[proofplan] The proof tests the Hamiltonian equation against all linear coordinate functionals on $\mathfrak g^*$. For a fixed $\eta \in \mathfrak g$, the linear function $F_\eta(\mu)=\mu(\eta)$ has differential $d(F_\eta)_\mu=\eta$ under the canonical identification $T_\mu^*\mathfrak g^* \cong \mathfrak g$. The Lie-Poisson bracket then computes the scalar derivative of $\mu(t)(\eta)$, and the definition of the coadjoint action converts this scalar identity into the vector equation in $\mathfrak g^*$. The converse follows by reversing the same calculation and using the chain rule for arbitrary smooth observables. [/proofplan] [step:Compute the differential of the linear test observable] Fix $\eta \in \mathfrak g$. Define the linear test observable $F_\eta:\mathfrak g^* \to \mathbb R$ by \begin{align*} F_\eta(\nu) = \nu(\eta) \end{align*} for every $\nu \in \mathfrak g^*$. For $\mu \in \mathfrak g^*$ and $\alpha \in T_\mu\mathfrak g^* \cong \mathfrak g^*$, the directional derivative is \begin{align*} d(F_\eta)_\mu(\alpha) = \alpha(\eta). \end{align*} Thus, under the canonical identification $T_\mu^*\mathfrak g^* \cong (\mathfrak g^*)^* \cong \mathfrak g$, we have \begin{align*} d(F_\eta)_\mu = \eta. \end{align*} [guided] Fix $\eta \in \mathfrak g$. The point of introducing $F_\eta$ is that evaluating $\mu(t)$ on every $\eta$ determines the curve $\mu(t)$ in the dual [vector space](/page/Vector%20Space) $\mathfrak g^*$. Define the map $F_\eta:\mathfrak g^* \to \mathbb R$ by \begin{align*} F_\eta(\nu) = \nu(\eta) \end{align*} for each $\nu \in \mathfrak g^*$. We now compute its differential. Since $\mathfrak g^*$ is a finite-dimensional vector space, its tangent space at $\mu$ is canonically identified with $\mathfrak g^*$. Therefore a tangent vector at $\mu$ is an element $\alpha \in \mathfrak g^*$. By the definition of the differential, \begin{align*} d(F_\eta)_\mu(\alpha) = \frac{d}{ds}\bigg|_{s=0} F_\eta(\mu+s\alpha). \end{align*} Using the definition of $F_\eta$, this becomes \begin{align*} d(F_\eta)_\mu(\alpha) = \frac{d}{ds}\bigg|_{s=0}(\mu+s\alpha)(\eta). \end{align*} Linearity of each covector gives \begin{align*} (\mu+s\alpha)(\eta) = \mu(\eta)+s\alpha(\eta). \end{align*} Differentiating at $s=0$ gives \begin{align*} d(F_\eta)_\mu(\alpha) = \alpha(\eta). \end{align*} Thus $d(F_\eta)_\mu$ is precisely the functional on $\mathfrak g^*$ given by evaluation at $\eta$. Under the canonical finite-dimensional identification $(\mathfrak g^*)^* \cong \mathfrak g$, this functional corresponds to $\eta$, so \begin{align*} d(F_\eta)_\mu = \eta. \end{align*} [/guided] [/step] [step:Test the Hamiltonian identity against every element of $\mathfrak g$] Assume first that $\mu:I \to \mathfrak g^*$ is a Hamiltonian trajectory for $h$. For the test observable $F_\eta$ and for each $t \in I$, the Hamiltonian convention gives \begin{align*} \frac{d}{dt}F_\eta(\mu(t)) = \{h,F_\eta\}_{\mathfrak g^*}(\mu(t)). \end{align*} The left-hand side is \begin{align*} \frac{d}{dt}F_\eta(\mu(t)) = \frac{d}{dt}\mu(t)(\eta) = \dot{\mu}(t)(\eta). \end{align*} Using $d(F_\eta)_{\mu(t)}=\eta$, the right-hand side is \begin{align*} \{h,F_\eta\}_{\mathfrak g^*}(\mu(t)) = \mu(t)([dh_{\mu(t)},\eta]). \end{align*} By the definition of $\operatorname{ad}^*$, \begin{align*} (-\operatorname{ad}_{dh_{\mu(t)}}^*\mu(t))(\eta) = \mu(t)([dh_{\mu(t)},\eta]). \end{align*} Therefore \begin{align*} \dot{\mu}(t)(\eta) = (-\operatorname{ad}_{dh_{\mu(t)}}^*\mu(t))(\eta) \end{align*} for every $\eta \in \mathfrak g$ and every $t \in I$. [/step] [step:Conclude the vector equation in $\mathfrak g^*$] For each fixed $t \in I$, both $\dot{\mu}(t)$ and $-\operatorname{ad}_{dh_{\mu(t)}}^*\mu(t)$ are elements of $\mathfrak g^*$. The preceding step shows that these two linear functionals have the same value on every $\eta \in \mathfrak g$. Hence they are equal in $\mathfrak g^*$: \begin{align*} \dot{\mu}(t) = -\operatorname{ad}_{dh_{\mu(t)}}^*\mu(t). \end{align*} Since $t \in I$ was arbitrary, this proves the Lie-Poisson Hamilton equation along the whole trajectory. [/step] [step:Recover the Hamiltonian identity from the vector equation] Conversely, suppose $\mu:I \to \mathfrak g^*$ is smooth and satisfies \begin{align*} \dot{\mu}(t) = -\operatorname{ad}_{dh_{\mu(t)}}^*\mu(t) \end{align*} for every $t \in I$. Let $F \in C^\infty(\mathfrak g^*)$ be arbitrary. By the chain rule on the finite-dimensional vector space $\mathfrak g^*$, \begin{align*} \frac{d}{dt}F(\mu(t)) = dF_{\mu(t)}(\dot{\mu}(t)). \end{align*} Under the identification $T_{\mu(t)}^*\mathfrak g^* \cong \mathfrak g$, write $dF_{\mu(t)} \in \mathfrak g$. Substituting the assumed equation gives \begin{align*} dF_{\mu(t)}(\dot{\mu}(t)) = (-\operatorname{ad}_{dh_{\mu(t)}}^*\mu(t))(dF_{\mu(t)}). \end{align*} Using the definition of $\operatorname{ad}^*$, this equals \begin{align*} (-\operatorname{ad}_{dh_{\mu(t)}}^*\mu(t))(dF_{\mu(t)}) = \mu(t)([dh_{\mu(t)},dF_{\mu(t)}]). \end{align*} By the definition of the Lie-Poisson bracket, \begin{align*} \mu(t)([dh_{\mu(t)},dF_{\mu(t)}]) = \{h,F\}_{\mathfrak g^*}(\mu(t)). \end{align*} Therefore \begin{align*} \frac{d}{dt}F(\mu(t)) = \{h,F\}_{\mathfrak g^*}(\mu(t)). \end{align*} Since $F$ was arbitrary, $\mu$ is a Hamiltonian trajectory for $h$ with the stated convention. This completes the proof. [/step]