Bott–Samelson bimodules carry delta and nabla support filtrations

Write ${\theta }_s^0:=R{\otimes }_{R^s}(-)$ for the unshifted balanced tensor functor; Soergel's shifted reflection functor, whose recursions [F4] records, is ${\theta }_s=R[1]{\otimes }_{R^s}(-)$. Since $B_s=R{\otimes }_{R^s}R(1)=R{\otimes }_{R^s}R\{-1\}$ and shifts may be moved across a balanced tensor product, $B_s{\otimes }_{R}N\cong {\theta }_s^0(N)\{-1\}\cong {\theta }_s(N)$ for every graded bimodule $N$, and associativity of ${\otimes }_R$ with $B_{(i_2,\dots ,i_r)}$ gives $B_{\underset{‾}{i}}\cong {\theta }_{i_1}^0\left({\theta }_{i_2}^0\right(\cdots {\theta }_{i_r}^0(R)\cdots \left)\right)\{-r\};$ the empty word gives $B_{\mathrm{\varnothing }}=R=R_e$, which is its own $\mathrm{\Delta }$-flag and $\mathrm{\nabla }$-flag, and an internal shift $M\mapsto M\{k\}$ preserves the flag categories because it changes only internal degrees and not supports.

One-step calculus: a standard bimodule $R_x$ is free of rank one as a right $R$-module, so the functor $R_x\{a\}{\otimes }_R-$ is exact, and applying it to the two short exact sequences of [F1], whose total space is $B_s$, and identifying the outer terms by [F3] gives the two exact sequences $0\to R_{xs}\{a+1\}\to R_x\{a\}{\otimes }_R{B}_s\to R_x\{a-1\}\to 0,$ $0\to R_x\{a+1\}\to R_x\{a\}{\otimes }_R{B}_s\to R_{xs}\{a-1\}\to 0,$ whose outer terms are standard bimodules with graph indices $x$ and $xs$ and shifts differing by one; the first sequence exhibits the $xs$-graph as a sub-bimodule and the second exhibits the $x$-graph as a sub-bimodule of the same total space, so $R_x\{a\}{\otimes }_R{B}_s$ has a two-term flag in either order. Taking opposites with [F5] turns these two sequences into the corresponding two sequences for ${\theta }_s(R_x\{a\})=B_s{\otimes }_R{R}_x\{a\}$, whose outer terms have the graph indices $x$ and $sx$ and again differ by one in shift.

Membership: by [F4] the functor $B_s{\otimes }_R-$ carries $F_{\mathrm{\Delta }}$ into $F_{\mathrm{\Delta }}$ and $F_{\mathrm{\nabla }}$ into $F_{\mathrm{\nabla }}$, and by step 1.1 it differs from the unshifted balanced tensor functor ${\theta }_s^0=R{\otimes }_{R^s}-$ by the internal shift $\{-1\}$, which preserves both categories because it changes only degrees, so ${\theta }_s^0$ carries each flag category into itself. Induction on the number of letters in the normal form of step 1.1 therefore gives $B_{\underset{‾}{i}}\in F_{\mathrm{\Delta }}\cap F_{\mathrm{\nabla }}$ together with a $\mathrm{\Delta }$-flag and a $\mathrm{\nabla }$-flag; the empty word is the single layer $R_e=R$ of step 1.1.

Quotient and shift bookkeeping: by the recursions of [F4], read in the direction of the contribution of an old quotient, a flag quotient of $M$ with graph $y$ and index $e$ contributes to the $\mathrm{\Delta }$-flag of $B_s{\otimes }_{R}M$ the two quotients with graph $sy$ and index $e$, and with graph $y$ and index $e-1$ when $\mathrm{\ell }(y)>\mathrm{\ell }(sy)$ respectively index $e+1$ when $\mathrm{\ell }(sy)>\mathrm{\ell }(y)$; in the $\mathrm{\nabla }$-chart the same pair of recursions holds with the two signs exchanged, which is the dual pair of [F4]. Iterating this along the $r$ letters of $\underset{‾}{i}$, each flag quotient of $B_{\underset{‾}{i}}$ arises from a choice, at every letter $s_{i_j}$, of keeping the graph or multiplying it by $s_{i_j}$ on the left, so its graph is the subexpression product $s_{i_{j_1}}\cdots s_{i_{j_m}}$ of the $m$ kept letters, and its normalized index is unchanged at each graph-changing branch and changes by $\pm 1$ only at a graph-preserving branch, with the sign specified above. The recursion is applied from the rightmost letter to the leftmost letter, so the final graph is the subexpression product in its original order. For the unnormalized generator degree $a=\mathrm{\ell }(y)-e$ in the $\mathrm{\Delta }$-chart or $a=-\mathrm{\ell }(y)-e$ in the $\mathrm{\nabla }$-chart, each branch changes $a$ by $\pm 1$, since $\mathrm{\ell }(sy)-\mathrm{\ell }(y)=\pm 1$; the $\mathrm{\nabla }$-chart is governed by the dual pair of recursions of [F4]. The two branches at a letter give different graphs, since $sx\mathrm{\ne }x$ for a simple reflection $s$, so the number of occurrences of a graph $x$ equals the number of those subexpressions of $\underset{‾}{i}$ whose product is $x$; for the one-letter word the two charts give the two displayed pairs of the statement.

Right tensors: let $M\in F_{\mathrm{\Delta }}$. By [F5] $M^{\mathrm{o}\mathrm{p}}\in F_{\mathrm{\Delta }}$, so $B_s{\otimes }_R{M}^{\mathrm{o}\mathrm{p}}\in F_{\mathrm{\Delta }}$ by [F4]; applying [F5] again, $M{\otimes }_R{B}_s\cong (B_s{\otimes }_R{M}^{\mathrm{o}\mathrm{p}}{)}^{\mathrm{o}\mathrm{p}}$ lies in $F_{\mathrm{\Delta }}$, since the opposite functor preserves the $\mathrm{\Delta }$-flag category. The same argument with the roles of $\mathrm{\Delta }$ and $\mathrm{\nabla }$ exchanged, using that the opposite functor preserves the $\mathrm{\nabla }$-flag category as well, gives the closure clause for $F_{\mathrm{\nabla }}$.

Freeness and rank: [F2] gives that $B_{\underset{‾}{i}}$ is finite free of rank $2^r$ on each side, and each layer $R_x\{a\}$ is free of rank one on each side; the support of $B_{\underset{‾}{i}}$, a finite union of graphs, is contained in $Gr(S_n)$ because every flag quotient is a standard bimodule of a permutation, and the internal shifts change only degrees. ∎