Tensoring a direct sum of Hodge-Riemann bimodules with B_s preserves the signs

Unshifted orthogonal blocks and their scalars. By [F1] the given unshifted decomposition of $B$ has orthogonal isotypic components, with the normalized perfect $B_z$ form tensored with a nonsingular real symmetric multiplicity matrix. Orthogonally diagonalize these finite matrices to obtain orthogonal unshifted copies $B_z$ with form $c_{\alpha }⟨-,-{⟩}_z$, where $c_{\alpha }\mathrm{\ne }0$. In each copy its bottom vector has degree $-\mathrm{\ell }(z)$ and is primitive, since ${\rho }^{\mathrm{\ell }(z)+1}$ vanishes by the degree bounds. Its normalized Lefschetz value is positive by [F1]. Ambient HR on $B$ says the value on this primitive vector has sign $(-1{)}^{(d-\mathrm{\ell }(z))\mathrm{/}2}$. Hence $c_{\alpha }$ has exactly that sign. This proves the claimed scalar comparison for every copy, including components which do not attain the bottom degree of $B$. No grading shifts are introduced.

Tensoring preserves the orthogonal decomposition and stability. Tensoring the chosen decomposition of $B$ with $B_s$ gives an orthogonal decomposition $B{B}_s={⨁}_{\alpha }{B}_{z_{\alpha }}^R{B}_s$ for the induced form, because the induced form on a direct sum is the orthogonal sum of the induced forms and the isotypic components remain mutually orthogonal. The deformed operator $L_{\zeta }$ is built from left multiplication by $\rho$ on the first factor and middle multiplication on the tensor position, so it preserves each block $B_z^R{B}_s$: the projections onto the blocks are bimodule maps commuting with left multiplication on the first factor, and middle multiplication by $\rho$ acts inside the balanced tensor position and commutes with the outer actions; hence the restriction of $L_{\zeta }$ to each block is the operator of the same name on that block.

Primitive signs on the blocks. On each block $B_z^R{B}_s$ satisfying $\mathrm{H}\mathrm{R}(z,s{)}_{\zeta }$, the primitive space in degree $-i$ is definite with sign $(-1{)}^{(\mathrm{\ell }(z)+1-i)\mathrm{/}2}$ by [F2], relative to the induced form normalized with bottom value $+1$; the actual restricted form differs from that normalization by the scalar of step 1.1, which is the sign $(-1{)}^{(d-\mathrm{\ell }(z))\mathrm{/}2}$ times a positive number, and a positive scalar changes no definiteness sign. The product of the two signs is $(-1{)}^{(d-\mathrm{\ell }(z)+\mathrm{\ell }(z)+1-i)\mathrm{/}2}=(-1{)}^{(d+1-i)\mathrm{/}2}$, independent of $z$; hence each block's primitive signs are the common standard signs of $B{B}_s$, whose lowest degree is $-d-1$: the bottom of each block is $-\mathrm{\ell }(z)-1$, and the largest occurring $\mathrm{\ell }(z)$ equals $d$ in the unshifted sum.

The descent summands at $\zeta =0$ and conclusion. If $\zeta =0$ the hypothesis excludes summands with $zs<z$, so every block $B_z^R{B}_s$ is an ascent block or has $zs>z$; no HR statement for a shifted descent block is used. By [F3] the orthogonal sum of the blocks satisfies HR with the common standard signs of step 2.1, and the lowest-degree sign is positive by the normalization of the bottom: the total bottom is ${⨁}_{z\text{ at bottom}}{B}_z^{-\mathrm{\ell }(z)}{B}_s^{-1}$, and each block contributes a positive multiple of its bottom primitive form, so the assembled form is positive there. Hence $B{B}_s$ satisfies HR with the standard sign with respect to $L_{\zeta }$ and the induced form, and the displayed sign $(-1{)}^{(d+1-i)\mathrm{/}2}$ is exactly the standard sign of the tensor product; this completes the proof.