$\mu$-edges and left equivalence are transported by star operations

Recursion notation and mixed descents. For $a<b$ put ${\pi }_{a,b}:=p_{a,b}\mathrm{/}v$, put ${\pi }_{a,a}:=v^{-1}$, and put ${\pi }_{a,b}:=0$ if $a\le ̸b$; then ${\pi }_{a,b}\in \mathbb{Z}[v]$ and ${\pi }_{a,b}(0)=\mu (a,b)$. If $a<b$, $as>a$, and $bs<b$, coefficient comparison in $C_b{H}_s=v^{-1}{C}_b$ gives $p_{a,b}=v{p}_{as,b}$, hence ${\pi }_{a,b}=v{\pi }_{as,b}$; therefore $\mu (a,b)=0$ unless $as=b$, when it is $1$. The right recursion from [F3], when $as<a$, $bs<b$, and $a\mathrm{\ne }bs$, reads ${\pi }_{a,b}={\pi }_{as,bs}+({\pi }_{a,bs}-\mu (a,bs))\mathrm{/}v-{\sum }_{a<z<bs,zs<z}\mu (z,bs){\pi }_{a,z}$. In particular, if $a\le ̸bs$, it gives ${\pi }_{a,b}={\pi }_{as,bs}$.

Pairs in one right coset. A right $⟨s,t⟩$-coset meets $D_{ij}$ in two elements; the rank-two table shows these form a Bruhat cover, and their two star images form a Bruhat cover as well. By [F5], the $\mu$-coefficient is $1$ for both pairs. This proves (a) when $y,w$ belong to the same right coset.

Left-preorder transport. Let $I_s$ (respectively $I_t$) be the $A$-span of the $C_w$ with $ws<w$ (respectively $wt<w$), and put $J=I_s\cap I_t$. For each simple reflection $r$, if $rw<w$ then $C_r{C}_w=(v+v^{-1})C_w$; if $rw>w$, the multiplication formula expresses $C_r{C}_w$ as $C_{rw}$ plus lower terms $C_z$ that are nonzero off-diagonal left steps. In the second case the leading term is itself a left step. By [F4], every such step preserves each right descent of $w$, so $I_s$ and $I_t$ are stable under left multiplication by each $C_r$. Since $\mathrm{i}\mathrm{d}<r$ is a Bruhat cover, [F2, F5] give $C_r=H_r+v$; [F6] says the $H_r$ generate the algebra, so $I_s$ and $I_t$ are left ideals. By the basis property in [F2], $J$ is spanned by elements with both descents, and the quotient bases of $I_s\mathrm{/}J$ and $I_t\mathrm{/}J$ are indexed by $D_{ij}$ and $D_{ji}$. Right multiplication by $C_t$ maps $I_s$ into $I_t$ by the right multiplication formula in [F4], and maps $J$ into $J$ because $C_w{C}_t=(v+v^{-1})C_w$ for every both-descent basis vector; it is left-linear by associativity and induces $f:I_s\mathrm{/}J\to I_t\mathrm{/}J$. For $xs<x$, [F4] and $C_s=H_s+v$ from [F2, F5] give $C_x{H}_s=v^{-1}{C}_x$. If also $zs>z$, comparison of the $H_z$ coefficients using [F6] gives $p_{z,x}=v{p}_{zs,x}$. If $zs=x$, then $z=xs$ and $\mu (z,x)=1$; if $zs<x$, the right side lies in $v^2\mathbb{Z}[v]$ and $\mu (z,x)=0$; if $zs\le ̸x$, both coefficients vanish by support. Thus in this mixed-descent situation $\mu (z,x)$ can be nonzero only for $z=xs$. In a right $⟨s,t⟩$-coset with shortest representative $\tilde{w}$, its two $D_{ij}$ elements are $x=\tilde{w}s$ and $x=\tilde{w}ts$. For $x=\tilde{w}s$, the right-ascent formula for $C_x{C}_t$ has leading term $C_{xt}=C_{\tilde{w}st}$. A correction index surviving modulo $J$ has $zs>z$, so the mixed-descent identity forces $z=xs=\tilde{w}$. But $\tilde{w}t>\tilde{w}$, contradicting the condition $zt<z$ on correction indices; hence no correction survives and $f([C_x])=[C_{x^{\ast }}]$. For $x=\tilde{w}ts$, the leading term $C_{xt}$ has both descents and dies in $J$; any correction surviving modulo $J$ must have $zs>z$, so the identity forces $z=xs=\tilde{w}t=x^{\ast }$, with coefficient $1$ because $x^{\ast }s=x$ is a Bruhat cover. Thus again $f([C_x])=[C_{x^{\ast }}]$. Exchanging $s,t$ gives the left-linear map $g:I_t\mathrm{/}J\to I_s\mathrm{/}J$ induced by right multiplication by $C_s$; it is well-defined since $C_w{C}_s=(v+v^{-1})C_w$ on $J$, and the exchanged two-case table sends each quotient basis vector to its star. Thus $gf$ and $fg$ fix every quotient basis vector, so these maps are inverse. Hence for every simple $r$ and $x,z\in D_{ij}$, left-linearity and comparison in the quotient bases give equality between the coefficient of $C_z$ in $C_r{C}_x$ and that of $C_{z^{\ast }}$ in $C_r{C}_{x^{\ast }}$. Along a left-preorder chain $x=w_0{\leftarrow }_L\cdots {\leftarrow }_L{w}_m=y$ with endpoints in $D_{ij}$, [F4] gives $R(y)\subseteq R(w_k)\subseteq R(x)$, so every intermediate has the same singleton descent set on $\{s,t\}$. In particular $J$ is an absorbing left ideal: a step cannot enter both descents and then return to $D_{ij}$. The chain stays in $D_{ij}$ and the coefficient equality transports every step; diagonal steps and zero-length chains transport as well. Applying $g$ gives the converse. Thus $x{\le }_{L}y$ iff $K_{ij}(x){\le }_L{K}_{ij}(y)$. This argument uses support and nonzero coefficients only, not positivity or Choice.

Right-preorder transport. By [F7], $K_{ij}(x){\le }_{R}x$ and $y{\le }_R{K}_{ij}(y)$, so transitivity gives $x{\le }_{R}y\Rightarrow K_{ij}(x){\le }_R{K}_{ij}(y)$. Applying the same implication to the inverse star $K_{ji}$ gives the converse.

Different cosets, stars moving by the same simple reflection. Let $a<b$ be a comparable pair with odd length difference in distinct right cosets. Suppose both stars move down by the same simple reflection, $a^{\ast }=as<a$ and $b^{\ast }=bs<b$. If $a\le ̸bs$, step 1.1 gives ${\pi }_{a,b}={\pi }_{as,bs}$. Otherwise $a<bs$ because the cosets differ. Parity gives $\mu (a,bs)=0$, and the right recursion gives ${\pi }_{a,b}\equiv {\pi }_{as,bs}+{\pi }_{at,bs}-{\sum }_{a<z<bs,zs<z}\mu (z,bs)\mu (a,z)(\mathrm{m}\mathrm{o}\mathrm{d}v)\mathrm{.}$ Here $bst<bs$ and $at>a$, so mixed descent gives ${\pi }_{a,bs}\mathrm{/}v={\pi }_{at,bs}$. Bruhat lifting gives $at\le bs$; distinct cosets make the inequality strict. The rank-two table gives $ats<at$. The only summand with nonzero constant term is $z=at$: if a summand has $zt>z$, mixed descent for $\mu (z,bs)$ forces $z=bst$, which has $zs>z$ and is excluded; hence $zt<z$, and mixed descent for $\mu (a,z)$ then forces $z=at$. Since $a⋖at$, $\mu (a,at)=1$, so this constant-term summand cancels ${\pi }_{at,bs}(0)$. Thus $\mu (a,b)=\mu (as,bs)$. If both stars move up by the same simple reflection, apply this calculation to the starred pair and use involutivity. When the common star multiplier gives opposite directions, mixed descent using the other generator rules out a nonzero coefficient, since each element of $D_{ij}$ has exactly one of the two right descents. These are the same-multiplier cases of Casselman's two-case calculation.

Different cosets, stars moving in opposite directions. Take elements $a,b$ in distinct cosets with $a^{\ast }=as<a$ and $b^{\ast }=bt>b$, without initially assuming $a<b$; coefficients outside Bruhat order are zero. We prove equality whenever either $\mu (a,b)$ or $\mu (as,bt)$ is nonzero. The rank-two table gives $bs<b<bt<bts$ and $ast<as<a<at$, with $asts>ast$ and $bst>bs$. If either $\mu (a,b)$ or $\mu (as,bt)$ is nonzero, then $as\le b$: this is immediate from $a<b$ in the first case; in the second, if $as\le ̸b$, step 1.1 with right descent $t$ gives ${\pi }_{as,bt}={\pi }_{ast,b}$, whose constant term is zero by mixed descent with $s$, because $asts$ lies in $a$'s right coset while $b$ lies in a different one. Since $as\le b$ and $ass=a>as$ while $bs<b$, Bruhat lifting also gives $a\le b$; equality is excluded by the distinct cosets. Either nonzero coefficient makes $\mathrm{\ell }(b)-\mathrm{\ell }(a)$ odd, since the two coefficient length differences differ by $2$. Thus $\mathrm{\ell }(b)-\mathrm{\ell }(as)$ is even, so $\mu (as,b)=0$, and mixed descent gives ${\pi }_{as,b}=v{\pi }_{a,b}$. The recursion of step 1.1 applied to $(as,bt)$ gives ${\pi }_{as,bt}={\pi }_{ast,b}+{\pi }_{as,b}\mathrm{/}v-{\sum }_{as<z<b,zt<z}\mu (z,b){\pi }_{as,z},$ hence modulo $v$ it is ${\pi }_{ast,b}+{\pi }_{a,b}-\sum \mu (z,b)\mu (as,z)$. This polynomial sum need not be empty, but its constant-term sum is zero. Indeed, if $\mu (z,b)\mathrm{\ne }0$, mixed descent with $s$ forces $zs<z$ unless $z=bs$; that exception has $bst>bs$, contrary to $zt<z$. If also $\mu (as,z)\mathrm{\ne }0$, mixed descent with $s$ and $ass>as$ forces $z=a$, but $at>a$, again contrary to $zt<z$. Also $\mu (ast,b)=0$: $asts>ast$, $bs<b$, and equality $asts=b$ would put $a$ and $b$ in the same right coset. Thus ${\pi }_{as,bt}(0)={\pi }_{a,b}(0)$. If neither coefficient is nonzero the equality is immediate; if either is nonzero this calculation proves the other is equal and nonzero. For the remaining original arrangement $a^{\ast }=at>a$, $b^{\ast }=bs<b$, apply the same calculation with $s,t$ exchanged to $A=at$ and $B=bs$. These satisfy $A^{\ast }=At=a<A$ and $B^{\ast }=Bs=b>B$ in distinct cosets. Its conditional nonzero hypothesis is precisely that either $\mu (at,bs)$ or $\mu (a,b)$ is nonzero, so the calculation establishes their equality and the required comparability even if $at\le bs$ was not initially known. If both vanish there is no edge to transport. Together with steps 1.2 and 2.1 this proves (a), including the symmetric convention for comparable orientations.

Cell equivalences. The left and right cell equivalences are mutual comparability. The left-preorder iff in step 1.3 and the right-preorder iff in step 1.4 therefore give both cell iff statements in (b); with (a) proved in step 3.1, all claims of the Statement follow.