The ring $K_T(G/B)$

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

Last update: 23 February 2013

The ring $K_T(G/B)$

Let $H$ and $\mathbb{Z}\left[X\right]$ be as in (1.5). The trivial representation of $H$ is defined by the homomorphism $1:H\to \mathbb{Z}$ given by $1\left(T_i\right)=1\text{.}$ The first of the maps

$$\begin{array}{ccccc}\mathbb{Z}\left[X\right]& \stackrel{\sim }{⟶}& \stackrel{\sim }{H}{T}_{w_0}& \stackrel{\sim }{⟶}& \stackrel{\sim }{H}{\otimes }_{H}1\\ f& ⟼& f{T}_{w_0}& ⟼& f\otimes 1\end{array}$$

is an $\stackrel{\sim }{H}\text{-module}$ isomorphism if the action of $\stackrel{\sim }{H}$ on $\mathbb{Z}\left[X\right]$ is given by

$$\begin{array}{cc}{T}_i·f=\frac{X^{{\alpha }_i}f-s_{i}f}{X^{{\alpha }_i}-1},\text{for}\hspace{0.17em}f\in \mathbb{Z}\left[X\right]\text{.}& \text{(2.1)}\end{array}$$

The group algebra of $P$ is

$$\begin{array}{cc}R=\mathbb{Z}\text{-span}\hspace{0.17em}\{e^{\lambda }\hspace{0.17em}\mid \hspace{0.17em}\lambda \in P\}\hspace{0.17em}\text{with}\hspace{0.17em}{e}^{\lambda }{e}^{\mu }=e^{\lambda +\mu },& \text{(2.2)}\end{array}$$

for $\lambda ,\mu \in P\text{.}$ Extend coefficients to $R$ so that ${\stackrel{\sim }{H}}_R=R{\otimes }_{\mathbb{Z}}\stackrel{\sim }{H}$ and $R\left[X\right]=R{\otimes }_{\mathbb{Z}}\mathbb{Z}\left[X\right]$ are $R\text{-algebras.}$ Define $K_T(G/B)$ to be the ${\stackrel{\sim }{H}}_R\text{-module}$

$$\begin{array}{cc}{K}_T(G/B)=R\text{-span}\hspace{0.17em}\{\left[{𝒪}_{X_w}\right]\hspace{0.17em}\mid \hspace{0.17em}w\in W\},& \text{(2.3)}\end{array}$$

so that the $\left[{𝒪}_{X_w}\right],$ $w\in W,$ are an $R\text{-basis}$ of $K_T(G/B),$ with ${\stackrel{\sim }{H}}_R\text{-action}$ given by

$$\begin{array}{cc}{X}^{\lambda }\left[{𝒪}_{X_1}\right]=e^{\lambda }\left[{𝒪}_{X_1}\right],\text{and}{T}_i\left[{𝒪}_{X_w}\right]=\{\begin{array}{cc}\left[{𝒪}_{X_{w{s}_i}}\right],& \text{if}\hspace{0.17em}w{s}_i>w,\\ \left[{𝒪}_{X_w}\right],& \text{if}\hspace{0.17em}w{s}_i<w\text{.}\end{array}& \text{(2.4)}\end{array}$$

If $R$ is an $R\left[X\right]\text{-module}$ via the $R\text{-algebra}$ homomorphism given by

$$\begin{array}{cc}\begin{array}{cccc}e:& R\left[X\right]& ⟶& R\\ & X^{\lambda }& ⟼& e^{\lambda }\end{array}& \text{(2.5)}\end{array}$$

then as ${\stackrel{\sim }{H}}_R\text{-modules,}$ $K_T(G/B)\cong {\stackrel{\sim }{H}}_R{\otimes }_{R\left[X\right]}{R}_e,$ where $R_e$ is the $R\text{-rank}$ 1 $R\left[X\right]\text{-module}$ determined by the homomorphism $e\text{.}$

Let $Q$ be the field of fractions of $R$ and let $\bar{Q}$ be the algebraic closure of $Q\text{.}$ For $w\in W$ let

$$\begin{array}{cc}{b}_w\text{in}\hspace{0.17em}\bar{Q}{\otimes }_R{K}_T(G/B)\text{be determined by}\hspace{0.17em}{X}^{\lambda }{b}_w=e^{w\lambda }{b}_w,\text{for}\hspace{0.17em}\lambda \in P\text{.}& \text{(2.6)}\end{array}$$

If the $b_w$ exist, then they are a $\bar{Q}\text{-basis}$ of $\bar{Q}{\otimes }_R{K}_T(G/B)$ since they are eigenvectors with distinct eigenvalues. If ${\tau }_i,$ $1\le i\le n,$ are the operators on $\bar{Q}{\otimes }_R{K}_T(G/B)$ given by

$$\begin{array}{cc}{\tau }_i=T_i-\frac{1}{1-X^{-{\alpha }_i}},\hspace{0.17em}\text{then}\hspace{0.17em}{b}_1=\left[{𝒪}_{X_1}\right]\hspace{0.17em}\text{and}\hspace{0.17em}{\tau }_i{b}_w=b_{w{s}_i},\text{for}\hspace{0.17em}w{s}_i>w,& \text{(2.7)}\end{array}$$

because, a direct computation with relation (1.3) gives that $X^{\lambda }{\tau }_i{b}_w={\tau }_i{X}^{s_i\lambda }{b}_w={\tau }_i{e}^{w{s}_i\lambda }{b}_w=e^{w{s}_i\lambda }{b}_{w{s}_i}\text{.}$ Thus the $b_w,w\in W,$ exist and the form of the $\tau \text{-operators}$ shows that, in fact, they form a $Q\text{-basis}$ of $\bar{Q}{\otimes }_R{K}_T(G/B)$ (it was not really necessary to extend coefficients all the way to $\bar{Q}\text{).}$ Eqs. (2.6) and (2.7) force

$$\underset{\underset{m_{ij}\text{factors}}{⏟}}{{\tau }_i{\tau }_j{\tau }_i\dots }=\underset{\underset{m_{ij}\text{factors}}{⏟}}{{\tau }_j{\tau }_i{\tau }_j\dots },\text{and the equality}\hspace{0.17em}{\tau }_i^2=\frac{1}{(X^{{\alpha }_i}-1)(X^{-{\alpha }_i}-1)}$$

is checked by direct computation using (1.3). Let ${\tau }_w={\tau }_{i_1}\dots {\tau }_{i_p}$ for a reduced word $w=s_{i_1}\dots s_{i_p}\text{.}$ Then, for $w\in W,$

$$\begin{array}{cc}{b}_w={\tau }_{w^{-1}}{b}_1,\left[{𝒪}_{X_w}\right]=T_{w^{-1}}\left[{𝒪}_{X_1}\right]\text{and we define}\left[{ℐ}_{X_w}\right]={\epsilon }_{w^{-1}}\left[{𝒪}_{X_1}\right],& \text{(2.8)}\end{array}$$

where ${\epsilon }_w$ is as in (1.11). In terms of geometry, $\left[{𝒪}_{X_w}\right]$ is the class of the structure sheaf of the Schubert variety $X_w$ in $G/B$ and, up to a sign, $\left[{ℐ}_{X_w}\right]$ is class of the sheaf ${ℐ}_{X_w}$ determined by the exact sequence $0\to {ℐ}_{X_w}\to {𝒪}_{X_w}\to {𝒪}_{\partial X_w}\to 0,$ where $\partial X_w={⨆}_{v<w}BvB$ (see [Mat2000, Theorem 2.1(ii)] and [LSe2003, Eq. (4)]). We are not aware of a good geometric characterization of the basis $\{\left[X^{-{\lambda }_w}\right]\hspace{0.17em}\mid \hspace{0.17em}w\in W\}$ of $K_T(G/B)$ which appears in the following theorem.

Theorem 2.9. Let ${\lambda }_w,$ $w\in W,$ be as defined in Theorem 1.7 and let $\left[X^{\lambda }\right]=X^{\lambda }\left[{𝒪}_{X_{w_0}}\right]=X^{\lambda }{T}_{w_0}\left[{𝒪}_{X_1}\right]$ for $\lambda \in P\text{.}$ Then the $\left[X^{-{\lambda }_w}\right],w\in W,$ form an $R\text{-basis}$ of $K_T(G/B)\text{.}$

		Proof.
	Up to constant multiples, $\left[{𝒪}_{X_{w_0}}\right]=T_{w_0}\left[{𝒪}_{X_1}\right]$ is determined by the property $$\begin{array}{cc}{T}_i\left[{𝒪}_{X_{w_0}}\right]=T_{w_0}\left[O_{X_1}\right],\text{for all}\hspace{0.17em}1\le i\le n\text{.}& \text{(2.10)}\end{array}$$ If constants $c_w\in Q$ are given by $$\left[{𝒪}_{X_{w_0}}\right]=\sum _{w\in W}{c}_w{b}_w,$$ then comparing coefficients of $b_{w{s}_i},$ for $w{s}_i>w,$ on each side of (2.10) yields a recurrence relation for the $c_w,$ $$c_w=c_{w{s}_i}\left(\frac{1}{1-e^{-w{\alpha }_i}}\right)\text{for}\hspace{0.17em}w{s}_i>w,$$ which implies $$\begin{array}{cc}{c}_{w_0{v}^{-1}}=\prod _{\alpha \in R\left(v\right)}\frac{1}{1-e^{w_0\alpha }}& \text{(2.11)}\end{array}$$ via (1.1) and the fact that $c_{w_0}=1\text{.}$ Thus, $$\left[X^{-{\lambda }_v}\right]=X^{-{\lambda }_v}\left[{𝒪}_{X_{w_0}}\right]=\sum _{w\in W}{c}_w{e}^{-w{\lambda }_v}{b}_w,$$ and if $C,M$ and $A$ are the $\mid W\mid \times \mid W\mid$ matrices given by $$C=\text{diag}\left(c_w\right),M=\left(e^{-w{\lambda }_v}\right),\text{and}A=\left(a_{zw}\right),\text{where}\hspace{0.17em}{b}_w=\sum _{z\in W}{a}_{zw}\left[{𝒪}_{X_z}\right],$$ then the transition matrix between the $X^{-{\lambda }_v}$ and the $\left[{𝒪}_{X_z}\right]$ is the product $ACM\text{.}$ By (2.8) and the definition of the ${\tau }_i,$ the matrix $A$ has determinant 1. Using the method of Steinberg [Ste1975] and subtracting row $e^{-s_{\alpha }w{\lambda }_v}$ from row $e^{-w{\lambda }_v}$ in the matrix $M$ allows one to conclude that $\text{det}\left(M\right)$ is divisible by $$\prod _{\alpha \in R^{+}}{(1-e^{-\alpha })}^{\mid W\mid /2}\text{and identifying}\prod _{w\in W}{e}^{-w{\lambda }_w}=\prod _{i=1}^n\prod _{s_{i}w<w}{e}^{-{\omega }_i}={\left(e^{-\rho }\right)}^{\mid W\mid /2}$$ as the lowest degree term determines $\text{det}\left(M\right)$ exactly. Thus, $$\text{det}\left(ACM\right)=1·\left(\prod _{w\in W}\prod _{\alpha \in R\left(w\right)}\frac{1}{1-e^{-\alpha }}\right){\left(e^{\rho }\prod _{\alpha \in R^{+}}(1-e^{-\alpha })\right)}^{\mid W\mid /2}={\left(e^{\rho }\right)}^{\mid W\mid /2}\text{.}$$ Since this is a unit in $R,$ the transition matrix between the $\left[{𝒪}_{X_w}\right]$ and the $X^{-{\lambda }_v}$ is invertible. $\square$

Theorem 2.12. The composite map

$$\begin{array}{cccccccc}\Phi :& R\left[X\right]& ⟶& {\stackrel{\sim }{H}}_R{T}_{w_0}& \hookrightarrow & {\stackrel{\sim }{H}}_R& ⟶& K_T(G/B)\\ & f& ⟼& f{T}_{w_0}& & h& ⟼& h\left[O_{X_1}\right]\end{array}$$

is surjective with kernel

$$\text{ker}\hspace{0.17em}\Phi =⟨f-e\left(f\right)\hspace{0.17em}\mid \hspace{0.17em}f\in R{\left[X\right]}^W⟩,$$

the ideal of the ring $R\left[X\right]$ generated by the elements $f-e\left(f\right)$ for $f\in R{\left[X\right]}^W\text{.}$ Hence

$$K_T(G/B)\cong \frac{R\left[X\right]}{⟨f-e\left(f\right)\hspace{0.17em}\mid \hspace{0.17em}f\in R{\left[X\right]}^W⟩}$$

has the structure of a ring.

		Proof.
	Since $\Phi \left(X^{\lambda }\right)=X^{\lambda }{T}_{w_0}\left[O_{X_1}\right]=X^{\lambda }\left[{𝒪}_{X_{w_0}}\right],$ it follows from Theorem 2.9 that $\Phi$ is surjective. Thus $K_T(G/B)\cong R\left[X\right]/\text{ker}\hspace{0.17em}\Phi \text{.}$ Let $I=⟨f-e\left(f\right)\hspace{0.17em}\mid \hspace{0.17em}f\in R{\left[X\right]}^W⟩\text{.}$ If $f\in R{\left[X\right]}^W$ then, for all $\lambda \in P,$ $$\begin{array}{ccc}\Phi \left(x^{\lambda }(f-e\left(f\right))\right)& =& X^{\lambda }(f-e\left(f\right))T_{w_0}\left[O_{X_1}\right]=X^{\lambda }{T}_{w_0}(f-e\left(f\right))\left[O_{X_1}\right]\\ & =& X^{\lambda }{T}_{w_0}(e\left(f\right)-e\left(f\right))\left[O_{X_1}\right]=0,\end{array}$$ since $f-e\left(f\right)\in Z\left({\stackrel{\sim }{H}}_R\right)\text{.}$ Thus $I\subseteq \text{ker}\hspace{0.17em}\Phi \text{.}$ The ring $K_T(G/B)=R\left[X\right]/\text{ker}\hspace{0.17em}\Phi$ is a free $R\text{-module}$ of rank $\mid W\mid$ and, by Theorem 1.7, so is $R\left[X\right]/I\text{.}$ Thus $\text{ker}\hspace{0.17em}\Phi =I\text{.}$ $\square$

Notes and References

This is an excerpt of the paper entitled Affine Hecke algebras and the Schubert calculus authored by Stephen Griffeth and Arun Ram. It was dedicated to Alain Lascoux.

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