots ,X_r]$ in characteristic zero, antisymmetrization identifies $H{H}^{\ast }(A,A)$ with ${\bigwedge }_A^{\ast }{\mathrm{Der}}_k(A)$. The map on a wedge of $p$ derivations is $\frac{1}{p!}{\sum }_{\sigma \in S_p}\mathrm{sgn}(\sigma ){\prod }_j{D}_{\sigma (j)}(a_j)$. The Hochschild cup product becomes the exterior product, and the left Gerstenhaber bracket becomes the left Schouten-Nijenhuis bracket. This assertion requires a smoothness hypothesis when generalized beyond polynomial rings.

A Poisson manifold has a skew bilinear bracket on smooth functions which is a derivation in each argument and satisfies the Jacobi identity. Locally a Poisson bivector $\Pi$ represents it byThe derivation property is automatic for this expression. On applying the Jacobi identity to coordinate functions, and then using the derivation property for arbitrary functions, the necessary and sufficient condition isEquivalently its Schouten-Nijenhuis bracket $[\Pi ,\Pi ]$ vanishes.

A symplectic manifold has a closed nondegenerate two-form $\omega$. Write its coefficient matrix as ${\Omega }_{ij}$ and set ${\Pi }^{ij}=({\Omega }^{-1}{)}^{ij}$. It is skew and defines the bracket above. Differentiating $\Pi \Omega =I$ gives ${\partial }_{\ell }\Pi =-\Pi ({\partial }_{\ell }\Omega )\Pi$. Substitution into the Jacobi condition giveswhere the last equality is $d\omega =0$. Thus every symplectic manifold is Poisson. In this convention the Hamiltonian vector field satisfies ${\iota }_{X_f}\omega =-df$, and $\{f,g\}=X_{g}f$; this will also be the convention below.

The converse fails because a Poisson bivector need not be invertible. On ${\mathbb{R}}^2$, take $\Pi =0$. Its bracket is identically zero and satisfies every Poisson axiom, but it cannot be the inverse of any nondegenerate two-form. This is an even-dimensional counterexample, so odd dimension is not the only obstruction.

A Poisson structure is a real bilinear operation on smooth functions on $M$,which is antisymmetric, obeys the Jacobi identity, and is a derivation in each argument. Explicitly,Antisymmetry supplies the corresponding Leibniz rule in the first argument. Equivalently it is a smooth Poisson bivector $\Pi$ with $\{f,g\}=\Pi (df,dg)$ and $[\Pi ,\Pi ]=0$ for the Schouten-Nijenhuis bracket. In local coordinates,The last equation is the coordinate Jacobi identity. Nondegeneracy is not part of the definition: a Poisson structure may have Casimir functions and symplectic leaves of different dimensions.

A Hamiltonian system consists of such a Poisson manifold, a Hamiltonian $H\in C^{\infty }(M)$, and the flow of its Hamiltonian vector field. Choose the conventionFor an observable with no explicit time dependence, $df/dt=\{f,H\}$. In particular $H$ is conserved because $\{H,H\}=0$. For the canonical Poisson bracket, this convention gives ${\dot{q}}_i=\partial H/\partial p_i$ and ${\dot{p}}_i=-\partial H/\partial q_i$.

A derivation of an algebra is a $k$-linear map $D:A\to A$ satisfying $D(ab)=D(a)b+aD(b)$. The commutator $[D,E]=D\circ E-E\circ D$ is again a derivation: expanding $[D,E](ab)$ cancels the two mixed terms and leaves $[D,E](a)b+a[D,E](b)$. The commutator on endomorphisms is bilinear, antisymmetric, and satisfies the Jacobi identity by cancellation of its twelve triple-composition terms. Therefore ${\mathrm{Der}}_k(A)$ is a Lie algebra.

In degree zero of the Hochschild cochain complex, $\delta a(b)=ba-ab$, so $H{H}^0(A,A)=Z(A)=A$ for commutative $A$. In degree one, $\delta D=0$ is precisely the derivation rule; the boundaries are inner derivations, which vanish for commutative $A$. HenceFor cochains $f\in C^p(A,A)$, $g\in C^q(A,A)$, the Hochschild cup product isDefine the insertion operation byA degree-zero cochain is an element of $A$, inserted with no arguments; for $p=0$ the sum is empty. For the Gerstenhaber bracket we use the left graded Leibniz rule convention, compatible with the unsigned Hochschild cup product just displayed:Both degree-zero inputs have bracket zero. Another common insertion convention writes $f\circ g-(-1{)}^{(p-1)(q-1)}g\circ f$; the two brackets differ by $(-1{)}^{(p-1)(q-1)}$. With an unsigned Hochschild cup product, that convention uses the corresponding right graded Leibniz rule. The distinction matters for a degree-two cochain bracketed with a function. Either consistent convention gives the same degree-one Lie bracket and the same derivation action on functions.

If $m(a,b)=ab$, this convention gives $\delta f=[m,f]$. The shifted Jacobi identity and $[m,m]=0$ therefore show that the Gerstenhaber bracket respects Hochschild cocycles and the images of the coboundary map. The Hochschild cup product and Gerstenhaber bracket induce operations on Hochschild cohomology. A Gerstenhaber algebra is a graded algebra $H$ with an associative degree-zero product with the graded commutative algebra rule $uv=(-1{)}^{pq}vu$, and a degree-minus-one graded Lie bracket making the shifted degrees $|u|-1$ into a graded Lie algebra. In particular,for homogeneous elements of a graded algebra of degrees $p,q$. The shifted Jacobi identity isThe Hochschild cup product does not make the cochains a graded commutative algebra in general, but does make their cohomology a graded commutative algebra; the insertion operation supplies the homotopy for this assertion and for the graded Leibniz rule. Thus these axioms describe the induced Gerstenhaber algebra, not a claim of a graded commutative algebra structure on the cochain multiplication itself.

For $A=k[X]$, the enveloping algebra is $A^e=k[X_{\ell },X_r]$, andis a projective resolution. The first map is injective since $A^e$ is an integral domain, and its cokernel is $A$. Applying ${\mathrm{Hom}}_{A^e}(-,A)$ gives a zero coboundary map. ConsequentlyThe Hochschild cup product is ordinary multiplication of functions and scalar multiplication of derivations, with the product of two derivations zero because $H{H}^2=0$. Every derivation is $f{\partial }_X$, since it is determined by its value on $X$. The Gerstenhaber bracket iswith all other orders fixed by graded antisymmetry. These formulas fully determine the Gerstenhaber algebra.

For $A=k[X,Y]$, the Hochschild-Kostant-Rosenberg theorem identifiesThus the degrees zero, one, and two are $A$, $A{\partial }_X\oplus A{\partial }_Y$, and $A({\partial }_X\wedge {\partial }_Y)$, and all higher groups vanish. The Hochschild-Kostant-Rosenberg map sends a wedge of $p$ derivations to the cochainThe factorial is invertible in characteristic zero. Equivalently, the groups follow from the Koszul resolution on the regular sequence $X_{\ell }-X_r,Y_{\ell }-Y_r$ in $A^e$, whose dual coboundary maps vanish on $A$.

The Hochschild cup product becomes the exterior product, and our Gerstenhaber bracket becomes the left Schouten-Nijenhuis bracket. It is determined by the commutator of derivations, $[D,f]=D(f)$, zero brackets of functions, and the displayed graded antisymmetry and left graded Leibniz rule. For explicit signs, put $\Pi ={\partial }_X\wedge {\partial }_Y$ and $D=F{\partial }_X+G{\partial }_Y$. ThenThe last bracket has degree three, whose exterior power is zero. For two derivations, the coefficient functions of their commutator give the remaining formula. This specifies the entire Gerstenhaber algebra; under the alternate insertion convention mentioned above, the first displayed bracket changes sign, together with the Leibniz convention. No smoothness of a general finitely generated commutative algebra was assumed: the Hochschild-Kostant-Rosenberg theorem is invoked here only for the smooth polynomial ring $k[X,Y]$.

Two Poisson structures are compatible when every linear combination ${\Pi }_1+\lambda {\Pi }_2$ is a Poisson structure. This family is a Poisson pencil. In terms of the Schouten-Nijenhuis bracket, compatibility means $[{\Pi }_1,{\Pi }_2]=0$, in addition to the two individual Jacobi identities. Constant Poisson bivectors are compatible. Over a real manifold, complex parameter values refer to the complexification of its function algebra.

For a polynomial ring $A$, a degree-$p$ polynomial polyvector field is an element of ${\bigwedge }_A^p{\mathrm{Der}}_k(A)$. Its product is the exterior product. Together with the Schouten-Nijenhuis bracket these spaces form a Gerstenhaber algebra.