The Elliptic Weyl character formula: The ring $\stackrel{˜}{Th}$

Last update: 02 March 2012

The ring $\stackrel{˜}{Th}$

Let ${\stackrel{\circ }{𝔥}}_{ℤ}$ be a free $ℤ-$module of rank $l$ with a positive definite symmetric bilinear form $\left(|\right):{\stackrel{\circ }{𝔥}}_{ℤ}×{\stackrel{\circ }{𝔥}}_{ℤ}\to ℤ.$ Let $𝔥∘ℝ = ℝ⊗ℤ𝔥∘ℤ and 𝔥∘ℂ = ℂ⊗ℤ𝔥∘ℤ$ and extend $\left(|\right)$ to a symmetric bilinear form on $(δ|δ) =0, (δ|𝔥∘ℂ)=0, (δ|Λ0)=1, 𝔥ℂ = ℂδ⊕ 𝔥∘ℂ⊕ ℂΛ0 by setting (𝔥∘ℂ|δ)=0, (𝔥∘ℂ|Λ0)=1, (Λ0|δ)=1, (Λ0|𝔥∘ℂ)=0, (Λ0|Λ0)=0.$ For $\beta \in {\stackrel{\circ }{𝔥}}_{ℤ}$ define ${t}_{\beta }:{𝔥}_{ℂ}\to {𝔥}_{ℂ}$ by $tβ(λ) = λ+mβ- (λ_+ 12mβ|β)δ, if λ=aδ+λ_+mΛ0, (RTh 1)$ with and $m\in ℂ.$ The motivation for this formula is ??????? Let For $\lambda =a\delta +\stackrel{_}{\lambda }+m{\Lambda }_{0}$ with define (see [Kac, (12.7.2) and (12.7.3)]) $Θλ = Θaδ+λ_+mΛ0 = e- (λ|λ)δ 2m ∑ β∈𝔥∘ℤ etβ(λ) (RTh 2) = e- 2am+(λ_|λ_)δ 2m ∑ β∈𝔥∘ℤ e aδ+λ_+mβ- (λ_+ 1 2 mβ|β)δ+mΛ0 = emΛ0 ∑ β∈𝔥∘ℤ e λ_+mβ- 1 2m ( (2λ_+mβ|mβ) +(λ_|λ_) )δ = emΛ0 ∑ β∈𝔥∘ℤ e λ_+mβ- 1 2m ( λ_+mβ|λ_+mβ )δ , (RTh 3)$ which (modulo $\delta$) is exactly the sum over translates of $\stackrel{_}{\lambda }$ by an $m-$dilate of the lattice ${\stackrel{\circ }{𝔥}}_{ℤ}.$ The expression ${\Theta }_{\lambda }$ is an element of where ${e}^{\lambda }$ are formal symbols indexed by $\lambda \in {𝔥}_{ℂ},$ infinite sums are allowed and If $a\in ℂ$ and $\beta \in {\stackrel{\circ }{𝔥}}_{ℤ}$ then $Θ aδ+λ_+mΛ0 = Θ λ_+mΛ0 and Θ λ_+mβ+mΛ0 = Θ λ_+mΛ0 .$ Write $(λ|λ) = ∥λ∥2 and q=e-δ so that Θ λ_+mΛ0 = emΛ0 ∑ β∈𝔥∘ℤ eλ_+mβ q 1 2m ∥λ+mβ∥2 , (RTh 4)$

Setting ${e}^{\lambda }{e}^{\mu }={e}^{\lambda +\mu }$ for $\lambda ,\mu \in {𝔥}_{ℂ},$ where $d λ+μ+mγ+(m+n)Λ0 λ+mΛ0, μ+nΛ0 = ∑ κ∈𝔥∘ℤ q 12 mn (m+n) ∥ 1mλ- 1nμ+ γ+(m+)κ ∥2 ,$ (see [Kac, Ex. 13.1] or [KP, §13.2]). This formula gives the product structure on the graded ${\stackrel{˜}{Th}}_{0}-$algebra WHAT IS THE RIGHT COMBINATORIAL DESCRIPTION OF THIS BASE RING? PUT THE ${G}_{ℤ}-$ACTION ON $\stackrel{˜}{Th}$ HERE?

Proof of the product formula for $\stackrel{˜}{Th}$.

 Proof. $Θμ1+m1Λ0 Θμ2+m2Λ0 =( em1Λ0 ∑ γ1∈𝔥∘ℤ e μ1+m1γ1- 1 2m1 ∥μ1+m1γ1∥2δ ) ( em2Λ0 ∑ γ2∈𝔥∘ℤ e μ2+m2γ2- 1 2m2 ∥μ2+m2γ2∥2δ ) =e(m1+m2)Λ0 ∑ γ1-γ2∈𝔥∘ℤ ∑ γ2∈𝔥∘ℤ e μ1+μ2+m1(γ1-γ2)+(m1+m2)γ2 e - 1 2(m1+m2) ∥μ1+μ2+m1γ1+m2γ2∥2δ ⋅e 1 2(m1+m2) ∥μ1+μ2+m1γ1+m2γ2∥2δ e - 1 2m1 ∥μ1+m1γ1∥2δ e - 1 2m2 ∥μ2+m2γ2∥2δ = ∑ γ1-γ2∈𝔥∘ℤ Θ μ1+μ2+m1(γ1-γ2)+(m1+m2)Λ0 e - (m1m2) 2(m1+m2) ∥ 1 m1 μ1- 1 m1 μ2+γ1-γ2∥2δ ,$ since $1 2(m1+m2) ∥μ1+μ2+m1γ1+m2γ2∥2 - 1 2m1 ∥μ1+m1γ1∥2 - 1 2m2 ∥μ2+m2γ2∥2 = 1 2(m1+m2) ∥μ1+m1γ1∥2 + 1 2(m1+m2) ∥μ2+m2γ2∥2 + 1 m1+m2 (μ1+m1γ1|μ2+m2γ2) - 1 2m1 ∥μ1+m1γ1∥2 - 1 2m2 ∥μ2+m2γ2∥2 =- m2 2m1m1+m2) ∥μ1+m1γ1∥2 - m1 2m2m1+m2) ∥μ2+m2γ2∥2 + 1 m1+m2 (μ1+m1γ1|μ2+m2γ2) =- m1m2 2(m1+m2) ∥ 1 m1 μ1+γ1∥2 - m1m2 2(m1+m2) ∥ 1 m2 μ2+γ2∥2 + m1m2 m1+m2 ( 1 m1 μ1+γ1| 1 m2 μ2+γ2 ) = -m1m2 2(m1+m2) ∥ 1 m1 μ1+γ1- 1 m2 μ2-γ2∥2 = -m1m2 2(m1+m2) ∥ 1 m1 μ1 - 1 m2 μ2 +γ1 -γ2∥2 .$ Thus $\square$

The subring ${\stackrel{˜}{Th}}^{{W}_{0}}$

The action of ${W}_{0}$ on $\stackrel{˜}{Th}$ induced by the action on given by $w{e}^{\lambda }={e}^{w\lambda }$ is $wΘλ =Θwλ, so that wΘλ_+mΛ0 = Θwλ_+mΛ0,$ for $w\in {W}_{0},$ $\stackrel{_}{\lambda }\in {\stackrel{\circ }{𝔥}}_{ℂ}$ and $m\in {ℤ}_{>0}.$ Let $Mλ = ∑ γ∈W0λ Θλ and Aμ = ∑ w∈W0 det(w)Θwμ.$ Let Since is a set of representatives of the ${W}_{0}-$orbits on and ${A}_{\mu }=0$ if $\mu \in {P}^{+}-{P}^{++},$ it follows (as in [KP, Prop. 4.3(d-e)]) that Because of the Weyl character formula and, since $P+ → P++ λ ↦ λ+ρ is a bijection,$ ${\stackrel{˜}{Th}}^{\mathrm{det}}$ is a free module of rank 1 over ${\stackrel{˜}{Th}}^{{W}_{0}},$ generated by ${A}_{\rho }.$ In other words, as ${\stackrel{˜}{Th}}^{{W}_{0}}-modules$ $Th˜W0 →∼ Th˜det f ↦ Aρf sλ ↦ Aλ+ρ$ BE SURE TO GET THE NORMALIZATION CONSTANT CORRECT ON THE LAST LINE!!!

Notes and References

These notes are taken from notes on the Elliptic Weyl character formula by Nora Ganter and Arun Ram.

References?