= Solution
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 \b[$\operatorname{Der}_k(A)$ is a <Lie algebra>].
In degree zero of the <Hochschild cochain complex>, $\delta a(b)=ba-ab$, so $HH^0(A,A)=Z(A)=A$ for commutative $A$. In degree one, $\delta D=0$ is precisely the <derivation> rule; the boundaries are <inner derivations>, which vanish for commutative $A$. Hence
$$
\boxed{HH^0(A,A)=A,\qquad HH^1(A,A)=\operatorname{Der}_k(A).}
$$
For cochains $f\in C^p(A,A)$, $g\in C^q(A,A)$, the <Hochschild cup product> is
$$
(f\smile g)(a_1,\ldots,a_{p+q})=f(a_1,\ldots,a_p)g(a_{p+1},\ldots,a_{p+q}).
$$
Define the insertion operation by
$$
f\circ g=\sum_{i=0}^{p-1}(-1)^{i(q-1)}f(a_1,\ldots,a_i,g(a_{i+1},\ldots,a_{i+q}),a_{i+q+1},\ldots,a_{p+q-1}).
$$
A 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:
$$
\boxed{[f,g]=(-1)^{(p-1)(q-1)}f\circ g-g\circ f.}
$$
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,
$$
[u,v]=-(-1)^{(p-1)(q-1)}[v,u],\qquad
[u,vw]=[u,v]w+(-1)^{(p-1)q}v[u,w]
$$
for <homogeneous elements of a graded algebra> of degrees $p,q$. The shifted <Jacobi identity> is
$$
[u,[v,w]]=[[u,v],w]+(-1)^{(p-1)(q-1)}[v,[u,w]].
$$
The <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 $A^e=k[X_\ell,X_r]$, and
$$
0\longrightarrow A^e\xrightarrow{\ X_\ell-X_r\ }A^e\longrightarrow A\longrightarrow0
$$
is a <projective resolution>. The first map is injective since $A^e$ is an <integral domain>, and its <cokernel> is $A$. Applying $\operatorname{Hom}_{A^e}(-,A)$ gives a zero <coboundary map>. Consequently
$$
\boxed{HH^*(k[X],k[X])=k[X]\otimes_k\Lambda(\partial_X),\qquad |\partial_X|=1.}
$$
The <Hochschild cup product> is ordinary multiplication of functions and scalar multiplication of <derivations>, with the product of two <derivations> zero because $HH^2=0$. Every <derivation> is $f\partial_X$, since it is determined by its value on $X$. The <Gerstenhaber bracket> is
$$
[f\partial_X,g\partial_X]=(fg'-gf')\partial_X,\qquad [f\partial_X,h]=fh',\qquad [h_1,h_2]=0,
$$
with all other orders fixed by graded antisymmetry. These formulas fully determine the <Gerstenhaber algebra>.
For $A=k[X,Y]$, the <Hochschild-Kostant-Rosenberg theorem> identifies
$$
\boxed{HH^*(A,A)=\bigwedge_A^*\operatorname{Der}_k(A)=A\otimes_k\Lambda(\partial_X,\partial_Y).}
$$
Thus 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 cochain
$$
\frac1{p!}\sum_{\sigma\in S_p}\operatorname{sgn}(\sigma)\prod_{j=1}^p D_{\sigma(j)}(a_j).
$$
The 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 $D=F\partial_X+G\partial_Y$. Then
$$
[h\Pi,f]=h(f_Y\partial_X-f_X\partial_Y),\qquad [D,h\Pi]=(D(h)-h(F_X+G_Y))\Pi,\qquad [h\Pi,j\Pi]=0.
$$
The 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> $k[X,Y]$.
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