Generalized Gelfand Pairs Associated to m-Step Nilpotent Lie Groups

Abstract

Let N be a nilpotent Lie group and K a compact subgroup of the automorphism group Aut(N) of N. It is well known that if $(K\ltimes N,K)$ is a Gelfand pair then N is at most 2-step nilpotent Lie group. This notion has been generalized to non-compact groups K. In this work, we exhibit a family $(K_m\ltimes N_m,K_m)$ of generalized Gelfand pairs, where $N_m$ is a $m+2$-step nilpotent Lie group and $K_m$ is isomorphic to ${\mathbb {R}}^{m+1}$.

Appendix

In the following, we denote by $E_{i,j}$ the matrix of $M_{m+3}({\mathbb {R}})$ that has 1 in the (i, j) position and 0 elsewhere.

Lemma 1

The map $\Phi :N_m\rightarrow M_m$ defined by

$$\begin{aligned}&\Phi (s_m,\ldots ,s_1,x,y,t)\\ {}&=\left( \begin{array}{cccccccc} 1&{}\quad 0&{}\quad \ldots &{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad (-1)^ms_m\\ x&{}\quad 1&{}\quad \ldots &{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad (-1)^{m-1}s_{m-1}+(-1)^{m}s_{m}x\\ \vdots &{}\quad \ddots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ \frac{x^{m-2}}{m-2!}&{}\quad \ldots &{}\quad x&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad s_2-s_3x+\sum \limits _{i=2}^{m-2}s_{2+i}\frac{(-x)^i}{i!}\\ \frac{x^{m-1}}{m-1!}&{}\quad \ldots &{}\quad \frac{x^2}{2!}&{}\quad x&{}\quad 1&{}\quad 0&{}\quad 0&{}\quad -s_1+s_2x-\sum \limits _{i=2}^{m-1}s_{1+i}\frac{(-x)^i}{i!}\\ \frac{x^{m+1}}{m+1!}&{}\quad \ldots &{}\quad \frac{x^4}{4!}&{}\quad \frac{x^3}{3!}&{}\quad \frac{x^2}{2!}&{}\quad 1&{}\quad x&{}\quad t+ \frac{xy}{2}-\sum \limits _{i=1}^ms_i\frac{(-x)^{i+1}}{i+1!}\\ \frac{x^m}{m!}&{}\quad \ldots &{}\quad \frac{x^3}{3!}&{}\quad \frac{x^2}{2!}&{}\quad x&{}\quad 0&{}\quad 1&{}\quad y-s_1x+\sum \limits _{i=2}^ms_i\frac{(-x)^i}{i!}\\ 0&{}\quad \ldots &{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 1 \end{array}\right) \end{aligned}$$

is an isomorphism of Lie groups.

Proof

The proof is by induction on m. For $(s'_m,s'_{m-1},\ldots ,s'_1,x',y',t')\in N_m$ we set

$$\begin{aligned} X'=\left( \begin{array}{c} x'\\ \frac{x'^2}{2}\\ \vdots \\ \frac{x'^{m-1}}{m-1!}\\ \frac{x'^{m+1}}{m+1!}\\ \frac{x'^m}{m!}\\ 0 \end{array}\right) , \ n'_{m-1}=(s'_{m-1},\ldots ,s'_1,x',y',t'). \end{aligned}$$

We assume that $\Phi _{m-1}:N_{m-1}\rightarrow M_{m-1}$ is an isomorphism of groups. Since $s_m$ defines an automorphism on $M_{m-1}$ and $s_m\circ \Phi _{m-1}=\Phi _{m-1}\circ s_m$, we have that

$$\begin{aligned} \Phi \left( 0,s_m\cdot n'_{m-1}\right)&=\left( \begin{array}{cc} 1 &{} 0\\ X' &{} \Phi _{m-1}(s_m\cdot (n'_{m-1})) \end{array}\right) \\ \\&=\left( \begin{array}{cc} 1 &{}\quad 0\\ X' &{}\quad s_m\cdot \Phi _{m-1}(n'_{m-1}) \end{array}\right) \\ \\&=\left( \begin{array}{cc} 1 &{}\quad 0\\ X' &{}\quad \Phi _{m-1}(n'_{m-1})+(-1)^{m-1}s_m\left( \begin{array}{cc} 0&{}X' \end{array}\right) \end{array}\right) \end{aligned}$$

where we have used the action of $s_m$ on $M_{m-1}$. So, as $\Phi (s_m,0,\ldots ,0)=I+(-1)^ms_m E_{1,m+3}$, we have that

$$\begin{aligned} \Phi \left( 0,s_m\cdot (s'_{m-1},\ldots ,s'_1,x',y',t')\right) \Phi (s_m,0,\ldots ,0)&=\left( \begin{array}{cc} 1 &{}\quad 0 \cdots 0 \ (-1)^ms_m\\ X' &{}\quad \Phi _{m-1}(n'_{m-1}) \end{array}\right) \nonumber \\&=\Phi (s_m,0,\ldots ,0) \nonumber \\&\quad \times \Phi (0,s'_{m-1}, \ldots ,s'_1,x',y',t'). \end{aligned}$$

(10)

Also, it is easy to see that

$$\begin{aligned}&\Phi (s_m,s_{m-1},\ldots ,s_1,x,y,t)=\Phi (0,s_{m-1},\ldots ,s_1,x,y,t)\Phi (s_m,0,\ldots ,0) \end{aligned}$$

(11)

$$\begin{aligned}&\Phi ((0,s_{m-1},\ldots ,s_1,x,y,t)(0,s'_{m-1},\ldots ,s'_1,x',y',t'))=\Phi (0,s_{m-1},\ldots ,s_1,x,y,t) \nonumber \\ {}&\Phi (0,s'_{m-1},\ldots ,s'_1,x',y',t') \end{aligned}$$

(12)

$$\begin{aligned}&\Phi (s_m+s'_m,0,\ldots ,0)=\Phi (s_m,0,\ldots ,0) \ \Phi (s'_m,0,\ldots ,0). \end{aligned}$$

(13)

So, by definition of semidirect product, (11), (12), (13), and (10), we have that

$$\begin{aligned}&\Phi \left( (s_m,\ldots ,s_1,x,y,t)(s'_m,\ldots ,s'_1,x',y',t')\right) \\&\quad =\Phi (s_m+s'_m,(s_{m-1},\ldots ,s_1,x,y,t)s_m\cdot (s'_{m-1},\ldots ,s'_1,x',y',t'))\\&\quad =\Phi (0,(s_{m-1},\ldots ,s_1,x,y,t)s_m\cdot (s'_{m-1},\ldots ,s'_1,x',y',t')) \ \Phi (s_m+s'_m,0,\ldots ,0)\\&\quad =\Phi (0,s_{m-1},\ldots ,s_1,x,y,t) \ \Phi (0,s_m\cdot (s'_{m-1},\ldots ,s'_1,x',y',t')) \ \Phi (s_m,0,\ldots ,0) \\&\qquad \times \Phi (s'_m,0,\ldots ,0)\\&\quad =\Phi (0,s_{m-1},\ldots ,s_1,x,y,t) \ \Phi (s_m,0,\ldots ,0) \\ {}&\quad \quad \Phi (0,s'_{m-1},\ldots ,s'_1,x',y',t') \ \Phi (s'_m,0,\ldots ,0)\\&\quad =\Phi (s_m,s_{m-1},\ldots ,s_1,x,y,t) \ \Phi (s'_m,s'_{m-1},\ldots ,s'_1,x',y',t'). \end{aligned}$$

$\square$