# Non-Noether symmetries in singular dynamical systems

Department of Theoretical Physics,A. Razmadze Institute of Mathematics,1 Aleksidze Street, Tbilisi 0193, Georgia
abstract. In the present paper geometric aspects of relationshipbetween non-Noether symmetries and conservation laws in Hamiltoniansystems is discussed. Case of irregular/constrained dynamical systemson presymplectic and Poisson manifolds is considered.
keywords. Non-Noether symmetry; Conservation laws; Constrained dynamics;
msc. 70H33, 70H06, 53Z05
Georgian Math. J. 10 (2003) 057-061

## Introduction

Noether's theorem associates conservation laws with particular continuous symmetries ofthe Lagrangian. According to the Hojman's theorem - there exists the definite correspondence betweennon-Noether symmetries and conserved quantities. In 1998 M. Lutzky showed that several integrals ofmotion might correspond to a single one-parameter group of non-Noether transformations. 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. ).In this case time evolution is governed by Hamilton's equationiXhω + 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 vectorfields and 1-forms given by $i$Xω + α= 0.The vector field is said to be Hamiltonian if it corresponds to exact formiXfω + df = 0.The Poisson bracket is defined as follows:{f , g} = Xf g = − Xg f = iXfiXgω.By introducing a bivector field $W$ satisfyingiXiYω = iW iXω ∧ iYω,Poisson bracket can be rewritten as{f , g} = iW df ∧ dg.It's easy to show thatiXiYLZω =i[Z,W] iXω ∧ iYω,where the bracket $[ · , · ]$ is actually a supercommutator,for an arbitrary bivector field$W = ∑s Vs ∧ Us$ we have[X,W] = s[X,Vs] ∧ Us+ sVs ∧ [X,Us]Equation (6) is based on the following useful property of the Lie derivativeLXiWω = i[X,W]ω +iWLXω.Indeed, for an arbitrary bivector field$W = ∑s Vs ∧ Us$ we haveLXiWω = LXsiVs ∧ Usω =LXs iUsiVsω= s i[X,Us]iVsω +s iUsi[X,Vs]ω +siUsiVsLXω =i[X,W]ω + iWLXωwhere $L$Z denotes the Lie derivative along the vector field $Z$.According to Liouville's theorem Hamiltonian vector fieldpreserves $ω$LXfω = 0;therefore it commutes with $W$:[Xf ,W] = 0.In the local coordinates $z$s where$ω = ∑rsωrsdz$r ∧ zs bivector field$W$ has the following form$W = ∑rsWrs∂∂z$r∂zs where$Wrs$ is matrix inverted to $ωrs$.

## Case of regular Lagrangian systems

We can say that a group of transformations$g(z) = ezL$E generated by the vectorfield $E$ maps the space of solutions of equation onto itself ifiXhg*(ω) + g*(dh) = 0For $X$h satisfyingiXhω + dh = 0Hamilton's equation.It's easy to show that the vector field $E$ should satisfy$[E , X$h] = 0Indeed,iXhLEω + dLEh =LE(iXhω + dh) = 0since $[E,X$h] = 0. When $E$ is not Hamiltonian,the group of transformations $g(z) = ezL$E is non-Noethersymmetry (in a sense that it maps solutions onto solutions but does not preserve action).
theorem. (Lutzky, 1998) If the vector field $E$ generates non-Noether symmetry, then the following functions are constant along solutions:I(k) = iWk ωEk       k = 1...n,where $Wk$ and $ω$Ek are outerpowers of $W$ and $L$Eω.
proof. We have to prove that $I(k)$ is constant alongthe flow generated by the Hamiltonian. In other words, we should find that$L$XhI(k) = 0 isfulfilled. Let us consider$L$XhI(1)LXhI(1)= LXh(iWωE) =i[Xh , W]ωE+ iWLXhωE,where according to Liouville's theorem both terms$[X$h , W] = 0 andiWLXhLEω =iWLELXhω =0 since $[E , X$h] = 0 and $L$Xhω = 0 vanish.In the same manner one can verify that$L$XhI(k) = 0
remark. Theorem is valid for a larger class of generators $E$ .Namely, if $[E , X$h] = Xf where $X$f isan arbitrary Hamiltonian vector field, then $I(k)$ is still conserved. Such asymmetries map the solutions of the equation$i$Xhω + dh = 0on solutions ofiXhg*(ω) +d(g*h + f) = 0
remark. Discrete non-Noether symmetries give rise to the conservation of$I(k) = i$Wkg*(ω)kwhere $g$*(ω) is transformed $ω$.
remark. If $I(k)$ is a set of conserved quantitiesassociated 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$[X$h , E]. Namely it is easy to show by taking the Liederivative of (15) along vector field $E$ that{I(k) , f} = iWkωk[Xf , E] is fulfilled.As a result conserved quantities associated with Non-Noether symmetries form Lie algebra underthe Poisson bracket.
remark. If generator of symmetry satisfies Yang-Baxter equation$[[E[E , W]]W] = 0$ Lutzky's conservation laws are in involution ${Y(l) , Y(k)} = 0$

## 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" $u$s such that$i$usω = 0       s = 1,2 ... n − rank(ω),every Hamiltonian vector field isdefined up to linear combination of vectors $u$s. By identifying $X$fwith $X$f + sCsus, 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 satisfyiXiY ω =iW1,2 iXω ∧ iYω,theni(W1 − W2) iXω ∧ iYω = 0       ∀X,Yis fulfilled. It is possible only whenW1 − W 2 = svs ∧ uswhere $v$s are some vector fields and$i$usω = 0(in other words when $W$1 − W2 belongs to the class$0∗$)
theorem. If the non-Hamiltonian vector field $E$satisfies $[E , X$h] = 0 commutationrelation (generates non-Noether symmetry), then the functionsI (k)= iWkωEk        k = 1...rank(ω)(where $ω$ E = LEω) are constant along trajectories.
proof. Let's consider $I(1)$LXhI(1)= LXh(iWωE)= i[Xh , W]ωE +iWLXhωE = 0The second term vanishes since $[E , X$h] = 0 and$L$Xhω = 0. The first one iszero as far as $[X$h , W] = 0 and$[E , 0∗] = 0∗$ are satisfied. So$I (1)$ is conserved.Similarly one can show that $L$XhI(k) = 0 isfulfilled.
remark. $W$ is not unique, but $I(k)$ doesn't dependon choosing representative from the class $W∗$.
remark. Theorem is also valid for generators $E$ satisfying$[E , X$h] = Xf
example. Hamiltonian description of the relativistic particle leads to the following actionA = p0dx0 + spsdxswhere$p$0 = (p2 + m2)1/2with vanishing canonical Hamiltonian and degenerate 2-form defined byp0ω = s(psdps ∧ dx0 + p0dps ∧ dxs).$ω$ possesses the "null vector field"$i$uω = 0u = p0∂x0 + sps∂xs.One can check that the following non- Hamiltonian vector fieldE =p0x0∂x0+ p1x1∂x1 + ⋯ + pnxn∂xngenerates non-Noether symmetry. Indeed, $E$ satisfies$[E , X$h] = 0 because of$X$h = 0 and $[E,u] = u$.Corresponding integrals of motion are combinations of momenta:I(1) = spsI(2) = r > s prpsI(n) = spsThis example shows that the set of conserved quantities can be obtained from a singleone-parameter group of non-Noether transformations.
acknowledgements. Author is grateful to Z. Giunashvili and M. Maziashvili forconstructive discussions and particularly grateful to George Jorjadze for invaluable help.This work was supported by INTAS (00-00561)and Scholarship from World Federation of Scientists.

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