[proofplan]
The forward implication is immediate from the displayed moment map identity: each contraction one-form is the [exterior derivative](/theorems/1525) of the corresponding Hamiltonian component. For the converse, choose a finite basis of the [Lie algebra](/page/Lie%20Algebra) and choose primitives for the finitely many exact one-forms associated to that basis. Extending those primitives linearly in the Lie algebra variable produces scalar functions $H_\xi$, and evaluating $H_\xi$ pointwise defines a $\mathfrak g^*$-valued smooth map. The final verification is that this construction is linear in $\xi$, smooth in $p$, and satisfies the prescribed sign convention $d\mu^\xi=\iota_{\xi_M}\theta$.
[/proofplan]
[step:Derive exactness from the moment map identity]
Assume there exists a smooth map
\begin{align*}
\mu:M\to\mathfrak g^*
\end{align*}
such that, for every $\xi\in\mathfrak g$, the function $\mu^\xi:M\to\mathbb R$ defined by $\mu^\xi(p)=\mu(p)(\xi)$ satisfies
\begin{align*}
d\mu^\xi=\iota_{\xi_M}\theta.
\end{align*}
Fix $\xi\in\mathfrak g$. The one-form $\iota_{\xi_M}\theta\in\Omega^1(M)$ is the exterior derivative of the smooth function $\mu^\xi$, hence it is exact. Therefore its de Rham cohomology class vanishes:
\begin{align*}
[\iota_{\xi_M}\theta]=0\quad\text{in }H^1_{\mathrm{dR}}(M).
\end{align*}
Since $\xi\in\mathfrak g$ was arbitrary, the displayed vanishing holds for every $\xi\in\mathfrak g$.
[/step]
[step:Choose primitives on a finite basis of the Lie algebra]
Conversely, assume that
\begin{align*}
[\iota_{\xi_M}\theta]=0\quad\text{in }H^1_{\mathrm{dR}}(M)
\end{align*}
for every $\xi\in\mathfrak g$. Since $\mathfrak g$ is finite-dimensional, let $m=\dim\mathfrak g$ and choose a basis
\begin{align*}
\xi_1,\dots,\xi_m\in\mathfrak g.
\end{align*}
For each index $i\in\{1,\dots,m\}$, the hypothesis applied to $\xi_i$ says that the one-form $\iota_{(\xi_i)_M}\theta$ is exact. Hence there exists a smooth function
\begin{align*}
H_i:M\to\mathbb R
\end{align*}
such that
\begin{align*}
dH_i=\iota_{(\xi_i)_M}\theta.
\end{align*}
[guided]
We now use the finite-dimensionality of $\mathfrak g$. Let
\begin{align*}
m=\dim\mathfrak g.
\end{align*}
Choose a basis
\begin{align*}
\xi_1,\dots,\xi_m\in\mathfrak g.
\end{align*}
The goal is to build a single map $\mu:M\to\mathfrak g^*$, but a covector on $\mathfrak g$ is determined by its values on a basis. Thus it is enough to first prescribe the functions that will become the components of $\mu$ along the basis vectors $\xi_i$.
For each $i\in\{1,\dots,m\}$, the assumed cohomology condition applied to $\xi_i$ is
\begin{align*}
[\iota_{(\xi_i)_M}\theta]=0\quad\text{in }H^1_{\mathrm{dR}}(M).
\end{align*}
The meaning of this vanishing in degree-one de Rham cohomology is that the closed one-form $\iota_{(\xi_i)_M}\theta$ is exact. Therefore there exists a smooth function
\begin{align*}
H_i:M\to\mathbb R
\end{align*}
with
\begin{align*}
dH_i=\iota_{(\xi_i)_M}\theta.
\end{align*}
These primitives are not unique: adding a constant to any $H_i$ gives another valid primitive. This non-uniqueness is harmless because the theorem asks only for existence of a moment map and imposes no normalization or equivariance condition.
[/guided]
[/step]
[step:Extend the primitives linearly in the Lie algebra variable]
For each $\xi\in\mathfrak g$, write its unique coordinate expansion in the chosen basis as
\begin{align*}
\xi=\sum_{i=1}^m a_i(\xi)\xi_i,
\end{align*}
where $a_i:\mathfrak g\to\mathbb R$ is the $i$th coordinate functional associated to the basis. Define a smooth function
\begin{align*}
H_\xi:M&\to\mathbb R
\end{align*}
by
\begin{align*}
H_\xi(p)=\sum_{i=1}^m a_i(\xi)H_i(p).
\end{align*}
For every $\xi\in\mathfrak g$, the fundamental vector field depends linearly on $\xi$, so
\begin{align*}
\xi_M=\sum_{i=1}^m a_i(\xi)(\xi_i)_M.
\end{align*}
Using linearity of the exterior derivative, contraction, and the two-form $\theta$, we obtain
\begin{align*}
dH_\xi=\sum_{i=1}^m a_i(\xi)dH_i.
\end{align*}
Substituting the defining identities for the $H_i$ gives
\begin{align*}
dH_\xi=\sum_{i=1}^m a_i(\xi)\iota_{(\xi_i)_M}\theta.
\end{align*}
By linearity of contraction in the vector field argument,
\begin{align*}
\sum_{i=1}^m a_i(\xi)\iota_{(\xi_i)_M}\theta=\iota_{\xi_M}\theta.
\end{align*}
Thus
\begin{align*}
dH_\xi=\iota_{\xi_M}\theta.
\end{align*}
[/step]
[step:Define the moment map and verify smoothness]
Define a map
\begin{align*}
\mu:M\to\mathfrak g^*
\end{align*}
by declaring that, for every $p\in M$ and every $\xi\in\mathfrak g$,
\begin{align*}
\mu(p)(\xi)=H_\xi(p).
\end{align*}
For fixed $p\in M$, the map $\xi\mapsto H_\xi(p)$ is linear because
\begin{align*}
H_\xi(p)=\sum_{i=1}^m a_i(\xi)H_i(p)
\end{align*}
and each $a_i:\mathfrak g\to\mathbb R$ is linear. Hence $\mu(p)\in\mathfrak g^*$.
Let $\xi_1^*,\dots,\xi_m^*\in\mathfrak g^*$ denote the basis dual to $\xi_1,\dots,\xi_m$. Then, for every $p\in M$,
\begin{align*}
\mu(p)=\sum_{i=1}^m H_i(p)\xi_i^*.
\end{align*}
Since each coefficient function $H_i:M\to\mathbb R$ is smooth and $\mathfrak g^*$ is finite-dimensional, this formula proves that $\mu:M\to\mathfrak g^*$ is smooth.
[/step]
[step:Identify the Hamiltonian components with the chosen primitives]
Fix $\xi\in\mathfrak g$. The function $\mu^\xi:M\to\mathbb R$ associated to $\mu$ is defined by
\begin{align*}
\mu^\xi(p)=\mu(p)(\xi).
\end{align*}
By the definition of $\mu$, this equals
\begin{align*}
\mu^\xi(p)=H_\xi(p)
\end{align*}
for every $p\in M$. Therefore $\mu^\xi=H_\xi$ as smooth functions on $M$. From the identity proved above,
\begin{align*}
d\mu^\xi=dH_\xi=\iota_{\xi_M}\theta.
\end{align*}
Since $\xi\in\mathfrak g$ was arbitrary, the map $\mu:M\to\mathfrak g^*$ satisfies the required moment map identity for every Lie algebra element. This proves the converse implication and completes the proof.
[/step]
