Consider a directed coxeter diagram $\vec{\Gamma}$, i.e. a finite graph where each edge is decorated with one of the integer weights $\big\{3,4,6\big\}$ and those edges with weights $4$ or $6$ are assigned orientations. Let $p$ be a vertex of $\vec{\Gamma}$ attached to a leaf $q$ and let $m= m_{p,q}$ be the weight of the corresponding edge. Let $\mathring{\vec{\Gamma}}$ be the induced coxeter diagram obtained by removing the leaf $q$.

Let $G$ and $\mathring{G}$ be the Kac-Moody groups associated to $\vec{\Gamma}$ and $\mathring{\vec{\Gamma}}$ respectively. Choose Borel subgroups $B$ and $\mathring{B}$ of $G$ and $\mathring{G}$ and let $N$ and $\mathring{N}$ denote their associated unipotent radicals.

Question: Is there a representation $\mathcal{H}_{p,m}$ of $\mathring{N}$ such that $N \cong \mathcal{H}_{p,m} \rtimes \mathring{N}$ ?

  • $\begingroup$ The tag 'lie-groups' here seems inappropriate, since Lie groups don't in general have an intrinsic Jordan decomposition (in particular, "unipotent radical" isn't generally defined). Maybe 'kac-moody-algebras' or 'algebraic-groups'? So far there isn't a tag for Kac-Moody groups. $\endgroup$ – Jim Humphreys Mar 10 '16 at 15:59
  • $\begingroup$ Also, your notion of "Coxeter diagram" is much more limited than the usual definition of "Coxeter graph" (or "Coxeter matrix") and creates ambiguity in passing to Lie algebras or groups: non-isomorphic ones may have isomorphic Coxeter groups attached. $\endgroup$ – Jim Humphreys Mar 15 '16 at 18:55
  • $\begingroup$ I realise this is more narrow. However, once orientations are assigned to the type $B_2$ and type $G_2$ edges the corresponding unipotent radical is well defined --- in fact it can be constructed directly by generators and relations alone without recourse to either the Lie algebra or the enveloping Kac-Moody group. best, A. Leverkühn $\endgroup$ – A. Leverkuhn Mar 15 '16 at 19:13

To A. Leverkühn,

This is not a general answer to your question but rather a nice confirmation in the case of $\vec{\Gamma} = B_n $. The dynkin diagram $B_n$ has two leaves --- one joined by a type $A_2$-bond (weight $3$ by your convention) and another joined by a type $B_2$-bond (weight $4$ by your convention). So there ought to be two distinct semi-direct product factorisations of the unipotent radical $N_{B_n}$ according to your conjecture.

As I noted in your other posting the Kac-Moody group $G_{B_n}$ in this case is the symplectic group $\text{Sp}_{2n} \big( \Bbb{C} \big)$ whose unipotent radical $N_{B_n}$ can be identified as the group of all $2n \times 2n$ invertible complex matrices having the following block-decomposition

\begin{equation} \left( \begin{array}{c|c} A & A \dot{w} S \\ \hline \Bbb{O} & \dot{w}A^{-T} \dot{w} \\ \end{array} \right) \end{equation}

where $A$ is any $n \times n$ unipotent matrix (i.e. upper-triangular with $1$'s on the diagonal), $S$ is any $n \times n$ symmetric matrix, and $\dot{w}$ is the $n \times n$ permutation matrix of $w \in S_n$ defined by $w(i) = n+1 -i$ for $1 \leq i \leq n$.

Consider first the case of the leaf attached by the $B_2$-bond. Let $H$ denote the abelian (indeed additive) subgroup of $N_{B_n}$ consisting of matrices of the form

\begin{equation} \left( \begin{array}{c|c} \Bbb{I} & \dot{w} S \\ \hline \Bbb{O} & \Bbb{I} \\ \end{array} \right) \end{equation}

It's easy to see that $H$ is normal in $N_{B_n}$; furthermore it is abelian and isomorphic to the vector space $\mathcal{H}$ consisting of all symmetric complex matrices. The quotient $N_{B_n} \big/ H$ is isomorphic to the group $N_{A_{n-1}}$ of all $n \times n$ unipotent complex matrices which acts linearly on $\mathcal{H}$ by

\begin{equation} A \cdot S \ = \ \dot{w} \, A \,\dot{w} \, S \, \dot{w} \, A^{T} \, \dot{w} \end{equation}

and consequently we have the semi-direct product factorisation $N_{B_n} \cong \mathcal{H} \rtimes N_{A_{n-1}}$ expressed as

\begin{equation} \big(S_1,A_1\big) \circ \big(S_2,A_2 \big) \ = \ \Big(S_2 + A_2^{-1} \cdot S_1 \, , \, A_1A_2 \Big) \end{equation}

Consider now the leaf with the type $A_2$-bond. In view of your conjecture we should expect a semi-direct factorisation involving $N_{B_{n-1}}$ instead of $N_{A_{n-1}}$. To effect this decomposition let's embed $\text{Sp}_{2n-2}\big( \Bbb{C} \big)$ into the central $(2n-2) \times (2n-2)$ block of $\text{Sp}_{2n} \big( \Bbb{C} \big)$, namely

\begin{equation} \left( \begin{array}{c|c|c} 1 & & \\ \hline & \text{Sp}_{2n-2}\big( \Bbb{C} \big) & \\ \hline & & 1 \end{array} \right) \end{equation}

Let $\acute{H}$ be the subgroup of $N_{B_n}$ consisting of matrices having the block-form

\begin{equation} \left( \begin{array}{c|c} U & V \\ \hline \Bbb{O} & \dot{w}U^{-T}\dot{w} \\ \end{array} \right) \end{equation}


\begin{equation} U \ = \ \left( \begin{array}{c|l} 1 & \vec{u}^{ \ \scriptscriptstyle T} \\ \hline \vec{0} & \Bbb{I} \\ \end{array} \right) \quad \text{and} \quad V \ = \ \left( \begin{array}{l|c} \vec{v}^{\ \scriptscriptstyle T} & v \\ \hline \Bbb{O} & \vec{v} \\ \end{array} \right) \end{equation}

for any pair of vectors $\vec{u}$ and $\vec{v}$ in $\Bbb{C}^{n-1}$ and any $v \in \Bbb{C}$. Again $\acute{H}$ is a normal (although non-abelian), it trivially intersects $N_{B_{n-1}}$ with respect to the embedding of $\text{Sp}_{2n-2}\big( \Bbb{C} \big)$ in $\text{Sp}_{2n}\big( \Bbb{C} \big)$, and $ N_{B_n} = \acute{H} \, N_{B_{n-1}}$. Let $\acute{\mathcal{H}}$ denote the $\log$ of $\acute{H}$ so to speak --- the vector space $\Bbb{C}^{n-1} \oplus \Bbb{C}^{n-1} \oplus \Bbb{C}$ consisting of all tuples $\big(\vec{u}, \vec{v}, v \big)$. The conjugation action of $N_{B_{n-1}}$ on $\acute{H}$ induces a linear action of $N_{B_{n-1}}$ on $\acute{\mathcal{H}}$ and so we get a second semi-direct product factorisation $N_{B_n} \cong \acute{\mathcal{H}} \rtimes N_{B_{n-1}}$ as anticipated with the caveat however that although $\acute{\mathcal{H}}$ is a representation of $N_{B_{n-1}}$ its group structure does not coincide with its additive structure as a vector space.

yours as always, Ines


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