Finite descent parabolics, parabolic factorization, the Steinberg inclusion-exclusion identity, and rational growth

Let $J\subseteq S$ and $w\in W$ with $J\subseteq D_R(w)$. Write $w=dv$ with $d\in W^J$, $v\in W_J$ as in [F2], so $\mathrm{\ell }(ws)=\mathrm{\ell }(d)+\mathrm{\ell }(vs)$ for every $s\in J$; since $\mathrm{\ell }(ws)<\mathrm{\ell }(w)=\mathrm{\ell }(d)+\mathrm{\ell }(v)$ by [F1], we get $\mathrm{\ell }(vs)<\mathrm{\ell }(v)$ for all $s\in J$. Now $D_L(x)=D_R(x^{-1})$ for every $x\in W$, because $\mathrm{\ell }(x^{-1})=\mathrm{\ell }(x)$ and $(xs{)}^{-1}=s{x}^{-1}$ turn $\mathrm{\ell }(xs)<\mathrm{\ell }(x)$ into $\mathrm{\ell }(s{x}^{-1})<\mathrm{\ell }(x^{-1})$; hence $D_L(v^{-1})=D_R(v)\supseteq J$, and [F3] applied to $v^{-1}\in W_J$ gives that $W_J$ is finite and $v^{-1}=w_0(J)$; since $w_0(J{)}^2=1$ by [F4], $v=w_0(J)$. Symmetrically, if $J\subseteq D_L(w)$, write $w=ud$ with $u\in W_J$, $d\in {}^{J}W$ as in [F2]; then $sw=(su)d$ for $s\in J$, so $\mathrm{\ell }(sw)=\mathrm{\ell }(su)+\mathrm{\ell }(d)<\mathrm{\ell }(u)+\mathrm{\ell }(d)=\mathrm{\ell }(w)$ gives $\mathrm{\ell }(su)<\mathrm{\ell }(u)$ for all $s\in J$, and [F3] applied to $u$ shows that $W_J$ is finite. In particular $W_{D_L(w)}$ and $W_{D_R(w)}$ are finite for every $w$, and taking $J=S$ shows that if some $x$ has $D_L(x)=S$ or $D_R(x)=S$ then $W=W_S$ is finite.

Let $W_J$ be finite with longest element $w_0(J)$ as in [F4]. For $s\in J$ the element $s$ lies in $W_J$, so $\mathrm{\ell }(s{w}_0(J))=\mathrm{\ell }(w_0(J))-\mathrm{\ell }(s)=\mathrm{\ell }(w_0(J))-1<\mathrm{\ell }(w_0(J))$ and likewise $\mathrm{\ell }(w_0(J)s)<\mathrm{\ell }(w_0(J))$: thus $J\subseteq D_L(w_0(J))\cap D_R(w_0(J))$. Conversely let $s\in S\setminus J$ and suppose $\mathrm{\ell }(s{w}_0(J))<\mathrm{\ell }(w_0(J))$. Fix a reduced expression $w_0(J)=s_1\cdots s_k$ with all $s_i\in J$ (possible by [F5], since $W_J=⟨J⟩$); by [F13] $s{w}_0(J)=s_1\cdots \widehat{s_i}\cdots s_k$ is a product of elements of $J$, hence lies in $W_J$, so $s=w_0(J)\cdot (s{w}_0(J){)}^{-1}\in W_J$ and therefore $s\in W_J\cap S=J$ by [F5], a contradiction. Since $\mathrm{\ell }(s{w}_0(J))-\mathrm{\ell }(w_0(J))\in \{\pm 1\}$ by [F1], it follows that $\mathrm{\ell }(s{w}_0(J))>\mathrm{\ell }(w_0(J))$, i.e. $s\mathrm{\notin }{D}_L(w_0(J))$. The right-handed version uses the right form of [F13] with the same computation; hence $D_L(w_0(J))=D_R(w_0(J))=J$. In particular, if $W$ is finite then $D_L(w_0)=D_R(w_0)=S$ for $w_0=w_0(S)$.

Fix $J\subseteq S$. By [F2] the maps $W^J\times W_J\to W$, $(d,v)\mapsto dv$, and $W_J\times {}^{J}W\to W$, $(u,d)\mapsto ud$, are bijections along which $\mathrm{\ell }$ is additive. For each $n\in \mathbb{N}$, the map $(d,v)\mapsto dv$ restricts to a bijection from the finite disjoint union ${⨆}_{i=0}^n\{d\in W^J:\mathrm{\ell }(d)=i\}\times \{v\in W_J:\mathrm{\ell }(v)=n-i\}$ onto $\{w\in W:\mathrm{\ell }(w)=n\}$; finiteness follows from [F7, F15]. Its cardinality is exactly the Cauchy-product coefficient $[t^n](P_{W^J}{P}_{W_J})$, so coefficientwise equality gives $P_W=P_{W^J}{P}_{W_J}$. The same argument with the factorization $w=ud$ gives $P_W=P_{{}^{J}W}{P}_{W_J}$. If $S=\mathrm{\varnothing }$, then $W=\{1\}$ and both series and the empty product are $1$. Otherwise, in the disconnected case [F6] gives an isomorphism $W_{S_1}\times \cdots \times W_{S_k}\to W$ carrying $(w_1,\dots ,w_k)$ to $w_1\cdots w_k$ with additive length. Iterating the same coefficient argument over the finite set of components gives $P_W={\prod }_i{P}_{W_{S_i}}$.

Fix $I\subseteq J\subseteq S$ and for $L\subseteq S$ put $E_L:=\{w\in W:D_R(w)=L\}$, a set partition of $W$ indexed by the finite power set $\mathcal{P}(S)$ by [F11], so that $P_{W^{S\setminus K}}={\sum }_{L\subseteq K}{P}_{E_L}$ for every $K\subseteq S$ by [F7] and the displayed definition in statement (3), while $D_I^J={\sum }_{I\subseteq L\subseteq J}{P}_{E_L}$. Substituting the first display into the sum of the statement gives ${\sum }_{J\setminus I\subseteq K\subseteq J}(-1{)}^{\mathrm{\mid }J\setminus K\mathrm{\mid }}{\sum }_{L\subseteq K}{P}_{E_L}={\sum }_{L\subseteq J}{P}_{E_L}c(L)$ with $c(L):={\sum }_{K:(J\setminus I)\cup L\subseteq K\subseteq J}(-1{)}^{\mathrm{\mid }J\setminus K\mathrm{\mid }}$, a finite interchange licensed by [F8, F10]. Writing $K=J\setminus N$ with $N\subseteq J$, the condition $K\supseteq (J\setminus I)\cup L$ becomes $N\subseteq M:=J\setminus ((J\setminus I)\cup L)$ and $\mathrm{\mid }J\setminus K\mathrm{\mid }=\mathrm{\mid }N\mathrm{\mid }$, so $c(L)={\sum }_{N\subseteq M}(-1{)}^{\mathrm{\mid }N\mathrm{\mid }}$; when $M\mathrm{\ne }\mathrm{\varnothing }$ fix $a\in M$ and the map $N\mapsto N\mathrm{△}\{a\}$ is a fixed-point-free involution of the subsets of $M$ that negates $(-1{)}^{\mathrm{\mid }N\mathrm{\mid }}$, so $c(L)=0$, and when $M=\mathrm{\varnothing }$ the sum is the single term $1$. Now $M=\mathrm{\varnothing }$ means $(J\setminus I)\cup L=J$, and since $L\subseteq J$ this holds exactly when $I\subseteq L$: each $x\in I$ lies in $J=(J\setminus I)\cup L$ but not in $J\setminus I$, hence in $L$, while $I\subseteq L$ gives $(J\setminus I)\cup L\supseteq (J\setminus I)\cup I=J$. Together with $L\subseteq J$ this is exactly $I\subseteq L\subseteq J$, so the whole sum equals ${\sum }_{I\subseteq L\subseteq J}{P}_{E_L}=D_I^J$, which is the asserted identity.

By step 1.2, if $W$ is finite then some element (namely $w_0$) has all left descents and all right descents; by step 1.1, if some element has all descents then $W$ is finite. Hence $W$ is finite if and only if some $x$ satisfies $D_L(x)=S$, if and only if some $x$ satisfies $D_R(x)=S$; so if $W$ is infinite then no element has all descents. Now $D_S^S(t)={\sum }_{\{w:D_R(w)=S\}}{t}^{\mathrm{\ell }(w)}$. If $W$ is infinite the index set is empty, so $D_S^S=0$. If $W$ is finite and $D_R(w)=S$, apply the factorization argument of step 1.1 with $J=S$: write $w=dv$ with $d\in W^S$ and $v\in W_S$. Since $W_S=W$, the set $w{W}_S$ is all of $W$, so its unique minimal representative is $d=1$ and $v=w$; the argument in step 1.1 then gives $v^{-1}=w_0$, so $w=v=w_0$; conversely $D_R(w_0)=S$ by step 1.2. Hence $D_S^S(t)=t^{\mathrm{\ell }(w_0)}$ in the finite case.

Take $I=J=S$ in step 1.4; since $S\setminus S=\mathrm{\varnothing }$, this gives $D_S^S(t)={\sum }_{K\subseteq S}(-1{)}^{\mathrm{\mid }S\setminus K\mathrm{\mid }}{P}_{W^{S\setminus K}}(t)={\sum }_{K\subseteq S}(-1{)}^{\mathrm{\mid }K\mathrm{\mid }}{P}_{W^K}(t)$ after the substitution $K\mapsto S\setminus K$. By step 1.3 $P_W=P_{W^K}{P}_{W_K}$ for every $K\subseteq S$, and $P_{W_K}(0)=1$ because $W_K$ contains the identity, so [F7, F9] gives $P_{W^K}=P_W\mathrm{/}{P}_{W_K}$ in $\mathbb{Z}\mathrm{⟦}t\mathrm{⟧}$. Substituting and dividing the resulting identity by the unit $P_W$ yields ${\sum }_{K\subseteq S}\frac{(-1{)}^{\mathrm{\mid }K\mathrm{\mid }}}{P_{W_K}(t)}=\frac{D_S^S(t)}{P_W(t)}$, and step 2.1 evaluates the right side as $t^{\mathrm{\ell }(w_0)}\mathrm{/}{P}_W(t)$ when $W$ is finite and as $0$ when $W$ is infinite, which is the asserted two-case identity. In the finite case, $w\mapsto w_{0}w$ is a bijection of $W$ with $\mathrm{\ell }(w_{0}w)=\mathrm{\ell }(w_0)-\mathrm{\ell }(w)$ by [F4], so summing over $w$ gives $P_W(t)={\sum }_w{t}^{\mathrm{\ell }(w_0)-\mathrm{\ell }(w)}=t^N{P}_W(t^{-1})$ for $N=\mathrm{\ell }(w_0)$, i.e. $t^N{P}_W(t^{-1})=P_W(t)$ as polynomials; hence $t^N\mathrm{/}{P}_W(t)=1\mathrm{/}{P}_W(t^{-1})$ as rational functions and the identity rewrites as $1\mathrm{/}{P}_W(t^{-1})={\sum }_{K\subseteq S}(-1{)}^{\mathrm{\mid }K\mathrm{\mid }}\mathrm{/}{P}_{W_K}(t)$.

By [F14], induct on rank, proving rationality over $\mathbb{Z}$ in the sense of [F12]. At rank zero $W=\{1\}$ and $P_W=1\mathrm{/}1$. Suppose the assertion holds through rank $n$, and let $\mathrm{\mid }S\mathrm{\mid }=n+1$. For every proper $K⊊S$, [F5] identifies $P_{W_K}$ with the intrinsic series of $(W_K,K)$, so write $P_{W_K}=U_K\mathrm{/}{V}_K$ with integer polynomials and $V_K(0)=\pm 1$. Its constant term is $1$ by [F7], hence $U_K(0)=V_K(0)=\pm 1$ and its reciprocal is $V_K\mathrm{/}{U}_K$. A finite sum of these signed reciprocals is rational over $\mathbb{Z}$: put it over the common denominator ${\prod }_{K\mathrm{\ne }S}{U}_K$, which has constant term $\pm 1$. In the infinite case step 3.1 gives $1\mathrm{/}{P}_W=(-1{)}^{\mathrm{\mid }S\mathrm{\mid }+1}{\sum }_{K\mathrm{\ne }S}(-1{)}^{\mathrm{\mid }K\mathrm{\mid }}\mathrm{/}{P}_{W_K}$. This rational series has constant term $1$, so its reciprocal $P_W$ is rational by [F12]. In the finite case the same step gives $P_{W}D=t^N-(-1{)}^{\mathrm{\mid }S\mathrm{\mid }}$ for $D={\sum }_{K\mathrm{\ne }S}(-1{)}^{\mathrm{\mid }K\mathrm{\mid }}\mathrm{/}{P}_{W_K}$. Choose $s_0\in S$; pairing $K$ with $K\mathrm{△}\{s_0\}$ cancels the full alternating subset sum, so $D(0)={\sum }_{K\mathrm{\ne }S}(-1{)}^{\mathrm{\mid }K\mathrm{\mid }}=-(-1{)}^{\mathrm{\mid }S\mathrm{\mid }}=\pm 1$. Thus $1\mathrm{/}D$ is rational over $\mathbb{Z}$ by [F12], and so is $P_W=(t^N-(-1{)}^{\mathrm{\mid }S\mathrm{\mid }})\mathrm{/}D$. This proves the induction, both recursions, and in particular rationality in $\mathbb{Q}(t)$. The single finite-set selection and all unique factorizations use no choice principle.