All-class Artin lifting for a curved BV calculus

We first give the Artin exactness argument used throughout. Every flat module M over an Artin local A is free, including infinite-rank M. Lift a k-basis of $M\mathrm{/}{m}_{A}M$ to obtain a map from a free module onto M: its cokernel Q satisfies $Q=m_{A}Q$, hence $Q=m_A^{r}Q=0$ for r with $m_A^r=0$. Flatness gives $To{r}_1^A(M,k)=0$, so the kernel K of that map satisfies $K\mathrm{/}{m}_{A}K=0$ and is zero by the same nilpotent argument. For a bounded flat complex $C_A$ with $C_A$ tensor $k=C_k$, split the central complex into its cohomology representatives H, boundaries B and complementary L, with $d_k:L^n$ isomorphic to $B^{n+1}$. Lift those bases to free A-modules. The B component of $d_A$ on L is 1 plus an $m_A$-valued map, hence invertible by a finite geometric series. Set $B^{\mathrm{\prime }}=d_A(L)$; projection B' to B is an isomorphism, so $C_A=B^{\mathrm{\prime }}$ direct-sum L direct-sum H as graded modules and d:L to B' is an isomorphism. Replace h in H by $h-(d\mathrm{\mid }L{)}^{-1}p{r}_{B^{\mathrm{\prime }}}dh$. Its differential has no B' component; applying $d^2=0$ shows that differential is again in this graph of H. Thus $C_A$ is a direct sum of a contractible complex and a complex (H tensor A,D) with D valued in $m_A$. This is a literal Gaussian cancellation; all its inverse matrices are finite nilpotent series.

If $H^n(C_A)$ to $H^n(C_k)$ is surjective in every degree, choose closed lifts of the central H basis in the residual complex. They are the columns of an A-linear map U on H tensor A with $U=1$ modulo $m_A$ and $DU=0$. The inverse of U is a finite nilpotent geometric series, so $D=0$. Consequently $H^n(C_A)=H^n(C_k)$ tensor A is free and the actual base-change maps are isomorphisms: tensoring the constructed direct-sum cancellation with any A to B preserves its contraction and its zero residual differential. These conclusions apply to $(G_A,\breve{d})$. To see that it is a complex, put $w=\mathrm{\ell }+y$ and use F1: $(\bar{\mathrm{\partial }}+\mathrm{\Delta }{)}^2=[w,-]$, whereas its graded anticommutator with wedge by the odd w is wedge by (barpartial w+Delta w) minus [w,-]. Here barpartial w+Delta $w=0$ and w wedge $w=0$. Thus breve $d^2=0$. Its central cohomology is finite by F3 and the finite bounds. We obtain finite free corrected-total cohomology and its canonical base change at every Artin A.

Products of formal coefficients introduce no extra flatness premise. For a flat $M_A$ write it, noncanonically, as A tensor V by step 1.1. Since A is finite dimensional over $k$, $M_A[\text{\hspace{0.17em}⁣}[h]\text{\hspace{0.17em}⁣}]$ is A tensor $V[\text{\hspace{0.17em}⁣}[h]\text{\hspace{0.17em}⁣}]$, hence is A-free; its Laurent localization is A-flat as well. For an Artin map A to B, B is finitely presented as an A-module. A finite presentation of B, followed by coefficientwise products of the resulting exact module sequences, proves $M_A[\text{\hspace{0.17em}⁣}[h]\text{\hspace{0.17em}⁣}]$ ${\otimes }_{A}B$ is $M_B[\text{\hspace{0.17em}⁣}[h]\text{\hspace{0.17em}⁣}]$; localizing gives the Laurent statement. The product sequences are exact because coefficientwise primitives can be chosen. Thus a small extension gives actual short exact sequences of these coefficient modules with kernel $I{\otimes }_k{M}_k[\text{\hspace{0.17em}⁣}[h]\text{\hspace{0.17em}⁣}]$, or its Laurent version. No flatness of a global $\mathrm{\Lambda }[\text{\hspace{0.17em}⁣}[h]\text{\hspace{0.17em}⁣}]$ module is used.

Define $D_h=\bar{\mathrm{\partial }}+h\mathrm{\Delta }+h^{-1}\mathrm{\ell }$ wedge+y wedge on $G_A((h))$. With $w_h=h^{-1}\mathrm{\ell }+y$, the same calculation as step 2.1 gives $D_h^2=0$: $(\bar{\mathrm{\partial }}+h\mathrm{\Delta }{)}^2=[\mathrm{\ell }+h$ y,-], its anticommutator with wedge $w_h$ subtracts $h[w_h,-]$, and the remaining multiplication is zero by F1. The diagonal map $\rho (\theta )=h^{-p-1}\theta$ for theta in $G_A^{p,q}$ is an isomorphism of complexes from $D_h$ to breve d. Indeed the barpartial and y terms keep p, the h Delta term changes p to p+1, and the $h^{-1}\mathrm{\ell }$ term changes p to p-1; in each case the two total h exponents agree. The p-range is finite, so rho and its inverse preserve Laurent series. This is the integral-power version of the source scaling $\sigma (\theta )=u^{q-p-2}\theta ,u^2=h$, followed in total degree m by $u^{-m}$. Cohomology for an h-independent differential is coefficientwise, so $H^n(G_A((h)),D_h)$ is naturally $H^n(G_A,\breve{d})$((h)). step 2.1 therefore makes its $H^0$ base change surjective over a small extension. The long exact sequence then makes $H^1(G_k((h)),{\bar{\mathrm{\partial }}}_k+h{\mathrm{\Delta }}_k)$ tensor I inject into $H^1(G_A((h)),D_h)$.

We prove the precise central Rees injection needed to remove negative powers. Central E1 degeneration implies that every ${\bar{\mathrm{\partial }}}_k$-closed homogeneous a has a finite Rees lift Z(h) with $Z(0)=a$ and $({\bar{\mathrm{\partial }}}_k+h{\mathrm{\Delta }}_k)Z=0$: start in its column and solve barpartial $a_{p+1}=-\mathrm{\Delta }{a}_p$ successively. Each obstruction is the corresponding spectral differential and vanishes by F3. There are only finitely many columns; an initial exact difference in the leading representative can be removed by a Rees boundary. Sum this construction over the finitely many columns in a fixed total degree. If $hz=({\bar{\mathrm{\partial }}}_k+h{\mathrm{\Delta }}_k)b$ in the power-series complex, b(0) is ${\bar{\mathrm{\partial }}}_k$-closed. Subtract its closed Rees lift from b and divide by h, which is injective on coefficient series. It follows that z is already a power-series boundary. Thus the cohomology has no h-torsion. Iterating shows that a power-series cycle which is exact in the Laurent complex is exact in the power-series complex. This remains true after tensoring the finite vector space I. No assertion that separate regularity or a formal inverse automatically removes negative powers has been used.

We now establish the column base change needed for arbitrary classical classes, without assuming that they have semi-classical enhancements. Let phi be ANY classical solution over A and put $i=i_{\varphi }$ on forms. Set $Q=\mathrm{\partial }+\bar{\mathrm{\partial }}+i_{\mathrm{\ell }}$. From F2 and barpartial $\omega =i_y\omega$ one has $Q\kappa =\kappa$ breve d: in the predifferential rule the term $(-1{)}^{\mathrm{\mid }\theta \mathrm{\mid }}{i}_{\theta }{i}_y$ equals $i_y{i}_{\theta }$, so it contributes exactly kappa(y wedge theta). Hence $Q^2=0$ and its total cohomology has the free base change of step 2.1, with degree reindexing by d. The Cartan commutators in F2 give the finite conjugation $e^{-i_{\varphi }}Q{e}^{i_{\varphi }}=\mathrm{\partial }+\bar{\mathrm{\partial }}+L_{\varphi }+i_{\mathrm{\ell }+\bar{\mathrm{\partial }}\varphi +[\varphi ,\varphi ]\mathrm{/}2}=\mathrm{\partial }+{\bar{\mathrm{\partial }}}_{\varphi }\mathrm{.}$ Indeed the first commutator is $L_{\varphi }+i_{\bar{\mathrm{\partial }}\varphi }$; the second is $i_{[}\varphi ,\varphi ]$; all higher commutators vanish because contractions commute with $i_{\varphi }$. Thus $(A_A,\mathrm{\partial },{\bar{\mathrm{\partial }}}_{\varphi })$ is an actual double complex. The anticommutation with partial follows from the Cartan formula and ${\mathrm{\partial }}^2=0$, and the remaining square is zero by the conjugation. This is the precise sign-compatible Cartan comparison, with $e^{+i_{\varphi }}$ on the right.

For Phi in $m_A{G}_A^0[\text{\hspace{0.17em}⁣}[h]\text{\hspace{0.17em}⁣}]$, put $F_h(\mathrm{\Phi })=\bar{\mathrm{\partial }}$ Phi+h Delta Phi+[Phi,Phi]/2+ell+h y. The BV identity and the product derivation give $D_h\exp (\mathrm{\Phi }\mathrm{/}h)=h^{-1}{F}_h(\mathrm{\Phi })$ wedge exp(Phi/h). This follows by $\mathrm{\Delta }({\mathrm{\Phi }}^n)=n{\mathrm{\Phi }}^{n-1}\mathrm{\Delta }\mathrm{\Phi }+n(n-1){\mathrm{\Phi }}^{n-2}[\mathrm{\Phi },\mathrm{\Phi }]\mathrm{/}2$, since Phi has even total degree. The exponential is finite in the nilpotent $m_A$ direction and is a Laurent series with a finite lower bound; its inverse is exp(-Phi/h). Hence the quantum equation $F_h(\mathrm{\Phi })=0$ is equivalent to $D_h\exp (\mathrm{\Phi }\mathrm{/}h)=0$, without inferring the equation merely from the vanishing of an adjoint operator.

We give the finite-length Hodge argument explicitly. For any bounded flat complex $K_A$ with central cohomology dimension $h_q$, induction along a composition series of A using its short exact coefficient complexes gives ${\mathrm{length}}_k{H}^q(K_A)\le {\mathrm{length}}_k(A)h_q$; the long exact sequence proves this bound even when the terms have infinite rank. It also proves finiteness of these cohomology groups. Apply this to each column of the double complex in step 3.3. Its total cohomology is free and base-change compatible by that step, so its length in total degree n is ${\mathrm{length}}_k(A)$ times the central total cohomology dimension. The central E1 degeneration identifies that dimension with the sum of the central column dimensions in total degree n. On the other hand the finite column spectral sequence gives length $H^n(\text{total})\le {\sum }_{i+q=n}\mathrm{length}{H}^q(\text{column}i)$. Combining with the preceding upper bounds makes the two endpoints equal. Every deficit is nonnegative, so every column reaches its maximal length, and every spectral differential is zero. For a small quotient A to B the long exact column sequence has middle length equal to length $H^q(\text{column}B)+{\dim }_k(I)h_q$. Equality forces I tensor $H^q(\text{column}k)$ to inject and $H^q(\text{column}A)$ to surject onto $H^q(\text{column}B)$. In particular each column surjects onto its central cohomology. Lift a central basis; nilpotent Nakayama gives a surjection $A_q^h$ onto the column cohomology, and equality of lengths makes it an isomorphism. Thus all column cohomology is finite free. For ANY Artin base change A to B, the canonical map from this free cohomology tensored with B to the B-column cohomology reduces to the identity on the central basis. Between free modules of equal rank it is invertible by nilpotent Nakayama. This proves actual column base change, not merely equality of dimensions.

The top column required here is $A_A^{d,\ast }$, not an unqualified column of index zero. For f in $m_A{G}_A^{0,0}$ put ${\omega }_f=e^f$ contracted with omega and ${\kappa }_f(\theta )=\theta$ contracted with ${\omega }_f$. Its inverse is obtained by multiplication by $e^{-f}$. The BV identity, Delta $f=[f,f]=0$ and $[e^f,\theta ]=e^f[f,\theta ]$ give $\mathrm{\partial }{\kappa }_f(\theta )={\kappa }_f(\mathrm{\Delta }\theta +[f,\theta ])$. Since $\mathrm{\partial }{\omega }_f=0$ by top degree, $L_{\varphi }{\omega }_f=\mathrm{\partial }(i_{\varphi }{\omega }_f)$. Using barpartial $\omega =i_y\omega$ now gives ${\bar{\mathrm{\partial }}}_{\varphi }{\omega }_f={\kappa }_f(\bar{\mathrm{\partial }}f+[\varphi ,f]+y+\mathrm{\Delta }\varphi )\mathrm{.}$ The sign [f,phi]=[phi,f] is positive because both have total degree zero and the bracket has degree one. This identity proves that the semi-classical second equation is equivalent to the actual closedness of the top volume omega_f.

Let A to B be small and let ${\mathrm{\Phi }}_B$ be ANY quantum solution. Choose a coefficientwise lift tildePhi in $m_A{G}_A^0[\text{\hspace{0.17em}⁣}[h]\text{\hspace{0.17em}⁣}]$. Its defect $c=F_h(\tilde{\mathrm{\Phi }})$ lies in I tensor $G_k^1[\text{\hspace{0.17em}⁣}[h]\text{\hspace{0.17em}⁣}]$. The equality $D_h\exp (\tilde{\mathrm{\Phi }}\mathrm{/}h)=c\mathrm{/}h$ is exact because $I{m}_A=0$, so multiplying c/h by the exponential changes nothing. Applying $D_h$ again makes c/h a central Laurent cycle. Its image in the A-complex is a boundary. The $H^1$ injection in step 3.1 therefore makes it a boundary in the I-valued central Laurent complex. Multiplying its Laurent primitive by h shows c exact there as well. step 3.2 supplies a POWER-SERIES primitive gamma in I tensor $G_k^0[\text{\hspace{0.17em}⁣}[h]\text{\hspace{0.17em}⁣}]$ with $({\bar{\mathrm{\partial }}}_k+h{\mathrm{\Delta }}_k)\gamma =c$. Then ${\mathrm{\Phi }}_A=\tilde{\mathrm{\Phi }}-\gamma$ is a quantum solution: every cross bracket contains $I{m}_A$, every quadratic correction contains $I^2$, and its remaining defect is $c-({\bar{\mathrm{\partial }}}_k+h{\mathrm{\Delta }}_k)\gamma =0$. It reduces to the originally given ${\mathrm{\Phi }}_B$. This proves all-element quantum lifting across a small extension.

Let A to B be a surjection, let $({\varphi }_B,f_B)$ be a semi-classical pair, and let ${\varphi }_A$ be ANY prescribed classical lift of ${\varphi }_B$. Apply step 4.2 to that actual ${\varphi }_A$. The map $H^0(A_A^{d,\ast },{\bar{\mathrm{\partial }}}_{{\varphi }_A})$ to $H^0(A_B^{d,\ast },{\bar{\mathrm{\partial }}}_{{\varphi }_B})$ is surjective. Since q is nonnegative there are NO degree-minus-one boundaries in these columns, so $H^0$ is the space of closed sections itself. Consequently the closed volume ${\omega }_{f_B}$ has a closed lift $et{a}_A$ restricting EXACTLY to ${\omega }_{f_B}$, not only to its cohomology class. Write $g={\kappa }^{-1}(et{a}_A)$ in $G_A^{0,0}$. Its reduction is $e^{f_B}$, and its central reduction is1, so $g=1+v$ with v in $m_A{G}_A^{0,0}$. The finite logarithm $f_A=\log (1+v)$ lies in that maximal-ideal module, reduces to $f_B$, and satisfies $e^{f_A}=g$. step 4.3 now says $({\varphi }_A,f_A)$ is semi-classical. This proves smoothness of the forgetful map while respecting the given classical lift; no other classical element is substituted for it.

Call a quantum solution obtuse if its $h^0,G^{0,0}$ coefficient is zero. An obtuse ${\mathrm{\Phi }}_B$ has a quantum lift by step 5.1. Let z be the $h^0,G^{0,0}$ coefficient of that lift; it lies in I tensor $G_k^{0,0}$. The $h^0,G^{0,1}$ part of its quantum equation says barpartial $z+[z,{\mathrm{\Phi }}_0^{-1,1}]=0$. The bracket term vanishes by $I{m}_A=0$, so barpartial $z=0$; also Delta $z=0$ by the p-range. Subtracting z preserves the whole quantum equation and makes the lift obtuse. A semi-classical pair (phi,f) gives the obtuse quantum solution $\mathrm{\Phi }=\varphi +hf$: Delta $f=[f,f]=0$ by the p-range, and its two remaining coefficients are exactly the two semi-classical equations. Conversely, from ANY obtuse quantum lift extract ${\varphi }^{\mathrm{\prime }}={\mathrm{\Phi }}_{h^0}^{-1,1}$ and $f^{\mathrm{\prime }}={\mathrm{\Phi }}_{h^1}^{0,0}$. The $h^0,G^{-1,2}$ equation gives the classical equation for phi'; its only other possible bracket terms involve the absent apex. The $h^1,G^{0,1}$ equation gives the semi-classical equation for (phi',f'); its other possible bracket terms again involve that apex. Thus every given semi-classical pair lifts across a small extension.

Every Artin surjection factors into small ones by a composition series of its kernel as a module over the source local ring; its simple factors are killed by the maximal ideal. Thus step 6.1 makes the semi-classical functor smooth for every surjection. Given ANY classical phi_B, apply step 5.2 to B to k and the central pair (0,0) to obtain a semi-classical enhancement of precisely phi_B. Lift that pair across the given surjection A to B by step 6.1 and the small factorization, and then forget f. The resulting classical phi_A restricts to the ORIGINAL phi_B as an element. This proves the all-class assertion, not merely existence of one formal smoothing or smoothness after replacing a class by a preferred representative.