Right-descent Soergel bimodules descend to simple invariants under the character hypothesis

A genuine bimodule section. Put $U=M{\otimes }_{S^s}S$, so $M{B}_s=U(1)$. The actual bimodule decomposition of [F1] gives $U\cong M\oplus M(-2)$. The multiplication counit $c:U\to M$, $c(b\otimes r)=br$, is an $S$-bimodule map. The bottom degree of $U$ is the single line $M^{-\mathrm{\ell }(x)}\otimes 1$, on which $c$ is nonzero. Only the unshifted summand $M$ in this decomposition has that degree, so $c$ restricts to a nonzero degree-zero endomorphism of that summand. By [F1] it is a nonzero scalar, hence invertible. Compose its inverse with the actual bimodule inclusion to obtain a degree-zero bimodule section $\sigma :M\to U$. On the bottom line $\sigma (b)=b\otimes 1$, because $c\sigma =1$ and that line in $U$ is one dimensional.

The twisted counit. Define $c_s:U\to M_s$ by $c_s(b\otimes r)=bs(r)$. It is balanced over $S^s$, left $S$-linear and right-$s$-semilinear. Since the section $\sigma$ in step 1.1 is genuinely right-$S$-linear, $t=c_s\sigma :M\to M$ is degree zero, left $S$-linear and right-$s$-semilinear. It is the identity on the bottom line by step 1.1. This argument uses the actual bimodule summand, rather than the generally non-right-linear map $b\mapsto b\otimes 1$.

The involution property of $t$. Since $t$ is right-$s$-semilinear and degree zero, $t^2$ is an ordinary degree-zero $S$-bimodule endomorphism of $M$, hence a scalar by ${\mathrm{End}}^0(M)=\mathbb{R}$; its scalar is $1$ because $t^2$ acts as the identity on the one-dimensional bottom degree by step 2.1. Hence $t^2={\mathrm{i}\mathrm{d}}_M$.

The descent isomorphism. For $u\in U$ one has $c_s(u)-c(u)\in M{\alpha }_s$, because $s(r)-r\in {\alpha }_{s}S$ for every $r\in S$; consequently $t(m)-m\in M{\alpha }_s$ for every $m\in M$. Define $N:=\{m\in M:t(m)=m\}$. Every $m$ splits as $m=\frac{1}{2}(m+t(m))+\frac{1}{2}(m-t(m))$ into an invariant and an anti-invariant part, and an anti-invariant $m$ (that is, $t(m)=-m$) lies in $M{\alpha }_s$, say $m=n{\alpha }_s$; since right multiplication by ${\alpha }_s$ is injective by [F2] and $t(m)=-m$ with $s({\alpha }_s)=-{\alpha }_s$ gives $t(n{\alpha }_s)=t(n)s({\alpha }_s)=-t(n){\alpha }_s$ and hence $t(n)=n$, we get $n\in N$ and the anti-invariant part of $m$ is $N{\alpha }_s$. Therefore $M=N\oplus N{\alpha }_s$ as $(S,S^s)$-bimodules, and the multiplication map induces an isomorphism $N{\otimes }_{S^s}S\to M$; this proves the descent isomorphism.

The matrix of $L_{\zeta }$. Use the proved descent $M=N{\otimes }_{S^s}S$ to write $M{B}_s=N{\otimes }_{S^s}S{\otimes }_{S^s}S(1)$. Splitting the middle factor by $S=S^s\oplus ({\alpha }_s\mathrm{/}2)S^s$ gives the bimodule isomorphism $M{B}_s\cong M(1)\oplus M(-1)$, with structure maps ${\iota }_1,{\iota }_2,{\pi }_1,{\pi }_2$ of [F2] satisfying ${\pi }_a{\iota }_b={\delta }_{ab}$. In this basis, middle multiplication by $\rho$ has entries ${\pi }_1(\rho )$, ${\pi }_1({\alpha }_s\rho )\mathrm{/}2$, $\rho ({\alpha }_s^{\vee })$ and ${\mathrm{\partial }}_s({\alpha }_s\rho )\mathrm{/}2$, and the operators of multiplication by $\rho$ on the two outer factors contribute the same diagonal $\rho \cdot (-)$ on $M(1)$ and $M(-1)$. All entries except the off-diagonal $\rho ({\alpha }_s^{\vee })$ have positive polynomial degree in $\rho$ and vanish after the right reduction; the off-diagonal entry has the constant term $\rho ({\alpha }_s^{\vee })$ and contributes the degree-two map $\zeta \rho ({\alpha }_s^{\vee }):M(1)\to M(-1)$; hence, after reduction, $L_{\zeta }$ is the displayed matrix.

Hard Lefschetz for $\zeta >0$. Assume $\mathrm{h}\mathrm{L}(x)$, so the reduced module $\bar{M}$ is a direct sum of centered $\rho$-strings $u_0,\dots ,u_m$ with $\mathrm{deg}{u}_j=-m+2j$ and $\rho u_j=u_{j+1}$; on each string, left multiplication by $\rho$ is the standard nilpotent Jordan block. In the two-dimensional middle factor with basis $v_{-},v_{+}$ of degrees $-1,+1$ and $e(v_{-})=\zeta \rho ({\alpha }_s^{\vee })v_{+}$, $e(v_{+})=0$, the operator $e=L_{\zeta }$ on the string tensor ${⨁}_j{u}_j\otimes \{v_{-},v_{+}\}$ acts by $e^k(u_j{v}_{-})=u_{j+k}{v}_{-}+k{u}_{j+k-1}{v}_{+}$ and $e^k(u_j{v}_{+})=u_{j+k}{v}_{+}$ (indices out of range zero), after normalizing the nonzero scalar $\zeta \rho ({\alpha }_s^{\vee })$ to $1$ by rescaling $v_{+}$; for $0\le k\le m-1$ of the compatible parity $k\equiv m+1(\mathrm{m}\mathrm{o}\mathrm{d}2)$, in centered degree $-k$ the source basis $u_{(m+1-k)\mathrm{/}2}{v}_{-},u_{(m-1-k)\mathrm{/}2}{v}_{+}$ and the corresponding target basis exhibit $e^k$ as a matrix with diagonal entries $1$ and $1$, lower-left entry $k$, and upper-right entry $0$, which is invertible; at $k=m+1$ the map sends $u_0{v}_{-}$ to $(m+1)u_m{v}_{+}$, nonzero by the normalization $\zeta \rho ({\alpha }_s^{\vee })>0$. The operator on each string is thus an isomorphism in every centered degree, and $L_{\zeta }$ on the direct sum of strings satisfies hL. Positivity of $\zeta \rho ({\alpha }_s^{\vee })$ uses $\zeta >0$ and strict dominance of $\rho$.

Failure at $\zeta =0$ and conclusion. At $\zeta =0$ the matrix of step 5.1 is diagonal with respect to the decomposition $M{B}_s=M(1)\oplus M(-1)$. The bottom class $b\in {\bar{M}}^{-\mathrm{\ell }(x)}$ sits in the first summand in degree $-\mathrm{\ell }(x)-1$ and satisfies ${\rho }^{\mathrm{\ell }(x)+1}b=0$ because ${\bar{M}}^{\mathrm{\ell }(x)+2}=0$; the opposite class in $M(-1)$ of degree $\mathrm{\ell }(x)+1$ is nonzero by self-duality and dimension symmetry. Hence ${\rho }^{\mathrm{\ell }(x)+1}$ is not injective on $\overline{M{B}_s}$ in the centered grading and centered hL fails at $\zeta =0$, even though the two individual shifted copies of $\bar{M}$ have their own strings centered at $\pm 1$. Steps 1.1-5.1 prove the descent isomorphism and the matrix, step 6.1 the positive-parameter hard Lefschetz, and this step the failure at zero; no singular-category classification or character conjecture beyond $\mathrm{S}(x)$ is used.