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.
What you seem to want to compute is
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.