The $R$-coefficient recursion, support, degree bounds and inversion

Descent recursion. Put $\alpha :=v-v^{-1}$. If $sx>x$, concatenating reduced words gives $H_s{H}_x=H_{sx}$. If $sx<x$, then $x=s(sx)$ is reduced and $H_s{H}_x=H_s^2{H}_{sx}=H_{sx}+(v^{-1}-v)H_x$ by the quadratic relation. For a left descent $sw<w$, write $w=s(sw)$ and apply bar to its reduced product: $\overline{H_w}=(H_s+\alpha ){\sum }_x{r}_{x,sw}{H}_x$. In the expansion, the coefficient correction $(v^{-1}-v)r_{x,sw}$ from the $sx<x$ terms cancels the $\alpha r_{x,sw}$ term; reindexing the remaining $H_{sx}$ terms gives $\overline{H_w}={\sum }_y{r}_{sy,sw}{H}_y+\alpha {\sum }_{sy>y}{r}_{y,sw}{H}_y$. Comparing coefficients in the standard basis proves (a).

Inverse-index symmetry. Apply $\mathrm{♭}$ to $\overline{H_w}={\sum }_y{r}_{y,w}{H}_y$. Since $\mathrm{♭}$ is $A$-linear, the result is ${\sum }_y{r}_{y,w}{H}_{y^{-1}}$; because $\mathrm{♭}$ commutes with bar and $\mathrm{♭}(H_w)=H_{w^{-1}}$, it is also $\overline{H_{w^{-1}}}={\sum }_x{r}_{x,w^{-1}}{H}_x$. Comparing coefficients at $H_{y^{-1}}$ gives $r_{y,w}=r_{y^{-1},w^{-1}}$.

Support and diagonal. Induct on $\mathrm{\ell }(w)$, with $r_{\mathrm{i}\mathrm{d},\mathrm{i}\mathrm{d}}=1$ and all other coefficients in the identity column zero. If $sy<y$ and $r_{y,w}\mathrm{\ne }0$, (a) gives $r_{sy,sw}\mathrm{\ne }0$, so induction gives $sy\le sw$; take a reduced expression for $sw$ and a reduced subword for $sy$. Prefixing $s$ gives a subword for $s(sy)=y$ in the reduced expression $w=s(sw)$; it is reduced because its length is $1+\mathrm{\ell }(sy)=\mathrm{\ell }(y)$. Hence $y\le w$. If $sy>y$ and $r_{y,w}\mathrm{\ne }0$, at least one of $r_{sy,sw}$ and $r_{y,sw}$ is nonzero. In the first case induction gives $sy\le sw$, and $y<sy$ implies $y\le w$; in the second it gives $y\le sw<w$. Thus the support is contained in $[\mathrm{i}\mathrm{d},w]$. Taking $y=w$ in (a) gives $r_{w,w}=r_{sw,sw}=1$ for a left descent, completing the induction (and the $n=1$ case is the identity base).

Bar symmetry. Induct on $\mathrm{\ell }(w)$ using (a), with the identity column as base. If $sy<y$, then $r_{y,w}=r_{sy,sw}$ and $\mathrm{s}\mathrm{g}\mathrm{n}(sy)\mathrm{s}\mathrm{g}\mathrm{n}(sw)=\mathrm{s}\mathrm{g}\mathrm{n}(y)\mathrm{s}\mathrm{g}\mathrm{n}(w)$, so the induction identity for the smaller column proves the first formula in (d). If $sy>y$, set $\epsilon =\mathrm{s}\mathrm{g}\mathrm{n}(y)\mathrm{s}\mathrm{g}\mathrm{n}(w)$; induction gives $\overline{r_{sy,sw}}=\epsilon r_{sy,sw}$ and $\overline{r_{y,sw}}=-\epsilon r_{y,sw}$, while $\bar{\alpha }=-\alpha$. Applying bar to (a) therefore gives $\overline{r_{y,w}}=\epsilon r_{y,w}$.

Degree, parity, and extreme coefficients. Induct on $\mathrm{\ell }(w)$ for $y\le w$ and write $d=\mathrm{\ell }(w)-\mathrm{\ell }(y)$. If $sy<y$, then $sy\le sw$: indeed $sy\le w$ by $sy<y\le w$, and the first lifting implication in [F3] applied to $sy$ gives $sy\le sw$. Now (a) identifies $r_{y,w}=r_{sy,sw}$, whose length difference is $d$; induction gives the asserted parity, range of degrees, and both extreme coefficients because $\mathrm{s}\mathrm{g}\mathrm{n}(sy)\mathrm{s}\mathrm{g}\mathrm{n}(sw)=\mathrm{s}\mathrm{g}\mathrm{n}(y)\mathrm{s}\mathrm{g}\mathrm{n}(w)$. If $sy>y$, the same lifting implication applied to $y\le w$ gives $y\le sw$, so $r_{y,sw}$ has length difference $d-1$. Its product with $\alpha$ has exponents between $-d$ and $d$, all congruent to $d$ modulo $2$; its top term is $v^d$ and its bottom term is $-\mathrm{s}\mathrm{g}\mathrm{n}(y)\mathrm{s}\mathrm{g}\mathrm{n}(sw)v^{-d}=\mathrm{s}\mathrm{g}\mathrm{n}(y)\mathrm{s}\mathrm{g}\mathrm{n}(w)v^{-d}$. The other term $r_{sy,sw}$ in (a) is zero unless $sy\le sw$ by (b), and when nonzero its length difference is $d-2$, so it has the same parity and lies strictly between those two extreme degrees. This proves (c), including the endpoint $d=0$ through the first case.

Matrix inversion. Apply bar to $\overline{H_z}={\sum }_y{r}_{y,z}{H}_y$ and use bar squared equal to the identity to obtain $H_z={\sum }_{x,y}{r}_{x,y}\overline{r_{y,z}}{H}_x$. Standard-basis independence gives ${\sum }_y{r}_{x,y}\overline{r_{y,z}}={\delta }_{x,z}$. Applying coefficient bar to this equality gives $\bar{R}R=1$ as well as $R\bar{R}=1$; all sums are finite because $S_n$ is finite. By (b), coefficients outside Bruhat order vanish, including when the interval is empty. The inductions use the identity permutation as their length-zero base, and all arguments are choice-free.