Undeformed Lefschetz and Hodge-Riemann restrict to the next Soergel summand

Restriction of hard Lefschetz (a). The inclusion $i:B_{xs}^R\to M$ and the projection $\pi :M\to B_{xs}^R$ of the summand are degree-zero bimodule maps, and they commute with left multiplication by $\rho$, since left multiplication by a central element is a bimodule endomorphism; after right reduction they remain a split inclusion and projection of the finite-dimensional real spaces. Hence the operator $L_0^i$ on $\bar{M}$ is block diagonal with respect to the decomposition of $\bar{M}$ into the summand and its complement, and each block is the corresponding power of the restricted operator. If $L_0^i$ is an isomorphism, then its inverse commutes with the block decomposition, so each diagonal block is invertible; in particular the block ${\overline{B_{xs}^R}}^{-i}\to {\overline{B_{xs}^R}}^i$ is an isomorphism. This holds for every $i\ge 0$, which is $\mathrm{h}\mathrm{L}(xs)$.

The converse under the character hypotheses (b). Assume $\mathrm{S}(<xs)$ and $\mathrm{h}\mathrm{L}(y)$ for all $y<xs$. By [F1] the decomposition $M=B_{xs}^R\oplus {⨁}_{y<xs}{B}_y^{R,a_y}$ has no shifts, and every lower label $y<xs$ satisfies $\mathrm{S}(y)$; the module $B_{xs}^R$ is self-dual with symmetric Betti numbers, and by the comparison lemma the same holds for every lower summand. If $\mathrm{h}\mathrm{L}(xs)$ holds, the block of $L_0^i$ on $B_{xs}^R$ is invertible; by hypothesis each lower block $B_y^R$ is invertible (hL$(y)$), and no other summands occur; therefore every diagonal block is invertible, so $L_0^i$ is invertible, which is $\mathrm{h}\mathrm{L}(x,s{)}_0$. This proves the equivalence.

Restriction of HR (c). Assume $\mathrm{H}\mathrm{R}\mathrm{b}\mathrm{a}\mathrm{r}(x,s{)}_0$, that is, HR for the restriction of the intersection form over a fixed reduced-word embedding of $\underset{‾}{x}$ and a fixed induced form of $M$; the summand $B_{xs}^R$ has symmetric Betti numbers by self-duality, and the inclusion is $L_0$-stable. Applying the stable-subspace lemma [F2] to $V=\overline{B_{xs}^R}$ inside $\bar{M}$ gives that the restricted operator and form satisfy hL and HR with the ambient signs; since the lowest degree of $B_{xs}^R$ equals the lowest degree $-\mathrm{\ell }(xs)$ of $M$, the ambient standard sign is the standard sign of $B_{xs}^R$. Once $\mathrm{S}(xs)$ is known, [F3] identifies the restricted form with a positive multiple of the normalized intrinsic form, and $\mathrm{H}\mathrm{R}(xs)$ and $\mathrm{H}\mathrm{R}\mathrm{b}\mathrm{a}\mathrm{r}(xs)$ follow with no change of signs.

General stable subspaces (d). For an $L_0$-stable graded subspace with symmetric Betti numbers, [F2] gives hL and HR with the ambient signs. If instead a graded decomposition $H=V\oplus W$ has both $V$ and $W$ $L_0$-stable and $H$ satisfies hL, then each $L_0^i:H^{-i}\to H^i$ is block diagonal. Its injectivity and surjectivity give injectivity and surjectivity on each block separately, hence hL on $V$, as in step 1.1. Merely being a graded direct summand is insufficient: in a length-one Lefschetz string, the stable top line has zero degree $-1$ part and a nonzero degree $1$ part. A positive deformation $L_{\zeta }$ with $\zeta >0$ preserves a summand only when the summand's projector commutes with the middle-position operator, which is not automatic; when it does not, the restriction of $L_{\zeta }$ to the summand is not defined as an operator on the reduced summand, so no positive-parameter restriction statement is made. In particular no step identifies the restriction of $L_{\zeta }$ with left multiplication by $\rho$. This proves (d) and completes the lemma.