Superintegrability in a non-conformally-at space

(Joint work with Ernie Kalnins and Willard Miller) School of Mathematics and Statistics University of New South Wales ANU, September 2011

Outline Background What is a superintegrable system Extending the classical superintegrable TTW systems Extending the quantum superintegrable TTW systems

What is a Superintegrable Systems? What is a Superintegrable System? Second Order Constants Tremblay-Turbiner-Winternitz (TTW) system By classical superintegrable system is a Liouville integrable natural Hamiltonian on a 2n-dimensional phase space of the form n H = g ij p i p j + V (x 1,..., x n) i,j=1 admitting 2n 1 functionally independent globally dened functions of the coordindates and momenta {L 0 = H, L 1,..., L 2n 2} that are constants of the motion. If {, } is the usual Poisson bracket, H is integrable if and superintegrable if, in addition, {L i, L j } = 0 for i = 0... n 1. {H, L i } = 0 for i = 0... 2n 2. Here we also ask that the the constants are polynomial in the momenta. For example, second order constants would have the form n L i = a jk (i) p jp k + W (i) (x 1,..., x n). j,k=1

An Example Background What is a Superintegrable System? Second Order Constants Tremblay-Turbiner-Winternitz (TTW) system The Hamiltonian H = p 2 x + p 2 y + α(x 2 + y 2 ) + β x 2 + γ y 2 has two constants that are second order polynomials in the momenta L 1 = p 2 x + αx 2 + β x 2, L2 = (xpy ypx)2 + β y 2 That is, {H, L 1} = {H, L 2} = 0. x 2 + γ x2 y 2. If we dene R = {L 1, L 2}, the Poisson algebra closes quadratically, {R, L 1} = 8L 2 1 8HL 1 + 16αL 2, {R, L 2} = 16L 1L 2 + 8HL 2 16(β + γ)l 1 + 16β. (This is TTW with k = 1.)

Killing Tensors Background What is a Superintegrable System? Second Order Constants Tremblay-Turbiner-Winternitz (TTW) system The space of functions polynomial in the momenta is isomorphic to the space of symmetric contravariant tensors. If H 0 and L 0 are the parts of H and L quadratic and polynomial of degree r in the momenta, then {H 0, L 0} = 0 is equivalent to where [, ] is the Schouten bracket and g = g ij x i [g, K] = 0 and K = a i 1 i r x j x i1 So K is a second rank (symmetric) Killing tensor. x ir.

Tremblay-Turbiner-Winternitz (TTW) system What is a Superintegrable System? Second Order Constants Tremblay-Turbiner-Winternitz (TTW) system Tremblay, Turbiner and Winternitz (2009) introduced a family of potentials V TTW = αr 2 + β r 2 cos 2 (kθ) + γ r 2 sin 2 (kθ). For k = 1, 2, 3 these were known systems (Smorodinsky-Winternitz, rational BC 2 model, Wolfes model). For all k N +, Tremblay, Turbiner and Winternitz showed that the quantum systems are exactly solvable and the classical systems have closed trajectories. demonstrated superintegrability for k = 1, 2, 3, 4. conjectured that the classical and quantum systems are superintegrable. Since then KMP 2009: Rational k classical TTW systems are superintegrable. KKM 2009: Extended treatment of classical TTW systems. KKM 2010: Rational k quantum TTW systems are superintegrable. Results can be extended to higher dimensions.

Generalized TTW Background Generalized TTW Subgroup separable classical Hamiltonians A 4D example The classical TTW Hamiltonian for k = 1 can be written in Cartesian coordinates as and in polar coordinates as H TTW = p 2 x + p 2 y + α(x 2 + y 2 ) + β1 x 2 + β2 y 2 H TTW = p 2 r + αr 2 + 1 r 2 A natural generalization to 3 dimensions is or in polar coordinates, ( p 2 θ + β1 cos 2 θ + ) β2 sin 2 θ H = p 2 x + p 2 y + p 2 z + α(x 2 + y 2 + z 2 ) + β1 z 2 + β2 x 2 + β3 y 2 H = p 2 r + αr 2 + 1 r 2 ( p 2 θ1 + β1 cos 2 + 1 ( θ 1 sin 2 p 2 θ2 θ + 1 This pattern will continue in higher dimensions. β2 cos 2 θ 2 + β3 sin 2 θ 2 ))

Generalized TTW Background Generalized TTW Subgroup separable classical Hamiltonians A 4D example The classical TTW Hamiltonian for arbitrary k is written in polar coordinates as H TTW = pr 2 + αr 2 + 1 ( ) r 2 p 2 β 1 θ + cos 2 (kθ) + β 2 sin 2 (kθ) and so we generalize this as H = p 2 r +αr 2 + 1 r 2 and so on to higher dimensions. ( p 2 θ1 + β 1 cos 2 (k 1θ 1) + 1 sin 2 (k 1θ 1) Inside each pair of brackets is a constant of the motion. ( )) p 2 θ2 + β 2 cos 2 (k + β 3 2θ 2) sin 2 (k 2θ 2) Next we consider some machinery for dealing with a Hamiltonian having this `nested' structure. This is a special case of a more general approach for orthogonal separable coordinates. [J. Phys. A: Math. Theor. 43 (2010) 092001]

Subgroup separable classical Hamiltonians Generalized TTW Subgroup separable classical Hamiltonians A 4D example Consider the n functionally independent functions of q i and p i (i = 1,..., n), H = L 1 = p 2 1 + V 1(q 1) + f 1(q 1)L 2, L 2 = p 2 2 + V 2(q 2) + f 2(q 2)L 3,. L n 1 = p 2 n 1 + V n 1(q n 1) + f n 1(q n 1)L n, L n = p 2 n + V n(q n). The coordinates q i separate the Hamilton-Jacobi equation for H. If p i = S q i then ( ) 2 S L i = + V i (q i ) + f i (q i )L i+1 = λ i q i is an equation in q i alone. (For convenience, dene L n+1 = 0.)

Additional constants Background Generalized TTW Subgroup separable classical Hamiltonians A 4D example It is easy to check that H is integrable, that is, {L i, L j } = 0 i, j = 1,..., n. We would like to nd more constants of the motion to show superintegrability. For each i = 1,..., n 1 try to nd M i (q i, L 1, L 2,..., L n) and N i (q i+1, L 1, L 2,..., L n) satisfying i i {H, M i } = f k (q k ) and {H, N i } = f k (q k ). k=1 k=1 Then {H, N i M i } = 0 and hence L i = N i M i is in involution with H.

Additional constants Background Generalized TTW Subgroup separable classical Hamiltonians A 4D example The M i and N i satisfy 2p i M i q i = f i (q i ) and 2p i+1 N i q i+1 = 1. The equation can be solved by writing p i in terms of L i and integrating f i (q i )dq i M i = 2 L i V i (q i ) f i (q i )L i+1 dq i+1 N i = 2. L i+1 V i+1 (q i+1 ) f i+1 (q i+1 )L i+2 It's straightforward to check that {L i, L i 1} = 1 for i > 1 and {L i, L j} = 0 for j i 1. and deduce set {L 1,..., L n, L 1,..., L n 1} is functionally independent. They are a set of action-angle variables.

Four-dimensional TTW Background Generalized TTW Subgroup separable classical Hamiltonians A 4D example We now consider a 4-dimensional generalization of the TTW system. H = L 1 = p 2 r + αr 2 + L2 r 2 L 2 = p 2 θ1 + β 1 cos 2 (k 1θ 1) + L 3 sin 2 (k 1θ 1) L 3 = p 2 θ2 + β 2 cos 2 (k 2θ 2) + L 4 sin 2 (k 2θ 2) L 4 = p 2 θ3 + β 3 cos 2 (k 3θ 3) + β 4 sin 2 (k 3θ 3). This system is superintegrable for all positive rational k 1, k 2 and k 3.

Four-dimensional TTW Background Generalized TTW Subgroup separable classical Hamiltonians A 4D example The corresponding metric g = e 0 e 0 + e 1 e 1 + e 2 e 2 + e 3 e 3, e 0 = dr, e 1 = rdθ 1, e 2 = r sin(k 1θ 1)dθ 2, e 3 = r sin(k 1θ 1) sin(k 2θ 2)dθ 3. is not conformally at since W abcd W abcd = 4(k2 1 k 2 2 ) 2 3r 4 sin 4 (k 1θ 1). W abcd is the Weyl conformal curvature tensor. The metric g is conformally at if and only if k 2 1 = k 2 2. Next, illustrate the superintegrability with k 1 = 2, k 2 = k 3 = 1. The general case is similar.

Four-dimensional TTW (k 1, k 2, k 3 = 2, 1, 1) Generalized TTW Subgroup separable classical Hamiltonians A 4D example We need M 1(r, H, L 2, L 3, L 4) and N 1(θ 1, H, L 2, L 3, L 4) such that and so M 1 = {H, M 1} = 1 r 2 and {H, N 1} = 1 r 2 dr, N 1 = 2r H 2 αr 2 L r 2 2 L 2 dθ 1 β1 cos 2 (k 1 θ1) L 3 sin 2 (k 1 θ1) which gives where sinh B 1 = i sinh A 1 = i M 1 = B1 4 A 1, N 1 =, L 2 4k 1 L2 H 2L 2 r 2 H 2 cosh B 1 = 2 L 2p r 4αL 2 r H 2 4αL 2 L 2 cos(2k 1θ 1) + L 3 β 1 (β1 L 2 L 3) 2 4L 2L 3 cosh A 1 = L2 sin(2k 1θ 1)p θ1 (β1 L 2 L 3) 2 4L 2L 3

Four-dimensional TTW (k 1, k 2, k 3 = 2, 1, 1) Generalized TTW Subgroup separable classical Hamiltonians A 4D example sinh(8 L 2(N 1 M 1)) = sinh(a 1 2B 1) = 2 cosh A 1 sinh B 1 cosh B 1 + 2 sinh A 1 cosh 2 B 1 sinh A 1 = (( ) ) 4iL 2 H 2L2 sin(4θ1) 2(L2 cos(4θ1) + L3 β1) r 2 p θ1 p r + r r 2 pr 2 (H 2 4αL 2) (β 1 L 2 L 3) 2 4L 2L 3 i (H 2 4αL 2)(L 2 cos(4θ 1) + L 3 β 1) (H 2 4αL 2) (β 1 L 2 L 3) 2 4L 2L 3 = 4iL 2L 1 ih 2 (L 3 β 1) (H 2 4αL 2) (β 1 L 2 L 3) 2 4L 2L 3 Everything apart from L 1 is known to be a constant of the motion and hence ( ) L 1 = H 2L2 sin(4θ1) 2(L2 cos(4θ1) + L3 β1) r 2 p θ1 p r + r r 2 pr 2 1 4 (H2 αl 2) cos(4θ 1) is a constant quartic in he momenta.

Four-dimensional TTW (k 1, k 2, k 3 = 2, 1, 1) Generalized TTW Subgroup separable classical Hamiltonians A 4D example Next M 2(θ 1, H, L 2, L 3, L 4) and N 2(θ 2, H, L 2, L 3, L 4) must satisfy and so {H, M 2} = M 2 = N 2 = 1 r 2 sin 2 (k 1θ 1) 2 sin 2 (k 1θ 1) L 2 2 L 3 and {H, N 2} = dθ 2 dθ 1 1 r 2 sin 2 (k 1θ 1). β1 cos 2 (k 1 θ1) L 3 sin 2 (k 1 θ1) β2 cos 2 (k 2 θ2) L 4 sin 2 (k 2 θ2), where M 2 = B 2 4k 1 L3 and N 2 = A 2 4k 2 L3 sinh B 2 = 2L3cosec2 (k 1θ 1) + β 1 L 2 L 3 4β1L2 (L 3 L 2 β 1) 2 2i L 3 cot(k 1θ 1)p θ1 cosh B 2 = 4β1L 2 (L 3 L 2 β 1) 2 sinh A 2 = i L 3 cos(2k 2θ 2) + β 3 β 2 L3 sin(2k 2θ 2)p θ2 cosh A 2 = (β2. (β2 β 3 L 3) 2 4β 3L 3 β 3 L 3) 2 4β 3L 3

Four-dimensional TTW (k 1, k 2, k 3 = 2, 1, 1) Generalized TTW Subgroup separable classical Hamiltonians A 4D example From this we can nd another quartic constant L 2 = 2(L 3 cos(2θ 2) + L 4 β 2) cot(2θ 1) sin(2θ 2)p θ1 p θ2 + ((β 2 L 3 L 4) 2 4L 3L 4)cosec 2 (2θ 1) sin 2 (2θ 2)(2L 3cosec 2 (2θ 1) + β 1 L 2 L 3)p 2 θ2 and then M 3(θ 2, H, L 2, L 3, L 4) and N 3(θ 3, H, L 2, L 3, L 4) lead to a cubic constant L 3 = 2(L 4 cos(2θ 3) + β 4 β 3) cot(θ 2)p θ2 (2L 4cosec 2 (θ 2) + β 2 L 3 L 4) sin(2θ 3)p θ3 So we have 7 functionally independent polynomial constants in total and hence the system is superintegrable. The same process works for any rational k 1, k 2 and k 3.

Four-dimensional Quantum TTW The quantum version has four mutually commuting dierential operators, H = 2 r + 3 r r ω2 r 2 + L1 r 2 L 1 = 2 θ1 + 2k1 cot(k1θ1) θ1 + L 2 = 2 θ2 + k2 cot(k2θ2) θ2 + L 3 = 2 θ3 + β 3 cos 2 (k 3θ 3) + β 4 sin 2 (k 3θ 3) β 1 cos 2 (k + L 2 1θ 1) sin 2 (k 1θ 1) β 2 cos 2 (k + L 3 2θ 2) sin 2 (k 2θ 2) The Hamiltonian is of the form H = 2 + V 0 where 2 is the Laplacian on a four-dimensional manifold with metric g = e 0 e 0 + e 1 e 1 + e 2 e 2 + e 3 e 3, e 0 = dr, e 1 = rdθ 1, e 2 = r sin(k 1θ 1)dθ 2, e 3 = r sin(k 1θ 1) sin(k 2θ 2)dθ 3.

Four-dimensional Quantum TTW The equation HΨ = EΨ is separable in the coordinate r, θ 1, θ 2, θ 3 but for reasons that will be explained, we add some extra terms. H = L 0 = r 2 + 3 r r ω2 r 2 + L1 r 2 + 1 k2 1 r 2 L 1 = 2 θ1 + 2k1 cot(k1θ1) β 1 θ1 + cos 2 (k + L 2 1θ 1) sin 2 (k + k2 1 k2 2 1θ 1) 4 sin 2 (k 1θ 1) L 2 = 2 θ2 + k2 cot(k2θ2) β 2 θ2 + cos 2 (k + L 3 2θ 2) sin 2 (k 2θ 2) L 3 = 2 θ3 + β 3 cos 2 (k 3θ 3) + β 4 sin 2 (k 3θ 3) HΨ = EΨ remains separable with the additional terms.

Four-dimensional Quantum TTW For the metric g, the scalar curvature is and if we dene R = 6 r 2 + k2 1 ( 6 r 2 2 r 2 sin 2 (k 1θ 1) ) + W = 3W abcd W abcd = 2(k2 1 k 2 2 ) r 2 sin 2 (k 1θ 1). 2k 2 2 r 2 sin 2 (k 1θ 1) then H = 2 + V 0 1 6 R 1 24 W The metric g is conformally at if and only if k 2 1 = k 2 2.

Four-dimensional Quantum TTW We look for solutions of the form HΨ = EΨ, Ψ = Ψ 0(r)Ψ 1(θ 1)Ψ 2(θ 2)Ψ 3(θ 3), and L 3Ψ 3 = l 3Ψ 3, L 2Ψ 2Ψ 3 = l 2Ψ 2Ψ 3, L 1Ψ 1Ψ 2Ψ 3 = l 1Ψ 1Ψ 2Ψ 3. For the angular equations we will nd the equation ( 1 u 4 + α2 1 sin 2 y + 4 ) β2 cos 2 + (2n + α + β + 1)2 u = 0. y which has solution u(y) = (sin y) α+ 1 2 (cos y) β+ 1 2 P (α,β) n (cos 2y) where P n (α,β) (x) is a Jacobi function.

Four-dimensional Quantum TTW With the replacements β 3 = k 2 3 ( ) ( ) 1 4 a2 4, β 4 = k3 2 1 4 a2 3, the separated θ 3 equation is ( ( k 2 1 L 3Ψ 3(θ 3) = Ψ 3 4 3 (θ 3) + ( 1 4) a2 cos 2 (k + k2 3 4 ) 3) a2 3θ 3) sin 2 Ψ 3(θ 3) = l 3Ψ 3(θ 3) (k 3θ 3) which has solutions Ψ a 3,a 4 3,n 3 (θ 3) = (sin(k 3θ 3)) a 3+ 1 2 (cos(k 3θ 3)) a 4+ 1 2 P (a 3,a 4 ) n 3 (cos(2k 3θ 3)) and eigenvalues L 3Ψ a 3,a 4 3,n 3 (θ 3) = l 3Ψ a 3,a 4 3,n 3 (θ 3), l 3 = k 2 3 (2n 3 + a 3 + a 4 + 1) 2.

Four-dimensional Quantum TTW The separated θ 2 equation L 2Ψ 2(θ 2) = l 2Ψ 2(θ 2) is ( Ψ 2 (θ 2) + k 2 cot(k 2θ 2)Ψ 2(θ 2) + which we transform with β 2 cos 2 (k 2θ 2) + l 3 sin 2 (k 2θ 2) Ψ 2(θ 2) = (sin(k 2θ 2)) 1 2 ψ(θ 2) to absorb the rst derivative term to give ( ψ 2 (θ 2) + β 2 cos 2 (k 2θ 2) + l3 + 1 4 k2 2 sin 2 (k 2θ 2) + 1 4 k2 2 l 2 ) Ψ 2(θ 2) = l 2Ψ 2(θ 2) ) ψ 2(θ 2) = 0 and we make the replacements ( ) β 2 = k2 2 1 4 a2 2, l 3 + 1 ( ) 4 k2 2 = k2 2 1 4 A2 2 A 2 = k3 k 2 (2n 3 + a 3 + a 4 + 1)

Four-dimensional Quantum TTW The separated θ 2 equation becomes ( 1 k2 2 ψ 2 4 (θ 2) + a2 1 2 cos 2 (k + 4 ) A2 2 2θ 2) sin 2 (k + 1 2θ 2) 4 l2 k2 2 ψ 2(θ 2) = 0 where 1 4 l2 k 2 2 = (2n 2+a 2+A 2+1) 2 l 2 = k 2 2 Λ 2(Λ 2+1), Λ 2 = 2n 2+a 2+A 2+ 1 2 and has solution Ψ A 2,a 2 2,n 2 (θ 2) = (sin(k 2θ 2)) A2 (cos(k 2θ 2)) a 2+ 1 2 P (A 2,a 2 ) n 2 (cos(2k 2θ 2)).

Four-dimensional Quantum TTW The separated θ 1 equation L 1Ψ 1(θ 1) = l 1Ψ 1(θ 1) is ( Ψ 1 (θ 1)+2k 1 cot(k 1θ 1)Ψ 1(θ 1)+ which we transform with to absorb the rst order term to give k 2 1 ψ 1 (θ 1) + and we make the replacements ( β 1 = k1 2 1 4 a2 1 β 1 cos 2 (k + l2 + 1 4 (k2 1 k2 2 ) 1θ 1) sin 2 (k 1θ 1) Ψ 1(θ 1) = (sin(k 1θ 1)) 1 ψ(θ 1) ( 1 4 a2 1 cos 2 (k + l2 + 1 4 (k2 1 k2 2 ) 1θ 1) sin 2 + 1 l1 (k 1θ 1) k1 2 ), l 2 + 1 4 (k2 1 k 2 2 ) = k 2 1 A 1 = k2 k 1 (2n 2 + A 2 + a 2 + 1) ) ) Ψ 1(θ 1) = l 1Ψ 1(θ 1), ψ 1(θ 1) = 0 ( ) 1 4 A2 1

Four-dimensional Quantum TTW The separate θ 1 equation becomes where 1 l1 k 2 1 ( 1 k1 2 2 θ1 ψ1(θ1) + 4 a2 1 1 cos 2 (k + 4 A2 1 1θ 1) sin 2 (k + 1 l1 1θ 1) k 2 1 ) ψ 1(θ 1) = 0 = (2n 1 + A 1 + a 1 + 1) 2 l 1 = k 2 1 Λ 1(Λ 1 + 2), Λ 1 = 2n 1 + a 1 + A 1 and has solutions Ψ A 1,a 1 1,n 1 (θ 1) = (sin(k 1θ 1)) A 1 1 2 (cos(k 1θ 1)) a 1+ 1 2 P (A 1,a 1 ) n 1 (cos(2k 1θ 1))

Four-dimensional Quantum TTW Finally, the separated radial equation is HΨ 0(r) = r 2 Ψ 0(r) + 3 ( r r Ψ0(r) + ω 2 r 2 + l1 ) k2 1 + 1 r 2 Ψ 0(r) = EΨ 0(r). which has solution Ψ A 0 0,n 0 (r) = e ωr2 2 r A 0 1 L (A 0 ) n 0 (ωr 2 ) where L A n (x) is a Laguerre functions and A 2 0 = k 2 1 l 1, A 0 = k 1(2n 1 + a 1 + A 1 + 1), E = ω(4n 0 + 2A 0 + 2).

Four-dimensional Quantum TTW Now, putting these together we nd E = ω(4n 0 + 2A 0 + 2) A 0 = k 1(2n 1 + a 1 + A 1 + 1) A 1 = k2 k 1 (2n 2 + A 2 + a 2 + 1) A 2 = k3 k 2 (2n 3 + a 3 + a 4 + 1) E = 2ω(2n 0+2k 1n 1+2k 2n 2+2k 3n 3+k 1a 1+k 2a 2+k 3a 3+k 3a 4+k 1+k 2+k 3+1) for a solution of the form Ψ n0,n 1,n 2,n 3 = Ψ A 0 0,n 0 (r)ψ A 1,a 1 1,n 1 (θ 1)Ψ A 2,a 2 2,n 2 (θ 2)Ψ a 3,a 4 3,n 3 (θ 3). Our aim is to use special function identities to raise and lower the n i while preserving E and produce new operators commuting with H.

Raising and lowering operators for TTW solutions Some examples... 0 n 0 = 1 A0 r + (2n 0 + A 0 + 1)ω + 1 A2 0 r r 2 K + A 0 0 n 0 = 1 + A0 r + (2n 0 + A 0 + 1)ω + 1 A2 0 r r 2 K A 0 These raising and lower n 0 and A 0 simultaneously. K + A 0 0 n 0 Ψ A 0 n 0 = 2(n 0 + 1)(n 0 + A 0)Ψ A 0 2 n 0 +1 K A 0 0 n 0 Ψ A 0 n 0 = 2ω 2 Ψ A 0+2 n 0 1

Raising and lowering operators for TTW solutions For the angular functions, we can use Jacobi function identities to make operators that raising and lower n alone. J n + (N + 1) sin(2kθ) = θ 1 ( (N + 1)(N + 1 c d) cos(2kθ) 2k 2 (N + 1)(c d) + a 2 b 2) Jn (N 1) sin(2kθ) = θ 1 ( (N 1)(N 1 + c + d) cos(2kθ) 2k 2 + (N 1)(c d) + a 2 b 2) N = 2n + a + b + 1, Θ (a,b) n = sin a+c (kθ) cos b+d (kθ)p n (a,b) cos(2kθ) J + n Θ (a,b) n = 2(n + 1)(n + a + b + 1)Θ (a,b) n+1 J n Θ (a,b) n = 2(n + a)(n + b)θ (a,b) n 1

Raising and lowering operators for TTW solutions We can also use Jacobi function identities to make operators that raising and lower n and a simultaneously. Again, for we have K n + a (1 a) cos(kθ) = θ 2(n(n + a + b + 1) + a(a + b)) k sin(kθ) (1 a)(a c) (1 a)(a + c + b + d) sin 2 (kθ) Kn a (1 + a) cos(kθ) = θ 2n(n + a + b + 1) k sin(kθ) (1 + a)(a + c) (1 + a)(a + c + b + d) + sin 2 (kθ) Θ (a,b) n = sin a+c (kθ) cos b+d (kθ)p n (a,b) cos(2kθ) K n + a Θ (a,b) n = 2(n + 1)(n + a)θn+1 a 2,b Kn a Θ (a,b) n = 2(n + a + b + 1)(n + b)θ a+2,b n 1

Raising and lowering operators for TTW solutions We now consider k 1 = p 1/q 1 for relatively prime p 1 and q 1 and form the operator Ξ + 01 = K A 0+2(p 1 1) 0 n 0 (p 1 1) K A 0 0 n 0 J 1 + n 1 +q 1 1 J 1 + n 1 which has the eect on a basis function of and so n 0 n 0 p 1, n 1 n 1 + q 1, A 0 A 0 + 2p 1. E = 2ω(2n 0 + 2k 1n 1 + ) 2ω(2(n 0 p 1) + 2k 1(n 1 + q 1) + ) = 2ω(2n 0 + 2k 1n 1 + ), that is, E is unchanged. A similar lowering operator is Ξ 01 = K + A 0 2(p 1 1) 0 n 0 +(p 1 1) K + A 0 0 n 0 J 1 n 1 (q 1 1) J 1 n 1.

Additional symmetries of generalized TTW The transformation n 1 n 1 A 1 a 1 1 while holding E constant has the eect of changing the sign of A 0. It is straightforward to check from the explicit expressions for the operators that L + 01 = Ξ + 01 + Ξ 01 and L 01 = Ξ+ 01 Ξ 01 A 0 are polynomials in E, A 2 0 and A 2 1. Since A 2 0 = k 2 1 l 1 and A 2 1 = 1 4 k2 2 l 2 k 2 1 we can replace E, A 2 0 and A 2 1 with a second order dierential operators where ever they appear in these expressions. E.g. for k 1, k 2, k 3 = 2, 1, 1 these are operators that commute with H of orders 5 and 6 that are algebraically independent of the second order operators.

Additional symmetries of generalized TTW For the case k 1, k 2, k 3 = 2, 1, 1, L + 01 is a 5th order operator. ( L + 01 = 2 r 3 r + 6 r 4 ( + + ) ( A 2 0 A2 1 + cos(4θ 1) 2r 3 r + sin(4θ 1) 4r 4 θ1 + 1 + 5 cos(4θ ) 1) 2r 4 A0 4 ( ) ( 1 r + 2 r r 2 E A1 2 sin(4θ1 ) + θ1 + cos(4θ ) 1) + 1 E 2 16 4 8 r + sin(4θ 1) 4r 2 θ1 + 3 cos(4θ ) ( ) 1) + 1 2r 2 E A0 2 + 10 r 3 r + 4 r 2 2 r 6 r 4 A1 2 cos(4θ 1) 4r ( sin(4θ1 ) r θ1 + 3 cos(4θ 1) + 2 a1 2 r r ( + sin(4θ 1) 4r 2 r 2 θ1 + 13 sin(4θ 1) r 5 sin(4θ 1) 4r 2 θ1 (6 cos(4θ 1) + 5 4a 2 1 ) ) 2r 2 E 4r 3 r θ1 (4 cos(4θ 1) + 1) 2r 2 r 2 + 13 + 27 cos(4θ 1) 4a 2 1 2r 3 r 5 sin(4θ 1) r 4 θ1 25 cos(4θ 1) + 20 12a 2 ) 1 2r 4 A0 2 + 11 sin(4θ 1) 4r 2 r 2 (11 8a 2 θ1 + 1 + 14 cos(4θ 1)) 2r 2 r 2 23 sin(4θ 1) 4r 3 r θ1 26 cos(4θ 1) + 23 20 a1 2 2r 3 r 21 sin(4θ 1) 4r 4 θ1 3(10 cos(4θ 1) + 7 4a 2 1 ) 2r 4 with the replacements E H, A 2 0 k2 1 L 1, A 2 1 (k2 2 4L 2)/(4k 2 1 ).

Additional symmetries of generalized TTW The operators L ± 01 are not of the lowest possible order. We can attempt to nd operators M 01 ± such that [L 1, M 01] ± = L ± 01, [H, M ± 01] = 0, [L 2, M ± 01] = 0, [L 3, M ± 01] = 0. For M01 the solution (when acting on a basis function) is M01 = 1 ( L ) 01 4q 1 A 0(A 0 + p + L + 01 S1(H, L2) + 1) A 0(A 0 p 1) A 2 0 p2 1 where k 1 = p 1/q 1 with p 1 and q 1 relatively prime and S 1(H, L 2) is an polynomial in H and L 2 to be determined.

Additional symmetries of generalized TTW From M01 = 1 ( 4q 1 L 01 A 0(A 0 + p 1) + L + 01 A 0(A 0 p 1) ) + S1(H, L2) A 2 0, p2 1 multiplying both sides by A 2 0 p 2 1 gives 4q 1(A 2 0 p1)m 2 01 = L 01 (A0 p1) + L+ 01(A 0 + p 1) + S 1(H, L 2) A 0 A 0 and so S 1(H, L 2) can be determined by setting A 0 = ±p 1. For example for p 1 = 2 and q 1 = 1, S 1(H, L 2) = (H2 4ω)(A 2 1 a 2 1) 16 and M 01 + turns out to be a polynomial in E and even powers of A0 and A1.

Further work Background The commutator algebra of the symmetry operators can be derived using similar methods (see e.g. SIGMA 7 (2011) 031). This provides examples of irreducible Killing tensors that naturally lead to symmetries of a modied Laplacian. Why does this work?