Non-Noether symmetries in singular dynamical systems

Noether's theorem associates conservation laws with particular continuous symmetries of the Lagrangian. According to the Hojman's theorem [1]-[3] there exists the definite correspondence between non-Noether symmetries and conserved quantities. In 1998 M. Lutzky showed that several integrals of motion might correspond to a single one-parameter group of non-Noether transformations [4]. In the present paper, the extension of Hojman-Lutzky theorem to singular dynamical systems is considered.

First of all let us recall some basic knowledge of description of the regular dynamical systems (see, e. g. [5]). In this case time evolution is governed by Hamilton's equation $$ i_{X_h}ω + dh = 0, $$ where $ω$ is the closed ($dω = 0$) and non-degenerate ($i_Xω = 0 ⇒ X = 0$) 2-form, $h$ is the Hamiltonian and $i_Xω$ denotes contraction of $X$ with $ω$. Since $ω$ is non-degenerate, this gives rise to an isomorphism between the vector fields and 1-forms given by $i_Xω + α= 0$. The vector field is said to be Hamiltonian if it corresponds to exact form $$ i_{X_f}ω + df = 0. $$ The Poisson bracket is defined as follows: $$ \{f , g\} = X_f g = − X_g f = i_{X_f} i_{X_g}ω. $$ By introducing a bivector field $W$ satisfying $$ i_Xi_Yω = i_W i_Xω ∧ i_Yω, $$ Poisson bracket can be rewritten as $$ \{f , g\} = i_W df ∧ dg. $$ It's easy to show that $$ i_Xi_YL_Zω = i_{[Z,W]} i_Xω ∧ i_Yω, $$ where the bracket $[ · , · ]$ is actually a supercommutator, for an arbitrary bivector field $ W = \stackrev{∑}{s} V^s ∧ U^s $ we have $$ [X,W] = \stackrev{∑}{s}[X,V^s] ∧ U^s + \stackrev{∑}{s}V^s ∧ [X,U^s] $$ Equation (6) is based on the following useful property of the Lie derivative $$ L_Xi_Wω = i_{[X,W]}ω + i_WL_Xω. $$ Indeed, for an arbitrary bivector field $W = \stackrev{∑}{s} V^s ∧ U^s $ we have $$ L_Xi_Wω = L_X\stackrev{∑}{s}i_{V^s ∧ U^s}ω = L_X\stackrev{∑}{s} i_{U^s}i_{V^s}ω\\ = \stackrev{∑}{s} i_{[X,U^s]}i_{V^s}ω + \stackrev{∑}{s} i_{U^s}i_{[X,V^s]}ω + \stackrev{∑}{s}i_{U^s}i_{V^s}L_Xω = i_{[X,W]}ω + i_WL_Xω $$ where $L_Z$ denotes the Lie derivative along the vector field $Z$. According to Liouville's theorem Hamiltonian vector field preserves $ω$ $$ L_{X_f}ω = 0; $$ therefore it commutes with $W$: $$ [X_f ,W] = 0. $$ In the local coordinates $ z_s $ where $ω = \stackrev{∑}{rs}ω^{rs}dz_r ∧ z_s$ bivector field $W$ has the following form $W = \stackrev{∑}{rs}W^{rs}\frac{∂}{∂z_r} ∧ \frac{∂}{∂z_s}$ where $W^{rs}$ is matrix inverted to $ω^{rs}$.

We can say that a group of transformations $g(z) = e^{zL_E}$ generated by the vector field $E$ maps the space of solutions of equation onto itself if $$ i_{X_h}g_{*}(ω) + g_{*}(dh) = 0 $$ For $X_h$ satisfying $$ i_{X_h}ω + dh = 0 $$ Hamilton's equation. It's easy to show that the vector field $E$ should satisfy $[E , X_h] = 0$ Indeed, $$ i_{X_h}L_Eω + dL_Eh = L_E(i_{X_h}ω + dh) = 0 $$ since $[E,X_h] = 0$. When $E$ is not Hamiltonian, the group of transformations $g(z) = e^{zL_E}$ is non-Noether symmetry (in a sense that it maps solutions onto solutions but does not preserve action).

The singular Lagrangian (Lagrangian with vanishing Hessian) leads to degenerate 2-form $ω$ and we no longer have isomorphism between vector fields and 1-forms. Since there exists a set of "null vectors" $u_s$ such that $i_{u_s}ω = 0       s = 1,2 ... n − rank(ω),$ every Hamiltonian vector field is defined up to linear combination of vectors $u_s$. By identifying $X_f$ with $X_f + \stackrev{∑}{s}C_su_s,$ we can introduce equivalence class $ X_f^{∗}$ (then all $u_s$ belong to $0^{∗}$ ). The bivector field $W$ is also far from being unique, but if $W_1$ and $W_2$ both satisfy $$ i_Xi_Y ω = i_{W_{1,2}} i_Xω ∧ i_Yω, $$ then $$ i_{(W_1 − W_2)} i_Xω ∧ i_Yω = 0       ∀X,Y $$ is fulfilled. It is possible only when $$ W_1 − W _2 = \stackrev{∑}{s}v_s ∧ u_s $$ where $v_s$ are some vector fields and $i_{u_s}ω = 0$ (in other words when $ W_1 − W_2$ belongs to the class $0^{∗}$)