Hard Lefschetz for the deformed operator on B_x B_s, in the three cases

Case (1), positive parameter. Assume $xs<x$, $\mathrm{S}(x)$ and $\mathrm{h}\mathrm{L}(x)$. By [F1] the reduced operator is the matrix with diagonal blocks $\rho \cdot (-)$ on $B_x(1)$ and $B_x(-1)$ and lower off-diagonal block $\zeta \rho ({\alpha }_s^{\vee })$, a scalar multiple of the identity; since $\rho \cdot (-)$ satisfies hL on $B_x$, the $\rho$-action completes to an ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$-action and, after rescaling by $\zeta \mathrm{\ne }0$, the matrix describes the action of the raising operator on the tensor product of $B_x$ with the two-dimensional standard representation; hence its powers are isomorphisms in the centered degrees and $\mathrm{h}\mathrm{L}(x,s{)}_{\zeta }$ holds, which is case (1).

Case (2), setting and form identity. Assume $\zeta >0$, $xs>x$, $\mathrm{S}(\le x)$, $\mathrm{H}\mathrm{R}(z,t)$ for all $z<x$ with $zt>z$, $\mathrm{H}\mathrm{R}\mathrm{b}\mathrm{a}\mathrm{r}(<x,s{)}_{\zeta }$ and $\mathrm{H}\mathrm{R}(x)$. By [F2] the scalars ${\gamma }_i$ for $1\le i\le m$ are strictly positive and ${\gamma }_{m+1}=(x^{-1}\rho )({\alpha }_s^{\vee })+\zeta \rho ({\alpha }_s^{\vee })>0$, and with $\phi :M\to {}^1{F}_x{B}_s\oplus B_x(1)$ the first differential of $F_x{F}_s$ one has $⟨b,L_{\zeta }{b}^{\mathrm{\prime }}{⟩}_{\bar{M}}=⟨\bar{\phi }(b),\bar{\phi }(b^{\mathrm{\prime }}){⟩}_{\gamma }$ for all $b,b^{\mathrm{\prime }}\in \bar{M}$; by [F3] the reduced $\phi$ is injective in internal degrees $\le \mathrm{\ell }(x)$.

Failure at $\zeta =0$ for a descent, without $\mathrm{S}(x)$. Put $m=\mathrm{\ell }(x)$ and choose a reduced expression $\underset{‾}{x}=\underset{‾}{y}s$, where $y=xs$. By [F9], tensoring the split inclusion $B_x\subseteq \mathrm{B}\mathrm{S}(\underset{‾}{x})$ with $B_s$ and then reducing on the right gives an $L_0$-equivariant split inclusion $\bar{M}\subseteq \overline{\mathrm{B}\mathrm{S}(\underset{‾}{x})B_s}$. The rank-one splitting gives the ambient bimodule decomposition $\mathrm{B}\mathrm{S}(\underset{‾}{x})B_s=\mathrm{B}\mathrm{S}(\underset{‾}{y})B_s{B}_s\cong \mathrm{B}\mathrm{S}(\underset{‾}{x})(1)\oplus \mathrm{B}\mathrm{S}(\underset{‾}{x})(-1)$, which commutes with left multiplication by $\rho$. The reduction of $\mathrm{B}\mathrm{S}(\underset{‾}{x})$ is supported in $[-m,m]$, so $L_0^{m+1}$ vanishes on both shifted copies, and therefore on $\bar{M}$. A homogeneous right basis of $B_x$ has extremal degrees $-m,m$; tensoring with the right basis of $B_s$ adds $-1,1$ to these degrees. Hence ${\bar{M}}^{-m-1}$ and ${\bar{M}}^{m+1}$ are both nonzero. The centered map $L_0^{m+1}:{\bar{M}}^{-m-1}\to {\bar{M}}^{m+1}$ is thus zero with nonzero source and target, proving the unconditional failure statement.

The operator commutes with $\phi$. Writing $\phi =(d_1,d_2)$ with $d_1$ into ${}^1{F}_x{B}_s$ and $d_2$ into $B_x(1)$, the deleted-word factorization of [F2] shows that $d_1$ commutes with $L_{\zeta }$ and that $d_2(L_{\zeta }b)=\rho \cdot d_2(b)+d_2(b)\cdot \zeta \rho$. After right reduction the last right multiple vanishes. Thus, with $L$ equal to $L_{\zeta }$ on the reduced first summand and $\rho \cdot (-)$ on the reduced second summand, $\bar{\phi }(L_{\zeta }b)=L\bar{\phi }(b)$ for every $b\in \bar{M}$. No such equality with this $L$ is asserted on unreduced $M$.

The target satisfies HR. Consider the graded space $W:=\overline{{}^1{F}_x(-1)B_s}\oplus \overline{B_x}$ with the operator $L$ of step 2.1 and the form $⟨-,-{⟩}_{\gamma }$. The two summands are orthogonal for the induced form, $B_x$ satisfies hL and HR by $\mathrm{H}\mathrm{R}(x)$, and ${}^1{F}_x(-1)B_s$ satisfies hL and HR with the standard sign: its summands $B_z$ with $m_{z,1}>0$ and $z<x$ satisfy $\mathrm{S}(z)$ and, for every such $z$, the hypothesis $\mathrm{H}\mathrm{R}(z,s{)}_{\zeta }$ (which $\mathrm{H}\mathrm{R}\mathrm{b}\mathrm{a}\mathrm{r}(<x,s{)}_{\zeta }$ supplies, including descents since $\zeta >0$), so the tensoring lemma [F6] applies to ${}^1{F}_x(-1)={⨁}_{z<x}{B}_z^{m_{z,1}}$ with the positivity of the rescalings; the induced signs of ${}^1{F}_x(-1)$ furnished by [F5] match the standard sign. Hence $W$ satisfies HR with the standard sign, and the reduced map $\bar{\phi }$ sends $\bar{M}$ into $W(1)$, since right reduction of the first differential target is $\overline{{}^1{F}_x{B}_s}\oplus \overline{B_x}(1)=W(1)$.

Case (2) concluded. Apply the weak-Lefschetz substitute [F4] to $\bar{\phi }:\bar{M}\to W(1)$: its three hypotheses hold by steps 1.2, 2.1 and 3.1 (injectivity in degrees $\le \mathrm{\ell }(x)$ contains injectivity in degrees $\le -1$, the form identity is the one of [F2], and $W$ satisfies HR by step 3.1). Hence $L_{\zeta }^k:{\bar{M}}^{-k}\to {\bar{M}}^k$ is injective for all $k\ge 0$; since $\bar{M}$ has symmetric dimensions by self-duality, equal dimensions give isomorphisms, which is $\mathrm{h}\mathrm{L}(x,s{)}_{\zeta }$ and proves case (2).

Case (3), the separated differential. Assume $\zeta =0$, $xs>x$, $\mathrm{S}(\le x)$, the ascent HR hypothesis for lower elements, $\mathrm{H}\mathrm{R}(x)$ and $\mathrm{h}\mathrm{L}(z)$ for all $z<xs$. Use the same positive scalars ${\gamma }_1,\dots ,{\gamma }_{m+1}$ with $\zeta =0$ and the same form identity as in [F2] with $\phi$ the first differential of $F_x{F}_s$. Decompose ${}^1{F}_x(-1)=B^{\uparrow }\oplus B^{\downarrow }$ orthogonally according to $zs>z$ and $zs<z$ as in step 3.1, retaining the multiplicities $m_{z,1}$. To prove the required tensor bound, take any $z\le x$. If $zs>z$, [F8] makes $B_z{B}_s$ perverse without assuming $S(zs)$, so $B_z{F}_s=[B_z{B}_s\to B_z(1)]$ satisfies the perverse lower bound. If $zs<z$, the descent contraction in [F7] gives $B_z{F}_s\simeq B_z(-1)$, which also satisfies that bound. By linearity, the $k$-th term of $F_x$ is a sum of $B_z(k)$ with $z\le x$. Tensor its stupid-filtration triangles with $F_s$: each quotient is a sum of $(B_z{F}_s)(k)[-k]$, whose internal shifts and cohomological indices both increase by $k$, preserving the lower bound. Finite descending induction and extension closure in the radical/perverse lemma give $F_x{F}_s\in p{K}_{\ge 0}^b$. The offending degree-one block is ${\tau }_{\le -2}(B^{\downarrow }{B}_s(1))=B^{\downarrow }(2)$. In the radical quotient it cannot receive a differential from the perverse degree-zero source: its shifted simple classes differ. The minimal model has no such shift in degree one, so the quotient differential out of this block is injective (otherwise that quotient simple would survive in degree one). Lift its quotient left inverse and invert the resulting identity-plus-radical composite, as in the radical lemma, to obtain a genuine split injection. Gaussian elimination cancels this forward contractible block without changing the degree-zero term. The resulting homotopy-equivalent complex has first differential $d=(d_1,d_2,d_3):B_x{B}_s\to B_x(1)\oplus B^{\uparrow }{B}_s(1)\oplus B^{\downarrow }$, all components commuting with left multiplication by $\rho$, and $\bar{d}$ injective in degrees $\le \mathrm{\ell }(x)$ by [F3].

Case (3), the two claims. Fix $0\mathrm{\ne }b\in {\bar{M}}^{-k}$ with $k\ge 0$, so that $\bar{d}(b)\mathrm{\ne }0$ by the injectivity just recalled. If ${\bar{d}}_3(b)\mathrm{\ne }0$, then ${\rho }^k(d_3(b))=d_3({\rho }^{k}b)$ by commutation, and ${\rho }^k(d_3(b))\mathrm{\ne }0$ because ${\bar{d}}_3(b)$ lies in $B^{\downarrow }$, each of whose indecomposable summands $B_z$ has $z<xs$ and satisfies $\mathrm{h}\mathrm{L}(z)$ by hypothesis, so left multiplication by $\rho$ satisfies hL on $B^{\downarrow }$; hence ${\rho }^{k}b\mathrm{\ne }0$. If ${\bar{d}}_3(b)=0$, put $V=\ker ({\bar{d}}_3)\subseteq \bar{M}$ and $W=\overline{B_x}\oplus \overline{B^{\uparrow }{B}_s}$ with the restricted forms and the operator $\rho \cdot (-)$, write ${\phi }_V:V\to W(1)$ for the restriction of $({\bar{d}}_1,{\bar{d}}_2)$: it commutes with $\rho$, it is injective in degrees $\le -1$ (indeed in degrees $\le \mathrm{\ell }(x)$), and it is form-compatible as follows: in the original degree-one target the descent block is $B^{\downarrow }(2)\oplus B^{\downarrow }$, pairing the two shifted pieces. The first piece has zero self-pairing: its invariant form map is $B^{\downarrow }(2)\to D(B^{\downarrow }(2))\cong B^{\downarrow }(-2)$, a negative-degree map between unshifted perverse summands, so it vanishes by the proved perverse Hom bound, while on $V$ the surviving $B^{\downarrow }$ coordinate $d_3$ is zero. Consequently both its cross-pairings and its surviving self-pairing contribute zero on $V$, so $⟨b,\rho b^{\mathrm{\prime }}{⟩}_V=⟨{\phi }_V(b),{\phi }_V(b^{\mathrm{\prime }}){⟩}_W$; and $W$ satisfies HR with the standard sign, since $B_x$ does by $\mathrm{H}\mathrm{R}(x)$ and $B^{\uparrow }{B}_s$ does by the tensoring lemma [F6] applied with $\zeta =0$ to the ascent summands. The weak-Lefschetz substitute [F4] gives ${\rho }^k$ injective on $V^{-k}$, whence ${\rho }^{k}b\mathrm{\ne }0$. Both claims show that ${\rho }^k$ has trivial kernel on ${\bar{M}}^{-k}$; by symmetric dimensions of $\bar{M}$ it is an isomorphism, which is $\mathrm{h}\mathrm{L}(x,s)$.

Conclusion and caveats. Steps 1.1 and 1.3 treat the descent cases, step 4.1 case (2) and step 5.1 case (3); no step assumes HR for the unseparated zero-parameter descent target, and no step invokes any singular-category descent theorem or any statement beyond the suppliers listed. This proves the three-case theorem and the failure statement.