## The ring $ℤ{\left[P\right]}^{W}$

Last update: 27 July 2012

## The ring $ℤ{\left[P\right]}^{W}$

$ℤ{\left[P\right]}^{W}$ is a polynomial ring, i.e. there are algebraically independent elements ${e}_{1},...,{e}_{n}\in ℤ{\left[P\right]}^{W}$ such that $ℤ[P]W = ℤ[e1,...,en].$

 Proof. Let ${e}_{1},...,{e}_{n}\in 𝕂{\left[P\right]}^{W}$ be such that $ei = xωi + ( lower terms in dominance order ).$ If $\lambda ={l}_{1}{\omega }_{1}+\cdots +{l}_{n}{\omega }_{n}\in {P}^{+}$ then $e1l1 e2l2 ⋯ enln = xλ + ( lower terms in dominance order ).$ Thus $\left\{{e}_{1}^{{l}_{1}}\cdots {e}_{n}^{{l}_{n}}\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}{l}_{1},...,{l}_{n}\in {ℤ}_{\ge 0}\right\}$ 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 $𝒜- = { Hα, Hα-δ | α∈R+ } where Hα = { x∈ℝn | ⟨x, α∨⟩ = 0 }, Hα-δ = { x∈ℝn | ⟨x, α∨⟩ = -1 }. (ZPW 1)$ 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 $λw = w-1 ∑siw $Hα1+α2 Hα1 Hα2 Hα1+2α2 Hα1+α2-δ Hα1-δ Hα1-δ Hα2-δ Hα1+2α2-δ C s1C s2C s1s2C s2s1C s1s2s1C s2s1s2C λ1 λs1 λs2s1 λs2 λs1s2 λs1s2s1 λs2s1s2 λs1s2s1s2 The arrangement 𝒜-$

$ℤ\left[P\right]$ is a free $ℤ{\left[P\right]}^{W}$ module of rank $|W|$ with basis $\left\{{x}^{{\lambda }_{w}}\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}w\in W\right\}.$

 Proof. The proof is accomplished by establishing three facts: Let be a family of elements of $ℤ\left[P\right].$ Then $\mathrm{det}\left(z{f}_{y}\right)$ is divisible by $\prod _{\alpha \in {R}^{+}}{\left(1-{x}^{-\alpha }\right)}^{\frac{|W|}{2}}.$ $\mathrm{det}{\left(z{x}^{{\lambda }_{y}}\right)}_{z,y\in W}={\left(\prod _{\alpha >0}{x}^{\rho }\left(1-{x}^{-\alpha }\right)\right)}^{\frac{|W|}{2}}.$ If $f\in ℤ\left[P\right]$ then there is a unique solution to the equation $∑w∈W awxλw=f, with aw∈ℤ[P]W.$ For each $\alpha \in {R}^{+}$ subtract row $z{f}_{y}$ from row ${s}_{\alpha }z{f}_{y}$. Then this row is divisible by $\left(1-{x}^{-\alpha }\right).$ Since there are $\frac{|W|}{2}$ pairs of rows $\left(z{f}_{y},{s}_{\alpha }z{f}_{y}\right)$ the whole determinant is divisible by ${\left(1-{x}^{-\alpha }\right)}^{\frac{|W|}{2}}.$ For $\alpha ,\beta \in {R}^{+}$ the factors $\left(1-{x}^{-\alpha }\right)$ and $\left(1-{X}^{-\beta }\right)$ are coprime, and so $\mathrm{det}\left(z{f}_{y}\right)$ is divisible by $\prod _{\alpha \in {R}^{+}}{\left(1-{x}^{-\alpha }\right)}^{\frac{|W|}{2}}.$ 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 $∑ αi∈R(y) ωi.$ 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 $\mathrm{det}\left(z{x}^{{\lambda }_{y}}\right)$ is equal to $∏z∈W zxλz = ∏z∈W ∏ i siz Since ${s}_{i}\mathrm{det}\left(z{x}^{{\lambda }_{y}}\right)=-\mathrm{det}\left(z{x}^{{\lambda }_{y}}\right)$ the lowest term of $\mathrm{det}\left(z{x}^{{\lambda }_{y}}\right)$ is ${w}_{0}{\left({x}^{\rho }\right)}^{\frac{|W|}{2}}={\left({x}^{\rho }\right)}^{-\frac{|W|}{2}}.$ These are the same as the highest and lowest terms of $\left({x}^{\rho }\prod _{\alpha \in {R}^{+}}\left(1-{x}^{-\alpha }\right)\right)$ and so (2) follows from (1). Assume that ${a}_{y}\in ℤ{\left[P\right]}^{W}$ 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 $|W|$ equations $∑y∈W (zxλy)ay = zf, z∈W.$ 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 ${a}_{y}\in ℤ{\left[P\right]}^{W}.$ 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 $ℤ\left[P\right].$ Since each determinant is an alternating function (an element of Fock space), the quotient is an element of $ℤ{\left[P\right]}^{W}.$ $\square$

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 $\left\{{e}^{\epsilon \left(w\right)}\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}w\in W\right\}$ as a basis, where $ε(w) = w-1 (ρw) where ρw = ∑i∈D(w)wi, D(w) = {i | siw There is an explicit expansion formula $eλ = ∑w∈W χλ,w eε(w).$

 Proof. $eλ+μ = ∑w∈W χλ,w eμ+ε(w) for all μ∈P.$ We shall use this for a set of $|W|$ values $\mu$ below. Recall that $Δw0: R(T) → R(T)W eλ ↦ ch(λ) the Reynolds/Demazure operator.$ Then $Δw0 (eλ+μ) = ∑w∈W χλ,w Δw0( eμ+ε(w) ).$ So let $μ ∈ {-ε(yw0)-ρ | y∈W}$ and we get a $|W|×|W|$ matrix $( ch( e ε(w) - ε(yw0) -ρ ) ) y,w∈W$ which Hulsurkar proved to be unipotent. The diagonal entries are $\mathrm{sgn}\left(w\right).$ By definition $ρw0w = ρ-ρw and so w-1 ρw0w = w-1ρ - w-1ρw w-1 w0-1 w0 ρw0w = -w-1 w0-1 ρw0w = -ε(w0w).$ So $-ε(w0w) = w-1ρ - ε(w).$ So $ε(w) - ε(w0w) = w-1ρ.$ But one can look at these elements as follows: Remove the strips ${x∈ℝn | ⟨x,α∨⟩<1} Hα1 Hα2 Hβ 0$ 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 $ew = w-1λw = w-1 ∏i∈D(w) ewi where D(w) = {i | siw

$det(uev)u,v∈W = ∏α>0 ( eα2 - e-α2 ) nα$ where $nαi = | {v∈W | v-1αi<0} |$ and $nwαi = nαi.$

Let ${f}_{v}\in ℤ\left[X\right].$ Then $det(ufv) is divisible by ∏α>0 ( eα2 - e-α2 ) nα .$

If $f\in ℤ\left[X\right]$ then there is a unique solution to $∑awew = f with aw ∈ ℤ[X]W .$ 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 $\mathrm{det}\left(u{f}_{v}\right)$ is divisible by $\prod _{\alpha }{\left(1-{e}^{-\alpha }\right)}^{{n}_{\alpha }}.$ $\square$

 Proof of Theorem 2.2. The top coefficient of $\mathrm{det}\left(u{e}_{v}\right)$ is $∏v∈W vev = ∏v∈W λv = ∏v∈W ∏i∈D(v) ewi = ∏i=1n ewini.$ The top coefficient of $\prod _{\alpha }{\left(1-{e}^{\alpha }\right)}^{{n}_{\alpha }}$ is ${e}^{\sum _{\alpha >0}{n}_{\alpha }\alpha }.$ Now $si( ∑α>0 nαα ) = ∑α>0 nαα - 2nαi αi.$ So $( ∑α>0 nαα , αi∨ ) = 2nαi = 2ni.$ So $∑α>0 nαα = ∑i=1n 2niwi.$ $\square$

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

Example for ${S}_{2}$. $W={1,s1} z1=x1$ $z1 = 1 e1 = 1 zs1 = z1 es1 = x2 D = det 1 x2 1 x1 = x1-x2$ and $1 x2 1 x1 -1 = 1 x1-x2 x1 -x2 -1 1 .$

Example for ${S}_{3}$. $W = { 1,s1, s2, s1s2, s2s1, s1s2s1 } z1 = x1 z2 = x1x2$ $z1 = 1 e1 = 1 zs1 = z1=x1 es1 = x2 zs2 = z2 = x1x2 es2 = x1x3 zs1s2 = z1=x1 es1s2 = x3 zs2s1 = z2 = x1x2 es2s1 = x2x3 zs1s2s1 = z1z2 = x12x2 es1s2s1 = x2x32$ $D = 1 x2 x1x3 x3 x2x3 x2x32 1 x1 x2x3 x3 x1x3 x1x32 1 x3 x1x2 x2 x2x3 x22x3 1 x3 x1x2 x1 x1x3 x12x3 1 x1 x1x3 x2 x1x2 x1x22 1 x2 x2x3 x1 x1x2 x12x2 1 s1 s2 s1s2 s2s1 s1s2s1$ This is divisible by $(x2-x1)3 (x3-x2)3 (x3-x1)3$ and both things have top term ${x}_{1}^{6}{x}_{2}^{3}.$

In the proof of Theorem 2.4 we get $av = det uew uf uf uf vth column det(uew) = ∏α>0 ( eα2 - e-α2 ) -nα det uew uf uf uf$ 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.