presentation for a nilpotent group associated to the square of a coxeter element

Question

This question is related to one asked earlier about inductive presentations of unipotent radicals in Kac-Moody groups.

Let $\Gamma$ be a coxeter diagram --- i.e. an unoriented graph with $r$ vertices whose edges are labeled by the integer weights $\big\{3,4,6 \big\}$. Let $\vec{\Gamma}$ be a directed version of $\Gamma$ where each edge with weight $4$ or $6$ is assigned an orientation. Let $W$ and $G$ respectively denote the coxeter group and the Kac-Moody group associated to $\Gamma$ and $\vec{\Gamma}$; in addition choose an opposing pair of Borel subgroups $B_{\pm}$ of $G$ and let $N_{\pm}$ denote the corresponding unipotent radicals. Given $w \in W$ let $\dot{w}$ denote a lifting of $w$ to $G$ and consider the subgroup $N(w) := N \cap \big( \dot{w} \big)^{-1} N_{-} \big( \dot{w} \big)$ of $N$; it is a finite dimensional nilpotent group which, as a variety, is isomorphic to $\Bbb{C}^{l(w)}$ where $l(w)$ is the length of $w$.

Question Let $c = \sigma_1 \dots \sigma_r$ be a choice of coxeter element of $W$ and consider $N\big(c^2 \big)$. Is there way to read off a presentation of this group from $\Gamma$; in a manner analagous to the Chevalley presentation of $N$ (but with added relations) ?

regards, A. Leverkühn

p.s. This question only becomes interesting in the case that $l \big(c^2 \big) = 2 \cdot l(c)$ otherwise $N \big( c^2 \big)$ will be abelian. For example type $\Gamma = A_r$ is not interesting while type $\Gamma = D_r$ for $r \geq 4$ is. Is it possible that $N \big( c^2 \big) \cong \ N\big( c \big) \rtimes N \big( c \big)$ when this length condition is satisfied ?

Answer 1

This is not an answer to your main question but response to your post script remark. In general it is not the case that $N \big(c^2 \big) \cong N(c) \rtimes N(c)$. Consider the case of $\vec{\Gamma} = B_2$. The Kac-Moody group in this case is the symplectic group $\text{Sp}_{4}\big( \Bbb{C} \big)$; namely all $4 \times 4$ invertible complex matrices $g$ for which

\begin{equation} g^T \, \left( \begin{array}{r|c} \Bbb{O} & \dot{w} \\ \hline -\dot{w} & \Bbb{O} \\ \end{array} \right) \, g \ = \ \left( \begin{array}{r|c} \Bbb{O} & \dot{w} \\ \hline -\dot{w} & \Bbb{O} \\ \end{array} \right) \end{equation}

where $\dot{w}$ is the $2 \times 2$ permutation matrix of the transposition $w = \big(1 \, 2 \big)$ in $S_2$. The corresponding unipotent radical $N$ can be expressed as the group of all $4 \times 4$ matrices $g$ possessing a block-decomposition of the form

\begin{equation} \left( \begin{array}{cc|cr} 1 &a &ax + z &az + y \\ 0 &1 &x &z \\ \hline 0 &0 &1 &-a \\ 0 &0 &0 &1 \end{array} \right) \end{equation}

for any choice or complex parameters $a$, $x$, $y$, and $z$. The square-coxeter subgroup $N \big(c^2 \big)$ in this case coincides with $N$. On the other hand $N(c)$ will be the two-dimensional subgroup consisting of all $4 \times 4$ matrices with block-form

\begin{equation} \left( \begin{array}{cc|cr} 1 &0 &z &0\\ 0 &1 &x &z \\ \hline 0 &0 &1 &0 \\ 0 &0 &0 &1 \end{array} \right) \end{equation}

One can check that $N(c)$ is not normal in $N$ which immediately rules out your conjecture. However the subgroup $H$ consisting of matrices of the form

\begin{equation} \left( \begin{array}{cc|cr} 1 &0 &z &y \\ 0 &1 &x &z \\ \hline 0 &0 &1 &0 \\ 0 &0 &0 &1 \end{array} \right) \end{equation}

is normal in $N$ and indeed there is a semi-direct product decomposition $N \cong H \rtimes \Bbb{C}$ where we view $\Bbb{C}$ as an additive group; which is related to a question you raised in another posting.

yours, Ines