Bott–Samelson bimodules carry delta and nabla support filtrations

We prove flag preservation and the multiplicity recursion together. For the empty word $R$ has its single graph flag. Shifts preserve supports. The generator tensor functors are exact by [F2], and [F1] gives the two graph sequences for each $B_s{R}_y$; their graph indices are $y,sy$. Each sequence splits on the right, since its quotient is right-free. The issue for a longer flag is placing the two graphs in the required length order, which we resolve next.

Refine a $\mathrm{\Delta }$ length flag at each integer length $i$ into the left-descent indices $x$ with $\mathrm{\ell }(x)=i$, $sx<x$, and the left-ascent indices of that length. Group the descent part at length $i$ with the ascent part at length $i-1$ into one block. The length layers are sums of graphs, so these refinements exist. Distinct graph indices of the same length do not differ by a reflection, by length parity. If a descent index $x$ of length $i$ and ascent index $y$ of length $i-1$ differ by a reflection, strong exchange gives $y$ by one deletion in a reduced expression $x=s\underset{‾}{v}$. If the first letter is deleted, $y=sx$; if a later letter is deleted, the remaining expression begins with $s$ and has exactly $i-1=\mathrm{\ell }(y)$ letters, so it is reduced and $sy<y$, contradicting the ascent condition. Thus in this block the only potentially nonsplit extensions are between $R_x$ and $R_{sx}$. The cocycle calculation of [F4], applied to matrices of graph generators, then decomposes the block into the independent two-graph pairs after restricting its left action to $R^s$.

Here is why the restriction in the last sentence splits even the paired extensions. After twisting the graph action, the two left actions on the anti-invariant coordinate ${\alpha }_s$ are $a$ and $-a$ for a nonzero right-linear coordinate $a$, and the extension matrix is $\left(\begin{array}{cc}a& h\\ 0& -a\end{array}\right)$ in a right-free splitting. Its square is $a^{2}I$, with zero off-diagonal entry. All left coordinates fixed by $s$ already have identical diagonal action and zero off-diagonal obstruction in the graph-extension cocycle of [F4]. These coordinates together with ${\alpha }_s^2$ generate $R^s$. Thus every element of $R^s$ acts by the common scalar on the paired extension, and any right-linear splitting is also left-$R^s$-linear. With multiple copies the same matrix calculation applies with $h$ a rectangular matrix. Consequently every restricted block is the direct sum of its graph layers, retaining their homogeneous degrees.

Tensor a restricted block back with $R{\otimes }_{R^s}(-)(1)$. Each graph generator of degree $a$ now contributes the two rank-one quotients of [F1], with generator degrees $a+1$ on the descent graph $x$ and $a-1$ on the ascent graph $sx$. This follows directly from the two rank-one sequences: the high-length graph is the submodule in the $\mathrm{\Delta }$ order, and its generator is raised by one; the low-length quotient is lowered by one. Both restricted indices $x,sx$ give this same pair of degrees because their restricted actions coincide. Arranging the resulting blocks in decreasing length therefore produces a genuine $\mathrm{\Delta }$ flag. If $x>sx$, converting generator degrees to normalized $\mathrm{\Delta }$ indices gives $(B_{s}M:{\mathrm{\Delta }}_x(d))=(M:{\mathrm{\Delta }}_x(d+1))+(M:{\mathrm{\Delta }}_{sx}(d))$ and $(B_{s}M:{\mathrm{\Delta }}_{sx}(d))=(M:{\mathrm{\Delta }}_x(d))+(M:{\mathrm{\Delta }}_{sx}(d-1))$. This also proves preservation for an arbitrary $\mathrm{\Delta }$-flagged $M$.

For a $\mathrm{\nabla }$ flag use the same half-length blocks in increasing length order. The reflection test of step 1.2 and restriction splitting of step 2.1 remain identical. The low graph is now the submodule in the appropriate rank-one sequence and has generator degree $a+1$; the high graph quotient has degree $a-1$. Since a ${\mathrm{\nabla }}_y(d)$ generator has degree $-\mathrm{\ell }(y)-d$, this gives $(B_{s}M:{\mathrm{\nabla }}_x(d))=(M:{\mathrm{\nabla }}_x(d-1))+(M:{\mathrm{\nabla }}_{sx}(d))$ and $(B_{s}M:{\mathrm{\nabla }}_{sx}(d))=(M:{\mathrm{\nabla }}_x(d))+(M:{\mathrm{\nabla }}_{sx}(d+1))$. This proves $\mathrm{\nabla }$ preservation independently, rather than assuming that taking opposites exchanges the two flag categories.

By [F5] taking opposites reverses tensor order and preserves each length flag category separately. Hence left closure from steps 3.1–3.2 implies right closure. Iterating left closure from the empty word proves both flags for every word. At each tensoring, every graph occurrence branches into $y$ and $sy$ exactly once, with generator degree changing by $\pm 1$. Applying letters from right to left yields exactly the subexpressions in their original order. The normalized indices follow the two recursions just proved; in the one-letter case [F1] gives exactly the two pairs in the Statement.

Finite freeness and rank $2^r$ follow from [F2]. Every graph in either flag is a subexpression graph by step 4.1, so its support is contained in $Gr(S_n)$. The closure statements, shifts and multiplicities are established in steps 3.1–4.1, completing all the assertions without an imported closure theorem.