## The ring ${K}_{T}\left(G/B\right)$

Last update: 23 February 2013

## The ring ${K}_{T}\left(G/B\right)$

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

$ℤ[X] ⟶∼ H∼Tw0 ⟶∼ H∼⊗H1 f ⟼ fTw0 ⟼ f⊗1$

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

$Ti·f= Xαif-sif Xαi-1 ,for f∈ℤ[X]. (2.1)$

The group algebra of $P$ is

$R=ℤ-span {eλ ∣ λ∈P} with eλeμ= eλ+μ, (2.2)$

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

$KT(G/B)=R -span {[𝒪Xw] ∣ w∈W}, (2.3)$

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

$Xλ[𝒪X1]= eλ[𝒪X1], andTi [𝒪Xw]= { [𝒪Xwsi], if wsi>w, [𝒪Xw], if wsi

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

$e: R[X] ⟶ R Xλ ⟼ eλ (2.5)$

then as ${\stackrel{\sim }{H}}_{R}\text{-modules,}$ ${K}_{T}\left(G/B\right)\cong {\stackrel{\sim }{H}}_{R}{\otimes }_{R\left[X\right]}{R}_{e},$ 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 $\stackrel{‾}{Q}$ be the algebraic closure of $Q\text{.}$ For $w\in W$ let

$bwin Q‾ ⊗RKT(G/B) be determined by Xλbw= ewλbw,for λ∈P. (2.6)$

If the ${b}_{w}$ exist, then they are a $\stackrel{‾}{Q}\text{-basis}$ of $\stackrel{‾}{Q}{\otimes }_{R}{K}_{T}\left(G/B\right)$ since they are eigenvectors with distinct eigenvalues. If ${\tau }_{i},$ $1\le i\le n,$ are the operators on $\stackrel{‾}{Q}{\otimes }_{R}{K}_{T}\left(G/B\right)$ given by

$τi=Ti- 11-X-αi, then b1= [𝒪X1] and τibw=bwsi, for wsi>w, (2.7)$

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 $\stackrel{‾}{Q}{\otimes }_{R}{K}_{T}\left(G/B\right)$ (it was not really necessary to extend coefficients all the way to $\stackrel{‾}{Q}\text{).}$ Eqs. (2.6) and (2.7) force

$τiτjτi… ⏟mijfactors = τjτiτj… ⏟mijfactors ,and the equality τi2= 1 (Xαi-1) (X-αi-1)$

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 $w\in W,$

$bw=τw-1b1 ,[𝒪Xw]= Tw-1[𝒪X1] and we define[ℐXw] =εw-1[𝒪X1], (2.8)$

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 (see [Mat2000, Theorem 2.1(ii)] and [LSe2003, Eq. (4)]). We are not aware of a good geometric characterization of the basis $\left\{\left[{X}^{-{\lambda }_{w}}\right] \mid w\in W\right\}$ of ${K}_{T}\left(G/B\right)$ 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 ${K}_{T}\left(G/B\right)\text{.}$

 Proof. Up to constant multiples, $\left[{𝒪}_{{X}_{{w}_{0}}}\right]={T}_{{w}_{0}}\left[{𝒪}_{{X}_{1}}\right]$ is determined by the property $Ti[𝒪Xw0] =Tw0[OX1] ,for all 1≤i≤n. (2.10)$ If constants ${c}_{w}\in Q$ are given by $[𝒪Xw0]= ∑w∈Wcwbw,$ 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 ${c}_{w},$ $cw=cwsi (11-e-wαi) for wsi>w,$ which implies $cw0v-1= ∏α∈R(v) 11-ew0α (2.11)$ via (1.1) and the fact that ${c}_{{w}_{0}}=1\text{.}$ Thus, $[X-λv]= X-λv [𝒪Xw0]= ∑w∈Wcw e-wλvbw,$ and if $C,M$ and $A$ are the $\mid W\mid ×\mid W\mid$ matrices given by $C=diag(cw),M= (e-wλv), andA=(azw), where bw=∑z∈W azw[𝒪Xz],$ 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 $∏α∈R+ (1-e-α)∣W∣/2 and identifying∏w∈W e-wλw= ∏i=1n ∏siw as the lowest degree term determines $\text{det}\left(M\right)$ exactly. Thus, $det(ACM)=1· ( ∏w∈W ∏α∈R(w) 11-e-α ) ( eρ∏α∈R+ (1-e-α) ) ∣W∣/2 =(eρ)∣W∣/2.$ 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

$Φ: R[X] ⟶ H∼RTw0 ↪ H∼R ⟶ KT(G/B) f ⟼ fTw0 h ⟼ h[OX1]$

is surjective with kernel

$ker Φ= ⟨ f-e(f) ∣ f∈R[X]W ⟩ ,$

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

$KT(G/B)≅ R[X] ⟨ f-e(f) ∣ f∈R[X]W ⟩$

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 ${K}_{T}\left(G/B\right)\cong R\left[X\right]/\text{ker} \Phi \text{.}$ Let $I=⟨f-e\left(f\right) \mid f\in R{\left[X\right]}^{W}⟩\text{.}$ If $f\in R{\left[X\right]}^{W}$ then, for all $\lambda \in P,$ $Φ(xλ(f-e(f))) = Xλ(f-e(f)) Tw0[OX1] =XλTw0 (f-e(f)) [OX1] = XλTw0 (e(f)-e(f)) [OX1]=0,$ since $f-e\left(f\right)\in Z\left({\stackrel{\sim }{H}}_{R}\right)\text{.}$ Thus $I\subseteq \text{ker} \Phi \text{.}$ The ring ${K}_{T}\left(G/B\right)=R\left[X\right]/\text{ker} \Phi$ 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 $\text{ker} \Phi =I\text{.}$ $\square$

## 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.