The $R$- and Kazhdan–Lusztig recursions on a small singular interval

The full interval. The displayed word for $w$ has inversion length $4$, so it is reduced. Its reduced subwords of lengths $0,1,2,3,4$ give respectively $1234$; $1324,2134,1243$; $1342,1423,2314,3124,2143$; $1432,2413,3142,3214$; and $3412$. The four excluded elements $1234,2134,1243,2143$ have no reduced subword $s_2$, so they are not above $b$; the other ten are. For an explicit order check, write $a=1342=s_2{s}_3$, $c=1423=s_3{s}_2$, $d=2314=s_1{s}_2$, $f=3124=s_2{s}_1$, and $A=1432=s_2{s}_3{s}_2$, $C=2413=s_1{s}_3{s}_2$, $D=3142=s_2{s}_1{s}_3$, $F=3214=s_2{s}_1{s}_2$. The reduced-subword criterion gives the intermediate covers $a<A,D$; $c<A,C$; $d<C,F$; $f<D,F$; each rank-two element is also above $b=s_2$, and each rank-three element is below $w$. These are all cover incidences between adjacent ranks, so all other comparisons are their transitive consequences.

The correction interval. Left $s_2$ gives $s_{2}w=2413$ and $s_{2}b=1234$. Step 1.1 gives $[b,2413]=\{b,1423,2314,2413\}$. Their left $s_2$ products are respectively $1234,1432,3214,3412$, of lengths $0,3,3,4$, whereas the original lengths are $1,2,2,3$. Thus only $b$ has that descent, and $\mu (b,2413)=0$ by its even length gap. In particular $1342$ is excluded: $R_{1342}(3,2)=1<2=R_{2413}(3,2)$.

All the $R$-coefficients. By [F9], noncomparable pairs have $r_{y,z}=0$, diagonal entries are $1$, and every comparable coefficient is nonzero because its leading term is $v^{\mathrm{\ell }(z)-\mathrm{\ell }(y)}$. For a cover, the degree range and parity in [F9] leave only the terms $v$ and $-v^{-1}$, so $r_{y,z}=\alpha$. For a comparable pair of gap two, induct on $\mathrm{\ell }(y)$ and choose a left descent $s$ of $z$. If $sy<y$, the recursion gives $r_{y,z}=r_{sy,sz}$; this coefficient is nonzero, so [F9] implies $sy\le sz$, and induction gives $r_{y,z}={\alpha }^2$. If $sy>y$, then $r_{sy,sz}=0$: its indices have equal length, and equality would force $y=z$. By the lifting implication in [F1], $y\le sz$, so the other recursion term is $\alpha r_{y,sz}={\alpha }^2$. Thus every gap-two pair in $S_4$ has coefficient ${\alpha }^2$. The only gap-three pair in $[b,w]$ is $(b,w)$. Since $s_2$ is a left descent of both, $r_{b,w}=r_{1234,2413}$. For $2413$, $s_1$ is a left descent, so $r_{1234,2413}=r_{2134,1423}+\alpha r_{1234,1423}$. The first term is zero because $1423=s_3{s}_2$ has no reduced subword $s_1$; the second is $\alpha \cdot {\alpha }^2$. Hence $r_{b,w}={\alpha }^3=v^3-3v+3v^{-1}-v^{-3}$. This determines every coefficient for pairs in the displayed interval.

The sole nonconstant polynomial. By [F3], all comparable pairs of gap at most two have $P=1$, so the only possibly nonconstant pair within the interval is $(b,w)$. To evaluate $P_{1234,2413}$, apply [F4] with left $s_1$ to $2413=s_1{s}_3{s}_2$, obtaining lower top $s_3{s}_2=1423$. No element below $1423$ has left $s_1$-descent: its subwords are $1234,1243,1324,1423$, with no inversion between the values $1,2$. Also $s_1\le ̸1423$. Hence $P_{1234,2413}=q{P}_{2134,1423}+P_{1234,1423}=0+1$. Now use [F4] at $(b,w,s_2)$: step 2.1 makes its correction sum empty, $c=1$, and $P_{b,2413}=1$, so $P_{b,w}=1+q$. Therefore $p_{b,w}=v^3+v$ and $\mu (b,w)=1$. Every other comparable distinct pair has $p_{y,z}=v^{\mathrm{\ell }(z)-\mathrm{\ell }(y)}$, so its nonzero $\mu$ occurs exactly on covers.

Inverse entries and both matrix products. Step 3.1 gives diagonal $p=1$, cover entries $v$, and gap-two entries $v^2$. Each gap-two interval in step 1.1 has two intermediate elements. Thus [F5] gives inverse entries $1,-v,v^2$ at gaps $0,1,2$. At the sole gap-three pair there are four elements at each intermediate rank, so $q_{b,w}^{\mathrm{\prime }}=-(v^3+v)-4(-v)v^2-4v^{2}v=-v^3-v$. This gives every entry of the inverse matrix, including every displayed value in part (d). For $Q^{\mathrm{\prime }}P$, the off-diagonal entries at gaps one and two are $-v+v=0$ and $v^2-2v^2+v^2=0$; at gap three the entry is $(v^3+v)-4v^3+4v^3-(v^3+v)=0$. For $P{Q}^{\mathrm{\prime }}$, these entries are $v-v=0$, $v^2-2v^2+v^2=0$, and $-(v^3+v)+4v^3-4v^3+(v^3+v)=0$. Diagonal entries are $1$ and noncomparable entries are zero by support, so both matrix products are the identity. Although all $p$-entries in this finite example are nonnegative, its cover inverse entries and $q_{b,w}^{\mathrm{\prime }}$ are negative. Every calculation is finite and uses no choice principle.