Reflection Groups

## Structure theorems for reflection groups

1. (Chevalley, Shephard-Todd) A finite group $W\subset GL\left({𝔥}^{*}\right)$ is generated by reflections if and only if $S 𝔥* W = ℂ I1,…,Ir$ where ${I}_{1},\dots ,{I}_{r}$ are algebraiclly independent and homogenous.
2. (Solomon) Let $W$ be a finite reflection group. Then $S𝔥* ⊗ Λ𝔥 W = ℂ I1,…,Ir ⊗ Λ dI1,…,dIr$ (see Benson page 86).
3. $S\left({𝔥}^{*}\right)\cong ℋ\otimes S{\left({𝔥}^{*}\right)}^{W}.$

1. $det 𝜕Ij 𝜕xj =λp, where p= ∏ α∈R+ α$ and $\lambda \in ℝ$, $\lambda \ne 0$.
2. $ℋ$ has basis $\left\{{h}_{w}={\Delta }_{w}^{*}\mid w\in W\right\}$ where Δw are the BGG operators and $deg\left({h}_{w}\right)=l\left(w\right).$

(Molien theorems)

1. $P S𝔥* ⊗ Λ𝔥 W ; q,t = 1 W ∑ w∈W det1+wq det1-wt$
2. $P S 𝔥* W ;t = 1 W ∑ w∈W 1 det 1-wt$

 Proof. $∑ j∈ℤ≥0 qj Tr w, Λj𝔥 = ∏ i=1 r 1+ λi-1 q = det 1+w-1q, 𝔥* .$ $∑ j∈ℤ≥0 tj Tr w,Sj𝔥 = 1 det 1-wt,𝔥* = ∏ i=1 r 1 1-λit .$ Now apply $1 W ∑ w∈W w$ to $S\left({𝔥}^{*}\right)\otimes \Lambda \left(𝔥\right)$: $P S𝔥* ⊗ Λ𝔥 W ; q,t = 1 W Trq,t ∑ w∈W w, S𝔥* ⊗ Λ𝔥 = 1 W ∑ w∈W Trq,t w, S𝔥* ⊗ Λ𝔥 .$ $\square$

$P S𝔥* ;t = ∏ i=1 r 1 1-t P S𝔥*W ;t = ∏ i=1 r 1 1-tdi P ℋ;t = ∏ i=1 r 1-tdi 1-t$ and $P S𝔥* ⊗ Λ𝔥 W ;q,t = ∏ i=1 r 1+qtdi-1 1-tdi .$

 Proof. (a) is clear. (b) follows from Chevalley's theorem. (c) follows from part (c) of the structure theorem since it implies $P S𝔥* ;t = Pℋ;t P S ( 𝔥* ) W ;t .$ (d) follows from Solomon's theorem. $\square$

Let

1. $∑ w∈W t dm(w) = ∏ i=1 r t χm∣di + di -1 .$
2. $∑ w∈W t dw = ∏ i=1 r t+di-1 .$
3. The number of reflections in $W$ is $\sum _{i=1}^{r}\left({d}_{i}-1\right)$.
4. $\left|W\right|=\prod _{i=1}^{r}{d}_{i}$.

 Proof. follows from (a) by putting $m=1$. follows from taking the coefficient of ${t}^{r-1}$ on both sides of the identity in (b). follows by putting $t=1$ in (b). (following Macdonald) Replace $q$ and $t$ with $t/\xi$, where $\xi =2\pi i/m$. Then $1 W ∑ w∈W detξ+qw detξ-tw = ∏ i=1 r ξdi +q t di-1 ξ di - t di$ from $1 W ∑ w∈W det 1+qw det 1-tw = ∏ i=1 r 1+q tdi-1 1- tdi .$ Now let $q=\left(1-t\right)X-1$. Then $det ξ-qw det ξ-tw = ∏ i=1 r ξ+λi 1-t X -1 ξ-λit .$ So, now take the limit as $t\to 1$. When we do this we get the result we want but we will need $∏ i=1 r di= W .$ To get this set $q=0$ and multiply by ${\left(1-t\right)}^{r}$ in the Molien formula to get $1 W ∑ w∈W 1-tr det 1-tw = ∏ i=1 r 1-t 1-tdi .$ Now set $t=1$. Then $LHS= 1 W 1+ ∑ w≠1 0 = 1 W and RHS= ∏ i=1 r 1 di .$ $\square$

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