the least action principle.

By the use of Eq. (\[xyzG2\]), it is shown that the Poisson bracket between $G(q,p)=\tilde{G}(x,y,z)$ and an arbitrary function $u(q,p)=\tilde{u}(x, y, z)$ vanishes such that $$\begin{aligned}
\{G, u\}_{\mbox{\tiny{PB}}} 
&=& \frac{1}{2}
\sum_{i, j=1}^{3} \frac{\partial (\tilde{G}, \tilde{u})}
{\partial (x_i, x_j)} \{x_i, x_j\}_{\mbox{\tiny{PB}}}
= \frac{1}{2} \sum_{i, j, k=1}^{3} \varepsilon_{ijk} 
\frac{\partial (\tilde{G}, \tilde{u})}{\partial (x_i, x_j)} 
\frac{\partial \tilde{G}}{\partial x_k}\nonumber \\
&=& \frac{\partial (\tilde{G}, \tilde{u}, \tilde{G})}{\partial (x, y, z)} = 0~.
\label{Gu=0}\end{aligned}$$ This means that $G$ is a constant. We can eliminate the constant by redefining $G$, and the resulting $\tilde{G}(x,y,z)=0$ can be regarded as a $\it constraint$, which is induced by enlarging the phase space from $(q, p)$ to $(x, y, z)$.

Here we give two comments on the induced constraint $\tilde{G}(x,y,z)=0$. First, in the case in which $\partial \tilde{G}/\partial z \ne 0$, we can solve $\tilde{G}(x,y,z)=0$ for $z$ and obtain $z=z(x,y)$. Because the condition $\partial \tilde{G}/\partial z= \{x, y\}_{\mbox{\tiny{PB}}} \ne 0$ also enables us to express $q$ and $p$ as functions of $x$ and $y$, the expression $z=z(x,y)$ can also be obtained by inserting $q=q(x,y)$ and $p=p(x,y)$ into $z=z(q,p)$. Therefore the implicit form of the constraint $\tilde{G}(x,y,z)=0$ has an equivalent explicit form $z=z(x,y)$, which clearly shows that $z$ is a 