The ring $\mathbb{Z}{\left[P\right]}^W$

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 27 July 2012

The ring $\mathbb{Z}{\left[P\right]}^W$

$\mathbb{Z}{\left[P\right]}^W$ is a polynomial ring, i.e. there are algebraically independent elements $e_1,...,e_n\in \mathbb{Z}{\left[P\right]}^W$ such that $$\mathbb{Z}{\left[P\right]}^W=\mathbb{Z}[e_1,...,e_n].$$

		Proof.
	Let $e_1,...,e_n\in 𝕂{\left[P\right]}^W$ be such that $$e_i=x^{{\omega }_i}+\left(lower\; terms\; in\; dominance\; order\right).$$ If $\lambda =l_1{\omega }_1+\cdots +l_n{\omega }_n\in P^{+}$ then $$e_1^{l_1}{e}_2^{l_2}\cdots e_n^{l_n}=x^{\lambda }+\left(lower\; terms\; in\; dominance\; order\right).$$ Thus $\{e_1^{l_1}\cdots e_n^{l_n}|l_1,...,l_n\in {\mathbb{Z}}_{\ge 0}\}$ is a basis of $𝕂{\left[P\right]}^W$ and the result follows. $\square$

The Shi arrangement ${𝒜}^{-}$ is the arrangement of (affine) hyperplanes given by $$\begin{array}{ll}{𝒜}^{-}=\{H_{\alpha },H_{\alpha -\delta }|\alpha \in R^{+}\}where\begin{array}{rcl}{H}_{\alpha }& =& \{x\in {ℝ}^n|⟨x,{\alpha }^{\vee }⟩=0\},\\ H_{\alpha -\delta }& =& \{x\in {ℝ}^n|⟨x,{\alpha }^{\vee }⟩=-1\}.\end{array}& (ZPW\; 1)\end{array}$$ Consider the partition of ${𝔥}_{ℝ}^{*}$ determined by the Shi arrangement. Each chamber $w^{-1}C,$ $w\in W,$ contains a unique region of ${𝒜}^{-}$ which is a cone, and the vertex of this cone is the point $$\begin{array}{ll}{\lambda }_w=w^{-1}\left(\sum _{s_{i}w<w}{\omega }_i\right).& (ZPW\; 2)\end{array}$$ $$\begin{array}{c}\begin{array}{c} \[ H_{{\alpha }_1+{\alpha }_2} \] \[ H_{{\alpha }_1} \] \[ H_{{\alpha }_2} \] \[ H_{{\alpha }_1+2{\alpha }_2} \] \[ H_{{\alpha }_1+{\alpha }_2-\delta } \] \[ H_{{\alpha }_1-\delta } \] \[ H_{{\alpha }_1-\delta } \] \[ H_{{\alpha }_2-\delta } \] \[ H_{{\alpha }_1+2{\alpha }_2-\delta } \] \[ C \] \[ s_{1}C \] \[ s_{2}C \] \[ s_1{s}_{2}C \] \[ s_2{s}_{1}C \] \[ s_1{s}_2{s}_{1}C \] \[ s_2{s}_1{s}_{2}C \] \[ {\lambda }_1 \] \[ {\lambda }_{s_1} \] \[ {\lambda }_{s_2{s}_1} \] \[ {\lambda }_{s_2} \] \[ {\lambda }_{s_1{s}_2} \] \[ {\lambda }_{s_1{s}_2{s}_1} \] \[ {\lambda }_{s_2{s}_1{s}_2} \] \[ {\lambda }_{s_1{s}_2{s}_1{s}_2} \]\end{array}\\ The\; arrangement{𝒜}^{-}\end{array}$$

$\mathbb{Z}\left[P\right]$ is a free $\mathbb{Z}{\left[P\right]}^W$ module of rank $\left|W\right|$ with basis $\left\{x^{{\lambda }_w}\right|w\in W\}.$

		Proof.
	The proof is accomplished by establishing three facts: Let $f_y, y\in W,$ be a family of elements of $\mathbb{Z}\left[P\right].$ Then $\det \left(z{f}_y\right)$ is divisible by $\prod _{\alpha \in R^{+}}{(1-x^{-\alpha })}^{\frac{\left|W\right|}{2}}.$ $\det {\left(z{x}^{{\lambda }_y}\right)}_{z,y\in W}={\left(\prod _{\alpha >0}{x}^{\rho }(1-x^{-\alpha })\right)}^{\frac{\left|W\right|}{2}}.$ If $f\in \mathbb{Z}\left[P\right]$ then there is a unique solution to the equation $$\sum _{w\in W}{a}_w{x}^{{\lambda }_w}=f,with{a}_w\in \mathbb{Z}{\left[P\right]}^W.$$ For each $\alpha \in R^{+}$ subtract row $z{f}_y$ from row $s_{\alpha }z{f}_y$. Then this row is divisible by $(1-x^{-\alpha }).$ Since there are $\frac{\left|W\right|}{2}$ pairs of rows $(z{f}_y,s_{\alpha }z{f}_y)$ the whole determinant is divisible by ${(1-x^{-\alpha })}^{\frac{\left|W\right|}{2}}.$ For $\alpha ,\beta \in R^{+}$ the factors $(1-x^{-\alpha })$ and $(1-X^{-\beta })$ are coprime, and so $\det \left(z{f}_y\right)$ is divisible by $\prod _{\alpha \in R^{+}}{(1-x^{-\alpha })}^{\frac{\left|W\right|}{2}}.$ Since $y{\lambda }_y$ is dominant, $y{\lambda }_y\ge zy{\lambda }_y.$ So all the entries in the $y^{\mathrm{th}}$ column are (weakly) less than the entry on the diagonal. If $y{\lambda }_y=zy{\lambda }_y$ then $z$ is in the stabilizer of $$\sum _{{\alpha }_i\in R\left(y\right)}{\omega }_i.$$ Thus $l\left(y^{-1}\right)=l\left(z^{-1}\right)+l\left(z{y}^{-1}\right).$ If $l\left(y\right)\le l\left(z\right)$ then $l\left(z{y}^{-1}\right)=0$ and so $z=y.$ Thus, if the rows are ordered so that the $y^{\mathrm{th}}$ row is above the $z^{\mathrm{th}}$ row when $l\left(y\right)\le l\left(z\right)$ then all terms above the diagonal are strictly less than the diagonal entry. Thus the top coefficient of $\det \left(z{x}^{{\lambda }_y}\right)$ is equal to $$\prod _{z\in W}z{x}^{{\lambda }_z}=\prod _{z\in W}\prod _{\genfrac{}{}{0ex}{}{i}{s_{i}z<z}}{x}^{{\omega }_i}=\prod _{i=1}^n{x}^{\frac{\left|W\right|}{2}{\omega }_i}={\left(x^{\rho }\right)}^{\frac{\left|W\right|}{2}}.$$ Since $s_i\det \left(z{x}^{{\lambda }_y}\right)=-\det \left(z{x}^{{\lambda }_y}\right)$ the lowest term of $\det \left(z{x}^{{\lambda }_y}\right)$ is $w_0{\left(x^{\rho }\right)}^{\frac{\left|W\right|}{2}}={\left(x^{\rho }\right)}^{-\frac{\left|W\right|}{2}}.$ These are the same as the highest and lowest terms of $\left(x^{\rho }\prod _{\alpha \in R^{+}}(1-x^{-\alpha })\right)$ and so (2) follows from (1). Assume that $a_y\in \mathbb{Z}{\left[P\right]}^W$ are solutions of the equation $\sum _{y\in W}{x}^{{\lambda }_y}{a}_y=f.$ Act on this equation by the elements of $W$ to obtain the system of $\left|W\right|$ equations $$\sum _{y\in W}\left(z{x}^{{\lambda }_y}\right)a_y=zf,z\in W.$$ By (1) the matrix ${\left(z{x}^{{\lambda }_y}\right)}_{z,y\in W}$ is invertible and so this system has a unique solution with $a_y\in \mathbb{Z}{\left[P\right]}^W.$ Cramer's rule provides an expression for $a_y$ as a quotient of two determinants. By (1) and (2) the denominator divides the numerator to give an element of $\mathbb{Z}\left[P\right].$ Since each determinant is an alternating function (an element of Fock space), the quotient is an element of $\mathbb{Z}{\left[P\right]}^W.$ $\square$

In [Sb2] Steinberg proves this type of result in full generality without the assumptions that $W$ acts irreducibly on ${𝔥}_{ℝ}^{*}$ and $L=P.$ Note also that the proof given above is sketchy, particularly in the aspect that the top coefficient of the determinant is what we have claimed it is. See [Sb2] for a proper treatment of this point.

From Verma at Magdeburg meeting August 1998

$R\left(T\right)=ℂ\left[P\right]$ ($P$ the weight lattice) is $R\left(G\right)=R{\left(T\right)}^W=ℂ{\left[P\right]}^W$ free with $\left\{e^{\epsilon \left(w\right)}\right|w\in W\}$ as a basis, where $$\epsilon \left(w\right)=w^{-1}\left({\rho }_w\right)where{\rho }_w=\sum _{i\in D\left(w\right)}{w}_i,D\left(w\right)=\left\{i\right|s_{i}w<w\}.$$ There is an explicit expansion formula $$e^{\lambda }=\sum _{w\in W}{\chi }_{\lambda ,w}{e}^{\epsilon \left(w\right)}.$$

		Proof.
	$$e^{\lambda +\mu }=\sum _{w\in W}{\chi }_{\lambda ,w}{e}^{\mu +\epsilon \left(w\right)}for\; all\mu \in P.$$ We shall use this for a set of $\left|W\right|$ values $\mu$ below. Recall that $$\begin{array}{rrcl}{\Delta }_{w_0}:& R\left(T\right)& \to & R{\left(T\right)}^W\\ & e^{\lambda }& \mapsto & \mathrm{ch}\left(\lambda \right)\end{array}the\; Reynolds/Demazure\; operator.$$ Then $${\Delta }_{w_0}\left(e^{\lambda +\mu }\right)=\sum _{w\in W}{\chi }_{\lambda ,w}{\Delta }_{w_0}\left(e^{\mu +\epsilon \left(w\right)}\right).$$ So let $$\mu \in \{-\epsilon \left(y{w}_0\right)-\rho |y\in W\}$$ and we get a $\left|W\right|\times \left|W\right|$ matrix $${\left(\mathrm{ch}\left(e^{\epsilon \left(w\right)-\epsilon \left(y{w}_0\right)-\rho }\right)\right)}_{y,w\in W}$$ which Hulsurkar proved to be unipotent. The diagonal entries are $\mathrm{sgn}\left(w\right).$ By definition $$\begin{array}{rcccl}{\rho }_{w_{0}w}& =& \rho -{\rho }_w& & and\; so\\ w^{-1}{\rho }_{w_{0}w}& =& w^{-1}\rho -w^{-1}{\rho }_w& & \\ w^{-1}{w}_0^{-1}{w}_0{\rho }_{w_{0}w}& =& -w^{-1}{w}_0^{-1}{\rho }_{w_{0}w}& =& -\epsilon \left(w_{0}w\right).\end{array}$$ So $$-\epsilon \left(w_{0}w\right)=w^{-1}\rho -\epsilon \left(w\right).$$ So $$\epsilon \left(w\right)-\epsilon \left(w_{0}w\right)=w^{-1}\rho .$$ But one can look at these elements as follows: Remove the strips $$\begin{array}{r}\{x\in {ℝ}^n|⟨x,{\alpha }^{\vee }⟩<1\}\\ \[ H_{{\alpha }_1} \] \[ H_{{\alpha }_2} \] \[ H_{\beta } \] \[ 0 \]\end{array}$$ The endpoints of the resulting cones in each chamber are the $\epsilon \left(w\right)$ (up to a possible multiplication by $w$ or $w^{-1}$). $\square$

This is not at all unrelated to

1. Steinberg, "On a theorem of Pittie", Topology 14 (1975), 173-177.

There he defines $$e_w=w^{-1}{\lambda }_w=w^{-1}\prod _{i\in D\left(w\right)}{e}^{w_i}whereD\left(w\right)=\left\{i\right|s_{i}w<w\}.$$

$$\det {\left(u{e}_v\right)}_{u,v\in W}=\prod _{\alpha >0}{(e^{\frac{\alpha }{2}}-e^{-\frac{\alpha }{2}})}^{n_{\alpha }}$$ where $$n_{{\alpha }_i}=\left|\{v\in W|v^{-1}{\alpha }_i<0\}\right|$$ and $$n_{w{\alpha }_i}=n_{{\alpha }_i}.$$

Let $f_v\in \mathbb{Z}\left[X\right].$ Then $$\det \left(u{f}_v\right)is\; divisible\; by\prod _{\alpha >0}{(e^{\frac{\alpha }{2}}-e^{-\frac{\alpha }{2}})}^{n_{\alpha }}.$$

If $f\in \mathbb{Z}\left[X\right]$ then there is a unique solution to $$\sum a_w{e}_w=fwith{a}_w\in \mathbb{Z}{\left[X\right]}^W.$$ Here $X$ is the weight lattice.

		Proof of Theorem 2.3.
	Subtract row $u{f}_v$ from row $s_{\alpha }u{f}_v.$ Then this row is divisible by $1-e^{-\alpha }.$ So $\det \left(u{f}_v\right)$ is divisible by $\prod _{\alpha }{(1-e^{-\alpha })}^{n_{\alpha }}.$ $\square$

		Proof of Theorem 2.2.
	The top coefficient of $\det \left(u{e}_v\right)$ is $$\prod _{v\in W}v{e}_v=\prod _{v\in W}{\lambda }_v=\prod _{v\in W}\prod _{i\in D\left(v\right)}{e}^{w_i}=\prod _{i=1}^n{e}^{w_i{n}_i}.$$ The top coefficient of $\prod _{\alpha }{(1-e^{\alpha })}^{n_{\alpha }}$ is $e^{\sum _{\alpha >0}{n}_{\alpha }\alpha }.$ Now $$s_i\left(\sum _{\alpha >0}{n}_{\alpha }\alpha \right)=\sum _{\alpha >0}{n}_{\alpha }\alpha -2n_{{\alpha }_i}{\alpha }_i.$$ So $$(\sum _{\alpha >0}{n}_{\alpha }\alpha ,{\alpha }_i^{\vee })=2n_{{\alpha }_i}=2n_i.$$ So $$\sum _{\alpha >0}{n}_{\alpha }\alpha =\sum _{i=1}^{n}2n_i{w}_i.$$ $\square$

		Proof of Theorem 2.4.
	The system $\sum \left(u{e}_v\right)a_v=uf$ has a unique solution with $a_v\in \mathbb{Z}{\left[X\right]}^W.$ So $\sum e_v{a}_v=f$ has a unique solution. $\square$

Example for $S_2$. $$W=\{1,s_1\}{z}_1=x_1$$ $$\begin{array}{rclcrcl}{z}_1& =& 1& & e_1& =& 1\\ z_{s_1}& =& z_1& & e_{s_1}& =& x_2\end{array}D=\det \left(\begin{array}{cc}1& x_2\\ 1& x_1\end{array}\right)=x_1-x_2$$ and $${\left(\begin{array}{cc}1& x_2\\ 1& x_1\end{array}\right)}^{-1}=\frac{1}{x_1-x_2}\left(\begin{array}{cc}{x}_1& -x_2\\ -1& 1\end{array}\right).$$

Example for $S_3$. $$W=\{1,s_1,s_2,s_1{s}_2,s_2{s}_1,s_1{s}_2{s}_1\}\begin{array}{rcl}{z}_1& =& x_1\\ z_2& =& x_1{x}_2\end{array}$$ $$\begin{array}{rclcrcl}{z}_1& =& 1& & e_1& =& 1\\ z_{s_1}& =& z_1=x_1& & e_{s_1}& =& x_2\\ z_{s_2}& =& z_2=x_1{x}_2& & e_{s_2}& =& x_1{x}_3\\ z_{s_1{s}_2}& =& z_1=x_1& & e_{s_1{s}_2}& =& x_3\\ z_{s_2{s}_1}& =& z_2=x_1{x}_2& & e_{s_2{s}_1}& =& x_2{x}_3\\ z_{s_1{s}_2{s}_1}& =& z_1{z}_2=x_1^2{x}_2& & e_{s_1{s}_2{s}_1}& =& x_2{x}_3^2\end{array}$$ $$D=\left(\begin{array}{cccccc}1& x_2& x_1{x}_3& x_3& x_2{x}_3& x_2{x}_3^2\\ 1& x_1& x_2{x}_3& x_3& x_1{x}_3& x_1{x}_3^2\\ 1& x_3& x_1{x}_2& x_2& x_2{x}_3& x_2^2{x}_3\\ 1& x_3& x_1{x}_2& x_1& x_1{x}_3& x_1^2{x}_3\\ 1& x_1& x_1{x}_3& x_2& x_1{x}_2& x_1{x}_2^2\\ 1& x_2& x_2{x}_3& x_1& x_1{x}_2& x_1^2{x}_2\end{array}\right)\begin{array}{l}1\\ s_1\\ s_2\\ s_1{s}_2\\ s_2{s}_1\\ s_1{s}_2{s}_1\end{array}$$ This is divisible by $${(x_2-x_1)}^3{(x_3-x_2)}^3{(x_3-x_1)}^3$$ and both things have top term $x_1^6{x}_2^3.$

In the proof of Theorem 2.4 we get $$a_v=\frac{\det \left(u{e}_w\stackrel{v^{\mathrm{th}}column}{\left(\begin{array}{r}\phantom{u}\phantom{f}\\ uf\\ \phantom{u}\phantom{f}\end{array}\right)}\right)}{\det \left(u{e}_w\right)}=\prod _{\alpha >0}{(e^{\frac{\alpha }{2}}-e^{-\frac{\alpha }{2}})}^{-n_{\alpha }}\det \left(u{e}_w\left(\begin{array}{r}\phantom{u}\phantom{f}\\ uf\\ \phantom{u}\phantom{f}\end{array}\right)\right)$$ from Cramer's Rule.

Notes and References

These notes are a retyping, into MathML, of the notes at http://researchers.ms.unimelb.edu.au/~aram@unimelb/Notes2005/zpnzpw7.18.05.pdf

§2 is taken from Verma at the Magdeburg meeting in August 1998.

References

[GL] S. Gaussent, P. Littelmann, LS galleries, the path model, and MV cycles, Duke Math. J. 127 (2005), no. 1, 35-88.

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