Existence and uniqueness of the parabolic Kazhdan–Lusztig bases

The coefficient matrices. Define $r_{x,y}\in A$ for $x,y\in {}^{J}W$ by $\overline{m_y}={\sum }_{x\le y}{r}_{x,y}{m}_x$, so $r_{x,y}=a_{x,y}$ for $x<y$, $r_{y,y}=1$, and $r_{x,y}=0$ when $x\le ̸y$; define $s_{x,y}$ analogously from $\overline{n_y}={\sum }_{x\le y}{s}_{x,y}{n}_x$. Applying bar to $\overline{m_y}={\sum }_{x\le y}{r}_{x,y}{m}_x$ and using $\overline{\stackrel{‾}{m_y}}=m_y$ gives $m_y={\sum }_{x\le y}\overline{r_{x,y}}\overline{m_x}={\sum }_{x^{\mathrm{\prime }}\le y}\left({\sum }_{x:x^{\mathrm{\prime }}\le x\le y}\overline{r_{x,y}}{r}_{x^{\mathrm{\prime }},x}\right)m_{x^{\mathrm{\prime }}}$; by [F1] the coefficient of $m_{x^{\mathrm{\prime }}}$ is ${\delta }_{x^{\mathrm{\prime }},y}$, that is, $R\bar{R}=I$ for the unitriangular matrix $R=(r_{x,y})$. A finite unitriangular matrix is invertible, so also $\bar{R}R=I$; the same identities hold for $S=(s_{x,y})$.

Construction of the coefficients. Fix $w\in {}^{J}W$ and increase $d=\mathrm{\ell }(w)-\mathrm{\ell }(x)$ over the finite lower ideal $I_w=\{x\in {}^{J}W:x\le w\}$ of [F3], starting with $c_{w,w}=1$. Suppose $x<w$ and $c_{y,w}$ has been constructed for every $y\in I_w$ with $x<y\le w$, satisfying $c_{y,w}={\sum }_{y\le z\le w}{r}_{y,z}\overline{c_{z,w}}$; note that all cited labels lie in $I_w$. Put $a_x:={\sum }_{x<y\le w}{r}_{x,y}\overline{c_{y,w}}$. Then $\overline{a_x}={\sum }_{x<y\le w}\overline{r_{x,y}}{c}_{y,w}={\sum }_{x<y\le z\le w}\overline{r_{x,y}}{r}_{y,z}\overline{c_{z,w}}=-{\sum }_{x<z\le w}{r}_{x,z}\overline{c_{z,w}}=-a_x$, where the middle equality substitutes the induction equations and the third uses $\bar{R}R=I$ from step 1.1: the sum over $x\le y\le z$ is zero for $x<z$, and its omitted diagonal term is $\overline{r_{x,x}}{r}_{x,z}=r_{x,z}$. Write $a_x={\sum }_{m\in \mathbb{Z}}{\gamma }_m{v}^m$; anti-invariance under $v\mapsto v^{-1}$ gives ${\gamma }_{-m}=-{\gamma }_m$, hence ${\gamma }_0=0$. Define $c_{x,w}:={\sum }_{m>0}{\gamma }_m{v}^m$, so $c_{x,w}\in v\mathbb{Z}[v]$ and $c_{x,w}-\overline{c_{x,w}}=a_x$. The induction is finite and uses no choice principle.

Bar invariance and the coefficient equations. Set $C_w:={\sum }_{y\in I_w}{c}_{y,w}{m}_y$. The coefficient of $m_x$ in $\overline{C_w}$ is ${\sum }_{x\le y\le w}{r}_{x,y}\overline{c_{y,w}}$. For $x=w$ this is $\overline{c_{w,w}}=1=c_{w,w}$; for $x<w$ it is $\overline{c_{x,w}}+a_x=c_{x,w}$ by step 2.1, and for $x\mathrm{\notin }{I}_w$ the support of $r_{x,y}$ makes the coefficient zero ($x\le y\le w$ would put $x\in I_w$). Hence $\overline{C_w}=C_w$ and the coefficient equation $c_{x,w}={\sum }_{x\le y\le w}{r}_{x,y}\overline{c_{y,w}}$ holds for every $x\le w\in {}^{J}W$; this proves the existence, the bar invariance and the triangular support $C_w\in m_w+{\sum }_{y<w}v\mathbb{Z}[v]m_y$ of (i),(ii). Repeating the identical construction with the matrix $S$ in place of $R$ produces $D_w\in n_w+{\sum }_{y<w}v\mathbb{Z}[v]n_y$ with $\overline{D_w}=D_w$.

Uniqueness. Suppose $h={\sum }_{y<w}{c}_y{m}_y\in M^J$ is bar-fixed with every $c_y\in v\mathbb{Z}[v]$. If $h\mathrm{\ne }0$, choose a Bruhat-maximal $x$ in its finite support $I_w\setminus \{w\}$. The coefficient of $m_x$ in $\bar{h}$ is $\overline{c_x}$: no $y>x$ is supported, and $r_{x,y}\mathrm{\ne }0$ requires $x\le y$. Since $h=\bar{h}$ and the $m_y$ are independent, $c_x=\overline{c_x}$; but bar maps $v\mathbb{Z}[v]$ into $v^{-1}\mathbb{Z}[v^{-1}]$, so the common value lies in $v\mathbb{Z}[v]\cap v^{-1}\mathbb{Z}[v^{-1}]=\{0\}$ by [F4], a contradiction. Hence $h=0$: the bar-fixed element with the prescribed triangular support is unique, and likewise in $N^J$.

Bases. The transition from $(m_x{)}_{x\in {}^{J}W}$ to $(C_x{)}_{x\in {}^{J}W}$ is unitriangular over the finite Bruhat order (each $C_x=m_x+{\sum }_{y<x}{c}_{y,x}{m}_y$), so it is invertible over $A$; hence $\{C_x\}$ is an $A$-basis of $M^J$. The same holds for $\{D_x\}$ in $N^J$. This proves the theorem for both modules.