Parabolic semi-regular duality and the reverse-column label map

All graded pieces and the crossing trace. A positive root has degree equal to the number of omitted simple roots in its support. Hence the full positive radical is ${⨁}_{d>0}{\mathfrak{g}}_d$, not generally ${\mathfrak{g}}_1$. The simple outside root vectors lie in ${\mathfrak{g}}_1$ and their negatives in ${\mathfrak{g}}_{-1}$; together with ${\mathfrak{g}}_0$ they generate the Lie algebra. For $H\in {\mathfrak{g}}_0$, ${\mathrm{tr}}_{{\mathfrak{g}}_{>0}}\mathrm{ad}H={\sum }_{\alpha \in {\mathrm{\Phi }}^{+}\setminus {\mathrm{\Phi }}_{\bar{J}}^{+}}\alpha (H)=\gamma (H)$ on the Cartan. Both sides vanish on the derived Levi, so this identity holds on all of ${\mathfrak{g}}_0$. In particular $\gamma$ is a Levi character.

Restricted coinduction and the two actions. Every component $N_{-r}$ is finite dimensional and $N_0=\mathbb{C}$. Give $N^{\mathrm{\#}}$ its regular transpose bimodule actions $(nf)(a)=f(an)$ and $(fn)(a)=f(na)$. PBW gives $N^{\mathrm{\#}}{\otimes }_{N}U=N^{\mathrm{\#}}\otimes B$, carrying its right $U$-action. On the other hand restrict a $B$-linear map $U\to {\mathbb{C}}_{\gamma }\otimes B$ to $N$ using $U=B\otimes N$. A homogeneous map of degree $d$ sends $N_{-r}$ into $B_{d-r}$, hence is zero for $r>d$. Consequently it has finite support in $r$, and the remaining domains are finite dimensional. This proves ${\underset{}{\mathrm{Hom}}}_B(U,{\mathbb{C}}_{\gamma }\otimes B)=N^{\mathrm{\#}}\otimes B$ even though $B_0$ is infinite dimensional. The left action is $(XG)(u)=G(uX)$. Its Lie relations follow from associativity in $U$; restriction to $N$ gives an algebraic action on $S$. The left $N$-action and right $B$-action already commute with the entire opposite action.

For homogeneous $X,Y$ of opposite degrees, take the $P$-block trace of $[\mathrm{ad}X,\mathrm{ad}Y]$. Its terms entirely within $P\mathfrak{g}$ form a commutator and have trace zero, leaving exactly $\omega (X,Y)$. The bracket has degree zero; its adjoint trace on ${\mathfrak{g}}_0$ is zero since that algebra is reductive, so step 1.1 gives $\omega (X,Y)=\gamma ([X,Y])$. For degrees whose sum is nonzero the crossing operators have nonzero degree and trace zero, and ${\gamma }_0([X,Y])=0$; degree-zero pairs have no crossing terms and $\gamma$ annihilates their bracket. This proves the displayed identity for all $X,Y$. For degrees $1,-1$, only ${\mathfrak{g}}_0$ crosses the boundary, giving the stated ${\mathfrak{g}}_0$ trace. For the root ${\mathfrak{s}\mathfrak{l}}_2$ example it is $2$, whereas the full trace on ${\mathfrak{s}\mathfrak{l}}_2$ is $4$.

Normal ordering. Choose a basis $H_i$ of ${\mathfrak{g}}_0$. For $X\in {\mathfrak{g}}_1$, PBW uniquely defines maps $C_i^X,R_X:N\to N$ by $aX=Xa+{\sum }_i{H}_i{C}_i^X(a)+R_X(a)$. All these maps have degree $1$; higher negative degrees of $a$ are retained. On $f\otimes 1$ the left formulas are $H(f\otimes 1)=-(f\mathrm{ad}H)\otimes 1+f\otimes (H+\gamma (H))$ and $X(f\otimes 1)=f\otimes X+{\sum }_i(f{C}_i^X)\otimes (H_i+\gamma (H_i))+(f{R}_X)\otimes 1$. For $Y\in {\mathfrak{g}}_{-1}$ right multiplication gives $(f\otimes 1)Y=(f{L}_Y)\otimes 1$, where $L_Y$ is left multiplication on $N$. Comparing the two PBW expansions of $(Ya)X=Y(aX)$, with $[Y,X]={\sum }_i{c}_i{H}_i$, gives $C_i^X{L}_Y=L_Y{C}_i^X+c_i\mathrm{id}$ and $R_X{L}_Y=L_Y{R}_X+{\sum }_i{L}_{[Y,H_i]}{C}_i^X$.

Restricted endomorphisms and the inverse functor. Transposing a homogeneous left $N$-endomorphism of $N^{\mathrm{\#}}$ gives a homogeneous right $N$-endomorphism of $N$, because its homogeneous pieces are finite dimensional. Such an endomorphism is multiplication by its value at $1$. Transposition reverses composition, and summing finitely many degrees gives ${\underset{}{\mathrm{End}}}_N(N^{\mathrm{\#}})=N^{\mathrm{o}\mathrm{p}}$. For unrestricted maps this argument produces a product of degrees: in the polynomial example $x$ acts locally nilpotently on the dual basis, so every power series in $x$ acts, and these are all the endomorphisms. Its $\zeta$-locally finite subspace under commutator is exactly the polynomial subalgebra; unrestricted completion is therefore explicitly excluded.

Substitute these two identities into $X((f\otimes 1)Y)-(X(f\otimes 1))Y$. All terms with a nonconstant operator on $N$ cancel; the remaining scalar is the difference $\gamma ([X,Y])-{\mathrm{tr}}_{{\mathfrak{g}}_0}(\mathrm{ad}X\mathrm{ad}Y)$, multiplied by $f\otimes 1$, and is zero by step 2.1. For $H,Y$ the cancellation is $[\mathrm{ad}H,L_Y]=L_{[H,Y]}$. The same relations on $f\otimes b$ follow by induction on a word in the generators ${\mathfrak{g}}_0,{\mathfrak{g}}_1$ of $B$: commute right multiplication by the next generator past $Y$, whose bracket lies in ${\mathfrak{g}}_0$ or ${\mathfrak{g}}_{-1}$. Since $U$ is generated by ${\mathfrak{g}}_0,{\mathfrak{g}}_{\pm 1}$, both actions commute everywhere. Thus higher radical degrees require no extra assumption; their relations are already relations in the coinduction and tensor models, and their crossing traces were checked in step 2.1.

Left PBW, without a dimension shortcut. The map $U{\otimes }_N{N}^{\mathrm{\#}}\to S$ is left multiplication on the inclusion $f\mapsto f\otimes 1$. Its PBW domain is $B\otimes N^{\mathrm{\#}}$. For a homogeneous $f$ filter first by its nonnegative dual degree and then by the ordinary PBW length of the $B$-word. In the formula for $X$, every correction decreases the dual degree; in the formula for $H$, corrections preserve that degree but decrease the $B$-length. Thus the map on the associated successive quotients is precisely the vector-space interchange $b\otimes f\mapsto f\otimes b$. Inducting over the dual degree and, at fixed degree, over PBW length proves surjectivity. A finite linear relation has a largest pair of filtration indices; its leading symbols are independent under the interchange, proving injectivity. Each expression involves finitely many words, so the argument needs neither finite dimension of $B_0$ nor a completed series.

From the right PBW model $T(M)=N^{\mathrm{\#}}{\otimes }_{N}M$, so free finite rank becomes cofree finite rank. From the left PBW model $H(K)={\underset{}{\mathrm{Hom}}}_N(N^{\mathrm{\#}},K)$, which is free finite rank by step 2.3. Tensor-Hom adjunction supplies the unit and counit. On each shifted rank-one summand they are the identity under the endomorphism identification and the evaluation map, respectively, so both are isomorphisms. A short exact sequence of graded free finite-rank objects splits over $N$ since its quotient is graded free. A sequence of graded cofree objects splits over $N$ since its subobject is graded injective: $N^{\mathrm{\#}}$ is the graded coinduction of the field, and its Hom functor is the exact dual of the corresponding homogeneous component. Hence $T,H$ preserve the exact sequences at issue.

Duality and standards. Restricted duality with the principal antipode gives an exact anti-equivalence from finite cofree to finite free objects; the transpose of that antipode identifies the two regular transpose actions on $N^{\mathrm{\#}}$. Finite homogeneous components give biduality, so composing with $T$ gives $\mathrm{\Phi }$ with inverse $H\circ (-{)}^{\mathrm{\#}}$; no involutivity of $T$ is assumed. For $M=U{\otimes }_{B}E$, tensor cancellation gives $T(M)=S{\otimes }_{B}E=N^{\mathrm{\#}}\otimes E$. On the augmentation vector $\epsilon \otimes E$, the Levi action in step 2.2 is the original $E$-action plus $\gamma$, since $\epsilon \mathrm{ad}H=0$. Dualizing makes the generator space ${\mathbb{C}}_{-\gamma }\otimes E^{\ast }$. The positive radical annihilates this generator space by degree, and the negative regular action makes the dual free on it. The induced PBW map is therefore an isomorphism $\mathrm{\Phi }(M)=U{\otimes }_B({\mathbb{C}}_{-\gamma }\otimes E^{\ast })$. This is a standard for the same parabolic.

A finite standard flag is graded free over $N$ of finite rank, since its standard sections are such and every extension splits over $N$. Applying $\mathrm{\Phi }$ gives a reversed finite standard flag by step 6.1. Conversely $H\circ (-{)}^{\mathrm{\#}}$ sends a standard back to a standard: the transformation $E\mapsto {\mathbb{C}}_{-\gamma }\otimes E^{\ast }$ is involutive, and the unit and counit give its inverse. Induction on flag length thus restricts the equivalence precisely to the standard-flag categories. In category $\mathcal{O}$, $\zeta$ is semisimple; a finite generating set gives finitely many eigenvalue cosets modulo $\mathbb{Z}$. Choose a grading origin on each coset. Each standard has finite-dimensional graded components, because its finite Levi space has scalar $\zeta$-action and the graded pieces of $N$ are finite dimensional; the same holds for finite flags. Weight restriction and degree restriction therefore agree here. All $U$-morphisms preserve the actual $\zeta$-eigenvalues. Changing origins just shifts gradings, so the construction descends to the ungraded flag category using the $\zeta$-locally finite internal Hom. On outputs the eigenvalue cosets are shifted by $\gamma (\zeta )$ and negated by the principal dual; this is part of the construction, not an extra integrality assumption on highest weights.

Reverse-column shifts. The automorphism $c$ preserves the upper Borel and carries $\bar{\mathfrak{p}}$ to $\mathfrak{p}$. If the source Levi highest weight is $\lambda$, that of $E^{\ast }$ is $-w_{\bar{J}}\lambda$, so step 6.1 gives $\nu =-w_{\bar{J}}\lambda -2{\rho }_{\bar{\mathfrak{u}}}$ before applying $c^{\ast }$. Pullback by $c$ sends a highest weight to $-w_0\nu$, giving the displayed weight map. A target block of length $r$ starting after $b$ coordinates has reversed source offset $b^{\mathrm{\prime }}=n-b-r$. The map $w_0{w}_{\bar{J}}$ preserves its position $k=1,\dots ,r$ inside the column. The value of $2{\rho }_{\bar{\mathfrak{u}}}$ on the source block is $n-2b^{\mathrm{\prime }}-r=b-b^{\mathrm{\prime }}$, hence the value of $2w_0{\rho }_{\bar{\mathfrak{u}}}$ on the target block is $b-b^{\mathrm{\prime }}$. The source coordinate $a_i+b^{\mathrm{\prime }}+k-1$ is consequently transported to $a_i+b^{\mathrm{\prime }}+k-1+b-b^{\mathrm{\prime }}=a_i+i-1$. Thus $\mathrm{\Psi }(\bar{\mathrm{\Delta }}(A))=\mathrm{\Delta }(A)$ including the scalar trace-centre coordinate.

With ${\rho }_i=(n+1-2i)\mathrm{/}2$, the shifted coordinate is ${\lambda }_{A,i}+{\rho }_i=a_i+(n-1)\mathrm{/}2$. Reversing columns preserves its multiset, so corresponding source and target standards belong to matching central-character summands; extensions of standards in one summand remain in it, and hence so do the images under $\mathrm{\Psi }$. At fixed total weight, each reversed-column prefix is the complementary suffix of the original reading; subtracting from the total changes the direction of each prefix-dominance inequality. This reverses the tableau order. Neither connectedness of a central summand nor an equivalence on arbitrary objects is used.

Covariant equivalence. Restricted BGG duality is exact and contravariant, sends target standards to costandards and preserves the target central summands. Composing it with the contravariant equivalence $\mathrm{\Psi }$ gives the covariant exact equivalence $F=D\mathrm{\Psi }$ and $F(\bar{\mathrm{\Delta }}(A))=\mathrm{\nabla }(A)$. The label order reversal comes from step 8.1, whereas applying $D$ changes the variance and the kind of flag. This proves all the stated algebraic and filtered-category assertions with their restricted-Hom boundary.