Parabolic Category O and Fock-Space Bases — Examples

The bar matrix. Each natural-factor bar fixes its two coordinate vectors. The only nonconstant quasi-$R$ term acting on this tensor square is $(s^{-1}-s)E_0\otimes F_0$; it takes $M_{10}$ to $(s^{-1}-s)M_{01}$ and kills the other three monomials. Higher terms vanish because $E_0^2=F_0^2=0$ on a natural factor. This gives the displayed $B_{\psi }$ in the mixed order $(M_{01},M_{10})$.

The four families. A vector $M_{10}+c{M}_{01}$ is bar fixed exactly when $c-\bar{c}=s^{-1}-s$. The two triangular charts give $c=-s$ for the $s\mathbb{Z}[s]$-chart and $c=s^{-1}$ for the $s^{-1}\mathbb{Z}[s^{-1}]$-chart, that is $L_{10}=M_{10}-s{M}_{01}$ and $T_{10}=M_{10}+s^{-1}{M}_{01}$. The quasi-canonical family is defined by the inverse transition matrix, giving $P_{01}=M_{01}+s{M}_{10}$ and $P_{10}=M_{10}$; the reverse reading gives ${\tilde{L}}_{01}={\tilde{M}}_{01}-s{\tilde{M}}_{10}$. Since the off-diagonal entries are $-s$, $s^{-1}$ and $s$, the four families are pairwise different over $\mathbb{Q}(s)$.

The Gram matrices and delta identities. The defining forms have labelled monomial Gram matrices equal to the inverse ordinary-bar matrix, so $S=B_{\psi }^{-1}=\left(\begin{array}{cc}1& s-s^{-1}\\ 0& 1\end{array}\right)$ for both forms, as displayed. Substituting the coefficient matrices of step 2.1 and multiplying the resulting $4\times 4$ matrices (which are the identity on the $00$- and $11$-labelled directions) gives ${\bar{P}}^{t}SL=I$ and ${\tilde{L}}^{t}ST=I$ exactly; these are the delta identities $⟨P_A,L_B⟩={\delta }_{AB}$ and $C({\tilde{L}}_A,T_B)={\delta }_{AB}$ of [F1].

The form rescaling. The mixed-weight vector in the highest constituent is $T_{10}=F_0{M}_{00}$; its norm under the normalized bilinear form $C_0$ is $[2{]}_s=s+s^{-1}$, and by [F3] the library polarization gives $B_8(T_{10},T_{10})=s^{-1}[2{]}_s=1+t^2$ at $t=s^{-1}$. The projected dual-canonical vector is $D_{01}=\pi L_{01}=T_{10}\mathrm{/}(s+s^{-1})$, and the library form-dual upper vector is its rescaled $G_{\mathrm{u}\mathrm{p}}=s{D}_{01}$, which checks the exact rescaling of [F3] without asserting that the source $D$-family itself is the library upper family.

Conclusion. In this four-dimensional model the ordinary bar matrix is $B_{\psi }$, both Gram matrices are $S=B_{\psi }^{-1}$, the displayed $L,T,P,\tilde{L}$ vectors are the four distinct source families with the stated delta pairings, and the highest-component rescaling checks $B_8(T_{10},T_{10})=1+t^2$, $D_{01}=T_{10}\mathrm{/}(s+s^{-1})$ and $G_{\mathrm{u}\mathrm{p}}=s{D}_{01}$.

The two families at label $10$. In the mixed order the bar fixes $M_{01}$ and sends $M_{10}$ to $M_{10}+(s^{-1}-s)M_{01}$. A vector $M_{10}+c{M}_{01}$ is fixed exactly when $c-\bar{c}=s^{-1}-s$. The positive chart gives $c=-s$ and the negative chart gives $c=s^{-1}$; uniqueness follows because a bar-fixed polynomial in either $s\mathbb{Z}[s]$ or $s^{-1}\mathbb{Z}[s^{-1}]$ is zero. Thus $L_{01}=T_{01}=M_{01}$, $L_{10}=M_{10}-s{M}_{01}$ and $T_{10}=M_{10}+s^{-1}{M}_{01}$. Inverting the $L$ chart gives $M_{10}=L_{10}+s{L}_{01}$, so the transposed inverse chart is $P_{01}=M_{01}+s{M}_{10}$ and $P_{10}=M_{10}$. Linear independence of the two monomials proves $P_{10}\mathrm{\ne }{T}_{10}$.

The rectified highest-label coincidence. At label $01$ the same computations give $P_{01}=M_{01}+s{M}_{10}=s{T}_{10}$, so the two families do meet at the rectified highest label with the power $a=1$ of [F2]; this is an identity between specific vectors, not an identification of the two labelled families.

Conclusion. Since $P_{10}\mathrm{\ne }{T}_{10}$, the canonical and quasi-canonical families are different over $\mathbb{Q}(s)$; the dual-canonical and monomial families differ too, because $L_{10}=M_{10}-s{M}_{01}\mathrm{\ne }{M}_{10}$. The highest-submodule coincidence $P_{01}=s{T}_{10}$ is exactly the rectified normalization of [F2], and at $s=1$ it does not make the whole families interchangeable.

The moves. Apply the Levi tensor decomposition to each one-box column and add the two valid positions: tensoring with the natural module moves an entry $0$ to $1$ in either position, giving $f_0{M}_{00}=M_{01}+M_{10}$; there is no entry $0$ in $M_{11}$, giving $f_0{M}_{11}=0$; and the unique entry $0$ in either mixed monomial changes to $1$, giving $f_0{M}_{01}=f_0{M}_{10}=M_{11}$. The dual functor $e_0$ is computed verbatim with the dual natural module and gives the displayed reverse moves.

The commutator. In the ordered basis $(M_{00},M_{01},M_{10},M_{11})$ the matrices of $f_0$ and $e_0$ are $f_0=\left(\begin{array}{cccc}0& 0& 0& 0\\ 1& 0& 0& 0\\ 1& 0& 0& 0\\ 0& 1& 1& 0\end{array}\right),e_0=\left(\begin{array}{cccc}0& 1& 1& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{array}\right),$ and direct multiplication gives $[e_0,f_0]=\mathrm{diag}(2,0,0,-2)$, the ${\mathfrak{s}\mathfrak{l}}_2$-Cartan action on the Fock basis.

The actual simple classes. Here $\mathfrak{g}={\mathfrak{g}\mathfrak{l}}_2$ and the parabolic is the Borel. The tuple formula gives highest weights $(0,1),(0,2),(1,1),(1,2)$ for $00,01,10,11$, with root highest weights $-1,-2,0,-1$. On a Verma vector of root highest weight $m$, induction using [F3] gives $e{f}^{k}v=k(m-k+1)f^{k-1}v$. For $m=-1,-2$ every coefficient for $k>0$ is nonzero, so a nonzero weight submodule contains $v$ and the Verma is simple. For $m=0$, the span of $f^{k}v$, $k\ge 1$, is the simple Verma of highest weight $(0,2)$, and the quotient is the one-dimensional simple of weight $(1,1)$. Thus $L_{00}=M_{00}$, $L_{01}=M_{01}$, $[L_{10}]=[M_{10}]-[M_{01}]$ and $L_{11}=M_{11}$. Both singular endpoint blocks have one label, so BGG reciprocity makes $L_{00}=P_{00}$ and $L_{11}=P_{11}$ projective-injective. In the mixed block reciprocity gives $[P_{01}]=[M_{01}]+[M_{10}]$ and $[P_{10}]=[M_{10}]$.

Crystal and head/socle comparison. The moves in step 1.1 give $[f_0{P}_{00}]=[e_0{P}_{11}]=[P_{01}]$; these objects are projective, so finite decomposition into projective covers and independence of their classes identify both with $P_{01}$. They are self-dual because the endpoint simples are self-dual and the residue functors commute with duality. Hence they have simple head and socle $L_{01}$ and class $2[L_{01}]+[L_{10}]$. Also $f_0{L}_{01}=L_{11}$ and $e_0{L}_{01}=L_{00}$ by the standard moves, while exactness gives $[f_0{L}_{10}]=[e_0{L}_{10}]=0$, so these two translates vanish. The endpoint outward translates vanish as well. Independently, the source signature cancels $-+$ pairs, leaving the component $00\to 01\to 11$ and isolated $10$. These direct rank-one categorical computations establish precisely its head, socle and vanishing labels without using a general tensor-to-simple crystal identification.

Conclusion. On the four standard monomials the residue functors act by the displayed moves, their commutator is $\mathrm{diag}(2,0,0,-2)$, the source crystal is $00\to 01\to 11$ with isolated $10$, and the categorical translates have the corresponding unique heads and socles; all computations use the actual generalized standards and exact functors.

The two matrices. Finite Levi Weyl cancellation with $W_J=\{e,s_1\}$ and $W_{\bar{J}}=\{e,s_2\}$, together with all ordinary $S_3$ Kazhdan-Lusztig values at $1$ being $1$, gives the target standard/simple matrix $D$ and the opposite matrix $N$ independently on the displayed labels; subtracting the scalar $(2,2,2)$ reduces the comparison to the actual regular $S_3$ example of the block, and the target dominant labels appear in the reverse tableau order while the reverse-side dominant labels appear in the displayed order.

The semiregular shift and the functor values. Applying the explicit block-offset shift of [F1] to each of the three highest weights verifies the label computation: the shifted coordinates return the original column readings, so $F\bar{M}(A)=\mathrm{\nabla }(A)$ and the central-character blocks match. Opposite BGG then gives the entrywise tilting-standard matrix $N^t$ through $F\bar{P}(A)=T(A)$.

The proper ideal and the quotient. Restricting to $\mathrm{\Gamma }=\{A_0,A_1\}$, the invariant tiltings have rows $\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$, while the projective-corner $\mathrm{Hom}$ quotient of [F3] has standard/simple rows $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$; this is the opposite highest-weight Serre quotient by the complementary label, not the same opposite down-set subcategory. Simples and canonical classes are unchanged ambient ones, while projectives and quasi-canonical classes use the standard-flag truncation of the ambient projectives.

Conclusion. In this finite ${\mathfrak{g}\mathfrak{l}}_3$ example the semiregular label shift, the functor values $F\bar{M}=\mathrm{\nabla }$ and $F\bar{P}=T$, the opposite BGG tilting matrix $N^t$, the distinction from the target projective rows $D^t$, and the two-label ideal quotient data are all verified; the three Fock families correspond to simples, projectives and tiltings exactly as stated.