3
$\begingroup$

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 $N \cong \mathcal{H}_{p,m} \rtimes \mathring{N}$ ?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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}

where

\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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.