The type-A character recursion under simple Soergel tensoring

Flag membership (1): by [F4] the functor $B_s{\otimes }_R-$ carries $F_{\mathrm{\Delta }}$ into $F_{\mathrm{\Delta }}$ and $F_{\mathrm{\nabla }}$ into $F_{\mathrm{\nabla }}$ (the closure clause of each of the two imported propositions recorded there), and by [F5] it is exact and preserves finite freeness, so $B_s{\otimes }_{R}M\in F_{\mathrm{\Delta }}\cap F_{\mathrm{\nabla }}$ with finitely many flag quotients. For the right tensor, the opposite identification $(M{\otimes }_R{B}_s{)}^{\mathrm{o}\mathrm{p}}\cong B_s^{\mathrm{o}\mathrm{p}}{\otimes }_R{M}^{\mathrm{o}\mathrm{p}}$ together with $B_s^{\mathrm{o}\mathrm{p}}\cong B_s$ expresses $M{\otimes }_R{B}_s$ as the opposite of $B_s{\otimes }_R{M}^{\mathrm{o}\mathrm{p}}$; the opposite functor preserves each of the two flag categories, as recorded in Standard graph bimodules, support filtrations and characters, and $M^{\mathrm{o}\mathrm{p}}$ again lies in both with the same shift data, so the left-closure clause applied to $M^{\mathrm{o}\mathrm{p}}$ gives $M{\otimes }_R{B}_s\in F_{\mathrm{\Delta }}\cap F_{\mathrm{\nabla }}$. Exactness and preservation of finite freeness for $-{\otimes }_R{B}_s$ hold by [F5] because $B_s$ is free of rank two as a left $R$-module.

Multiplicities in the two charts: for $\mathrm{\ell }(x)>\mathrm{\ell }(sx)$ the imported recursions of [F4] express the $\mathrm{\Delta }$-multiplicities of $B_s{\otimes }_{R}B$ in terms of those of $B$, and the matching Hecke formulas of [F4] are the expansion of ${\tilde{T}}_s{\tilde{T}}_x$ and ${\tilde{T}}_s{\tilde{T}}_{sx}$ in the standard basis of [F3]: since ${\tilde{T}}_x=v^{\mathrm{\ell }(x)}{T}_x$, the left-multiplication rule is $T_s{T}_x=T_{sx}$ when $\mathrm{\ell }(sx)=\mathrm{\ell }(x)+1$ and $T_s{T}_x=(q-1)T_x+q{T}_{sx}$ when $\mathrm{\ell }(sx)=\mathrm{\ell }(x)-1$, so ${\tilde{T}}_s{\tilde{T}}_x={\tilde{T}}_{sx}$ in the ascent case and ${\tilde{T}}_s{\tilde{T}}_x={\tilde{T}}_{sx}+(v^{-1}-v){\tilde{T}}_x$ in the descent case, with coefficient one on ${\tilde{T}}_{sx}$; this coefficient one is exactly what the recursions of [F4] assert, and the diagonal term $(v^{-1}-v){\tilde{T}}_x$ is the shift by one of the diagonal that the recursion moves.

Base case: for $M=R$ the two flags of [F2] give $h_{\mathrm{\Delta }}(B_s)=v^1{\tilde{T}}_e+v^0{\tilde{T}}_s=v+v{T}_s$ and $h_{\mathrm{\nabla }}(B_s)=v^1{\tilde{T}}_e+v^0{\tilde{T}}_s=v+v{T}_s$, both equal to $H_s\cdot 1=H_s$; and $h_{\mathrm{\Delta }}(R)=h_{\mathrm{\nabla }}(R)={\tilde{T}}_e=1$.

Shift rule: ${\mathrm{\Delta }}_x(d)\{k\}=R_x\{\mathrm{\ell }(x)-d+k\}={\mathrm{\Delta }}_x(d-k)$ and ${\mathrm{\nabla }}_x(d)\{k\}={\mathrm{\nabla }}_x(d-k)$, so the multiplicity of ${\mathrm{\Delta }}_x(d)$ in $M\{k\}$ equals that of ${\mathrm{\Delta }}_x(d+k)$ in $M$, whence $h_{\mathrm{\Delta }}(M\{k\})={\sum }_d(M:{\mathrm{\Delta }}_x(d+k))v^d{\tilde{T}}_x=v^{-k}{h}_{\mathrm{\Delta }}(M)$; the same computation with the weights $v^{-d}$ gives $h_{\mathrm{\nabla }}(M\{k\})=v^k{h}_{\mathrm{\nabla }}(M)$.

Comparing coefficients in step 1.2 term by term shows $h_{\mathrm{\Delta }}(B_s{\otimes }_{R}B)={\sum }_{x,d}(B_s{\otimes }_{R}B:{\mathrm{\Delta }}_x(d))v^d{\tilde{T}}_x=({\tilde{T}}_s+v)h_{\mathrm{\Delta }}(B)=H_s{h}_{\mathrm{\Delta }}(B)$, so the $\mathrm{\Delta }$-recursion of claim (2) holds.

The dual chart is the same computation with the weights $v^{-d}$: the $\mathrm{\nabla }$-recursions of [F4] give $h_{\mathrm{\nabla }}(B_s{\otimes }_{R}B)=H_s{h}_{\mathrm{\nabla }}(B)$, since the two charts are exchanged by $d\mapsto -d$ in the displayed Hecke formulas of [F4].

Intrinsicness and conclusion: by The type-A support filtration multiplicities are intrinsic the multiplicities used in steps 2.1 to 3.1 depend only on the bimodules involved, so the displayed identities are identities between the characters of $M$, $B_i{\otimes }_{R}M$ and $M\{k\}$; claims (1) to (4) follow, the case $M=R$ of claim (4) being step 1.3. ∎