[proofplan] We first check that adding a constant to a Hamiltonian function does not change its Hamiltonian vector field, so the assignment from the quotient is well-defined. Injectivity follows because a Hamiltonian vector field vanishes exactly when the differential of its Hamiltonian vanishes, and on a connected manifold this forces the Hamiltonian to be constant. The image statement is then exactly the definition of Hamiltonian vector fields. Finally, Cartan's formula, the closedness of $\omega$, and the sign convention for the Poisson bracket give the commutator identity. [/proofplan]

[step:Show that adding constants does not change the Hamiltonian vector field] Let $H\in C^\infty(M)$ and let $c\in\mathbb R$ be a constant, viewed as a constant smooth function on $M$. Since $d(H+c)=dH$, the defining equation for the Hamiltonian vector field gives \begin{align*} \iota_{X_{H+c}}\omega=d(H+c)=dH=\iota_{X_H}\omega. \end{align*} Because $\omega$ is nondegenerate at every point of $M$, contraction with $\omega$ determines a vector field uniquely. Hence $X_{H+c}=X_H$. Therefore the map \begin{align*} \Phi:C^\infty(M)/\mathbb R\to\mathfrak X(M),\qquad \Phi([H])=X_H \end{align*} is well-defined. The Poisson bracket also descends to the quotient. If $F,H\in C^\infty(M)$ and $a,b\in\mathbb R$, then $X_{F+a}=X_F$ and $X_{H+b}=X_H$, so \begin{align*} \{F+a,H+b\}=\omega(X_{F+a},X_{H+b})=\omega(X_F,X_H)=\{F,H\}. \end{align*} Thus the class $[\{F,H\}]$ depends only on $[F]$ and $[H]$. [/step]

[step:Prove injectivity by comparing two classes with the same Hamiltonian vector field]Suppose $F,H\in C^\infty(M)$ satisfy $\Phi([F])=\Phi([H])$. This means $X_F=X_H$ as vector fields on $M$. By the defining equations for $X_F$ and $X_H$, \begin{align*} d(F-H)=dF-dH=\iota_{X_F}\omega-\iota_{X_H}\omega=\iota_{X_F-X_H}\omega=0. \end{align*} Thus $F-H$ has zero differential at every point of $M$. Since $M$ is connected, every smooth function with zero differential is constant on $M$. Hence $F-H\in\mathbb R$ as an element of $C^\infty(M)$, so $[F]=[H]$ in $C^\infty(M)/\mathbb R$. This proves that $\Phi$ is injective.[/step]

[guided]To prove injectivity as a map of sets, we compare two inputs with the same output. Assume $F,H\in C^\infty(M)$ satisfy $\Phi([F])=\Phi([H])$. By definition of $\Phi$, this says that the Hamiltonian vector fields agree: \begin{align*} X_F=X_H. \end{align*} The Hamiltonian vector fields are defined by \begin{align*} \iota_{X_F}\omega=dF \end{align*} and \begin{align*} \iota_{X_H}\omega=dH. \end{align*} Subtracting these identities gives \begin{align*} d(F-H)=dF-dH=\iota_{X_F}\omega-\iota_{X_H}\omega=\iota_{X_F-X_H}\omega. \end{align*} Because $X_F=X_H$, the right-hand side is $\iota_0\omega=0$, and therefore \begin{align*} d(F-H)=0. \end{align*} Now $d(F-H)=0$ means that every [directional derivative](/page/Directional%20Derivative) of the smooth function $F-H:M\to\mathbb R$ vanishes at every point of $M$. On each connected coordinate neighbourhood this forces $F-H$ to be locally constant, because in local coordinates all first partial derivatives vanish. Since $M$ is connected, these local constants agree along overlapping coordinate neighbourhoods, so $F-H$ is constant on all of $M$. Thus $F$ and $H$ differ by a constant smooth function, which is exactly the condition that they represent the same class in the quotient $C^\infty(M)/\mathbb R$. Hence $[F]=[H]$, and this proves injectivity of $\Phi$.[/guided]

[step:Identify the image with the Hamiltonian vector fields] By definition, a vector field $X\in\mathfrak X(M)$ is Hamiltonian if there exists a smooth function $H\in C^\infty(M)$ such that $X=X_H$. Writing $\operatorname{im}\Phi$ for the image set of the map $\Phi$, we have \begin{align*} \operatorname{im}\Phi=\{X_H:H\in C^\infty(M)\}. \end{align*} This is exactly the space of Hamiltonian vector fields on $M$. [/step]

[step:Compute the contraction of the commutator with the symplectic form] Let $F,H\in C^\infty(M)$. For a smooth vector field $Y\in\mathfrak X(M)$, write $\mathcal L_Y$ for the Lie derivative along $Y$. We first record that Hamiltonian vector fields preserve $\omega$. Since $\omega$ is closed and $\iota_{X_F}\omega=dF$, Cartan's formula gives \begin{align*} \mathcal L_{X_F}\omega=d(\iota_{X_F}\omega)+\iota_{X_F}d\omega=d(dF)+\iota_{X_F}0=0. \end{align*} Using the commutator identity $[\mathcal L_Y,\iota_Z]=\iota_{[Y,Z]}$ for the Lie derivative and contraction operators, applied with $Y=X_F$ and $Z=X_H$, we obtain \begin{align*} \iota_{[X_F,X_H]}\omega=\mathcal L_{X_F}(\iota_{X_H}\omega)-\iota_{X_H}(\mathcal L_{X_F}\omega). \end{align*} Therefore \begin{align*} \iota_{[X_F,X_H]}\omega=\mathcal L_{X_F}(dH)-\iota_{X_H}0. \end{align*} For a smooth function $H:M\to\mathbb R$, the Lie derivative of $dH$ along $X_F$ is $d(X_FH)$, so \begin{align*} \iota_{[X_F,X_H]}\omega=d(X_FH). \end{align*} [/step]

[step:Convert the contraction identity into the stated bracket identity] By the Hamiltonian convention and the definition of the Poisson bracket, \begin{align*} X_FH=dH(X_F)=\omega(X_H,X_F)=-\omega(X_F,X_H)=-\{F,H\}. \end{align*} Therefore \begin{align*} \iota_{[X_F,X_H]}\omega=d(X_FH)=d(-\{F,H\})=-d\{F,H\}. \end{align*} Since $X_{\{F,H\}}$ is defined by \begin{align*} \iota_{X_{\{F,H\}}}\omega=d\{F,H\}, \end{align*} we have \begin{align*} \iota_{[X_F,X_H]}\omega=\iota_{-X_{\{F,H\}}}\omega. \end{align*} The nondegeneracy of $\omega$ implies equality of the vector fields: \begin{align*} [X_F,X_H]=-X_{\{F,H\}}. \end{align*} This proves the claimed compatibility between the Poisson bracket on $C^\infty(M)/\mathbb R$ and the commutator bracket on Hamiltonian vector fields. [/step]