Reflection Groups

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

Last updates: 20 October 2010

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 $$S{\left({𝔥}^{*}\right)}^W=ℂ\left[I_1,\dots ,I_r\right]$$ where $I_1,\dots ,I_r$ are algebraiclly independent and homogenous.
2. (Solomon) Let $W$ be a finite reflection group. Then $${\left(S\left({𝔥}_{*}\right)\otimes \Lambda \left(𝔥\right)\right)}^W=ℂ\left[I_1,\dots ,I_r\right]\otimes \Lambda \left(d{I}_1,\dots ,d{I}_r\right)$$ (see Benson page 86).
3. $S\left({𝔥}^{*}\right)\cong \mathscr{H}\otimes S{\left({𝔥}^{*}\right)}^W.$

Some additional remarks:

1. $$det\left(\frac{𝜕I_j}{𝜕x_j}\right)=\lambda p,\text{where}p=\prod _{\alpha \in R^{+}}\alpha$$ and $\lambda \in ℝ$, $\lambda \ne 0$.
2. $\mathscr{H}$ 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. $$P\left({\left(S\left({𝔥}^{*}\right)\otimes \Lambda \left(𝔥\right)\right)}^W;q,t\right)=\frac{1}{W}\sum _{w\in W}\frac{det\left(1+wq\right)}{det\left(1-wt\right)}$$
2. $$P\left(S{\left({𝔥}^{*}\right)}^W;t\right)=\frac{1}{W}\sum _{w\in W}\frac{1}{det\left(1-wt\right)}$$

		Proof.
	$$\sum _{j\in {\mathbb{Z}}_{\ge 0}}{q}^{j}Tr\left(w,{\Lambda }^j𝔥\right)=\prod _{i=1}^r\left(1+{\lambda }_i^{-1}q\right)=det\left(1+w^{-1}q,{𝔥}^{*}\right).$$ $$\sum _{j\in {\mathbb{Z}}_{\ge 0}}{t}^{j}Tr\left(w,S^j𝔥\right)=\frac{1}{det\left(1-wt,{𝔥}^{*}\right)}=\prod _{i=1}^r\frac{1}{\left(1-{\lambda }_{i}t\right)}.$$ Now apply $$\frac{1}{\left|W\right|}\sum _{w\in W}w$$ to $S\left({𝔥}^{*}\right)\otimes \Lambda \left(𝔥\right)$: $$P\left({\left(S\left({𝔥}^{*}\right)\otimes \Lambda \left(𝔥\right)\right)}^W;q,t\right)=\frac{1}{\left|W\right|}{Tr}_{q,t}\left(\sum _{w\in W}w,S\left({𝔥}^{*}\right)\otimes \Lambda \left(𝔥\right)\right)=\frac{1}{\left|W\right|}\sum _{w\in W}{Tr}_{q,t}\left(w,S\left({𝔥}^{*}\right)\otimes \Lambda \left(𝔥\right)\right).$$ $\square$

$$P\left(S\left({𝔥}^{*}\right);t\right)=\prod _{i=1}^r\frac{1}{1-t}P\left(S{\left({𝔥}^{*}\right)}^W;t\right)=\prod _{i=1}^r\frac{1}{1-t^{d_i}}P\left(\mathscr{H};t\right)=\prod _{i=1}^r\frac{1-t^{d_i}}{1-t}$$ and $$P\left({\left(S\left({𝔥}^{*}\right)\otimes \Lambda \left(𝔥\right)\right)}^W;q,t\right)=\prod _{i=1}^r\frac{1+q{t}^{d_i-1}}{1-t^{d_i}}.$$

		Proof.
	(a) is clear. (b) follows from Chevalley's theorem. (c) follows from part (c) of the structure theorem since it implies $$P\left(S\left({𝔥}^{*}\right);t\right)=P\left(\mathscr{H};t\right)P\left(S{\left({𝔥}^{*}\right)}^W;t\right).$$ (d) follows from Solomon's theorem. $\square$

Let $$\begin{array}{rcl}d\left(w\right)& =& dim\left(V^w\right)=\text{multiplicity of 1 as an eigenvalue of} w,\\ d_m\left(w\right)& =& \text{multiplicity of} e^{2\pi i/m} \text{as an eigenvalue of} w,\\ {\rm X}\left(m\mid d_i\right)& =& \left\{\begin{array}{ll}1,& \text{if} m \text{divides} d_i,\\ 0,& \text{if} m \text{does not divide} d_i.\end{array}\right.\end{array}$$

1. $$\sum _{w\in W}{t}^{d_m\left(w\right)}=\prod _{i=1}^r\left(t^{\chi \left(m\mid d_i\right)}+d_i-1\right).$$
2. $$\sum _{w\in W}{t}^{d\left(w\right)}=\prod _{i=1}^r\left(t+d_i-1\right).$$
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 $$\frac{1}{\left|W\right|}\sum _{w\in W}\frac{det\left(\xi +qw\right)}{det\left(\xi -tw\right)}=\prod _{i=1}^r\left(\frac{{\xi }^{d_i}+q{t}^{d_i-1}}{{\xi }^{d_i}-t^{d_i}}\right)$$ from $$\frac{1}{\left|W\right|}\sum _{w\in W}\frac{det\left(1+qw\right)}{det\left(1-tw\right)}=\prod _{i=1}^r\left(\frac{1+q{t}^{d_i-1}}{1-t^{d_i}}\right).$$ Now let $q=\left(1-t\right)X-1$. Then $$\frac{det\left(\xi -qw\right)}{det\left(\xi -tw\right)}=\prod _{i=1}^r\frac{\left(\xi +{\lambda }_i\left(\left(1-t\right)X-1\right)\right)}{\left(\xi -{\lambda }_{i}t\right)}.$$ So, now take the limit as $t\to 1$. When we do this we get the result we want but we will need $$\prod _{i=1}^r{d}_i=\left|W\right|.$$ To get this set $q=0$ and multiply by ${\left(1-t\right)}^r$ in the Molien formula to get $$\frac{1}{\left|W\right|}\sum _{w\in W}\frac{{\left(1-t\right)}^r}{det\left(1-tw\right)}=\prod _{i=1}^r\frac{1-t}{1-t^{d_i}}.$$ Now set $t=1$. Then $$\text{LHS}=\frac{1}{\left|W\right|}\left(1+\sum _{w\ne 1}0\right)=\frac{1}{\left|W\right|}\text{and}\text{RHS}=\prod _{i=1}^r\frac{1}{d_i}.$$ $\square$

References

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