where for $v\in\V(\R)$ we put $$\begin{array}{cccc}\delta_v:&\g(\R)&\rightarrow&\V(\R)\\ &x&\mapsto&\varrho_\R(x)(v)\end{array}$$ which is $\R$-linear, as $\varrho_\R={\rm Lie}(\rho)_\R$ is $\R$-linear. Thus for any $(v,f)\in \V(\R)\times \V(\R)^\vee$ and $x\in \g(\R)$ we have \_(v,f)(x)= \_(x)(v),f . We note that $\mu_\R$ is equivariant with respect to the natural action of $\G(\R)$, namely for $g\in\G(\R)$

$$\begin{aligned}
\nonumber
 \mu_\R\left(\rho_\R(g)(v),\rho^\vee(g)(f)\right))&=\left\langle\varrho_\R(x)(\rho_\R(g)(v)),\rho^\vee_\R(g)(f)\right\rangle\\& \nonumber =\left\langle\rho_\R(g^{-1})\circ\varrho_\R(x)\circ\rho_\R(g)(v),f\right\rangle\\&=\left\langle\varrho_\R({\rm Ad}(g^{-1})(x))(v),f\right\rangle \nonumber\\ &= \mu_\R(v,f)({\rm Ad}(g^{-1})(x))
 \label{equimoment}  \end{aligned}$$

where ${\rm Ad}$ denotes the adjoint action of $\G$ on $\g$. Hence it makes sense to consider the orbit space $\mu_\R^{-1}(0)/\G(\R)$.

Fourier transforms
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In this section we assume that $\R$ is a $\k$-Frobenius algebra with $1$-form $\lambda:\R\rightarrow\k$, $\V$ is a finite dimensional $\k$-vector space and $\rho:\G\rightarrow\GL(\V)$ a representation with infinitesimal action $\varrho:\g\rightarrow\gl(\V)$. In the following we make the identifications $\G(\R[\epsilon])\simeq\g(\R)\rtimes\G(\R)$ and $\V(\R[\epsilon])=\V(\R)\times\V(\R)$ where the second coordinate is the $\epsilon$-coordinate. We also use the simplified notations $$g\cdot y:=\rho_\R(g)(y),$$ and $$g\cdot f:= \rho^\vee_\R(g)(f)$$ with $g\in\G(\R)$, $y\in\V(\R)$ and $f\in \V^\vee(\R)$. By Remark \[GL\] (iii), the action of $\G(\R[\epsilon])$ on $\V(\R[\epsilon])$ reads

$$(x,g)\cdot(x_1,x_2)=(g\cdot x_1,\varrho_\R(x)(g\cdot x_1)+g\cdot x_2),
\label{actionfor}$$

for all $(x,g)\in\G(\R[\epsilon])$ and $(x_1,x_2)\in\V(\R[\epsilon])$. We will also need the following formula (the first equation follows from and the second one from equivariance of the moment map ) $$\left\langle \varrho_\R(x)(g\cdot x_1),g\cdot x_3\right\rangle=\mu_\R(g\cdot x_1,g\cdot x_3)(x)=\mu_\R(x_1,x_3)({\rm Ad}(g^{-1})(x)),
\label{momentfor}$$ for all 