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For $u^{\prime }=F(x,u)$, a point-symmetry generator $V=\xi {\partial }_x+\eta {\partial }_u$ must satisfyIt follows by applying the first prolongation to $u^{\prime }-F=0$ and restricting to its solution manifold.

A Lie point symmetry is a local one-parameter family of diffeomorphisms of $(t,\mathbf{x})$ that takes solution graphs of the differential equation to solution graphs wherever the transformed graph remains a graph over time. Its infinitesimal generator is $Y=\tau (t,\mathbf{x}){\partial }_t+{\sum }_i{\eta }_i(t,\mathbf{x}){\partial }_{x_i}$. Its first prolongation isFor a regular first-order equation the infinitesimal criterion is ${\mathrm{pr}}^{(1)}Y(\Delta )=0$ on $\Delta =0$; it says that the prolonged vector field is tangent to the equation submanifold.

With the constant nondegenerate canonical matrix $J$, define the Poisson bracket by $\{f,g\}=(\nabla f{)}^{T}J\nabla g$ and the Hamiltonian vector field by $V_f=J\nabla f$. Our convention gives $V_f[h]=\{h,f\}$. Applying the Jacobi identity to a test function $h$,soA first integral $F$ obeys $dF/dt=\{F,H\}=0$. Consequently $[V_F,V_H]=0$, and the local flows commute. The time-independent transformation generated by $V_F$, with $t$ unchanged, therefore maps integral curves of a vector field for $V_H$ to integral curves of a vector field: it is a Lie point symmetry.

Conversely, such a time-independent symmetry gives $[V_F,V_H]=0$, hence $V_{\{F,H\}}=0$. Nondegeneracy of $J$ makes $\{F,H\}$ constant on each connected component. At a fixed point ${\mathbf{x}}_0$, $J\nabla H({\mathbf{x}}_0)=0$ implies $\nabla H({\mathbf{x}}_0)=0$, so that constant is zero on its component. Thus the converse holds on a connected phase space containing a fixed point, or on each component that contains one.

The printed global converse needs this connectedness convention: a fixed point on one component does not control the constants on other components. For example take one component with $H=(q^2+p^2)/2$, $F=0$, and another with $H=p$, $F=q$. The Hamiltonian vector fields commute everywhere and there is a fixed point on the first component, but $F$ is not a first integral on the second. Even on a connected space the fixed-point condition matters: $H=p$, $F=q$ commute but $\dot{F}=1$.

Let $\Delta (x,u,u^{\prime })=u^{\prime }-F(x,u)$. Applying the first prolongation givesOn the solution manifold $\Delta =0$, one has $u^{\prime }=F$. The infinitesimal symmetry criterion is therefore exactly