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\begin{document}

\title{Noncrystallographic spherical unitary dual}

\maketitle

\section{Introduction}

Let $R$ be an arbitrary, possibly noncrystallographic root system
in a Euclidean space $V$ with Coxeter group $W$.  Thus $W$ is
either a Weyl group or of type $I_2(m)$ for $m=5$ or $m \geq 7$,
$H_3$, or $H_4$.  Fix a choice of positive roots $R^+$ and a basis
of simple roots $\al_1, \dots, \al_n$.

Fix a reduced expression for the longest element of $W$, $w_0=s_1
\cdots s_\ell$, where these indices denote the position of the
reflection in this expression and do not indicate which simple
reflection is being used.  This expression induces the ordering
$\be_1,\dots ,\be_\ell$ of the positive roots given by
$$\be_\ell=\al_\ell,\: \be_{\ell-1}=s_\ell \al_{\ell-1},\: \dots,\:
\be_1=s_\ell \cdots s_2 \al_1.$$

For each $\nu \in V$, define the following element of the group
algebra $\RR[W]$:
$$A(\nu)=(1+\langle \nu,\be_1^\vee\rangle s_1)\cdots
(1+\langle \nu,\be_\ell^\vee\rangle s_\ell).$$

\begin{lemma}
\label{le:indep}
$A(\nu)$ is independent of the choice of reduced expression for
$w_0$.
\end{lemma}
\begin{proof}
If $W$ is a Weyl group, this is well known (reference?).  Rogawski
defines a similar operator at the level of the affine Hecke
algebra and for {\it any} element in $W$ and views these as
elements in the principal series module $I(\nu)$ for the Hecke
algebra.  He mentions that if all the factors are invertible, then
in fact $A_w(\nu)$ is in the $w\nu$ weight space of $I(\nu)$. He
concludes that if $\nu$ is regular so the dimension of the $w\nu$
weight space is 1, then $A_w(\nu)$ is independent of the reduced
expression for $w$.

I have checked this for $W$ of type $I_2(m)$, $m=5,7$ using Maple
calculations.

I haven't yet tried proving this for $H_3$ and $H_4$.
\end{proof}

\begin{lemma}
If $w_0 \nu=-\nu$, then for the orthogonal form, $\rho$, of any
representation of $W$, $\rho(A(\nu))$ is symmetric.
\end{lemma}
\begin{proof}
First notice that since $w_0=w_0^{-1}$, using indices as described above,
\begin{align*}
\be_k^\vee&=s_\ell \cdots s_{k+1}\al_k^\vee \\
&=s_\ell \cdots s_{k+1}s_k \cdots s_1(s_1 s_2 \cdots s_k\al_k^\vee) \\
&=-w_0(s_1 s_2 \cdots s_{k-1}\al_k^\vee).
\end{align*}
Thus
$$\langle \nu,\be_k^\vee\rangle=\langle -w_0\nu,s_1 s_2 \cdots s_{k-1}\al_k^\vee\rangle
=\langle \nu,\tilde{\be}_{\ell-k+1}^\vee\rangle,$$
where $\tilde{\be}_1^\vee,\dots ,\tilde{\be}_\ell^\vee$ denotes
the ordering induced by $w_0=s_\ell \cdots s_1$.
This yields that
\begin{align*}
\rho(A(\nu))^T&=(1+\langle \nu,\be_\ell^\vee\rangle \rho(s_\ell))^T\cdots
(1+\langle \nu,\be_1^\vee\rangle \rho(s_1))^T \\
&=(1+\langle \nu,\be_\ell^\vee\rangle \rho(s_\ell))\cdots
(1+\langle \nu,\be_1^\vee\rangle \rho(s_1)) \\
&=(1+\langle \nu,\tilde{\be}_1^\vee\rangle \rho(s_\ell))\cdots
(1+\langle \nu,\tilde{\be}_\ell^\vee\rangle \rho(s_1)).
\end{align*}
Since this last expression is the operator with respect to the
reduced decomposition $w_0=s_\ell \cdots s_1$, this equals
$\rho(A(\nu))$ by Lemma~\ref{le:indep}.
\end{proof}

Let $V_0^+$ be the set of all dominant $\nu$ fixed by $-w_0$.

\section{Unitarity}

Include description of star structure on graded Hecke algebra.  Maybe this
should go before the lemmas above.

\begin{remark}
GOAL:
We wish to find all $\nu \in V_0^+$ so that $\rho(A(\nu))$ is
positive semi-definite (psd) for all unitary irreducible
representations $\rho$ of $W$.
\end{remark}

The following definition is modified slightly from Stembridge's notes.
\begin{definition}
A {\it cell} is a non-empty set obtained by selecting one of
$$\{\nu \in V \mid \langle \nu, \be^\vee\rangle >1\}, \;
\{\nu \in V \mid \langle \nu, \be^\vee\rangle =1\}, \;
\{\nu \in V \mid 0< \langle \nu, \be^\vee\rangle <1\}$$
for each positive root $\be$, and taking the intersection.
\end{definition}

\begin{remark}
ASSUMPTION:
We assume that the answer to whether $\rho(A(\nu))$ is psd is the
same for all $\nu$ in the same cell.  In other words, it suffices
to test $\rho(A(\nu))$ at one representative $\nu$ in each cell.
Proof????

This makes it possible to achieve the goal above through finitely
many calculations, except for the infinite nature of the dihedral
family.
\end{remark}


\section{The odd dihedral case}
We consider the case $I_2(m)$ where $m$ is odd.

Let $w_0=s_1 s_2 s_1 \cdots s_2 s_1$.  Since $w_0$ is a product of
an odd number of reflections, it is a reflection.  For
$\nu=\be_{\frac{m+1}{2}}$, $w_0(\nu)=-\nu$.  To see this, write
%\begin{alignat*}{2}
$$\be_{\frac{m+1}{2}}=s_1 s_2 \cdots s_i \al_j, \qquad \hbox{ where }
\left\{%
\begin{array}{ll}
    i=1, j=2, & \hbox{if $m=2k+1$ and $k$ is odd,} \\
    i=2, j=1, & \hbox{if $m=2k+1$ and $k$ is even,} \\
\end{array}%
\right.
$$
%\quad
%i&=1, j=2 &\quad &\hbox{if $m=2k+1$ and $k$ is odd, and} \\
%i&=2, j=1 &\quad &\hbox{if $m=2k+1$ and $k$ is even,}\end{alignat*}
so that
$$w_0\left(\be_{\frac{m+1}{2}}\right)=s_1 s_2 \cdots s_2 s_1(s_1 s_2 \cdots s_i \al_j)
=s_1 s_2 \cdots s_j \al_j=-s_1 s_2 \cdots s_i \al_j.$$ Thus the
only dominant $\nu$ for which $w_0(\nu)=-\nu$ are those that lie
on the central line in the fundamental chamber or equivalently,
the space $V_0^+$ of dominant $\nu$ fixed by $-w_0$ consists of
vectors of the form
$$\nu=a\be_{\frac{m+1}{2}}=a(\om_1+\om_2) \quad \hbox{for } a \in \RR.$$

This choice of expression for $w_0$ induces the ordering
$$\be_1=s_1 s_2 \cdots s_2 \al_1,\: \be_2=s_1 s_2 \cdots s_1 \al_2,\:
\dots,\: \be_{m-1}=s_1 \al_2,\: \be_m=\al_1.$$  By symmetry in the line
described above,
$$A(\nu)=(1+\langle \nu,\be_1^\vee\rangle s_1)(1+\langle \nu,\be_2^\vee\rangle s_2)\cdots
(1+\langle \nu,\be_2^\vee\rangle s_2)(1+\langle \nu,\be_1^\vee\rangle s_1),$$
where the indices now denote simple reflections.

Include a picture of the dominant cells for $I_2(m)$, $m$ odd, here.
Write $\nu=\nu_k$ if
$\langle \nu,\be_k^\vee\rangle = \langle \nu,\be_{m-k+1}^\vee\rangle = 1$
and $\nu=\nu_{ij}$ if
$\langle \nu,\be_i^\vee\rangle = \langle \nu,\be_j^\vee\rangle = 1$.

\begin{theorem}
For $W$ of type $I_2(m)$, $m$ odd, $\rho(A(\nu))$ is positive
semidefinite for all irreducible unitary representations $\rho$
of $W$ precisely when
$$\left\langle \nu,\be_{\frac{m+1}{2}}^\vee\right\rangle \leq 1
\hbox{ or } \nu = \nu_k \hbox{ for } 1 \leq k \leq \frac{m-1}{2}.$$
\end{theorem}
Notice that $\nu_k=\ga_{c,k}$ in the notation of~\cite{KR} (there
the positive roots $\be_k$ are numbered $0,1,\dots,m-1$).
\begin{proof}
It is easy to see that these conditions are necessary by considering
the case when $\rho$ is the sign representation,
$\rho_{sgn}(s_1)=\rho_{sgn}(s_2)=-1$.  Then
$$\rho_{sgn}(A(\nu))=(1-\langle \nu,\be_1^\vee\rangle)^2(1-\langle \nu,\be_2^\vee\rangle)^2
\cdots\left(1-\left\langle \nu,\be_{\frac{m-1}{2}}^\vee\right\rangle\right)^2
\left(1-\left\langle \nu,\be_{\frac{m+1}{2}}^\vee\right\rangle\right)$$
and $\rho_{sgn}(A(\nu))\geq 0$ precisely when the stated conditions hold.

Clearly $\rho_{triv}$ is psd for all dominant $\nu$ and $\rho(A(0))=1$ is
psd for all irreducible representations $\rho$ of $W$.

It remains to check that $\rho(A(\nu))$ is psd for each
2-dimensional representation $\rho$ at each $\nu_k$ and at the
representatives
$$\tfrac{1}{2}\left(\nu_{\frac{m-1}{2},\frac{m+1}{2}}+\nu_{\frac{m+1}{2},\frac{m+3}{2}}\right)$$
and
$$\tfrac{1}{3}\left(\nu_{\frac{m-1}{2},\frac{m+1}{2}}+\nu_{\frac{m+1}{2},\frac{m+3}{2}}\right)$$
for the one and two dimensional cells that contain points satisfying
$\left\langle \nu,\be_{\frac{m+1}{2}}^\vee\right\rangle \leq 1$.

Maple calculations support that this is true for $m=5$ and for the
reflection representation for $m=7$.  I do not see how to prove it
in general by hand using the form of the operator except in the
case of $\rho_{ref}$ and $\nu_1$ (the central character of the Steinberg
representation of the full algebra).  In this case, since
$\rho_{ref}(s_1)=\begin{pmatrix}
-1 & 0 \\ 0 & 1
\end{pmatrix}$, the first and last factors are
$(I+\rho_{ref}(s_1))=\begin{pmatrix}
0 & 0 \\ 0 & 2
\end{pmatrix}$,
so $\rho_{ref}(A(\nu))=\begin{pmatrix}
0 & 0 \\ 0 & 0
\end{pmatrix}$.
When $k$ is odd but not 1, there are two symmetrically placed factors
of the form $\begin{pmatrix}
0 & 0 \\ 0 & 2
\end{pmatrix}$
in the product, but these yield little information in general about the form
of the matrix for the product.
When $k$ is even, these factors are even more complicated since
$\rho_{ref}(s_1)=\begin{pmatrix}
-\cos(2\pi/m) & \sin(2\pi/m) \\ \sin(2\pi/m) & \cos(2\pi/m)
\end{pmatrix}$ and it does not seem at all obvious that even $\rho_{ref}(A(\nu))$ is psd.

I also have not yet considered explicit descriptions of the other 2 dimensional
representations (even in the m=7 case, where I forgot about the other one!).
Perhaps for the general result it would be better to consider the explicit
representation of the graded Hecke algebra directly.
\end{proof}

\section{The case of $H_3$}

Include proof that antichains or order ideals in the root poset
correspond to the biggest cells and perhaps list the antichains or
order ideals that can be used to describe minimal sets of defining
inequalities for the big cells.

Figure out some combinatorial way to describe all the other cells, maybe
by understanding Stembridge's description of coloring.

\section{The case of $H_4$}

\begin{thebibliography}{[KR]}

\bibitem[KR]{KR} C.~Kriloff and A.~Ram, \textit{Representations
of graded Hecke algebras}, Represent. Theory \textbf{6} (2002), 31--69.

\end{thebibliography}

\end{document}
