1$, then by (2) above, we must have $b=0$, and so the number of solutions of $ab=b$ is $(q-2)q^{2n+1}$.

\(ii) If $a_0=1$, $a_0'=0$, $A\neq 0$, then $b_0=b_0'=0$ and $B,B'$ can be anything, and the number of solutions of $ab=b$ is $q^{3n+1}(q^n-1)$.

\(iii) If $a_0=1$, $a_0'=0$, $A=0$ and $A'\neq 0$, then $b_0=0$ and $b_0', B, B'$ can be anything. In this case the number of solutions of $ab=b$ is $q^{2n+1}(q^n-1)$.

\(iv) If $a=1$, then $b$ can be anything, and so the number of solutions of $ab=b$ is $q^{2n+2}$.

\(v) If $a_0=1$, $a_0'\neq 0$, then $b_0=0$ and $B$ is determined, and the number of solutions of $ab=b$ is $(q-1)q^{3n+1}$.

The total number of orbits is then

$$\begin{aligned}
A_n(q)&=\frac{1}{(q-1)q^{2n+1}}\left((q-2)q^{2n+1}+q^{3n}(q^n-1)+q^{2n+1}(q^n-1)+q^{2n+2}+(q-1) q^{3n+1}\right)\\
&=2+q^n+\sum_{i=0}^{n-1}q^{i+n-1}+\sum_{i=0}^{n-1}q^i.\end{aligned}$$

We now compute the number of $\G(\R)$-orbits of $\mu_\R^{-1}(0)$. The moment map $\mu_\R:\V(\R)\times\D(\V)(\R)\rightarrow\D(\g(\k))(\R)$ is just the multiplication $\R\times\R\rightarrow\R$ and the action of $\G(\R)=\R^\times$ is given by $\lambda\cdot(a,b)=(\lambda a,\lambda^{-1} b)$. We compute the number $B_n(q)$ of $\R^\times$-orbits of $\{(a,b)\in\R\times\R\,|\, ab=0\}$.

Write an element $a\in\R$ in the form $a_0+A$ with $a_0\in\F_q$ and $A\in(t_1,\dots,t_n)$. The equation $ab=0$ is equivalent to the system

$$\begin{cases}a_0b_0=0,\\
a_0B+b_0A=0.\end{cases}$$ We have the following cases :

\(i) If $a_0\neq 0$ (resp. $b_0\neq 0$), then $b=0$ (resp. $a=0$) , and so there 