Introduction
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
iXhω + dh = 0,
where ω is the closed
(dω = 0) and non-degenerate
(iXω = 0 ⇒ X = 0) 2-form,
h is the Hamiltonian and
iXω 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 iXω + α= 0.
The vector field is said to be Hamiltonian if it corresponds to exact form
iXfω + df = 0.
The Poisson bracket is defined as follows:
{f , g} = Xf g = − Xg f = iXf
iXgω.
By introducing a bivector field W satisfying
iXiYω = iW iXω ∧ iYω,
Poisson bracket can be rewritten as
{f , g} = iW df ∧ dg.
It's easy to show that
iXiYLZω =
i[Z,W] iXω ∧ iYω,
where the bracket [ · , · ] is actually a supercommutator,
for an arbitrary bivector field
W = ∑s V^{s} ∧ U^{s} we have
[X,W] = ∑s[X,V^{s}] ∧ U^{s}
+ ∑sV^{s} ∧ [X,U^{s}]
Equation (6) is based on the following useful property of the Lie derivative
LXiWω = i[X,W]ω +
iWLXω.
Indeed, for an arbitrary bivector field
W = ∑s V^{s} ∧ U^{s} we have
LXiWω = LX∑siV^{s} ∧ U^{s}ω =
LX∑s iU^{s}iV^{s}ω
= ∑s i[X,U^{s}]iV^{s}ω +
∑s iU^{s}i[X,V^{s}]ω +
∑siU^{s}iV^{s}LXω =
i[X,W]ω + iWLXω
where LZ denotes the Lie derivative along the vector field Z.
According to Liouville's theorem Hamiltonian vector field
preserves ω
LXfω = 0;
therefore it commutes with W:
[Xf ,W] = 0.
In the local coordinates zs where
ω = ∑rsω^{rs}dzr ∧ zs bivector field
W has the following form
W = ∑rsW^{rs}∂∂zr ∧ ∂∂zs where
W^{rs} is matrix inverted to ω^{rs}.

Case of regular Lagrangian systems
We can say that a group of transformations
g(z) = e^{zLE} generated by the vector
field E maps the space of solutions of equation onto itself if
iXhg*(ω) + g*(dh) = 0
For Xh satisfying
iXhω + dh = 0
Hamilton's equation.
It's easy to show that the vector field E should satisfy
[E , Xh] = 0
Indeed,
iXhLEω + dLEh =
LE(iXhω + dh) = 0
since [E,Xh] = 0.
When E is not Hamiltonian,
the group of transformations g(z) = e^{zLE} is non-Noether
symmetry (in a sense that it maps solutions onto solutions but does not preserve action).

Theorem 1.
(Lutzky, 1998) If the vector field E generates non-Noether symmetry,
then the following functions are constant along solutions:
I^{(k)} = iW^{k} ωE^{k} k = 1...n,
where W^{k} and ωE^{k} are outer
powers of W and LEω.

Proof.
We have to prove that I^{(k)} is constant along
the flow generated by the Hamiltonian. In other words, we should find that
LXhI^{(k)} = 0 is
fulfilled. Let us consider
LXhI^{(1)}
LXhI^{(1)}
= LXh(iWωE) =
i[Xh , W]ωE
+ iWLXhωE,
where according to Liouville's theorem both terms
[Xh , W] = 0 and
iWLXhLEω =
iWLELXhω =
0
since [E , Xh] = 0 and
LXhω = 0 vanish.
In the same manner one can verify that
LXhI^{(k)} = 0
∎

Theorem is valid for a larger class of generators E .
Namely, if [E , Xh] = Xf where Xf is
an arbitrary Hamiltonian vector field, then I^{(k)} is still conserved. Such a
symmetries map the solutions of the equation
iXhω + dh = 0
on solutions of
iXhg*(ω) +
d(g*h + f) = 0

Discrete non-Noether symmetries give rise to the conservation of
I^{(k)} = iW^{k}g*(ω)^{k}
where g*(ω) is transformed ω.

If I^{(k)} is a set of conserved quantities
associated with E and f is any conserved quantity, then the set of functions
{I^{(k)} , f}
(which due to the Poisson theorem are integrals of motion) is associated with
[Xh , E]. Namely it is easy to show by taking the Lie
derivative of (15) along vector field E that
{I^{(k)} , f} = iW^{k}ω^{k}[Xf , E]
is fulfilled.
As a result conserved quantities associated with Non-Noether symmetries form Lie algebra under
the Poisson bracket.

If generator of symmetry satisfies Yang-Baxter equation
[[E[E , W]]W] = 0 Lutzky's conservation laws are in involution [7]
{Y^{(l)} , Y^{(k)}} = 0

Case of irregular Lagrangian systems
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" us such that
iusω = 0 s = 1,2 ... n − rank(ω),
every Hamiltonian vector field is
defined up to linear combination of vectors us. By identifying Xf
with Xf + ∑sCsus, we can introduce equivalence class
Xf^{∗} (then all us belong to
0^{∗} ).
The bivector field W is also far from being unique, but if
W1 and W2 both satisfy
iXiY ω =
iW1,2 iXω ∧ iYω,
then
i(W1 − W2) iXω ∧ iYω = 0 ∀X,Y
is fulfilled. It is possible only when
W1 − W2 = ∑svs ∧ us
where vs are some vector fields and
iusω = 0
(in other words when W1 − W2 belongs to the class
0^{∗})

Theorem 2.
If the non-Hamiltonian vector field E
satisfies [E , Xh^{∗}] = 0^{∗} commutation
relation (generates non-Noether symmetry), then the functions
I^{(k)}
= iW^{k}ωE^{k} k = 1...rank(ω)
(where ω E = LEω) are constant along trajectories.

Proof.
Let's consider I^{(1)}
LXh^{∗}I^{(1)}
= LXh^{∗}(iWωE)
= i[Xh^{∗} , W]ωE +
iWLXh^{∗}ωE = 0
The second term vanishes since [E , Xh^{∗}] = 0^{∗} and
LXh^{∗}ω = 0. The first one is
zero as far as [Xh^{∗} , W^{∗}] = 0^{∗} and
[E , 0^{∗}] = 0^{∗} are satisfied. So
I^{ (1)} is conserved.
Similarly one can show that LXhI^{(k)} = 0 is
fulfilled.
∎

W is not unique, but I^{(k)} doesn't depend
on choosing representative from the class W^{∗}.

Theorem is also valid for generators E satisfying
[E , Xh^{∗}] = Xf^{∗}

Example.
Hamiltonian description of the relativistic particle leads to the following action
A = ∫ p0dx0 +
∑spsdxs
where
p0 = (p^{2} + m^{2})^{1/2}
with vanishing canonical Hamiltonian and degenerate 2-form defined by
p0ω = ∑s(psdps ∧ dx0 + p0dps ∧ dxs).
ω possesses the "null vector field"
iuω = 0
u = p0∂∂x0 + ∑sps∂∂xs.
One can check that the following non- Hamiltonian vector field
E =p0x0∂∂x0
+ p1x1∂∂x1 + ⋯ + pnxn∂∂xn
generates non-Noether symmetry. Indeed, E satisfies
[E , Xh^{∗}] = 0^{∗} because of
Xh^{∗} = 0^{∗} and [E,u] = u.
Corresponding integrals of motion are combinations of momenta:
I^{(1)} = ∑sps
I^{(2)} = ∑r > s prps
⋯
I^{(n)} = ∏sps
This example shows that the set of conserved quantities can be obtained from a single
one-parameter group of non-Noether transformations.