## 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?