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Locally write the real analytic hypersurface as $\Sigma =\{F=0\}$ with $dF\ne 0$. The conormal bundle is spanned by $dF$. The surface is a characteristic hypersurface at $p$ precisely when the principal symbol vanishes on that conormal:Only the matrix $(a_{ij}+a_{ji})/2$, which is a symmetric matrix, contributes to this expression.

If $H{|}_L$ is constant, then $dH(v)=0$ for every $v\in TL$. By the definition of the Hamiltonian vector field,Since $L$ is Lagrangian, $(TL{)}^{\omega }=TL$, so $X_H$ is tangent to $L$. Its Hamiltonian flow preserves $L$. This is the Hamiltonian flow preserves a constant-level Lagrangian principle.

The cotangent lift is functorial:Since ${\varphi }_{t+s}={\varphi }_t\circ {\varphi }_s$, it follows thatThus $({\psi }_t)$ is a flow with infinitesimal vector field $V_{\#}$ as stated in the question.

Finally, a cotangent lift sends the conormal bundle $N^{\ast }Y$ to $N^{\ast }f(Y)$. Explicitly, if $\xi$ annihilates $T_{x}Y$, then $(d{f}_x^{-1}{)}^{\ast }\xi$ annihilates $T_{f(x)}f(Y)$. Since ${\varphi }_t(Y)=Y$, we haveso the conormal bundle is invariant.