# odd length Chevalley relations (in rank two)

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:

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

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

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

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

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

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

Now we compute the two relevant length five products:

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

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

$$\cdots \ \ \text{and} \ \ \cdots$$

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

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

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

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

together with the one-parameter subgroups

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

The length five products are then:

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

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

$$----- \ \text{and} \ -----$$

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

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

yours, Ines

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

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 ($*$)

$$\big(ab + ad + cd\big) \, \big(bcde\big) \ + \ \big(abcd \big) \, \big( bd + be + de \big) \ - \ \big(ade + abe + cde + abc \big) \, \big( bcd \big)$$

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

Suppose your system can be solved then we know

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

--- 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

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

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

best, A. Leverkühn