The Kazhdan–Lusztig polynomial descent recursion

The coefficient equation. Since $sw<w$, we have $s(sw)=w>sw$. By [F2], ${\underset{‾}{H}}_s{\underset{‾}{H}}_{sw}={\underset{‾}{H}}_w+{\sum }_{\begin{array}{c}z\le sw\\ sz<z\end{array}}\mu (z,sw){\underset{‾}{H}}_z\mathrm{.}$ By [F4], ${\underset{‾}{H}}_s$ is supported on $\{x:x\le s\}$, and [F5]'s reduced-subword characterization gives $[\mathrm{i}\mathrm{d},s]=\{\mathrm{i}\mathrm{d},s\}$. The constant-term and degree clauses in [F3] give $P_{\mathrm{i}\mathrm{d},s}=1$ and $p_{\mathrm{i}\mathrm{d},s}=v$, so ${\underset{‾}{H}}_s=H_s+v{H}_{\mathrm{i}\mathrm{d}}$. Expand each ${\underset{‾}{H}}_t={\sum }_{x\le t}{p}_{x,t}{H}_x$ using [F4]. Reduced words and the quadratic relation in [F1] give $H_s{H}_x=H_{sx}$ if $sx>x$; if $sx<x$, then $x=s(sx)$ is reduced and $H_s{H}_x=H_s^2{H}_{sx}=H_{sx}+(v^{-1}-v)H_x$. Thus the coefficient of $H_y$ on the left is $p_{sy,sw}+v^{-1}{p}_{y,sw}$ when $sy<y$, and $p_{sy,sw}+v{p}_{y,sw}$ when $sy>y$. The coefficient on the right is $p_{y,w}+{\sum }_{y\le z\le sw,sz<z}\mu (z,sw)p_{y,z}$. Therefore $p_{y,w}=\{\begin{array}{ll}{p}_{sy,sw}+v^{-1}{p}_{y,sw},& sy<y,\\ p_{sy,sw}+v{p}_{y,sw},& sy>y,\end{array}-{\sum }_{\begin{array}{c}y\le z\le sw\\ sz<z\end{array}}\mu (z,sw)p_{y,z}\mathrm{.}$

Convert to $q$-polynomials. Put $d:=\mathrm{\ell }(w)-\mathrm{\ell }(y)$. By [F5], $d\ge 0$. If $sy<y$, then $\mathrm{\ell }(sw)-\mathrm{\ell }(sy)=d$ and $\mathrm{\ell }(sw)-\mathrm{\ell }(y)=d-1$; after substituting $p_{x,z}=v^{\mathrm{\ell }(z)-\mathrm{\ell }(x)}{P}_{x,z}(v^{-2})$ in step 1.1 and dividing by $v^d$, the first two terms become $P_{sy,sw}(q)+q{P}_{y,sw}(q)$. If $sy>y$, then $\mathrm{\ell }(sw)-\mathrm{\ell }(sy)=d-2$, so they become $q{P}_{sy,sw}(q)+P_{y,sw}(q)$. These identities also hold when an index is outside the relevant Bruhat interval, using the zero convention for $P$. For a sum term, $\mathrm{\ell }(z)-\mathrm{\ell }(y)-d=-(\mathrm{\ell }(w)-\mathrm{\ell }(z))$, giving the factor $q^{(\mathrm{\ell }(w)-\mathrm{\ell }(z))\mathrm{/}2}$. Thus the two cases are the displayed formula with $c=1$ and $c=0$, respectively. If $\mu (z,sw)\mathrm{\ne }0$, then $\mathrm{\ell }(sw)-\mathrm{\ell }(z)$ is odd; since $\mathrm{\ell }(w)=\mathrm{\ell }(sw)+1$ by [F5], the exponent $(\mathrm{\ell }(w)-\mathrm{\ell }(z))\mathrm{/}2$ is an integer.

Right descents. If $ws<w$, inversion preserves Bruhat order by [F5], so $s{w}^{-1}<w^{-1}$ and $y^{-1}\le w^{-1}$. Apply the left formula to $y^{-1},w^{-1},s$. Replace every inverted index using [F4]; lengths and length differences are unchanged by [F5], while $sz<z$ becomes $zs<z$. This gives the right-descent recursion.