Product of moment maps I am trying to prove that if $G_{1}$ and $G_{2}$ act in a hamiltonian way in the symplectic manifold $(M,\omega)$, with moment maps $\mu_{1}$ and $\mu_{2}$, and if the actions commute, then $G_{1}\times G_{2}$ acts in a hamiltonian way in $(M,\omega)$ with moment map $\mu\colon M \longrightarrow \mathfrak{g}_{1}^{*}\oplus \mathfrak{g}_{2}^{*}$ such that $\mu(p)=(\mu_{1}(p),\mu_{2}(p))$.
I have defined the action as follows: $$\Phi\colon G_{1}\times G_{2}\longrightarrow \text{Sympl}(M,\omega)$$ such that $$\Phi(g,h):M\longrightarrow M$$ $$p\mapsto ((\psi_{2})_{h}\circ (\psi_{1})_{g})(p)$$ where $\psi_{1}$ and $\psi_{2}$ are the actions of $G_{1}$ and $G_{2}$, respectively.
To show that $\mu$ is a moment map, I first need to show that $d\mu^{(X,Y)}=\iota_{(X,Y)^{\#}}w$.
Here is my trouble. I am unable to compute $(X,Y)^{\#}$.
Let $p\in M$.
If $\varphi_{1}\colon G_{1} \longrightarrow M$, $$\varphi_{1}(g)=(\psi_{1})_{g}(p),$$
and $\varphi_{2}\colon G_{2}\longrightarrow M$, $$\varphi_{2} (h)=(\psi_{2})_{h}(p),$$
$X_{p}^{\#}$ is the image of $X_{e}$ under $d\varphi_{1}|_{e}$ and $Y_{p}^{\#}$ is the image of $Y_{e}$ under $d\varphi_{2}|_{e}$.
In order to compute $(X,Y)_{p}^{\#}$ I should compute the derivative in $e$ of $\varphi\colon G_{1}\times G_{2}\longrightarrow M$, such that $$\varphi(g,h)=(\psi_{2})_{h}((\psi_{1})_{g}(p)),$$ but I do not know how to do it.
Can anyone help me, please? Thanks in advance.
 A: What you seem to want to compute is

Expression 1
  $$\frac{d}{dt}|_{t=0}\psi_2(\exp(tY),\psi_1(\exp(tX),p)).$$  

We will be explicit in the notation, and set $\varphi_i^m(g):=\psi_i(g,m)$.
For $g \in G_2$, let $\nu_2 : M \to M$ be defined by $\nu_2^g(m):=\psi_2(g,m)$.
Proof of Expression 1:
By chain rule, we have
$$\frac{d}{dt}\psi_2(\exp(tY),\psi_1(\exp(tX),p))=\frac{d\varphi_2^{\psi_1(\exp(tX),p)}}{dg}|_{g=\exp(tY)}\frac{d}{dt}\exp(tY) +  \frac{d\nu_2^{\exp(tY)}}{dm}|_{m=\psi_1(\exp(tX),p)}\frac{d}{dt}\psi_1(\exp(tX),p)$$
And setting $t=0$, we note that $d(\nu_2^{\exp(tY)})=d(\nu_2^e)=d(id)=id$.
Therefore, at $t=0$, Expression 1 becomes
$$Y^{\sharp}_p +  X^{\sharp}_p.$$
End of proof of Expression 1.
Finally, let us check that this satisfies the moment map condition.
$\mu : M \to (\mathfrak{g}_1 \oplus \mathfrak{g}_2)^*=\mathfrak{g}_1^* \oplus \mathfrak{g}_2^*$ is a moment map if, for every $(X,Y) \in \mathfrak{g}_1 \oplus \mathfrak{g}_2$, the associated map
$\hat{\mu}(X,Y) : M \to \mathbb{R}$ given by $m \mapsto \langle \mu(m), X \rangle$ ($\langle, \rangle$ is the pairing of $\mathfrak{g}^*$ with $\mathfrak{g}$) satisfies
$$d(\hat{\mu}(X))(Z_m)=\omega_m((X,Y)^{\sharp}_m,Z_m)$$ for every $m \in M$, and every $Z \in \Gamma(TM)$ (here $\Gamma(TM)$ denotes smooth vector fields on $M$).
We just note that
$$\hat{\mu}(m)(X,Y)= \langle \mu(m)_1, X \rangle  + \langle \mu(m)_2, Y \rangle = \langle \mu_1(m), X\rangle  + \langle \mu_2(m), Y \rangle = \hat{\mu}_1(m)(X) + \hat{\mu}_2(m)(Y)$$ (here $\mu(m)_i \in \mathfrak{g}_i$ denotes the coordinate $i$ of $\mu(m)$).
In the first equality we use definition of $\hat{\mu}$, in the second equality, the definition of the moment map $\mu$, and in the third equality the definition of  $\hat{\mu}_1(m)$ and $\hat{\mu}_2(m)$.
Therefore,
$$d\hat{\mu}(m)(X,Y)=d\hat{\mu}_1(m)(X) + d\hat{\mu}_2(m)(Y)$$
Since $\mu_1$ and $\mu_2$ are moment maps, they satisfy 
$d\hat{\mu}_1(m)(X)(Z_m)= \omega_m(X^{\sharp}_m,Z_m)$ for $m \in M$, $Z \in \Gamma(TM)$ and $d\hat{\mu}_2(m)(Y)(Z_m)= \omega_m(Y^{\sharp}_m,Z_m)$ for $m \in M$, $Z \in \Gamma(TM)$.
Therefore,
$$d(\hat{\mu}(X))(Z_m)=\omega_m(X^{\sharp}_m,Z_m)+\omega_m(Y^{\sharp}_m,Z_m)=\omega_m((X,Y)^{\sharp}_m,Z_m)$$ for every $m \in M$, and every $Z \in \Gamma(TM)$, where we used linearity of $\omega_m$ in the first factor and Expression 1.
