Gan-Ginzburg and Skryabin inputs for the type-A pyramid good grading

Positive-weight transverse product. Choose the usual rowwise ${\mathfrak{s}\mathfrak{l}}_2$ triple. On $\mathrm{Hom}(R_b,R_a)$ the strings have highest $h$-weight $p_a+p_b-2k$, $1\le k\le \min (p_a,p_b)$. Since the pyramid grading element differs from $h$ by row scalars, their highest and lowest grading weights are $2(p_b-k)\ge 0$ and $-2(p_a-k)\le 0$. Thus the negative-degree part is injected by $\mathrm{ad}e$ and gives the direct-sum complement of [F1]. The differential of $\alpha (u,z)=uz{u}^{-1}$ at $(1,e)$ is $(x,y)\mapsto [x,e]+y$, an isomorphism. Use weight $-H$ on $\mathfrak{m}$ and weight $2-H$ on the translated slice and target coordinates; these are positive. In homogeneous coordinates of increasing weight, each component of $\alpha$ is an invertible linear block in the variables of that weight plus a polynomial in lower-weight variables. Successive substitution constructs its polynomial inverse. In particular derivative invertibility alone is not being used as a global inverse theorem.

Detection. The shifted operators $x-\chi (x)$ form a representation of the nilpotent Lie algebra $\mathfrak{m}$, because $\chi$ kills brackets. A nonzero central element has a nonzero kernel on any nonzero locally nilpotent module: apply its largest nonvanishing power to a nonzero vector. That kernel is stable, and the quotient Lie algebra acts on it with the same local nilpotence. Induction on $\dim \mathfrak{m}$ therefore gives a common nonzero invariant. The case $\mathfrak{m}=0$ is immediate.

The actual graded complex. PBW and [F3] give $\mathrm{gr}{Q}_{\chi }=\mathbb{C}[e+{\mathfrak{m}}^{\perp }]$, with no additional symbol factor. Step 1.1 identifies this with $\mathbb{C}[M_{\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{p}}]\otimes \mathbb{C}[S]$. Infinitesimal conjugation acts by invariant vector fields on the first factor and trivially on the slice factor. Giving the exterior dual of a grading-degree $-2r$ vector weight $2r$ makes the twisted Chevalley–Eilenberg differential filtration preserving. Its graded complex is the polynomial de Rham complex of $M_{\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{p}}$, tensored with $\mathbb{C}[S]$: invariant vector fields trivialize the tangent bundle and the bracket terms are precisely the de Rham differential. Polynomial exponential coordinates identify this unipotent group with affine space. On a polynomial form of positive total degree $t$, $t^{-1}{\iota }_{\mathrm{E}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{r}}$ is a contracting homotopy; constants are the only surviving degree-zero classes.

Filtered lifting and right freeness. A positive-degree cocycle in the actual complex has a graded leading term which is a boundary by step 2.1. Lift that boundary and subtract it, strictly lowering the integer filtration degree; boundedness below terminates this process. The same procedure lifts every graded invariant, using vanishing of graded degree-one cohomology to remove its differential. Hence $H^{>0}(\mathfrak{m},Q_{\chi })=0$ and $\mathrm{gr}W=\mathbb{C}[S]$, since $H^0(\mathfrak{m},Q_{\chi })=W$ by the defining invariant realization. Using the fixed PBW vector-space identification $U(\mathfrak{p})\to Q_{\chi }$, take homogeneous coordinate representatives in $U(\mathfrak{p})$ and form their ordered products there before projecting to $Q_{\chi }$. This never multiplies abstract classes in a left-ideal quotient. Their projected ordered monomials, acted on by the right $W$ bimodule action, span $Q_{\chi }$ by finite descending leading-term subtraction. A finite relation among them would give a relation between distinct monomials over $\mathrm{gr}W$; independence of polynomial monomials rules this out. Thus $Q_{\chi }$ is right $W$-free. All statements include the independent scalar coordinate of ${\mathfrak{g}\mathfrak{l}}_N$.

Transfer to every module without graded-flatness assumptions. The finite augmented cochain complex $0\to W\to C^0\to \cdots \to C^{\dim \mathfrak{m}}\to 0$, where $C^j=Q_{\chi }\otimes {\bigwedge }^j{\mathfrak{m}}^{\ast }$, is exact by step 3.1 and is a complex of right $W$-modules. Its differentials are right $W$-linear because the right action commutes with the twisted left $\mathfrak{m}$ action. Each term is right $W$-free. Starting at the last term, split each surjection onto the already-projective boundary module; descending induction makes every boundary projective and splits every short exact sequence, including $W\hookrightarrow C^0$. These splittings supply a right $W$-linear contraction of the augmented complex. Tensor the contraction with an arbitrary left $W$-module $X$. This remains a contraction, and the resulting complex is exactly the twisted Chevalley–Eilenberg complex of $Q_{\chi }{\otimes }_{W}X$. Consequently its positive cohomology vanishes and its degree-zero cohomology is $X$ via $x\mapsto 1_{\chi }\otimes x$. No filtration or flatness of $X$ is assumed. The splittings use exactly [F4].

The preceding steps prove the three assertions with the doubled even grading. In particular there is no degree $-1$ symplectic or Clifford contribution. If $e=0$, then $\mathfrak{m}=0$ and the product, augmentation and all cohomology identities reduce to their degree-zero identities.