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Expand both sides using the definition of a commutator: the middle terms $BAC$ cancel, leaving $ABC-BCA$. Thus commutation with a fixed operator is a derivation of an algebra. In particular, $[A,[B,C]]=[[A,B],C]+[B,[A,C]]$, which gives a short way to derive a Lie algebra representation from generator actions on an underlying algebra.

For a commutative $k$-algebra $A$, let $I$ be the kernel of multiplication $\mu :A{\otimes }_{k}A\to A$. The formula $a\mapsto [1\otimes a-a\otimes 1]$ is a derivation of an algebra into the quotient module $I/I^2$, with $A$ acting through the first factor. It induces an isomorphism from the module of Kähler differentials. An inverse is induced by $q(x\otimes y)=xdy$: the identity $q(rs)=\mu (r)q(s)+\mu (s)q(r)$ shows $q(I^2)=0$. For $z={\sum }_i{x}_i\otimes y_i\in I$, the equality ${\sum }_i{x}_i{y}_i=0$ gives $[z]={\sum }_i{x}_i[1\otimes y_i-y_i\otimes 1]$, proving that the two maps are inverse. For an affine variety, $I$ is the defining ideal of its diagonal morphism.

An algebraic derivation of an algebra over a field is a linear map satisfying the Leibniz rule for its multiplication: $D(xy)=D(x)y+xD(y)$. This definition makes sense even when the multiplication is not associative. A derivation of an algebra uses an associative product; a derivation of a Lie algebra uses the Lie bracket as its product.

Every real-linear derivation of an algebra $D:C^{\infty }(M)\to C^{\infty }(M)$ on a smooth manifold has the form $Dh=Z(h)$ for a unique smooth vector field $Z$. The product rule makes $D$ local: multiplying a function that vanishes near a point by a smooth cutoff function supported there gives zero derivative at that point. Locally write $h(x)=h(p)+{\sum }_i(x^i-x^i(p))h_i(x)$, where $h_i(p)={\partial }_{i}h(p)$. Applying $D$ at $p$ gives $Dh(p)={\sum }_{i}D(x^i)(p){\partial }_{i}h(p)$; the smooth coefficients $D(x^i)$ define the claimed vector field. No continuity hypothesis is needed. In dimension zero all such derivations vanish.

If $\theta$ has order at most one, set $c=\theta (1)$ and $\delta =\theta -m_c$. Then $\delta (1)=0$ and $[\delta ,m_a]=m_{\delta (a)}$, which is the Leibniz rule. Thus $\delta$ is a derivation of an algebra. Conversely any derivation satisfies this commutator identity. The sum is direct because a derivation that is multiplication must vanish at $1$.

Specify a commutative ground ring $k$ and a unital associative $k$-algebra $R$. For an $R$-bimodule $M$, the Hochschild cochain complex hasand coboundaryThe bimodule identities and associativity cancel the terms of ${\delta }^2$ in pairs. The Hochschild cohomology is $H{H}_k^q(R,M)=\ker {\delta }_q/\mathrm{im}{\delta }_{q-1}$; the paper's $H^q(R,R)$ uses the regular bimodule as coefficients.

Over a field, the bar resolution of an associative algebra explains its homological meaning. Put $R^e=R{\otimes }_k{R}^{\mathrm{op}}$. The degree-$q$ bar term is $R^{{\otimes }_k(q+2)}$, with the outer factors supplying the $R^e$ action, boundary obtained by multiplying adjacent factors with alternating signs, and augmentation multiplication $R\otimes R\to R$. Inserting $1$ at the left gives a $k$-linear contracting homotopy, proving exactness. The terms are free $R^e$-modules after choosing a $k$-basis for the inner tensor factors. Applying ${\mathrm{Hom}}_{R^e}(-,M)$ gives precisely the displayed cochain complex and thereforeThis ground-ring qualification in Hochschild cohomology matters for an arbitrary ring: one can take $k=\mathbb{Z}$, but the bar terms need not be projective over $R^e$. The ordinary cochain complex then computes relative Hochschild cohomology; the absolute Ext identification needs, for example, $R$ projective over $k$. Likewise the extension classification below applies to ground-ring-linearly split extensions. For algebras over a field these qualifications are automatic.

In degree zero, $(\delta m)(a)=am-ma$, so $H{H}^0(R,R)=Z(R)$. In degree one the cocycle equation isexactly the Leibniz rule for a derivation of an algebra. The coboundaries are the inner derivations $a\mapsto am-ma$. Thus the First Hochschild cohomology as outer derivations isIts significance is infinitesimal automorphisms modulo inner ones. Over the dual numbers $k[\epsilon ]/({\epsilon }^2)$, the map $a\mapsto a+\epsilon d(a)$ is multiplicative precisely when $d$ satisfies the Leibniz rule. Conjugation by $1+\epsilon m$ gives $a\mapsto a+\epsilon (ma-am)$, an inner derivation with the opposite sign convention. The quotient therefore measures infinitesimal symmetry not arising from conjugation.

Degree two controls extensions and changes of multiplication. A $k$-linear split square-zero extension of an algebra with prescribed kernel bimodule $M$ can be written as $E=R\oplus M$. Choose a unital section and let $f(a,b)$ be the difference between its product and its value on $ab$. ThenComputing both parenthesizations of a triple gives the associativity conditionnamely $\delta f=0$. A unital product has $f(1,a)=f(a,1)=0$, so one uses the normalized Hochschild cochain complex. Changing section by $g:R\to M$ gives $f\mapsto f+\delta g$. Conversely the displayed product for any normalized two-cocycle constructs such an extension, and a section change gives the corresponding extension isomorphism. This proves that second Hochschild cohomology classifies split square-zero extensions:The zero class is the split algebra extension with no extra product term. With $M=R$, this gives one meaning of the requested $H^2(R,R)$.

For a first-order associative deformation, retain the underlying $k$-module and put $a\ast b=ab+\epsilon f(a,b)$. Expanding $(a\ast b)\ast c=a\ast (b\ast c)$ modulo ${\epsilon }^2$ gives exactly $\delta f=0$. A coordinate change $T=1+\epsilon g$ transports the product to $T^{-1}(T(a)\ast T(b))$, whose first-order coefficient is $f+\delta g$. HenceThis classification is of infinitesimal deformations, not a guarantee that all classes extend to every order. For $a\ast b=ab+tf(a,b)+t^2{\mu }_2(a,b)$ modulo $t^3$, the next associativity equation isSubstitution of $\delta f=0$ shows the right-hand side is a three-cocycle. Its class in $H{H}^3(R,R)$ is the second-order obstruction to an associative deformation. It must vanish for a second-order extension to exist. At higher orders the same expansion gives further obstruction cocycles. If $H{H}^2(R,R)=0$, every first nonzero coefficient of a formal associative deformation is a coboundary and can be removed by a coordinate change. Repeating this order by order gives formal rigidity from vanishing second Hochschild cohomology, with the transformations converging in the $t$-adic topology. Vanishing of $H{H}^3$ instead removes these extension obstructions; these are distinct roles.

Examples make the two low degrees concrete. For $R=k[x]$ over a field, inner derivations vanish and every derivation is $h(x){\partial }_x$, determined by its value on $x$. The bimodule resolutionis exact: the middle tensor ring is the polynomial integral domain $k[u,v]$, multiplication identifies its quotient by $u-v$, and multiplication by $u-v$ is injective. Applying bimodule Hom gives zero differential. Consequently the Hochschild cohomology of a polynomial algebra in one variable isThis algebra has many infinitesimal automorphisms, while its ordinary associative multiplication is formally rigid.

For $R=k[x,y]$ in characteristic zero, the bilinear map $f(a,b)=({\partial }_{y}a)({\partial }_{x}b)$ is a two-cocycle: substitution in $\delta f$ and the two product rules cancel its four terms. It is not a coboundary, because every $\delta g$ on a commutative algebra is symmetric in its two inputs, whereas $f(y,x)-f(x,y)=1$. This class integrates to the Weyl deformation of a polynomial algebra in two variables, with $yx-xy=t$. The differential Ore extension construction supplies the ordered basis $x^i{y}^j$ over $k[t]$, so the $t$-adic completion has the fixed underlying module $k[x,y][[t]]$. Its first multiplication correction is exactly this $f$. This displays a genuinely noncommutative formal associative deformation detected by $H{H}^2$.

Finally, the Hochschild cup product $(f⌣g)(a_1,\dots ,a_{p+q})=f(a_1,\dots ,a_p)g(a_{p+1},\dots ,a_{p+q})$ descends to the graded cohomology algebra, and the Gerstenhaber bracket organizes the interaction of derivations and formal associative deformation cocycles. The calculations above explain why the first two positive degrees are especially useful: $H{H}^1$ measures outer infinitesimal symmetries, while $H{H}^2$ measures split extensions and infinitesimal associative structures, with degree three detecting the first obstruction to continuing them.

First establish the scalar Leibniz rule, since it is needed for an extension to all tensor fields. Suppose every component of $M$ has positive dimension. Computing $D(fgY)$ first with scalar $fg$ and then as $f(gY)$ givesAt any point, a smooth cutoff function times a coordinate vector field can be chosen nonzero there. Evaluating the displayed identity provesThus the scalar operator is a derivation of an algebra. Derivations of smooth functions are vector fields, so the scalar operator is differentiation along a unique vector field $Z$; no continuity assumption is needed. Indeed, if $h$ vanishes near $p$, take a smooth cutoff function $\chi$ equal to one near $p$ and supported where $h=0$. The identity $D(\chi h)=(D\chi )h+\chi Dh=0$ gives $Dh(p)=0$. Thus $D$ depends only on the local function germ, so local coordinate functions may be extended with cutoffs before applying it. The local identity $h(x)=h(p)+{\sum }_i(x^i-x^i(p))h_i(x)$ gives $Dh(p)={\sum }_{i}D(x^i)(p){\partial }_{i}h(p)$. Smooth local coefficients $D(x^i)$ define $Z$.

The vector-field operator is local as well. If $Y$ vanishes near $p$, choose a cutoff $h$ equal to one near $p$ with support where $Y=0$. Then $D(hY)=hDY+(Dh)Y$ gives $DY(p)=0$. We can therefore work with local frames without presuming a global frame.

For a differential one-form $\omega$, the only possible contraction-compatible definition isThe scalar product rule and the given vector-field rule show that this is linear over $C^{\infty }(M)$ in $Y$, so it defines a one-form. Its coefficients are smooth by evaluating the formula on local frame fields extended with cutoffs. It is real-linear in $\omega$, and direct substitution gives $D(f\omega )=fD\omega +(Df)\omega$.

The original PDF's hint has a transpose error. If $D{e}_i={\sum }_j{a}_{ij}{e}_j$ for $e_i=\partial /\partial x^i$, then evaluation on every $e_j$ forcesnot the untransposed coefficient array printed in the hint. For example, take $Df=0$, $D{e}_1=e_2$ and $D{e}_2=0$ on ${\mathbb{R}}^2$. The correct values are $D(d{x}^1)=0$, $D(d{x}^2)=-d{x}^1$. The printed hint instead makes the derivative of $d{x}^1(e_2)=0$ equal to $-1$.

There is also a genuine zero-dimensional edge case in the hypotheses: on a one-point manifold the vector-field space is zero, so the printed rule imposes no condition on the scalar map. Taking $Df=f$ satisfies that rule but cannot extend to a tensor derivation, since the product rule requires $D1=0$. Thus the extension theorem is valid on positive-dimensional manifolds as above, or in every dimension if the scalar product rule is added explicitly. The next two parts use these precise hypotheses.

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 $k[X,Y]$.

Define the Lie derivative of a differential form by ${\mathcal{L}}_X\alpha =(d/dt{)}_{t=0}{\varphi }_t^{\ast }\alpha$, using the local flow. The pullback of a differential form respects the wedge product of differential forms and commutes with the exterior derivative. Therefore ${\mathcal{L}}_X$ is a derivation of an algebra of degree zero and commutes with $d$.

Let $D_X={\iota }_{X}d+d{\iota }_X$, where ${\iota }_X$ is the interior product of a differential form. The graded product rules for $d$ and ${\iota }_X$ show that their anticommutator $D_X$ is also a degree-zero derivation of an algebra: the two mixed terms cancel. Since $d^2=0$, it commutes with $d$. On a function $f$, ${\iota }_{X}f=0$ and $D_{X}f={\iota }_{X}df=Xf={\mathcal{L}}_{X}f$. Hence $D_X(df)=d(Xf)={\mathcal{L}}_X(df)$.

In a manifold chart, every differential form is a sum of functions times products $d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. The two derivations agree on its generators $f$ and $d{x}^i$, and therefore agree on every such form. This proves Cartan's magic formula, including the case $p=0$:

For a product in a category of prevarieties, the projections $p_X,p_Y$ must induce a bijectionfor every prevariety $T$. If $Z$ and $Z^{\prime }$ both have this property, their projection pairs determine morphisms $Z\to Z^{\prime }$ and $Z^{\prime }\to Z$. Each composite has the same projections as the identity, so uniqueness in the universal property makes both composites identities. This gives the unique isomorphism respecting the projections.

Here the classical convention is that a prevariety is locally an affine variety, with a finite affine cover, and is irreducible; an algebraic variety is a separated prevariety, meaning that its diagonal morphism has closed image. The same separation argument works in the convention that also permits reducible varieties. For separated $X,Y$, under the rearrangement $(X\times Y{)}^2\cong X^2\times Y^2$, the diagonal isIt is closed, proving separation. For completeness, irreducibility in the classical convention follows from the affine calculation below: the products of affine charts are irreducible, and their pairwise intersections are nonempty since nonempty open subsets of each irreducible factor meet. An open cover by irreducible open sets with pairwise nonempty intersections has irreducible union. Thus the product is an algebraic variety.

To establish the coordinate ring of a product of affine varieties, write $A=k[X]=k[x_1,\dots ,x_r]/J$ and $B=k[Y]=k[y_1,\dots ,y_s]/L$. The algebrahas zero set $X\times Y$. To see directly that $(J,L)$ is prime when $X,Y$ are irreducible, take nonzero $F,G\in A{\otimes }_{k}B$. Express $F={\sum }_i{a}_i\otimes b_i$ with the $b_i$ linearly independent over $k$. The set of $x\in X$ for which $F_x={\sum }_i{a}_i(x)b_i$ is nonzero is a nonempty open set: it is the union of the nonvanishing loci of the $a_i$, and at least one $a_i$ is nonzero. The Hilbert Nullstellensatz guarantees that such a function cannot vanish at every point. The analogous set for $G$ is also nonempty; their intersection contains a point $x$ by irreducibility. Since $B$ is an integral domain, $F_x{G}_x\ne 0$, whence $FG\ne 0$. Thus $A{\otimes }_{k}B$ is an integral domain, and the Hilbert Nullstellensatz identifies $(J,L)$ with the defining ideal of its zero set. The resulting affine variety has the universal property: algebra maps from $A{\otimes }_{k}B$ to $\Gamma (T,{\mathcal{O}}_T)$ correspond to pairs of algebra maps from $A,B$, and these correspond to morphisms into $X,Y$. This dictionary holds for arbitrary $T$, since coordinate functions give the morphisms on affine charts and agree on overlaps. Uniqueness of the product now proves

For the final part put $T=A{\otimes }_{k}A$ and $\mu (x\otimes y)=xy$. Restriction to the diagonal morphism is $\mu$, so $I=\ker \mu$. Write $\delta (a)=1\otimes a-a\otimes 1$. The elements $\delta (a)$ generate $I$: quotienting by them identifies the two copies of $A$ and gives precisely the multiplication quotient $T/I\cong A$. The map $D(a)=[\delta (a)]$ is $k$-linear, vanishes on $k$, and takes values in $I/I^2$. Sinceit obeys $D(ab)=aD(b)+bD(a)$ for the stated first-factor action. Hence it is a derivation of an algebra. The universal property of the module of Kähler differentials supplies an $A$-linear map $\theta :{\Omega }_{A/k}^1\to I/I^2$ with $\theta (da)=D(a)$.

Define the $k$-linear map $q:T\to {\Omega }_{A/k}^1$ by $q(x\otimes y)=xdy$. It is $A$-linear for the first-factor action and, by the Leibniz rule, satisfiesTherefore $q(I^2)=0$ and restriction induces $\psi :I/I^2\to {\Omega }_{A/k}^1$. We have $\psi \theta (da)=da$, so $\psi \theta$ is the identity because the $da$ generate the module of Kähler differentials. Conversely, if $z={\sum }_i{x}_i\otimes y_i\in I$, then ${\sum }_i{x}_i{y}_i=0$ andThus the conormal module of the diagonal gives

For a ring homomorphism $A\to B$, the module of Kähler differentials ${\Omega }_{B/A}$ is the $B$-module generated by symbols $db$, subject toEquivalently, it represents $A$-linear derivations:For $A\to B\to C$, the Transitivity exact sequence for Kähler differentials isIf $B\to C=B/I$ is surjective, the Conormal exact sequence for Kähler differentials is

Let $L/K$ be a finite field extension. By the primitive element theorem, its maximal separable field extension $L_s/K$ is simple, and transitivity reduces the calculation to a simple algebraic extension. If $L=K(\alpha )$ with minimal polynomial $m_{\alpha }$, thenThus a separable simple extension has zero differentials. Conversely, if $L/K$ is not separable, the purely inseparable part has a generator whose minimal polynomial has zero formal derivative in positive characteristic, producing a nonzero differential. Hence

Now let $\mathrm{char}K=p>0$. If $L=K(a)$, $a^p\notin K$, and $a^{p^2}\in K$, then the minimal polynomial is $T^{p^2}-a^{p^2}$ and has zero derivative, soIf $L=K(a,b)$, where $a\notin K$, $b\notin K(a)$, and $a^p,b^p\in K$, then $[K(a):K]=[L:K(a)]=p$. Both defining equations have zero derivative, and

For the three morphisms, compute the Sheaf of relative Kähler differentials on coordinate rings.

If $\mathrm{char}k\ne 2$, this is $k[y]/(y)dy$ and its support is the origin $V(y)=V(x,y)$. If $\mathrm{char}k=2$, it is free of rank one and its support is all of $X$.

Its support is the origin $V(x)$ in every characteristic; the module is $k[x]/(x)$ unless $\mathrm{char}k=2$, when it is $k[x]/(x^2)$.

Its support is the entire component $V(x)\subset X$.