Minimal positive Rouquier complexes are linear under the lower Soergel induction

Base of the induction. [base] For $x=e$ the Rouquier complex is the unit $S$ concentrated in cohomological degree $0$, so ${}^0{F}_e=S=B_e$, there is no term of index $i\ge 1$, both perverse bounds hold trivially, and the only nonzero inverse coefficient is $g_{e,e}=1$; for $x=s_i$ the complex $F_{s_i}=[B_{s_i}^R\to S(1)]$ has ${}^0{F}_{s_i}=B_{s_i}$ and ${}^1{F}_{s_i}=S(1)=B_e^R(1)$ and no other term, so $m_{e,1}=1$ with $\mathrm{\ell }(s_i)-\mathrm{\ell }(e)=1\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}2)$, $B_{s_i}$ is perverse by $\mathrm{S}(s_i)$ and $S(1)$ is the shift of the perverse unit, so $F_{s_i}\in p{K}_{\ge 0}^b\cap p{K}_{\le 0}^b$ and $g_{s_i,s_i}=1$, $g_{e,s_i}=-v$.

Induction hypothesis and the last letter. [ih] Fix $x\mathrm{\ne }e$, write $\underset{‾}{x}=s_1\cdots s_m$ with $m\ge 1$ and put $y=s_1\cdots s_{m-1}$, $s=s_m$, so that $x=ys$ with $\mathrm{\ell }(x)=\mathrm{\ell }(y)+1$; the hypothesis $\mathrm{S}(\le x)$ contains $\mathrm{S}(\le y)$, and we assume the three assertions of the statement for $y$: ${}^0{F}_y=B_y$, ${}^i{F}_y={⨁}_{z<y}{B}_z(i{)}^{m_{z,i}^y}$ for $i\ge 1$, and $F_y\in p{K}_{\ge 0}^b\cap p{K}_{\le 0}^b$. By [F1], $F_x\cong F_y{F}_s$ in $K^b$.

Descent contraction. Let $z$ and a simple reflection $s$ satisfy $zs<z$ and $\mathrm{S}(z)$; then $B_z{F}_s\cong B_z(-1)$ in $K^b$. Indeed, by [F2] the two-term complex $B_z{F}_s$ has terms $B_z{B}_s=B_z(1)\oplus B_z(-1)$ in degree $0$ and $B_z(1)$ in degree $1$, with differential $d=(\delta ,\eta )$ where $\delta \in {\mathrm{End}}^0(B_z(1))={\mathrm{End}}^0(B_z)=\mathbb{R}$ by [F2] and [F6]. If $\delta =0$ then $B_z{F}_s$ is the direct sum of the two nonzero complexes $[B_z(1)\to 0]$ and $[B_z(-1)\stackrel{\eta }{\to }{B}_z(1)]$, contradicting the indecomposability of $B_z{F}_s$, which is the image of the indecomposable $B_z$ under the equivalence $-\otimes F_s$; hence $\delta \mathrm{\ne }0$ is an isomorphism, homological Gaussian elimination cancels the contractible two-term complex $B_z(1)\cong B_z(1)$, and the remainder is $B_z(-1)$ concentrated in degree $0$.

Top support and the Euler character. Every positive-degree term of the word complex omits at least one letter, so no such term has the $x$ graph by [F3]. The degree-zero term is its reduced word, whose top summand is exactly $B_x$ unshifted and once. Gaussian minimalization cannot remove it, since no target term has its label. Thus in the minimal complex $m_{x,0}^0=1$ and every other $m_{x,k}^i$ is zero, by actual support rather than alternating cancellation. Character additivity makes its Euler character unchanged by removing a contractible pair. The generator recursion and shift rule in [F3] give $\mathrm{ch}(F_z{F}_s)=\mathrm{ch}(F_z)({\underset{‾}{H}}_s-v)$ degree by degree. Iterating along the reduced word therefore gives Euler character $H_x$. Under the stated $S(\le x)$, expanding each actual summand character in the KL basis yields ${\sum }_{i,k}(-1{)}^i{m}_{w,k}^i{v}^k=g_{w,x}$. This equality is used only for the Euler coefficients, after the support proof of the top multiplicity.

Ascent perverse bound. Let $z\le y$ and a simple reflection $s$ satisfy $zs>z$; then $B_z{F}_s\in p{K}_{\ge 0}^b$. Indeed, $\mathrm{S}(u)$ holds for every $u<zs$, because $u<zs\le x$ implies $u<x$; so [F2] gives the unshifted decomposition $B_z{B}_s=B_{zs}\oplus {⨁}_{u<zs}{B}_u^{a_u}$, whose summands are perverse by $\mathrm{S}(zs)$ and $\mathrm{S}(u)$, while the second term $B_z(1)$ of $B_z{F}_s=[B_z{B}_s\to B_z(1)]$ is the internal shift of the perverse $B_z$ sitting in cohomological degree $1$; thus the chosen representative has p-split terms and satisfies ${\tau }_{<0}(B_z{B}_s)=0$ and ${\tau }_{<-1}(B_z(1))=0$, which is membership in $p{K}_{\ge 0}^b$ by [F5].

Shift propagation. Assume $\mathrm{S}(\le x)$ and let $F_x$ be minimal. If ${}^i{F}_x$ contains a summand $B_z(j)$ with $z<x$, then ${}^{i-1}{F}_x$ contains a summand $B_{z^{\mathrm{\prime }}}(j^{\mathrm{\prime }})$ with $z^{\mathrm{\prime }}>z$ and $j^{\mathrm{\prime }}<j$. First, components of the differential into summands of shift $k<j$ vanish: a degree-zero map $B_z(j)\to B_u(k)$ is an element of ${\mathrm{Hom}}^{k-j}(B_z,B_u)$, which is zero for $k<j$ since the graded Hom space has no negative-degree part by [F6]; components with $k=j$ are zero for $u\mathrm{\ne }z$ by [F6] and are scalars for $u=z$, and a nonzero scalar component would make $B_z(j)$ a contractible summand of the minimal complex, so all such components vanish as well; hence $d(B_z(j))$ lies in the sum of the summands of ${}^{i+1}{F}_x$ with shift $k>j$. Second, if a summand $B_u(k)$ of ${}^{i-1}{F}_x$ maps nontrivially to $B_z(j)$, then $k<j$: a degree-zero map $B_u(k)\to B_z(j)$ lies in ${\mathrm{Hom}}^{j-k}(B_u,B_z)$, which vanishes for $j<k$, and for $j=k$ it vanishes for $u\mathrm{\ne }z$ and is a scalar for $u=z$, the nonzero case again being excluded by minimality. Third, apply the support cutoff ${\mathrm{\Gamma }}_{\ge z\mathrm{/}>z}$ and reduce its free graph modules modulo $R_{+}$. Since $z<x$, its complex is contractible by [F5], hence this reduced finite-dimensional complex is exact. The chosen top layer ${\mathrm{\Delta }}_z(j)$ of $B_z(j)$ is a nonzero homogeneous basis vector. Every outgoing target has shift $k>j$, and its $z$ layer has shift $k+d$ with $d\ge 0$ by KL triangularity. A map from the chosen generator into it consequently has positive-degree coefficient, so vanishes after reduction. The chosen vector is a cycle and must therefore be an incoming boundary. Some incoming component has a nonzero constant coefficient onto it, so some $B_{z^{\mathrm{\prime }}}(j^{\mathrm{\prime }})$ in degree $i-1$ has a $z$ layer with $j^{\mathrm{\prime }}+d=j$. The underlying differential component is nonzero, and the preceding Hom estimate gives $j^{\mathrm{\prime }}<j$. Since $d=0$ for label $z$ itself, this requires $z^{\mathrm{\prime }}>z$. This proves shift propagation without using an Euler identity to recover multiplicities.

The perverse lower bound for $F_x$. Use the stupid filtration $w_{\ge k}{F}_y$ of the minimal complex $F_y$, with distinguished triangles $w_{\ge k+1}{F}_y\to w_{\ge k}{F}_y\to {}^k{F}_y[-k]\to$ and $w_{\ge \mathrm{\ell }(y)+1}{F}_y=0$; tensoring with the invertible $F_s$ preserves distinguished triangles. We prove $(w_{\ge k}{F}_y)F_s\in p{K}_{\ge 0}^b$ by descending induction on $k$: for $k=\mathrm{\ell }(y)+1$ the object is $0$; assume it for $k+1$. By [ih] and step 2.1, ${}^k{F}_y$ is a direct sum of summands $B_z(k)$ with $z\le y$ (with $z=y$ only for $k=0$), so ${(}^k{F}_y)F_s[-k]$ is a direct sum of copies of $(B_z{F}_s)(k)[-k]$. If $zs>z$ then $B_z{F}_s\in p{K}_{\ge 0}^b$ by step 4.1, and its internal shift $(k)$ and homological shift $[-k]$ place every summand $B_u(\mathrm{\ell })$ occurring in cohomological degree $i$ into degree $i+k$ with internal shift $\mathrm{\ell }+k\le i+k$, so $(B_z{F}_s)(k)[-k]\in p{K}_{\ge 0}^b$; if $zs<z$ then step 3.1 gives $B_z{F}_s\cong B_z(-1)$, so $(B_z{F}_s)(k)[-k]\cong B_z(k-1)[-k]$, a perverse object with internal shift $k-1$ in cohomological degree $k$, again in $p{K}_{\ge 0}^b$. Extension closure of $p{K}_{\ge 0}^b$ in the tensored triangles gives $(w_{\ge k}{F}_y)F_s\in p{K}_{\ge 0}^b$, and at $k=0$ we obtain $F_y{F}_s\in p{K}_{\ge 0}^b$; by step 2.1 this is $F_x\in p{K}_{\ge 0}^b$.

The top term and the perverse upper bound. If ${}^0{F}_x$ contained a summand $B_z(j)$ with $z<x$, then step 4.2 with $i=0$ would produce a summand of ${}^{-1}{F}_x=0$, which is impossible; hence every summand of ${}^0{F}_x$ has label $x$, and by step 3.2 ($m_{x,0}^0=1$ and $m_{x,k}^0=0$ for $k\mathrm{\ne }0$) we get ${}^0{F}_x=B_x$, which proves (1). Now suppose $i\ge 1$ and ${}^i{F}_x$ has a summand $B_z(k)$ with $k<i$; if $z=x$ this contradicts $m_{x,k}^i=0$ from step 3.2, so $z<x$, and step 4.2 produces a summand $B_{z^{\mathrm{\prime }}}(k^{\mathrm{\prime }})$ of ${}^{i-1}{F}_x$ with $k^{\mathrm{\prime }}<k<i$. This summand is again of the forbidden type: if $z^{\mathrm{\prime }}=x$ then $m_{x,k^{\mathrm{\prime }}}^{i-1}=0$ whenever $i-1\ge 1$, while for $i-1=0$ it contradicts ${}^0{F}_x=B_x$ because $k^{\mathrm{\prime }}<0$; otherwise $z^{\mathrm{\prime }}<x$ and step 4.2 applies again. Iterating this descent in the homological degree we reach a summand of ${}^0{F}_x$ with negative internal shift, contradicting ${}^0{F}_x=B_x$. Hence every summand of ${}^i{F}_x$ has internal shift $\ge i$ for $i\ge 1$, which together with $F_x\in p{K}_{\ge 0}^b$ from step 5.1 gives $F_x\in p{K}_{\le 0}^b$; since for $i\ge 1$ the label $x$ does not occur by step 3.2, we obtain ${}^i{F}_x={⨁}_{z<x}{B}_z(i{)}^{m_{z,i}}$ with $m_{z,i}:=m_{z,i}^i$, which is (2).

Inversion of the Kazhdan-Lusztig transition matrix. Let the two bases be related by ${\underset{‾}{H}}_w={\sum }_{y\le w}{h}_{y,w}{H}_y$ and $H_w={\sum }_{y\le w}{g}_{y,w}{\underset{‾}{H}}_y$ as in [F4]; substituting the first expansion into the second gives ${\sum }_{y\le z\le w}{h}_{y,z}{g}_{z,w}={\delta }_{y,w}$. Hence $g_{w,w}=1$ and, for $y<w$, $g_{y,w}=-{\sum }_{y<z\le w}{h}_{y,z}{g}_{z,w}$. We prove $g_{y,w}\in v^{\mathrm{\ell }(w)-\mathrm{\ell }(y)}\mathbb{Z}[v^{-2}]$ by descending induction on $\mathrm{\ell }(w)-\mathrm{\ell }(y)$: for $y=w$ this is $1\in v^0\mathbb{Z}[v^{-2}]$, and for $y<w$ each $h_{y,z}$ lies in $v^{\mathrm{\ell }(z)-\mathrm{\ell }(y)}\mathbb{Z}[v^{-2}]$ by [F4] and each $g_{z,w}$ with $y<z\le w$ lies in $v^{\mathrm{\ell }(w)-\mathrm{\ell }(z)}\mathbb{Z}[v^{-2}]$ by the hypothesis of the induction, so each product lies in $v^{\mathrm{\ell }(w)-\mathrm{\ell }(y)}\mathbb{Z}[v^{-2}]$ and the finite sum does too.

Parity and positivity of the inverse coefficients. By step 3.2 and step 6.1 the multiplicities appearing in $g_{z,x}={\sum }_i(-1{)}^i{m}_{z,i}{v}^i$ are exactly the multiplicities of the linear complex $F_x$, and by step 7.1 we have $g_{z,x}\in v^{\mathrm{\ell }(x)-\mathrm{\ell }(z)}\mathbb{Z}[v^{-2}]$; therefore the coefficient $(-1{)}^i{m}_{z,i}$ of $v^i$ can be nonzero only when $i\equiv \mathrm{\ell }(x)-\mathrm{\ell }(z)(\mathrm{m}\mathrm{o}\mathrm{d}2)$, which is the parity clause of (3). For such $i$ the sign $(-1{)}^{\mathrm{\ell }(x)-\mathrm{\ell }(z)+i}$ equals $+1$, so $(-1{)}^{\mathrm{\ell }(x)-\mathrm{\ell }(z)}{g}_{z,x}={\sum }_i(-1{)}^{\mathrm{\ell }(x)-\mathrm{\ell }(z)+i}{m}_{z,i}{v}^i\in {\mathbb{Z}}_{\ge 0}[v^{\pm 1}]$, and the coefficients are the nonnegative multiplicities $m_{z,i}$. This proves (3) and, with (1) and (2), completes the induction.