Non-Noether symmetries in singular dynamical systems George Chavchanidze Department of Theoretical Physics A. Razmadze Institute of Mathematics 1 Aleksidze Street Tbilisi 0193 Georgia Abstract

In the present paper geometric aspects of relationship between non-Noether symmetries and conservation laws in Hamiltonian systems is discussed. Case of irregular/constrained dynamical systems on presymplectic and Poisson manifolds is considered.

Keywords: Non-Noether symmetry, Conservation laws, Constrained dynamics

MSC 2000: 70H33, 70H06, 53Z05

Introduction

Noether's theorem associates conservation laws with particular continuous symmetries of the Lagrangian. According to the Hojman's theorem - 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 . 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 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 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 derivative LXiWω = i[X,W]ω + iWLXω. Indeed, for an arbitrary bivector field W = s Vs ∧ Us we have LXiWω = LXsiVs ∧ Usω = LXs iUsiVsω = s i[X,Us]iVsω + s iUsi[X,Vs]ω + siUsiVsLXω = 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 = rsωrsdzr ∧ zs bivector field W has the following form W = rsWrs∂zr∂zs where Wrs is matrix inverted to ωrs.

Case of regular Lagrangian systems

We can say that a group of transformations g(z) = ezLE 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) = ezLE 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) = iWk ωEk       k = 1...n, where Wk and ωEk 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) = iWkg*(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} = iWkωk[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  {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) = 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 = 0 The second term vanishes since [E , Xh] = 0 and LXhω = 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 = (p2 + m2)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.

Acknowledgements. Author is grateful to Z. Giunashvili and M. Maziashvili for constructive 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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