Wisconsin Bourbaki Seminar

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

Last update: 26 March 2014

Notes and References

This is an excerpt from notes of the Wisconsin Bourbaki Seminar, Fall 1993. The notes were written by Oliver Eng, Susan Hollingsworth, Mark Logan, Arun Ram and Louis Solomon.

§4

In the exercises below, $(W,S)$ designates a Coxeter system. One supposes that $S$ is finite and its cardinality is called the rank of $(W,S)\text{.}$ We identify $W$ with a subgroup of $GL\left(E\right)$ through $\sigma \text{.}$

1. Let $E^0$ be the orthgonal to $E$ with respect to the form $B_M\text{.}$ Show that $E^0$ is the radical of $E,$ and that $E/E^0$ is a direct sum of absolutely simple, pairwise nonisomorphic modules and that the number of them is the same as the number of connected componenets in the graph $S\text{.}$
   
   Solution. $E$ is a real vector space with basis $e_1,\dots ,e_n\text{.}$ $W$ acts on $E$ by $${\sigma }_s\left(e\right)=e-2B_M(e,e_s)e_s,$$ for each $s\in S\text{.}$ By definition $$E^0=\{v\in E\hspace{0.17em}|\hspace{0.17em}{B}_M(v,e)=0,\hspace{0.17em}\text{for all}\hspace{0.17em}e\in E\}\text{.}$$ We must show
   1. $E^0=\text{rad}\left(E\right),$ where $\text{rad}\left(E\right)$ is the intersection of all maximal submodules of $E\text{.}$
   2. $E/E^0$ is a direct sum of modules $L_j$ corresponding to the connected components ${\Gamma }_j$ of the graph $\Gamma \text{.}$
   3. The $L_j$ are simple.
   4. The $L_j$ are pairwise nonisomorphic.
   5. The $L_j$ are absolutely simple, i.e. upon extending to $ℂ$ they remain simple $W\text{-modules.}$
   1. Let $x\in E^0$ and let $M$ be a maximal submodule of $E\text{.}$ Since $x\in E^0,$ $$\sigma \left(x\right)=x-2B_M(x,e_s)e_s=x-0=x,$$ for all $s\in S,$ so $x$ is fixed by $W\text{.}$ Suppose $x\notin M\text{.}$ Then $M+ℝx$ is a submodule of $E\text{.}$ Since $M$ is maximal we must have that $M+ℝx=E\text{.}$ So for each basis element $e_i$ there exist $m_i\in M$ and $c_i\in ℝ$ such that $$e_i=m_i+c_{i}x\text{.}$$ Thus $e_i-c_i\left(x\right)\in M$ and ${\sigma }_i(e_i-c_{i}x)=-e_i-c_{i}x\in M\text{.}$ So $e_i-c_{i}x+(-e_i-c_{i}x)=e_i\in M\text{.}$ So $M=E$ which is clearly a contradiction. So $x\in M$ and we have that $E^0\subseteq \text{rad}\left(E\right)\text{.}$ Now let $x\in \text{rad}\left(E\right)\text{.}$ We will show that $B_M(x,e)=0$ by showing that ${\sigma }_s\left(x\right)=x$ for all $s\in S\text{.}$ Suppose that there is some $s_i$ such that ${\sigma }_{s_i}\left(x\right)\ne x\text{.}$ Then $\text{rad}\left(E\right)\supseteq {\sigma }_{s_i}\left(x\right)=x-2B_M(x,e_{s_i})e_{s_i}$ and $B_M(x,e_{s_i})\ne 0$ since ${\sigma }_{s_i}\left(x\right)\ne x\text{.}$ So $e_{s_i}\in \text{rad}\left(E\right)\text{.}$ Let ${\Gamma }_i$ be the connected component of $\Gamma$ containing $e_{s_i}$ If $i$ and $j$ are adjacent in ${\Gamma }_i$ then $B_M(e_i,e_j)\ne 0$ and since ${\sigma }_{s_i}\left(e_i\right)=e_i-2B_M(e_i,e_j)e_j$ we will get that $e_j\in \text{rad}\left(E\right)\text{.}$ In this way we get $e_j\in \text{rad}\left(E\right)$ for all $j\in {\Gamma }_i\text{.}$ If $\Gamma$ is connected then ${\Gamma }_i=\Gamma$ and we get that $\text{rad}\left(E\right)=E\text{.}$ This is a contradiction since because $E$ is finite dimensional there exist maximal submodules of $E\text{.}$ So $\Gamma$ is not connnected. For each $k,$ labeling the components of $\Gamma$ let $M_j$ be the span of the basis vectors $e_k$ such that $k\in {\Gamma }_j\text{.}$ Then $M_j$ is a submodule of $E\text{.}$ Let $P=\sum _{j\ne i}{M}_j\text{.}$ Then $P\ne E\text{.}$ So $P$ is contained in some maximal submodule $M\text{.}$ All basis vectors in ${\Gamma }_i$ are contained in $\text{rad}\left(E\right)$ and hence in $M\text{.}$ So $M=E,$ but this is a contradiction. So we have that ${\sigma }_{s_i}\left(x\right)=x$ for all $s_i\in S\text{.}$ So $B_M(e_i,x)=0$ for all $i\text{.}$ So $x\in \text{rad}\left(E\right)\text{.}$ So $\text{rad}\left(E\right)=E^0\text{.}$
   2. Let ${\Gamma }_1,\dots ,{\Gamma }_k$ be the connected components of the graph $\Gamma \text{.}$ Let $L_j\subseteq E/E^0$ be the span of the cosets $e_i+E^0$ for all $i\in {\Gamma }_j\text{.}$ Then $L_j$ is a submodule of $E/E^0$ and it is clear that $E/E^0$ is a direct sum of the $L_j\text{.}$
   3. Suppose that $X$ is a nonzero submodule of $L_j\text{.}$ Let $x+E^0\in X,$ $x\notin E^0\text{.}$ Note that ${\sigma }_{s_i}\left(x\right)=x$ and $B_M(x,e_i)=0$ for all $e_i\notin {\Gamma }_j\text{.}$ Since $x\notin E^0$ there exists $e_j\in {\Gamma }_j$ such that $B_M(x,e_j)\ne 0\text{.}$ Then $${\sigma }_j(x+E^0)=x-2B_M(x,e_j)e_j+E^0\in X\text{.}$$ So $e_j+E_0\in X\text{.}$ By a similar connectivity argument as above we have that $e_{\ell }\in X$ for all $\ell \in {\Gamma }_j\text{.}$ It follows that $X=L_j\text{.}$ Thus $L_j$ is an irreducible submodule of $E/E_0\text{.}$
   4. Suppose that $\varphi :L_j\to L_k$ is a module homomorphism. Let $e_j$ be a vertex in ${\Gamma }_j$ so that $e_j+E^0\in L_j\text{.}$ Then $\varphi (e_j+E^0)\in L_k$ and so ${\sigma }_{s_j}\left(\varphi (e_j+E^0)\right)=\varphi (e_j+E^0)\text{.}$ On the other hand $${\sigma }_{s_j}\left(\varphi (e_j+E^0)\right)=\varphi \left({\sigma }_{s_j}(e_j+E^0)\right)=\varphi (-e_j+E^0)=-\varphi (e_j+E^0)\text{.}$$ So $\varphi (e_j+E^0)=E^0$ for each vertex $e_j\in {\Gamma }_j\text{.}$ Since $B_M(e_j,e_j)=1\ne 0,$ $e_j\notin E^0$ and we have that $\varphi$ must be the zero map. Thus $L_j$ and $L_k$ are not isomorphic.
   5. Proposition 1, Chapt V §2, states that if $G$ is a group, $\rho$ an irreducible representation of $G$ on a vector space $V$ and if there exists $g\in G$ such that $\rho \left(g\right)$ is a pseudoreflection then $\rho$ is absolutely irreducible. Let $e_j$ be a vertex in ${\Gamma }_j$ and let $e_i$ be a vertex of ${\Gamma }_j$ which is connected to $e_j\text{.}$ Since $${\sigma }_{s_i}(e_j+E^0)-(e_j+E^0)=-B_M(e_i,e_j)e_i+E^0,$$ and since $B_M(e_i,e_j)\ne 0$ if $i$ is connected to $j$ in ${\Gamma }_j$ it follows that ${\sigma }_{s_i}-1\ne 0$ and that ${\sigma }_{s_i}$ is a pseudo reflection on $L_j\text{.}$ Thus it follows that $L_j$ is absolutely irreducible. $◻$
   
   Notes. It seems that it would be instructive to understand this explicitly in the cases of ${\stackrel{\sim }{A}}_1$ and ${\stackrel{\sim }{A}}_2\text{.}$
4. Suppose that $\text{Card}\left(S\right)=3\text{.}$ If $s\in S,$ let $a\left(s\right)=m(u,v),$ where $\{u,v\}=S-\left\{s\right\}\text{.}$ Let $A=\sum _{s\in S}1/a\left(s\right)\text{.}$
   1. If $A>1,$ show that $B_M$ is positive degenerate (in which case $W$ is finite).
   2. If $A=1,$ show that $B_M$ is positive degenerate.
   3. If $A<1,$ show that $B_M$ is non degenerate, and of signature $(2,1)\text{.}$
   Show that in case a), the order $q$ of $W$ is given by the formula $q=4/(A-1)\text{.}$
   
   Solution. Recall that $W$ acts on a vector space $E$ over $ℝ$ with basis $e_1,\dots ,e_n$ and that the form satisfies $$B_M(e_i,e_j)=-\text{cos}\left(\frac{\pi }{m(s_i,s_j)}\right)=-c_{ij}$$ where $m(s_i,s_i)=1$ and $m(s_i,s_j)\in \mathbb{Z},$ $m(s_i,s_j)\ge 2,$ $i\ne j\text{.}$ Then the matrix of the form is $$\begin{array}{ccc}B& =& \left(\begin{array}{ccc}1& -c_{12}& -c_{13}\\ -c_{12}& 1& -c_{23}\\ -c_{13}& -c_{23}& 1\end{array}\right),\text{and}\\ \text{det}\left(B\right)& =& 1-c_{12}^2-c_{13}^2-c_{23}^2-2c_{12}{c}_{13}{c}_{23}\text{.}\end{array}$$ The determinants of the principal minors are $1-c_{ij}^2,$ $i\ne j$ and since $c_{ij}\le 1$ we have that the determinants of the principal minors are all $\ge 0\text{.}$ Thus the form $B_M$ is positive if $\text{det}\left(B\right)\ge 0\text{.}$
   1. The possibilities such that $A=1/a+1/b+1/c>1$ are $$\begin{array}{ccccc}1/2& 1/2& x& & \text{det}\left(B\right)=1-0-0-{\text{cos}}^{2}x-2·0=1-{\text{cos}}^{2}x>0\\ 1/2& 1/2& 1/3& & \text{det}\left(B\right)=1-0-0-1/4-2·0=3/4\\ 1/2& 1/3& 1/4& & \text{det}\left(B\right)=1-0-1/4-1/2-2·0=1/4\\ 1/2& 1/3& 1/5& & \text{det}\left(B\right)=1-0-1/4-{\left(\frac{1+\sqrt{5}}{4}\right)}^2-2·0=\frac{3-\sqrt{5}}{8}\end{array}$$ where $x>0\text{.}$
   2. The possibilities such that $A=1/a+1/b+1/c=1$ are $$\begin{array}{ccccc}1/2& 1/2& 0& & \text{det}\left(B\right)=1-0-0-1-2·0=0\\ 1/2& 1/3& 1/6& & \text{det}\left(B\right)=1-0-1/4-3/4-2·0=0\\ 1/2& 1/4& 1/4& & \text{det}\left(B\right)=1-0-1/2-1/2-2·0=0\\ 1/3& 1/3& 1/3& & \text{det}\left(B\right)=1-1/4-1/4-1/4-2(1/2)(1/2)(1/2)=0\end{array}$$
   3. Given that $\text{cos}\hspace{0.17em}x$ is a decreasing function for $0\le x\le \pi$ it follows from b) that $\text{det}\hspace{0.17em}B<0\text{.}$ We show that $B_M$ has signature $(2,1)\text{.}$
      Case 1. Assume that $B_M(e_i,e_j)\ne -1$ for some $i,j,$ $i\ne j\text{.}$ Without loss of generality we may assume that $B_M(e_1,e_2)=-c_{12}\ne -1\text{.}$ Then if we let $$\begin{array}{ccc}{v}_1& =& e_1,\\ v_2& =& \alpha e_1+e_2,\\ v_3& =& \beta e_1+\gamma e_2+e_3,\end{array}$$ where $\alpha =c_{12}<1$ and $\beta$ and $\gamma$ are uniquely determined such that $B_M(v_i,v_j)=0\text{.}$ Since $$B_M(v_1,v_1)=B_M(e_1,e_1)=1,B_M(v_2,v_2)=1-{\alpha }^2>0,$$ and $\text{det}\left(B\right)<0$ it follows that $B_M$ has signature $(2,1)\text{.}$
      Case 2. $B_M(e_i,e_j)=-1$ for all $i\ne j\text{.}$ Then the matrix of $B_M$ is given by $$B=\left(\begin{array}{ccc}1& -1& -1\\ -1& 1& -1\\ -1& -1& 1\end{array}\right)\text{.}$$ One checks that $$\begin{array}{ccc}\left(\begin{array}{ccc}1& -1& 0\\ 1& 1& -2\\ 1& 1& 1\end{array}\right)\left(\begin{array}{ccc}1& -1& -1\\ -1& 1& -1\\ -1& -1& 1\end{array}\right)\left(\begin{array}{ccc}1& 1& 1\\ -1& 1& 1\\ 0& -2& 1\end{array}\right)& =& \left(\begin{array}{ccc}1& -1& 0\\ 1& 1& -2\\ 1& 1& 1\end{array}\right)\left(\begin{array}{ccc}2& 2& -1\\ -2& 2& -1\\ 0& -4& -1\end{array}\right)\\ & =& \left(\begin{array}{ccc}4& 0& 0\\ 0& 12& 0\\ 0& 0& -3\end{array}\right)\end{array}$$ Thus $B_M$ has signature $(2,1)\text{.}$
      We know that $⟨{\sigma }_{s_i},{\sigma }_{s_j}⟩$ is a dihedral group of order $2m(s_i,s_j)\text{.}$ Use the formula $$\sum _{I\subseteq S}{(-1)}^I\frac{\left|W\right|}{\left|W_I\right|}=1,$$ from Ex. 5 §3. Then we have that $$\begin{array}{cc}{(-1)}^3\frac{1}{\left|W\right|}+{(-1)}^2\frac{1}{2m(s_1,s_3)}& +{(-1)}^2\frac{1}{2m(s_2,s_3)}+{(-1)}^2\frac{1}{2m(s_1,s_2)}\\ & +{(-1)}^3\frac{1}{\left|W\right|}+(-1)\frac{1}{2}+(-1)\frac{1}{2}+(-1)\frac{1}{2}+1=\frac{1}{\left|W\right|}\text{.}\end{array}$$ So $$(1/2)A-(3/2)+1-\frac{1}{\left|W\right|}=\frac{1}{\left|W\right|}\text{.}$$ Thus $A/2-1/2=2/\left|W\right|$ and $\left|W\right|=(A-1)/4\text{.}$
6. 1. Let $m$ be an integer $\ge 2$ or $+\infty \text{.}$ Show that requiring $4{\text{cos}}^2\frac{\pi }{m}\in \mathbb{Z}$ is equivalent to requiring $m\in \{2,3,4,6,+\infty \}\text{.}$
   2. Let $\Gamma$ be a discrete subgroup of $E$ of rank $\text{dim}\hspace{0.17em}E\text{.}$ Show that, if $\Gamma$ is stable under $W,$ then the integers $m(s,s\prime )$ for $s\ne s\prime$ belong to the set $\{2,3,4,6,+\infty \}\text{.}$
   3. Suppose that $m(s,t)\in \{2,3,4,6,+\infty \}$ for $s\ne t\in S\text{.}$ A family ${\left(x_s\right)}_{s\in S}$ of positive real numbers is called radicielle if they satisfy the following conditions: $$\begin{array}{ccc}m(s,t)=3& ⟹& x_s=x_t\\ m(s,t)=4& ⟹& x_s=\sqrt{2}{x}_t& \text{or}& x_t=\sqrt{2}{x}_s\\ m(s,t)=6& ⟹& x_s=\sqrt{3}{x}_t& \text{or}& x_t=\sqrt{3}{x}_s\\ m(s,t)=+\infty & ⟹& x_s=x_t& \text{or}& x_s=2x_t& \text{or}& x_t=2x_s\text{.}\end{array}$$ If $\left(x_s\right)$ is such a family, one puts ${\alpha }_s=x_s{e}_s\text{.}$ Show that one has $${\sigma }_s\left({\alpha }_t\right)={\alpha }_t-n(s,t){\alpha }_s,\text{where}\hspace{0.17em}n(s,t)\in \mathbb{Z}\text{.}$$ Deduce that the discrete subgroup $\Gamma$ with base ${\left({\alpha }_s\right)}_{s\in S}$ is stable under $W\text{.}$
   4. With hypotheses as in c), suppose that the graph $(W,S)$ is a forest. Show that then there exists at most one radicielle family $\left(x_s\right)\text{.}$
   
   Solution. a) and b) were done in class.
   3. If $m(s,t)=2$ then $$\begin{array}{ccc}{\sigma }_s\left({\alpha }_t\right)& =& {\alpha }_t-2B_M(e_s,{\alpha }_t)e_s\\ & =& {\alpha }_t-2x_t{B}_M(e_s,e_t)e_s\\ & =& {\alpha }_t-2x_t(-\text{cos}(\pi /2))\\ & =& {\alpha }_t\end{array}$$ and so $n(s,t)=0\text{.}$
      If $m(s,t)=3$ then $$\begin{array}{ccc}{\sigma }_s\left({\alpha }_t\right)& =& {\alpha }_t-2x_t{B}_M(e_s,e_t)e_s\\ & =& {\alpha }_t-2x_t(-\text{cos}(\pi /3))\\ & =& {\alpha }_t-x_t{e}_s\\ & =& {\alpha }_t+x_t{e}_s\\ & =& {\alpha }_t+{\alpha }_s\end{array}$$ and so $n(s,t)=-1\text{.}$
      If $m(s,t)=4$ then $$\begin{array}{ccc}{\sigma }_s\left({\alpha }_t\right)& =& {\alpha }_t-2x_t(-\text{cos}(\pi /4))\\ & =& {\alpha }_t+\sqrt{2}{x}_t{e}_s\\ & =& {\alpha }_t+\{\begin{array}{c}\sqrt{2}\sqrt{2}{x}_s{e}_s\\ (\sqrt{2}/\sqrt{2})x_s{e}_s\end{array}\\ & =& {\alpha }_t+\{\begin{array}{c}2{\alpha }_s\\ {\alpha }_s\end{array}\end{array}$$ So $n(s,t)=-2$ or $n(s,t)=-1\text{.}$
      If $m(s,t)=6$ then $$\begin{array}{ccc}{\sigma }_s\left({\alpha }_t\right)& =& {\alpha }_t-2x_t(-\text{cos}(\pi /6))\\ & =& {\alpha }_t+\sqrt{3}{x}_t{e}_s\\ & =& {\alpha }_t+\{\begin{array}{c}\sqrt{3}\sqrt{3}{x}_s{e}_s\\ (\sqrt{3}/\sqrt{3})x_s{e}_s\end{array}\\ & =& {\alpha }_t+\{\begin{array}{c}3{\alpha }_s\\ {\alpha }_s\end{array}\end{array}$$ So $n(s,t)=-3$ or $n(s,t)=-1\text{.}$
      If $m(s,t)=\infty$ then $$\begin{array}{ccc}{\sigma }_s\left({\alpha }_t\right)& =& {\alpha }_t-2x_t(-\text{cos}(\pi /\infty ))\\ & =& {\alpha }_t+2x_t{e}_s\\ & =& {\alpha }_t+\{\begin{array}{cc}2{\alpha }_s,& \text{if}\hspace{0.17em}{x}_t=x_s,\\ {\alpha }_s,& \text{if}\hspace{0.17em}{x}_t=(1/2)x_s,\\ 4{\alpha }_s,& \text{if}\hspace{0.17em}{x}_t=2x_s\text{.}\end{array}\end{array}$$ So $n(s,t)=-2,-1,$ or $-4\text{.}$ Note that in these cases we have that $n(s,t)n(t,s)=4\text{.}$
   4. Since the Coxeter graph is a forest we may let $s_0$ be a pendant node. By induction we may assume that there is a radicelle family for the graph $S-\left\{s_0\right\}\text{.}$ Since $s_0$ is connected to a unique node of $S-\left\{s_0\right\}$ we can certainly extend according to the rules above.
      
      Notes. Suppose that $W$ is a finite reflection group with $\left\{e_s\right\}$ a simple system such that $\left|e_s\right|=1$ for all $s\in S\text{.}$ Define $\Phi =W{\left\{{\alpha }_s\right\}}_{s\in S}$ where the ${\alpha }_s$ are as given in c) above. We will show that $\Phi \cap ℝ\alpha =\{\pm \alpha \}$ for all $\alpha \in \Phi \text{.}$ Suppose the contrary. Then we may choose a simple system $\Delta$ such that ${\alpha }_s\in \Delta$ and such that $\Phi \cap {\alpha }_s\ne \{\pm {\alpha }_s\}\text{.}$ So there exists $w\in W,$ $t\in S$ and $c\in ℝ,$ $\left|c\right|\ne 1,$ such that $w{\alpha }_t=c{\alpha }_s\text{.}$ Then $$\left|x_t\right|=\left|{\alpha }_t\right|=\left|c\right|\left|{\alpha }_s\right|\ne \left|{\alpha }_s\right|=\left|x_s\right|\text{.}$$ So $t$ and $s$ cannot be joined in the Coxeter graph by a sequence of edges labeled 3. Let $A$ be the set of elements of $S$ which canot be joined to $s$ by edges labeled 3. Let $F$ be the free group on $S$ and let $N$ be the normal closure of the set of relators for the Coxeter system so that $F/N\cong W\text{.}$ Define a map $$\begin{array}{ccccccc}\varphi :& F& ⟶& \{\pm 1\}& x& ⟼& \{\begin{array}{cc}1,& \text{if}\hspace{0.17em}x\in A,\\ -1,& \text{if}\hspace{0.17em}x\notin A\text{.}\end{array}\end{array}$$ If $x\in S$ then $$\varphi \left(x^2\right)=\varphi {\left(x\right)}^2=1$$ and if $x,y\in S$ then $$\varphi \left({\left(xy\right)}^{m(x,y)}\right)=\{\begin{array}{cc}1,& \text{if}\hspace{0.17em}m(x,y)\hspace{0.17em}\text{is even,}\\ 1,& \text{if}\hspace{0.17em}m(x,y)=3\hspace{0.17em}\text{and}\hspace{0.17em}x,y\in A\\ 1,& \text{if}\hspace{0.17em}m(x,y)=3\hspace{0.17em}\text{and}\hspace{0.17em}x,y\notin A\text{.}\end{array}$$ So $\varphi$ factors through $N$ and thus $\varphi$ is a homomorphism from $W$ to $\{\pm 1\}\text{.}$ Since $\phi \left(s\right)=1$ and $\phi \left(t\right)=-1$ it follows that ${\sigma }_{a{\alpha }_s}$ and ${\sigma }_{{\alpha }_t}$ are not conjugate in $W\text{.}$ On the other hand if the form $B_M$ is invariant under $W$ we have that $$\begin{array}{ccc}w{\sigma }_{{\alpha }_s}{w}^{-1}\left(v\right)& =& w(w^{-1}v-2B_M(w^{-1}v,{\alpha }_s){\alpha }_s)\\ & =& v-2B_M(v,w{\alpha }_s)w{\alpha }_s\\ & =& {\sigma }_{w{\alpha }_s}\\ & =& {\sigma }_{{\alpha }_t}\text{.}\end{array}$$ So ${\sigma }_{{\alpha }_s}$ and ${\sigma }_{{\alpha }_t}$ are conjugate which is a contradiction. Therefore $\Phi \cap ℝ\alpha =\{\pm \alpha \}$ for all $\alpha \in \Phi \text{.}$

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