The Kazhdan–Lusztig Character and Multiplicity Theorem — Examples

The interval. The word $s_2{s}_1{s}_3{s}_2$ has length $4$ and is reduced, since $3412$ has exactly the four inversions $(3,1),(3,2),(4,1),(4,2)$. Its reduced subwords give: $1234$ at length $0$; $2134,1324,1243$ at length $1$; $1342,1423,2314,3124,2143$ at length $2$; $1432,2413,3142,3214$ at length $3$; and $3412$ at length $4$. The four elements $1234,2134,1243,2143$ contain no reduced subword equal to $s_2=1324$, so they are not above $b$ by the reduced-subword criterion; the remaining ten elements are exactly $[b,w]$.

All gaps at most two have $P=1$. Suppose $y<z$ and $d:=\mathrm{\ell }(z)-\mathrm{\ell }(y)\le 2$. By [F3] the degree of $P_{y,z}$ is at most $(d-1)\mathrm{/}2\le 1\mathrm{/}2$, hence at most $0$; together with constant term $1$ this forces $P_{y,z}=1$. If $d=0$ then $y=z$ and $P_{y,y}=1$; if $y$ and $z$ are incomparable, $P_{y,z}=0$ by [F3]. In particular $P_{b,2413}=1$ and $P_{1234,1423}=1$, both pairs having gap $2$.

The recursion at $(\mathrm{i}\mathrm{d},2413)$. Write $\mathrm{i}\mathrm{d}=1234$ and use $2413=s_1{s}_3{s}_2$, a reduced word with left descent $s_1$. With $s=s_1$ one has $s2413=1423$ and $s\mathrm{i}\mathrm{d}=2134$, so $c=0$ in [F4] and $P_{\mathrm{i}\mathrm{d},2413}(q)=q{P}_{2134,1423}(q)+P_{\mathrm{i}\mathrm{d},1423}(q)-\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\sum }_{\begin{array}{c}\mathrm{i}\mathrm{d}\le u\le 1423\\ s_{1}u<u,\mu (u,1423)\mathrm{\ne }0\end{array}}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\mu (u,1423)q^{(3-\mathrm{\ell }(u))\mathrm{/}2}{P}_{\mathrm{i}\mathrm{d},u}(q)\mathrm{.}$ The first summand vanishes: $1423=s_3{s}_2$ has no reduced subword $s_1$, so $2134\le ̸1423$ and $P_{2134,1423}=0$ by [F3] and [F5]. The correction sum is empty as well: $[\mathrm{i}\mathrm{d},1423]=\{1234,1324,1243,1423\}$ by step 1.1, and for each of these four elements left multiplication by $s_1$ swaps the values $1$ and $2$, raising the length from $0,1,1,2$ to $1,2,2,3$ respectively, so no element of the interval has the required left descent $s_1$. By step 1.2 the remaining term is $P_{\mathrm{i}\mathrm{d},1423}=1$, hence $P_{\mathrm{i}\mathrm{d},2413}=1$.

The main recursion. Apply [F4] to $b=1324$, $w=3412$ and the left descent $s=s_2$: $s_{2}w=2413$ and $s_{2}b=1234$, so $c=1$ because $s_{2}b$ has length $0<1=\mathrm{\ell }(b)$, and $P_{b,w}(q)=P_{\mathrm{i}\mathrm{d},2413}(q)+q{P}_{b,2413}(q)-\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\sum }_{\begin{array}{c}b\le u\le 2413\\ s_{2}u<u,\mu (u,2413)\mathrm{\ne }0\end{array}}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\mu (u,2413)q^{(4-\mathrm{\ell }(u))\mathrm{/}2}{P}_{b,u}(q)\mathrm{.}$ By step 1.1 the interval $[b,2413]$ is $\{1324,1423,2314,2413\}$; left multiplication by $s_2$ swaps the values $2$ and $3$, turning these into $1234,1432,3214,3412$ of lengths $0,3,3,4$, so only $u=b$ has the required descent, and there $\mu (b,2413)=0$ because $\mathrm{\ell }(2413)-\mathrm{\ell }(b)=2$ is even by [F3]. The correction sum is therefore empty; step 2.1 gives $P_{\mathrm{i}\mathrm{d},2413}=1$ and step 1.2 gives $P_{b,2413}=1$, so $P_{b,w}=1+q$, with $\mu (b,w)=1$ and $\mathrm{\ell }(w)-\mathrm{\ell }(b)=3$.

The comparison. The polynomial $P_{b,w}(q)=1+q$ has leading coefficient $1$ (the coefficient of $q^1$, its degree being $1$) while its value at $q=1$ is $2$. The multiplicity theorem [F1] computes the ordinary Verma-flag multiplicity as $(P(w\cdot \lambda ):\mathrm{\Delta }(b\cdot \lambda ))=P_{b,w}(1)=2$, with $0$ dominant regular integral in ${\mathfrak{s}\mathfrak{l}}_4$ (indeed $⟨\rho ,{\alpha }_i^{\vee }⟩=1\mathrm{\ne }0$ for every simple root and $0$ is integral). Hence the leading coefficient $1$ of the Kazhdan–Lusztig polynomial differs from the multiplicity $2$: the refuted claim's coefficientwise clause is false, and neither individual coefficient of this polynomial equals the ordinary multiplicity; its total value at one does. This is a genuine failure of the claim and not an artefact of undefined quantities: $b\le w$, all polynomials are classical Kazhdan–Lusztig polynomials, and the multiplicity is that of the ordinary block ${\mathcal{O}}_0$.

The grading clause. The remaining structural clause of the refuted claim is not supplied by the cited results either: the multiplicity of [F1] is the ordinary (ungraded) Verma-flag multiplicity, and the projective dictionary of [F2] is the ungraded reduction $V(P(w\cdot \lambda ))={\bar{B}}_{w^{-1}}$, whose auxiliary grading lives on the polynomial lift and is not a grading of ${\mathcal{O}}_{\lambda }$. Thus the character identity at $q=1$ cannot establish that the block is graded or Koszul; the graded and $\mathrm{E}\mathrm{x}\mathrm{t}$ refinements belong to the separate Koszul-duality frontier, and step 4.1 refutes only the identification of an individual coefficient with an ordinary multiplicity. It does not refute graded decomposition numbers or the possibility of a Koszul grading: those require additional constructions and proofs, and graded multiplicities may sum to the ordinary multiplicity. The stated conjunction is refuted by its false leading-coefficient clause, with witness $(y,w)=(1324,3412)$ in ${\mathfrak{s}\mathfrak{l}}_4$.

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.

The two matrix products. Write the displayed matrix as $B$. The product $AB$ has entries $(AB{)}_{y,w}={\sum }_{x:y\le x}{B}_{x,w}$, a sum over the interval $[y,\cdot ]$; reading rows $x$ of $B$ against the comparability lists of [F2]: for $y=123$ the six rows of $B$ sum to $(1,0,0,0,0,0)$; for $y=213$ the rows with $x\in \{213,231,312,321\}$ sum to $(0,1,0,0,0,0)$; for $y=132$ the rows with $x\in \{132,231,312,321\}$ sum to $(0,0,1,0,0,0)$; for $y=231$ the rows $231,321$ sum to $(0,0,0,1,0,0)$; for $y=312$ the rows $312,321$ sum to $(0,0,0,0,1,0)$; for $y=321$ the last row is $(0,0,0,0,0,1)$. Hence $AB=I$. Since $A$ is unitriangular by [F2], it has an integral two-sided inverse (the finite geometric series for $I+(A-I)$), so $AB=I$ implies $B=A^{-1}$ and $BA=I$. Equivalently, $(BA{)}_{y,w}={\sum }_{x:x\le w}{B}_{y,x}$ sum over the lower ideal of the column label, rather than over a prefix of the displayed order; for example the row-$123$ sums at $w=132$ and $w=312$ are $1-1=0$ and $1-1-1+1=0$ respectively.

The inverse coefficients. By [F3] the matrix of inverse Kazhdan–Lusztig coefficients is $Q^{\mathrm{\prime }}=P^{-1}$, and by [F4] its specialization at $v=1$ is $A^{-1}=B$; hence $q_{x,w}^{\mathrm{\prime }}(1)=B_{x,w}$ for all $x,w$, the entries of $B$ being integers, some negative. Substituting the rows of $B$ into the character formula of [F1], with the support $x\ge w$ of [F2], gives exactly the six displayed identities: row $123$ is $(1,-1,-1,1,1,-1)$, row $213$ is $(0,1,0,-1,-1,1)$ and row $132$ is $(0,0,1,-1,-1,1)$, row $231$ is $(1,0,-1)$ on the labels $231,312,321$, row $312$ is $(1,-1)$ on $312,321$, and row $321$ is $1$.

The $w=123$ row and the Euler characteristic. On the labels ordered by [F2], the first row of $B$ is $q_{123,x}^{\mathrm{\prime }}(1)=(-1{)}^{\mathrm{\ell }(x)}$: the values are $+1$ at $x=123$ (length $0$), $-1$ at $x=213,132$ (length $1$), $+1$ at $x=231,312$ (length $2$) and $-1$ at $x=321$ (length $3$). Hence the row-$123$ identity of step 2.1 is $\mathrm{c}\mathrm{h}L(0)={\sum }_x(-1{)}^{\mathrm{\ell }(x)}\mathrm{c}\mathrm{h}M(x\cdot 0)$, the BGG Euler-characteristic expression; by [F6] applied to the dominant integral weight $0$ this is the class of $L(0)$ in the Grothendieck group of the block, and by [F7] $\mathrm{c}\mathrm{h}L(0)=A(\rho )A(\rho {)}^{-1}=1$ is the character of the trivial module. This independent check confirms the signed inversion: all Kazhdan–Lusztig polynomials are $1$, yet the inverse coefficients are signed integers.