Multiplication by a generator in the Kazhdan–Lusztig basis

The left-ascent difference. Put $A_s:={\underset{‾}{H}}_s=H_s+v{H}_{\mathrm{i}\mathrm{d}}$ by [F2, F3]. We prove the formulas by induction on $\mathrm{\ell }(w)$, assuming the descent formula for all smaller upper indices. If $sx>x$, reduced concatenation gives $H_s{H}_x=H_{sx}$; if $sx<x$, write $x=s(sx)$ and use the quadratic relation to get $H_s{H}_x=H_{sx}+(v^{-1}-v)H_x$. Suppose $sw>w$ and set $C_w:={\underset{‾}{H}}_{sw}+{\sum }_{z:sz<z<w}\mu (z,w){\underset{‾}{H}}_z$. For any $x\le w$, fix a reduced expression for $w$ and a reduced subword for $x$. Since $sw>w$, prefixing $s$ gives a reduced expression for $sw$; when $sx>x$, prefixing $s$ to the subword gives a reduced subword for $sx$, and when $sx<x$, $sx<x\le w<sw$. Thus every standard-basis term of $A_s{\underset{‾}{H}}_w$ is indexed by an element $\le sw$. The same holds for $C_w$, since $w<sw$ and every $z<w$ satisfies $z<sw$. Only the leading term $H_w$ can produce $H_{sw}$ in $A_s{\underset{‾}{H}}_w$, with coefficient $1$: for $x<w$, the terms from $H_s{H}_x$ have length at most $\mathrm{\ell }(x)+1\le \mathrm{\ell }(w)<\mathrm{\ell }(sw)$. The leading term of ${\underset{‾}{H}}_{sw}$ gives coefficient $1$ in $C_w$. Thus $D_w:=A_s{\underset{‾}{H}}_w-C_w$ is supported strictly below $sw$. Using [F1], its coefficient at $H_y$ is $f_y=p_{sy,w}+\{\begin{array}{ll}{v}^{-1}{p}_{y,w},& sy<y,\\ v{p}_{y,w},& sy>y,\end{array}-p_{y,sw}-{\sum }_{\begin{array}{c}y\le z<w\\ sz<z\end{array}}\mu (z,w)p_{y,z},$ with $p_{a,b}=0$ when $a\le ̸b$.

Coefficients with $sy<y$. Let $y<sw$ and $sy<y$. The terms $p_{sy,w}$ and $p_{y,sw}$ are each either $0$ or in $v\mathbb{Z}[v]$, except when $sy=w$; that exception forces $y=sw$ and is excluded. If $y\le w$, then in fact $y<w$, since $y=w$ would give $sy=sw>w=y$, contrary to $sy<y$. The sum defining $f_y$ then contains its $z=y$ term $\mu (y,w)$, since $p_{y,y}=1$. By definition of $\mu$, $v^{-1}{p}_{y,w}-\mu (y,w)\in v\mathbb{Z}[v]$; every remaining sum term has $z>y$ and $p_{y,z}\in v\mathbb{Z}[v]$. If $y\le ̸w$, then $p_{y,w}=0$, $\mu (y,w)=0$, and the sum is empty. Thus $f_y\in v\mathbb{Z}[v]$ in both cases.

Coefficients with $sy>y$. For $y=w$, the coefficient is $f_w=v-p_{w,sw}=0$: the simple ascent makes $w<sw$ a Bruhat cover, so the degree bound and constant term of $P_{w,sw}$ give $p_{w,sw}=v$. Suppose $y<sw$ and $y\mathrm{\ne }w$. If $y\le ̸w$, then $sy\le ̸w$ (otherwise $y<sy\le w$), so $p_{sy,w}=p_{y,w}=0$ and the sum is empty; the remaining $p_{y,sw}$ is either zero or in $v\mathbb{Z}[v]$. Now assume $y\le w$. Fix a reduced expression for $w$ and a reduced subword for $y$. Since $sw>w$ and $sy>y$, prefixing $s$ gives a reduced expression for $sw$ and a reduced subword for $sy$, so $sy\le sw$; because $y\mathrm{\ne }w$, $sy<sw$. For each $z<w$ with $sz<z$, induction gives ${\underset{‾}{H}}_s{\underset{‾}{H}}_z=(v+v^{-1}){\underset{‾}{H}}_z$, or $(H_s-v^{-1}){\underset{‾}{H}}_z=0$. Comparing the coefficient of $H_y$ gives $p_{y,z}=v{p}_{sy,z}$. For every such $z$ with $y\le z$, the left-lifting clause in [F4] gives $sy\le z$; conversely $sy\le z$ implies $y<sy\le z$. Thus the sum in $f_y$ becomes $v{\sum }_{sy\le z<w,sz<z}\mu (z,w)p_{sy,z}$, and $f_y=v{f}_{sy}+v{p}_{sy,sw}-p_{y,sw}$. Since $s(sy)=y<sy<sw$, Step 2.1 gives $f_{sy}\in v\mathbb{Z}[v]$; the other two coefficients are also in $v\mathbb{Z}[v]$. Therefore $f_y\in v\mathbb{Z}[v]$.

Conclude the left-ascent formula. Both $A_s{\underset{‾}{H}}_w$ and $C_w$ are bar-invariant, since the $\mu$ coefficients are integers by [F3]. Thus $D_w$ is bar-invariant. By steps 1.1–3.1 it is a sum of lower standard basis elements with coefficients in $v\mathbb{Z}[v]$. Adding $D_w$ to ${\underset{‾}{H}}_{sw}$ would give another bar-invariant element in $H_{sw}+{\sum }_{y<sw}v\mathbb{Z}[v]H_y$; uniqueness in [F2] forces $D_w=0$. This proves the formula when $sw>w$.

Left descents. Suppose $sw<w$ and put $w^{\mathrm{\prime }}:=sw$. The ascent case for $w^{\mathrm{\prime }}$ gives ${\underset{‾}{H}}_s{\underset{‾}{H}}_{w^{\mathrm{\prime }}}={\underset{‾}{H}}_w+{\sum }_{z:sz<z<w^{\mathrm{\prime }}}\mu (z,w^{\mathrm{\prime }}){\underset{‾}{H}}_z$. The quadratic relation gives $(H_s-v^{-1}){\underset{‾}{H}}_s=0$. Apply $H_s-v^{-1}$ to this equality. Every $z$ in the sum has $z<w^{\mathrm{\prime }}<w$, so induction gives $(H_s-v^{-1}){\underset{‾}{H}}_z=0$. Thus $(H_s-v^{-1}){\underset{‾}{H}}_w=0$, and ${\underset{‾}{H}}_s{\underset{‾}{H}}_w=(H_s+v){\underset{‾}{H}}_w=(v+v^{-1}){\underset{‾}{H}}_w$. This also covers ${\underset{‾}{H}}_s^2$; the ascent sum is empty for $w=\mathrm{i}\mathrm{d}$ and for the stated length-at-most-one case.

Right multiplication. Apply the left formulas to $w^{-1}$ and then apply $\mathrm{♭}$. By [F5] it reverses the product and fixes ${\underset{‾}{H}}_s$; by inverse-index symmetry in [F2], it sends ${\underset{‾}{H}}_{w^{-1}}$ to ${\underset{‾}{H}}_w$. Bruhat inversion sends $sz<z<w^{-1}$ to $z{s}^{-1}<z^{-1}<w$, and $s^{-1}=s$; the coefficient is unchanged because $p_{z^{-1},w^{-1}}=p_{z,w}$ and $\mu$ is the coefficient of $v$ in that coefficient. This yields the asserted right formulas. The proof uses only finite Bruhat intervals and no choice principle.