The Kazhdan–Lusztig multiplicity matrix for the sl_3 principal block

The dot orbit. With $\rho ={\alpha }_1+{\alpha }_2$ for the root system $A_2$, the dot action $w\cdot 0=w(\rho )-\rho$ computes: $123\cdot 0=0$; $213\cdot 0=s_1(\rho )-\rho =-{\alpha }_1$; $132\cdot 0=s_2(\rho )-\rho =-{\alpha }_2$; $231\cdot 0=(s_1{s}_2)\cdot 0=s_1\cdot (-{\alpha }_2)=s_1(\rho -{\alpha }_2)-\rho =-2{\alpha }_1-{\alpha }_2$; $312\cdot 0=(s_2{s}_1)\cdot 0=-{\alpha }_1-2{\alpha }_2$; $321\cdot 0=w_0(\rho )-\rho =-2\rho$. The positive-root pairings of $\rho$ are $1,1,2$, and the negative-root pairings are their negatives; Weyl conjugation permutes these nonzero integral pairings, so all six weights are dot-regular and integral, so ${\mathcal{O}}_0$ is a regular integral block with the six labelled standards, simples and projectives.

The Kazhdan–Lusztig basis of $H_v(3)$. Put $X=H_{213}$, $Y=H_{132}$ and $A:=X+v$, $B:=Y+v$; by [F1] both $A$ and $B$ are bar-fixed, because $\overline{X+v}=(X+\alpha )+v^{-1}=X+v$ and likewise for $Y$. The reduced words give $H_{231}=XY$, $H_{312}=YX$ and $H_{321}=YXY$ in the one-based model, so $AB=H_{231}+v(H_{213}+H_{132})+v^2{H}_{123}$ and $BA=H_{312}+v(H_{132}+H_{213})+v^2{H}_{123}$. For the length-three element, expanding $(Y+v)(X+v)(Y+v)$ and replacing $Y^2$ by $1-\alpha Y$ gives $BAB=H_{321}+v(H_{231}+H_{312})+(1+v^2)H_{132}+v^2{H}_{213}+(v+v^3)H_{123}$, hence $BAB-B=H_{321}+v(H_{231}+H_{312})+v^2(H_{132}+H_{213})+v^3{H}_{123}$. Each of the six elements $H_{123}$, $A$, $B$, $AB$, $BA$, $BAB-B$ is bar-fixed and lies in its top standard basis element plus a $v\mathbb{Z}[v]$-span of strictly lower Bruhat elements (the lower indices are strictly lower by the rank description of [F4]); by the uniqueness in [F2] they are exactly ${\underset{‾}{H}}_{123},{\underset{‾}{H}}_{213},{\underset{‾}{H}}_{132},{\underset{‾}{H}}_{231},{\underset{‾}{H}}_{312},{\underset{‾}{H}}_{321}$.

All polynomials equal $1$, and the count. Reading the six expansions of step 1.2 against ${\underset{‾}{H}}_w={\sum }_{y\le w}{p}_{y,w}{H}_y$ gives $p_{y,w}=v^{\mathrm{\ell }(w)-\mathrm{\ell }(y)}$ whenever $y\le w$: the coefficients are $1$ on the diagonal, $v$ at gap one, $v^2$ at gap two and $v^3$ at the longest pair. Hence [F3] gives $P_{y,w}(q)=1$ for every one of the comparable pairs. By the rank description [F4] the comparable pairs are: the six diagonal pairs; for $y=123$ the five strict pairs; for each of $y=213,132$ the three strict pairs (the two rank-two elements and $321$); and for each of $y=231,312$ the pair with $321$; altogether $6+5+3+3+1+1=19$ pairs, and every one has $P_{y,w}=1$.

The multiplicity matrix. By [F5] and step 2.1, $[M(y\cdot 0):L(w\cdot 0)]=P_{y,w}(1)=1$ exactly when $y\le w$ and $0$ otherwise; this is the Bruhat zeta matrix of $S_3$ with rows $y$ and columns $w$. For the displayed instances: $y=123$ is the minimum of the Bruhat order, so $[M(0):L(w\cdot 0)]=1$ for all six $w$; $y=321$ is the maximum, so $[M(321\cdot 0):L(w\cdot 0)]=0$ unless $w=321$, where it is $1$; and for $y=213$ the comparable columns are $w=213,231,312,321$ by [F4], so $M(213\cdot 0)$ has exactly the four composition factors $L(213\cdot 0),L(231\cdot 0),L(312\cdot 0),L(321\cdot 0)$, each once.

The Verma-flag matrix. Applying BGG reciprocity [F6] with $\mu =w\cdot 0$ and $\nu =y\cdot 0$ gives $(P(w\cdot 0):\mathrm{\Delta }(y\cdot 0))=[M(y\cdot 0):L(w\cdot 0)]$, which by step 3.1 is $1$ exactly when $y\le w$ and $0$ otherwise; the interpretation of the left-hand multiplicity is the Verma-flag multiplicity of the labelled projective by [F7]. In particular $(P(123\cdot 0):\mathrm{\Delta }(y\cdot 0))=0$ for $y\mathrm{\ne }123$ (with value $1$ at $y=123$, since $123$ is the minimum and no other $y$ lies below it) and $(P(321\cdot 0):\mathrm{\Delta }(y\cdot 0))=1$ for all $y$ (all $y$ lie below the maximum $321$). So the projective-flag matrix is the same zeta matrix.