4
$\begingroup$

The unipotent radicals $\text{N}$ of the Borel subgroups of the complex algebraic groups of type $A_2$, $B_2$, and $G_2$ can each be abstractly presented using two one-parameter subgroups $x_1, x_2: \Bbb{C} \longrightarrow \text{N}$ subject to the following defining relations, namely:

\begin{equation} \begin{array}{llc} x_1(a)\, x_2(b) \, x_1(c) &= \ x_2 \Bigg( \displaystyle {bc \over {a+c}} \Bigg) \, x_1 \Big(a + b\Big) \, x_2 \Bigg( {ab \over {a+c}}\Bigg) &\text{for type $A_2$} \\ \\ x_1(a) \, x_2(b) \, x_1(c) \, x_2(d) &= \ x_2 \Bigg(\displaystyle {bc^2d \over {\pi_2}} \Bigg) \, x_1 \Bigg( {\pi_2 \over {\pi_1}} \Bigg) \, x_2 \Bigg( {\pi_1^2 \over \pi_2} \Bigg) \, x_1 \Bigg( {abc \over \pi_1} \Bigg)&\text{for type $B_2$} \\ \\ &\text{where} \ \pi_1 = ab + \big(a +c \big)d & \\ &\text{and} \ \pi_2 = a^2b + \big(a +c \big)^2d & \end{array} \end{equation}

For $G_2$ a six term relation is valid which converts a factorisation $x_1(a)x_2(b)x_1(c)x_2(d)x_1(e)x_2(f)$ into a product of the form $x_2(A)x_1(B)x_2(C)x_1(D)x_2(E)x_1(F)$ where $A$, $B$, $C$, $D$, $E$, and $F$ are subtraction-free rational expressions in the complex parameters $a$, $b$, $c$, $d$, $e$, and $f$ --- see the paper "Total positivity in Schubert varieties" by Arkady Berenstein and Andrei Zelevinsky.

Using a truncated version of the $\text{SL}_2 \Big( \Bbb{C} \Big)$ loop-group, namely $\text{SL}_2 \Big( \mathcal{L_d} \Big)$ where $\mathcal{L_d}$ is the Auslander ring $\Bbb{C}\big[ t \big] \Big/ \Big( t^d = 0 \Big)$ one can check the existence of similar relations of length $2d$ for $d \geq 1$; the cases of $d=1$ and $d=2$ correspond to types $B_2$ and $G_2$ respectively. Hint: interpret the abstract one-parameter groups $x_1$ an $x_2$ as the $\mathcal{L}_d$-valued $2 \times 2$ matrices

\begin{equation} x_1(a) = \begin{pmatrix} 1 & a \\ 0 & 1\end{pmatrix} \qquad x_2(a) = \begin{pmatrix} 1 & 0 \\ at & 1 \end{pmatrix} \end{equation}

These relations correspond to coxeter relations of length $2d$ within the group of dihedral symmetries of a regular $2d$-gon.

Question: Is there a length five relation of this kind ? If so, is there a nice representation of the corresponding nilpotent group ?

best, A. Leverkühn

$\endgroup$
3
$\begingroup$

Lieber Señor Leverkühn,

Instead of using the Auslander ring $\Bbb{C}\big[ t \big] \Big/ \Big( t^d = 0 \Big)$ use a two variable version $\mathcal{L}_{2+3} = \Bbb{C}\big[s,t \big]\Big/ \Big(s^3t^2 = s^2t^3 = 0 \Big)$ and form the truncated "double" loop group $\text{SL}_2 \Big( \mathcal{L}_{2+3} \Big)$. As in your construction let's define two one-parameter subgroups, namely

\begin{equation} x_1(a) := \begin{pmatrix} 1 & as \\ 0 & 1 \end{pmatrix}\quad \text{and} \quad x_2(a) := \begin{pmatrix} 1 & 0 \\ at & 1 \end{pmatrix}. \end{equation}

Now we compute the two relevant length five products:

\begin{equation} \begin{pmatrix} 1 & as \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ bt & 1 \end{pmatrix}\begin{pmatrix} 1 & cs \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ dt & 1 \end{pmatrix}\begin{pmatrix} 1 & es \\ 0 & 1 \end{pmatrix} \ = \end{equation}

\begin{equation} \begin{pmatrix} 1 + \big(ab + ad + cd \big)st + \big(abcd \big)s^2t^2 & \big(a + c + e \big)s + \big(ade + abe + cde + abc\big)s^2t \\ \big(b + d \big)t + \big(bcd \big)st^2 & 1 + \big( bc + be + de \big)st + \big( bcde \big)s^2t^2 \end{pmatrix} \end{equation}

\begin{equation} \cdots \ \ \text{and} \ \ \cdots \end{equation}

\begin{equation} \begin{pmatrix} 1 & 0 \\ At & 1 \end{pmatrix}\begin{pmatrix} 1 & Bs \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ Ct & 1 \end{pmatrix}\begin{pmatrix} 1 & Ds \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ Et & 1 \end{pmatrix}\ = \end{equation}

\begin{equation} \begin{matrix} 1 + \big( BC + BE + DE \big)st + \big( BCDE \big)s^2t^2 & \big(B + D \big)s + \big(BCD \big)s^2t \\ \big(A + C + E \big)t + \big(ADE + ABE + CDE + ABC \big)st^2 & 1 + \big(AB + AD + CD \big)st + \big( ABCD \big)s^2t^2 \end{matrix} \end{equation}

Now solve for $A$, $B$, $C$, $D$, and $E$ to obtain the conjectured "dihedral" relation.

yours always, Ines.


Ok, Here's a possible remedy which may avoid the inconsistency that you've discovered. Instead of $\mathcal{L}_{2+3}$ consider

\begin{equation} \mathcal{L}_{4} \ = \ \Bbb{C}\big[t \big]\Big/ \Big(t^4 = 0 \Big) \end{equation}

together with the one-parameter subgroups

\begin{equation} x_1(a) := \begin{pmatrix} 1 & at \\ 0 & 1 \end{pmatrix}\quad \text{and} \quad x_2(a) := \begin{pmatrix} 1 & 0 \\ at & 1 \end{pmatrix}. \end{equation}

The length five products are then:

\begin{equation} \begin{pmatrix} 1 & at \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ bt & 1 \end{pmatrix}\begin{pmatrix} 1 & ct \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ dt & 1 \end{pmatrix}\begin{pmatrix} 1 & et \\ 0 & 1 \end{pmatrix} \ = \end{equation}

\begin{equation} \begin{pmatrix} 1 + \big(ab + ad + cd \big)t^2 & \big(a + c + e \big)t + \big(ade + abe + cde + abc\big)t^3 \\ \big(b + d \big)t + \big(bcd \big)t^3 & 1 + \big( bc + be + de \big)t^2 \end{pmatrix} \end{equation}

\begin{equation} ----- \ \text{and} \ ----- \end{equation}

\begin{equation} \begin{pmatrix} 1 & 0 \\ At & 1 \end{pmatrix}\begin{pmatrix} 1 & Bt \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ Ct & 1 \end{pmatrix}\begin{pmatrix} 1 & Dt \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ Et & 1 \end{pmatrix}\ = \end{equation}

\begin{equation} \begin{pmatrix} 1 + \big( BC + BE + DE \big)t^2 & \big(B + D \big)t + \big(BCD \big)t^3 \\ \big(A + C + E \big)t + \big(ADE + ABE + CDE + ABC \big)t^3 & 1 + \big(AB + AD + CD \big)t^2 \end{pmatrix} \end{equation}

yours, Ines

$\endgroup$
  • $\begingroup$ sorry ... I couldn't fit the matrix parenthesis into the last line. $\endgroup$ – Ines Institoris Feb 4 '16 at 23:06
0
$\begingroup$

Your idea was promising but the system is inconsistent under the assumption that the initial factorisation parameters $a$, $b$, $c$, $d$, and $e$ are algebraically independent (which must be the case since they ought to be local coordinates on a 5-dimensional nilpotent group). Here's how to see inconsistency:

Note first the following polynomial identity ($*$)

\begin{equation} \big(ab + ad + cd\big) \, \big(bcde\big) \ + \ \big(abcd \big) \, \big( bd + be + de \big) \ - \ \big(ade + abe + cde + abc \big) \, \big( bcd \big) \end{equation}

\begin{equation} = \ \big( abcde \big)\big( b + d \big) \ = \ {\big(abcd \big) \, \big(bcde \big) \over {\big(bcd \big)}} \, \big(b +d \big) \end{equation}

Suppose your system can be solved then we know

\begin{equation} \begin{array}{ll} ab + ad + cd &= &BC + BE + DE \\ bd + be + de &= &AB + AD + CD \\ abcd &= &BCDE \\ bcde &= &ABCD \\ ade + abe + cde + abc &= &BCD \\ bcd &= &ADE + ABE + CDE + ABC \\ b + d &= &A + C + E \\ a + c + e &= &B + D \end{array} \end{equation}

--- Call this system of identities ($\dagger$).

The polynomial identity ($*$) is clearly valid when we replace $a$, $b$, $c$, $d$, and $e$ by $A$, $B$, $C$, $D$, and $E$ respectively --- call it ($**$). Moreover the left-hand (and right-hand) sides of both ($*$) and ($**$) are equal in view of the substitutions that ($\dagger$) permits. As long as $abcde \ne 0$ we may conclude that

\begin{equation} {b +d \over {bcd }} \ = \ {B + D \over {BCD}} \ = \ {a + c + e \over{ade + abe +cde +abc} }\end{equation}

which of course violates the independence of $a$, $b$, $c$, $d$, and $e$.

best, A. Leverkühn

$\endgroup$

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.