Formal Deformation Quantization and Hochschild Obstructions

On an overlap, write $J_i^a=\mathrm{\partial }{y}^a\mathrm{/}\mathrm{\partial }{x}^i$. By [F2], ${\mathrm{\partial }}_{x^i}f=J_i^a{\mathrm{\partial }}_{y^a}f$ and likewise for $g$. Then [F1] gives ${\pi }_x^{ij}({\mathrm{\partial }}_{x^i}f)({\mathrm{\partial }}_{x^j}g)={\pi }_x^{ij}{J}_i^a{J}_j^b({\mathrm{\partial }}_{y^a}f)({\mathrm{\partial }}_{y^b}g)={\pi }_y^{ab}({\mathrm{\partial }}_{y^a}f)({\mathrm{\partial }}_{y^b}g),$ so the local bracket is independent of the chart. Skew-symmetry and real-bilinearity follow from ${\pi }^{ij}=-{\pi }^{ji}$ and the coordinate formula.

The coordinate product rule gives $\{f,gh{\}}_{\pi }={\pi }^{ij}{f}_i(g_{j}h+g{h}_j)=\{f,g{\}}_{\pi }h+g\{f,h{\}}_{\pi }$. Skew-symmetry gives the derivation rule in the first entry as well. Hence the bracket is a biderivation.

In a chart, expand the cyclic Jacobiator $\{f,\{g,h{\}}_{\pi }{\}}_{\pi }+\{g,\{h,f{\}}_{\pi }{\}}_{\pi }+\{h,\{f,g{\}}_{\pi }{\}}_{\pi }$. Its first-derivative terms are $J^{ijk}{f}_i{g}_j{h}_k$. The second-derivative terms cancel in pairs: the terms with $g_{\mathrm{\ell }j}$ have coefficients ${\pi }^{i\mathrm{\ell }}{\pi }^{jk}$ and ${\pi }^{k\mathrm{\ell }}{\pi }^{ij}$, which cancel after interchanging $j,\mathrm{\ell }$ in the second term and using skew-symmetry; the terms with $h_{\mathrm{\ell }k}$ cancel after interchanging $\mathrm{\ell },k$; the terms with $f_{\mathrm{\ell }i}$ cancel after interchanging $\mathrm{\ell },i$. Thus the displayed Jacobiator formula holds for all smooth $f,g,h$.

If every $J^{ijk}$ vanishes, step 2.2 makes every local Jacobiator zero, so the glued bracket on $C^{\mathrm{\infty }}(M,\mathbb{R})$ satisfies Jacobi. Conversely, assume that global bracket satisfies Jacobi. Fix a point $p$ in a chart $x:U\to x(U)\subseteq {\mathbb{R}}^n$ and indices $i,j,k$. Choose $0<r<R$ with the closed coordinate ball of radius $R$ about $x(p)$ contained in $x(U)$. Define $\theta (t)=0$ for $t\le 0$ and $\theta (t)=e^{-1\mathrm{/}t}$ for $t>0$, and put $\chi (q):=\frac{\theta (R^2-\parallel q-x(p){\parallel }^2)}{\theta (R^2-\parallel q-x(p){\parallel }^2)+\theta (\parallel q-x(p){\parallel }^2-r^2)}\mathrm{.}$ The function $\theta$ is smooth at $0$ because all its right derivatives there tend to zero, and the denominator defining $\chi$ is everywhere positive. Thus $\chi$ is smooth, equals $1$ near $x(p)$, and vanishes outside the closed ball, which is compactly contained in $x(U)$. The functions $f,g,h$ defined on $U$ by $(\chi \circ x)x^i,(\chi \circ x)x^j,(\chi \circ x)x^k$ and extended by zero outside $U$ are therefore smooth global functions and agree with the corresponding coordinates near $p$. By step 2.2 their global Jacobiator at $p$ is $J^{ijk}(p)$, so global Jacobi forces $J^{ijk}(p)=0$. Since $p$ and the indices were arbitrary, every coefficient vanishes in every chart. The Jacobiator is formed from the chart-independent bracket of step 1.1, so the criterion is coordinate-independent. By [F3], the Schouten self-bracket vanishes exactly when these coefficients vanish, since $2$ is invertible over $\mathbb{R}$. This proves the claimed Poisson and Schouten criteria and justifies the bracket in General Poisson manifolds from bivector fields.

For any associative product, the commutator $[a,b{]}_{\star }=a\star b-b\star a$ satisfies the Jacobi identity. Expanding $[[f,g{]}_{\star },k{]}_{\star }$ gives $(f\star g)\star k-(g\star f)\star k-k\star (f\star g)+k\star (g\star f)$. After reassociating each term, this is $f\star g\star k-g\star f\star k-k\star f\star g+k\star g\star f$; the two cyclic rotations are $g\star k\star f-k\star g\star f-f\star g\star k+f\star k\star g$ and $k\star f\star g-f\star k\star g-g\star k\star f+g\star f\star k$. Their sum is zero by pairwise cancellation.

Associativity also gives the exact derivation identity $[f,g\star k{]}_{\star }=[f,g{]}_{\star }\star k+g\star [f,k{]}_{\star }$: expanding both sides and reassociating the four triple products leaves the same expression. Comparing coefficients of $\mathrm{\hslash }$ gives $\beta (f,gk)=\beta (f,g)k+g\beta (f,k)$.

Since $fg=gf$, the coefficient of $\mathrm{\hslash }$ in $[f,g{]}_{\star }$ is exactly $\beta (f,g)$. The coefficient of ${\mathrm{\hslash }}^2$ in the commutator Jacobi identity from step 1.1 is $\beta (\beta (f,g),k)+\beta (\beta (g,k),f)+\beta (\beta (k,f),g)=0$. Skew-symmetry of $\beta$ converts this to the Jacobi identity in [F2].

By definition, $\beta$ is real-bilinear and skew-symmetric. The derivation rule in step 1.2 and skew-symmetry give the derivation rule in the first entry; step 2.1 gives Jacobi. These are exactly the Poisson-algebra axioms in [F2].

Write $\varphi {\circ }_i\psi$ for insertion of a $q$-cochain into slot $i$ of a $p$-cochain, with slots numbered from $0$, so $\varphi \circ \psi ={\sum }_{i=0}^{p-1}(-1{)}^{i(q-1)}\varphi {\circ }_i\psi$. In $(\varphi \circ \psi )\circ \chi -\varphi \circ (\psi \circ \chi )$, terms where $\chi$ is inserted into the block occupied by $\psi$ cancel. If that block starts at slot $i$ and $\chi$ is inserted into its local slot $s$, the first expression has coefficient $(-1{)}^{i(q-1)+(i+s)(r-1)}$, while the second has coefficient $(-1{)}^{i(q+r-2)+s(r-1)}$; these exponents agree. Every remaining term is a pair of insertions into distinct original slots of $\varphi$. For original slots $i<j$, the coefficient with $\psi$ inserted first and $\chi$ second is $(-1{)}^{i(q-1)+(j+q-1)(r-1)}$; with the insertion order swapped it is $(-1{)}^{j(r-1)+i(q-1)}$. Their ratio is $(-1{)}^{(q-1)(r-1)}$. The same comparison for $j<i$ puts the earlier insertion first and gives that same ratio. Thus, with $\mathrm{\mid }\psi \mathrm{\mid }=q-1$ and $\mathrm{\mid }\chi \mathrm{\mid }=r-1$, $\mathrm{as}(\varphi ,\psi ,\chi )=(-1{)}^{\mathrm{\mid }\psi \mathrm{\mid }\mathrm{\mid }\chi \mathrm{\mid }}\mathrm{as}(\varphi ,\chi ,\psi ),\mathrm{as}(u,v,w):=(u\circ v)\circ w-u\circ (v\circ w)\mathrm{.}$

Since $m$ has shifted degree $1$, $[m,m{]}_G(a,b,c)=2((ab)c-a(bc))=0$ by associativity of $A$. Thus $\mathrm{\partial }=[m,-{]}_G$ has degree $+1$.

The bracket is graded skew by its defining formula. Put $e_{uv}=(-1{)}^{\mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }v\mathrm{\mid }}$. Expanding the graded Jacobi identity, the difference between its right and left sides is $\begin{array}{rl}{\displaystyle }& {\displaystyle \mathrm{as}(\varphi ,\psi ,\chi )-e_{\varphi \psi }\mathrm{as}(\psi ,\varphi ,\chi )+e_{\varphi \psi }{e}_{\varphi \chi }\mathrm{as}(\psi ,\chi ,\varphi )-e_{\varphi \psi }{e}_{\varphi \chi }{e}_{\psi \chi }\mathrm{as}(\chi ,\psi ,\varphi )}\\ {\displaystyle }& {\displaystyle +e_{\varphi \chi }{e}_{\psi \chi }\mathrm{as}(\chi ,\varphi ,\psi )-e_{\psi \chi }\mathrm{as}(\varphi ,\chi ,\psi )\mathrm{.}}\end{array}$ Step 1.1 gives $\mathrm{as}(\varphi ,\psi ,\chi )=e_{\psi \chi }\mathrm{as}(\varphi ,\chi ,\psi )$, $\mathrm{as}(\psi ,\varphi ,\chi )=e_{\varphi \chi }\mathrm{as}(\psi ,\chi ,\varphi )$, and $\mathrm{as}(\chi ,\psi ,\varphi )=e_{\varphi \psi }\mathrm{as}(\chi ,\varphi ,\psi )$. Substitution cancels the six terms in three pairs, proving graded Jacobi.

Apply the graded Jacobi identity from step 2.1 with the first two entries equal to $m$. It gives $2{\mathrm{\partial }}^2\varphi =[[m,m{]}_G,\varphi {]}_G=0$, so ${\mathrm{\partial }}^2=0$ over $\mathbb{R}$. Applying the same identity with entries $m,\varphi ,\psi$ gives $\mathrm{\partial }[\varphi ,\psi {]}_G=[\mathrm{\partial }\varphi ,\psi {]}_G+(-1{)}^{\mathrm{\mid }\varphi \mathrm{\mid }}[\varphi ,\mathrm{\partial }\psi {]}_G$. Hence these operations form a differential graded Lie algebra.

Substitute $m$ and $\varphi$ into the insertion formula in [F1]. The two outer insertions give $\varphi (a_1,\dots ,a_p)a_{p+1}$ and $(-1{)}^{p-1}{a}_1\varphi (a_2,\dots ,a_{p+1})$. The internal insertions give $-(-1{)}^{p-1}{\sum }_{i=1}^p(-1{)}^{i-1}\varphi (a_1,\dots ,a_{i-1},a_i{a}_{i+1},a_{i+2},\dots ,a_{p+1})$. Comparing this with the displayed formula for $\delta$ in [F1] gives $\mathrm{\partial }\varphi =(-1{)}^{p-1}\delta \varphi$. Since ${\mathrm{\partial }}^2=0$, this also proves ${\delta }^2=0$.

If a normalized cochain is evaluated on a tuple containing $1$, every insertion term vanishes: the unit either enters the inner cochain, which is normalized, or remains an input of the outer one. In $\delta \varphi$, the only potentially nonzero terms when $a_j=1$ are the two neighboring products (or the corresponding endpoint product and boundary term); they have identical cochain arguments and opposite signs, so cancel. Finally, finite compositions of local finite-order differential operators remain finite-order differential operators by the product and chain rules. Thus the normalized multidifferential cochains are closed under the differential and the bracket.

Let $m(f,g)=fg$ and $x={\sum }_{r=1}^{n-1}{\mathrm{\hslash }}^r{B}_r$, so $m+x$ is the truncated product. For arity-two cochains the insertion formula gives $\frac{1}{2}[m+x,m+x{]}_G(f,g,h)=((f{\star }_{<n}g){\star }_{<n}h)-f{\star }_{<n}(g{\star }_{<n}h)$. Thus its coefficient at order $n$, where there is no $B_n$ term, is precisely the displayed $O_n$; all lower coefficients vanish by the assumed partial associativity. The same insertion formula gives $\frac{1}{2}{\sum }_{i+j=n}[B_i,B_j{]}_G={\sum }_{i+j=n}{B}_i\circ B_j=O_n$, because the bracket of two 2-cochains is symmetric.

In the shifted Hochschild DGLA put $R=\mathrm{\partial }x+\frac{1}{2}[x,x{]}_G=\frac{1}{2}[m+x,m+x{]}_G$; the last equality uses $[m,m{]}_G=0$. The DGLA identities give the Bianchi identity $\mathrm{\partial }R+[x,R{]}_G=\frac{1}{2}\mathrm{\partial }[x,x{]}_G+[x,\mathrm{\partial }x{]}_G+\frac{1}{2}[x,[x,x{]}_G{]}_G=0.$ Indeed, the derivation rule and graded skew-symmetry make the first two terms cancel, and graded Jacobi gives $3[x,[x,x{]}_G{]}_G=0$, hence the last term vanishes over $\mathbb{R}$. By step 1.1, $R$ starts at order ${\mathrm{\hslash }}^n$ with coefficient $O_n$, while $x$ starts at order $\mathrm{\hslash }$; therefore the coefficient of ${\mathrm{\hslash }}^n$ in the Bianchi identity is $\mathrm{\partial }{O}_n=0$.

By [F3], $\mathrm{\partial }$ on 3-cochains is $\delta$, so $\delta O_n=0$. Each summand of $O_n$ is a composition of normalized bidifferential cochains. An input equal to $1$ either makes the inner cochain vanish or remains a unit input of the outer cochain; composition also preserves finite differential order. Thus $O_n$ lies in the stated normalized multidifferential subcomplex.

Add ${\mathrm{\hslash }}^n{B}_n$ to the truncated product. The order-$n$ terms in its associator that are linear in $B_n$ are $B_n(f,g)h+B_n(fg,h)-f{B}_n(g,h)-B_n(f,gh)=-(\delta B_n)(f,g,h)\mathrm{.}$ No term quadratic in $B_n$, or involving $B_n$ and a positive-order coefficient, contributes at order $n$, since those terms have order at least $2n$ or $n+1$. Hence the new order-$n$ associator coefficient is $O_n-\delta B_n$.

The lower associator coefficients already vanish. Therefore ${\star }_{<n}+{\mathrm{\hslash }}^n{B}_n$ is associative modulo ${\mathrm{\hslash }}^{n+1}$ exactly when $\delta B_n=O_n$. By step 2.2, $O_n$ is a cocycle in the normalized multidifferential complex; such a $B_n$ exists precisely when its cohomology class is zero. For $n=2$, the defining sum has only $(i,j)=(1,1)$, yielding $O_2=\frac{1}{2}[B_1,B_1{]}_G$. This proves the finite-order criterion without asserting that later obstruction classes vanish.

The coefficient of $\mathrm{\hslash }$ in associativity of $\star$ is $B_1(f,g)h+B_1(fg,h)-f{B}_1(g,h)-B_1(f,gh)=0$, which is $-(\delta B_1)(f,g,h)=0$. The same calculation applies to ${\star }^{\mathrm{\prime }}$, so both first coefficients are normalized 2-cocycles in the complex of [F4].

Comparing coefficients of $\mathrm{\hslash }$ in $T{T}^{-1}=\mathrm{id}$ gives $T^{-1}=\mathrm{id}-\mathrm{\hslash }{D}_1+O({\mathrm{\hslash }}^2)$.

Expand the defining transformation from [F2] through first order: $Tf\star Tg=fg+\mathrm{\hslash }(D_1(f)g+f{D}_1(g)+B_1(f,g))+O({\mathrm{\hslash }}^2)\mathrm{.}$ Applying $T^{-1}$ subtracts $\mathrm{\hslash }{D}_1(fg)$, hence $B_1^{\mathrm{\prime }}=B_1+f{D}_1(g)-D_1(fg)+D_1(f)g=B_1+\delta D_1$ by [F3].

Since $A$ is commutative, $\delta D_1(f,g)=f{D}_1(g)-D_1(fg)+D_1(f)g$ equals $\delta D_1(g,f)$. Thus the skew parts of $B_1$ and $B_1^{\mathrm{\prime }}$ agree, and their difference is the coboundary $\delta D_1$. With step 1.1 this proves equality of their classes in the cohomology of [F4].