EXPLICIT SYMMETRIES OF THE KEPLER HAMILTONIAN

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1 EXPLICIT SYMMETRIES OF THE KEPLER HAMILTONIAN HORST KNÖRRER AND EUGENE TRUBOWITZ To Richard Cushman on the occasion of his 65 th birthday Abstract. Using quaternions, we give explicit formulas for the global symmetries of the three dimensional Kepler problem. The regularizations of the Kepler problem that are based on the Hopf map and on stereographic projections, respectively, are interpreted in terms of these symmetries.. Introduction The motion of a point mass qt in the gravitation field of a mass that is fixed at the origin is described by the Kepler Hamiltonian Hq, p = 2 p 2 µ q Here, µ is determined by the gravitational constant and the central mass. Given an energy E, one regularizes the motion on the energy level hypersurface {q, p Hq, p = E} by introducing the eccentric anomaly as new independent variable. If E 0, the rescaled problem is the Hamiltonian flow of the Hamiltonian Hq, p = q p 2 2E 2µ on the hypersurface {q, p Hq, p = 0}. 2 E 2 E See [4], II.5.0b. Scaling q by the factor 2 E and p by the factor 2 E allows to reduce the discussion to the cases 2 E = or E = 0. That is, we consider the Hamiltonian flows of K ± q, p = q p 2 ± 2µ K 0 q, p = q p 2 2µ on the hypersurfaces {q, p K ± q, p = 0} and {q, p K 0 q, p = 0}, respectively. The case of negative energy that is, bounded orbits is described by K +, the case of positive energy by K. It is well known that, in three dimensions, this problem has SO4 symmetry for negative energy and SO +, 3 symmetry in the case of positive energy. For infinitesimal generators of the actions and the corresponding momentum maps see for example [4], II.3.2 or [5], I.3. We give explicit formulas for the actions of these groups on phase space R 3 R 3. Then we relate them to the regularizations of the Kepler problem by the Hopf map Kustaanheimo Stiefel [9] and by stereographic projection Györgyi [6] /Moser [] and Belbruno [2]/Osipov [2]. To motivate the construction, we first consider the two dimensional situation. Here we identify phase space R 2 R 2 with C C. The symmetry groups SO3 and SO +, 2 have double covers SU2 and SU,, respectively. We show that and Implicitly these actions are described in [], [3].

2 2 KNÖRRER AND TRUBOWITZ the symmetries are related to the action of these groups on the momentum variable q by fractional linear transformations. In three dimensions, we use the language of quaternions to describe the symmetries. In the case of negative energy, the symmetry group SO4 has as double cover SU2 SU2, which may be viewed as the group of pairs of quaternions of norm one. We identify R 3 with the set of all pure quaternions. The action of the group then is given by a formula similar to the two dimensional situation. In the case of negative energy, the symmetry group SO +, 3 has as double cover SL2, C. We describe SL2, C in terms of quaternions and again give the formula for the action on phase space. Then we select a point on the energy hypersurface {q, p K ± q, p = 0}. One gets a map from the symmetry group to the energy hypersurface by mapping a group element to the image of the chosen point under the action of the group element. We show that this map is the Kustaanheimo Stiefel regularization. The regularizations of Györgyi/Moser and Belbruno/Osipov are based on the stereographic projections from the three dimensional sphere, respectively hyperboloid. By viewing the sphere as homogenous space for SU2 SU2 and the hyperboloid as homogenous space for SL2, C, we get the relation between the Kustaanheimo Stiefel regularization and the Györgyi/Moser resp. Belbruno/Osipov regularizations described by Kummer [8]. 2. Two dimensions a b For an invertible complex 2 2 matrix A = and q, p C C set c d A q, p = c p + d 2 ap + b q, cp + d whenever it is defined. Observe that the second component is the standard action of GL2, C on the complex plane by fractional linear transformations. If det A =, the factor c p + d 2 is the complex conjugate of the inverse of the derivative of the fractional linear transformation p ap+b cp+d. Therefore, for A, A 2 SL2, C A A 2 q, p = A A 2 q, p whenever defined. Also, the action is symplectic with respect to the standard symplectic form Red p dq. Theorem 2.. Let q, p C C. Then K + A q, p = K+ q, p for all A SU2 K A q, p = K q, p for all A SU, { } a 0 K 0 A q, p = K0 q, p for all A P = a, c C, a 0 c a whenever defined. a b Proof. See also [5]. Let A = GL2, C. Then c d K ± A q, p + 2µ = c p + d 2 ap + b q 2 ± = q ap + b 2 ± cp + d 2 cp + d = q p 2 ± p 2 2

3 EXPLICIT SYMMETRIES OF THE KEPLER HAMILTONIAN 3 where p p 2 p = A If A SU2 then p 2 + p 2 2 = p 2 +, and if A SU, then p 2 p 2 2 = p 2. The proof for K 0 is similar. 3. Three dimensions, negative energy Recall that the quaternions are the four dimensional skew field over R H = {x 0 + x ı + x 2 j + x 3 k x 0,, x 3 R} with the associative multiplication characterized by For a quaternion we denote by u = u = u for all u H ı 2 = j 2 = k 2 = ıj = jı = k, jk = kj = ı, kı = ık = j u = x 0 + x ı + x 2 j + x 3 k u = x 0 x ı x 2 j x 3 k u = uu = x x2 + x2 2 + x2 3 the conjugate quaternion its norm Re u = x 0 = 2 u + u P u = x ı + x 2 j + x 3 k = 2 u u its real part its pure part R is considered as the subfield {x 0 x 0 R} of H. The multiplicative inverse of a non zero quaternion u is u = u u. It is easy to check that uv = v u for 2 u, v H. We identify R 3 with the space of pure quaternions H pure = {u H Re u = 0} = {u H u = Pu} = {u H u 2 R, u 2 0} by mapping q = q, q 2, q 3 R 3 to q ı+q 2 j +q 3 k H pure. The multiplicative group of quaternions of norm one is isomorphic to SU2. The product H unit = {u H u = } G + = H unit H unit with componentwise multiplication is a double cover of SO4, see for example [7], sec 3. For, 2 G + and q, p H pure H pure set whenever defined, 2 q, p = bp + a q bp + a, ap + bbp + a where a = 2 + 2, b =

4 4 KNÖRRER AND TRUBOWITZ Theorem 3.. Let, 2 G +. Then for all q, p H pure H pure. i, 2 q, p H pure H pure. ii, 2 2 q, p =, 2, 2 q, p for all, 2 G +. iii K +, 2 q, p = K + q, p. Furthermore the map q, p, 2 q, p is symplectic with respect to the symplectic form Re dq dp. The statements of the theorem apply whenever all expressions are defined. Proof. Observe that, by construction, a 2 + b 2 = and ab = ba, a b = b a. i Since q H pure we have q = q. Therefore bp + a q bp + a = bp + a q bp + a = bp + a q bp + a so that the first component of, 2 q, p lies in H pure. Similarly ap + bbp + a = bp + aap + b = bp + ap a + b = b p 2 a + bpb + ap a + ab = a p 2 b bp b apa ba = ap + bp b + a = ap + bbp + a so that ap + bbp + a = bp + a 2 ap + bbp + a = ap + bbp + a and ap + bbp + a H pure. ii By construction ap + bbp + a = [ 2 p + + p ][ 2 p + p ] Consequently = + p+ 2 p+ 2 ap + bbp + a + = ap + bbp + a = 2 p+ 2 2 p+ 2 p+ 2 Therefore the second component of, 2, 2 q, p is [ 2 p+ 2 + p+ 2 p+ 2 = ][ 2 [ 2 + p p + 2 p+ 2 = + p+ 2 2 p+ 2 2 which is the second component of, 2 2 q, p. p+ 2 ][ 2 p p + 2 ] p+ 2 ]

5 EXPLICIT SYMMETRIES OF THE KEPLER HAMILTONIAN 5 The derivative of the map p ap + bbp + a map at the point p is the linear Now u d ap + tu + b bp + tu + a dt t=0 = aubp + a ap + bbp + a bu bp + a bp + a 2[ a ap + bbp + a b ] = [ a ap + bbp + a b ] u bp + a = b 2 p 2 + a 2 + bpa + ap b a ap + bbp + a b = b 2 p 2 + a 2 a + a 2 bp + apa b a p 2 b 2 bp b 2 apa b ba b = a 2 a + a 2 bp + bp b 2 + bb a = bp + a So the derivative is the linear map u bp+a bp + a u bp + a on H 2 pure. Its inverse is the map u bp + a u bp + a ; and the adjoint of the inverse with respect to the bilinear form < u, v >= Re uv is the map u bp + a u bp + a. Thus the first component of q, p, 2 q, p is the adjoint of the derivative of the second component. This proves ii and also the fact that the map is symplectic. iii K +, 2 q, p + 2µ = bp + a 2 q 2 ap + b 2 bp + a 2 + = q 2 ap + b 2 + bp + a 2 = q 2 a 2 p 2 + b 2 + apb + bp a + b 2 p 2 + a 2 + ap b + bpa = q 2 p 2 + = K + q, p + 2µ For simplicity we assume from now on that 2µ =. Then the point ı, 0 lies on the energy hypersurface K+ 0. We consider the map ϕ + : {, 2 G + 2 } K + 0, 2, 2 ı, 0 = aıa, ba where, as before, a = 2 + 2, b = Under this parametrization, the angular momentum is L = Pq p = 4 ı 2 ı 2, and the eccentricity vector is q q + 2Pp L = 2 ı + 2 ı 2. Furthermore ϕ + maps the flow t e ıt, 2 e ıt on G + to the regularized Kepler flow the flow associated to the Hamiltonian K +. For the regularization of the Kepler problem, Kustaanheimo Stiefel [9] use the map that sends z, w C 2 \ {0} C 2 to q, p R 3 given by q = z σ 3 z, z σ z, z σ 2 z, p = Imz σ 3 z 2 w, Imz σ w, Imz σ 2 w where σ = 0 0, σ 2 = 0 ı ı 0, σ 3 = 0 0,

6 6 KNÖRRER AND TRUBOWITZ are the Pauli matrices. See [4] II.4.2 or [3] I.5.3. The first component of this map is the Hopf map hz = z 2 z 2 2, 2 Re z z 2, 2 Im z z 2 Its second component is 2 z 2 d dt hz + ıtw t=0. The Kustaanheimo Stiefelmap is the restriction of the map above to the set I = {z, w C 2 \{0} C 2 Re z w + z 2 w 2 = 0}. In the case of negative energy, one further restricts to I + = {z, w I z 2 + w 2 = }. We identify C 2 with H by mapping z, z 2 C 2 to z +k z 2 H. Under this identification I corresponds to and I + to Ĩ = {u, v H \ {0} H u v + v u = 0} Ĩ + = {u, v I u 2 + v 2 = } As before, we identify phase space R 3 R 3 with H pure H pure. With these identifications, the Kustaanheimo Stiefel map is KS : Ĩ H pure H pure u, v u ı u, v u Indeed, if u = z + kz 2, b = w + kw 2, then u ı u = z + kz 2 ı z z 2 k = z ı z + kz 2 ı z z ı z 2 k kz 2 ı z 2 k = z 2 ı + jz 2 z + z z 2 j j z 2 2 k = z 2 z 2 2 ı + 2 Rez z 2 j + 2 Imz z 2 k d corresponds to hz, and 2 z 2 dt hz + ıtw t=0 corresponds to d 2 u 2 dt u + tvı ı u + tvı = 2 u 2 vu + uv = u 2 vu = vu For similar descriptions of the Kustaanheimo Stiefel map, see [6] and [7]. Summarizing the discussion above, we see that the restriction of the Kustaanheimo Stiefel map to Ĩ+ is the same as the composition of ϕ + with the diffeomorphism Ĩ + {, 2 G + 2 } u, v u v, u + v Then a = = u, b = = v. Let S 3 = {x 0, x, x 2, x 3 R 4 x x2 3 = } be the three dimensional sphere. We denote by N =, 0, 0, 0 the north pole. Stereographic projection is the map from S 3 \ {N} to R 3 that sends x to sx = x 0 x, x 2, x 3. Its inverse is the map p = p, p 2, p 3 Bp = p 2 + p 2, 2p, 2p 2, 2p 3 For x S 3 we denote by T x S 3 = {y R 4 x y = 0} the tangent space to S 3 in the point x. The derivative of the stereographic projection in x is the map y y 0 y, y 2, y 3 + x 0 x 0 2 x, x 2, x 3 If p = sx, the adjoint of the derivative is the map R 3 T x S 3, q 2 Aq, p where Aq, p = 2 < p, q >, p, p 2, p 3 + p 2 + 0, q, q 2, q 3

7 EXPLICIT SYMMETRIES OF THE KEPLER HAMILTONIAN 7 The inverse of the adjoint is y s a x, y = x 0 y, y 2, y 3 + y 0 x, x 2, x 3 The stereographic projection is the starting point for the Moser regularization and for the Ligon Schaaf map 2. Let T S 3 = {x, y x S 3, y T x S 3, y = } be the unit tangent bundle of S 3. The Moser map is T S 3 \ T N S 3 R 3 R 3 x, y 2 s ax, y, sx We identify R 4 with H by mapping x 0, x, x 2, x 3 to x 0 + x ı + x 2 j + x 3 k. Then N corresponds to H and T S 3 to T S = {x, y H H x = y =, Re xy = 0} In this language the Moser map is since M : T S \ {, y Re y = 0} H pure H pure x, y 4 y y + 4 [xy x y], + x x 2 Rex Py + 2 ReyPx = 2 x + x y y + 2 x x y + y = y y + [xy x y] and Rex Px = x x 2 x x = x2 2x x 2 = +x x. The group G + acts on T S by, 2, x, y x 2, y 2. We show that the Moser map is equivariant with respect to the actions of G +. Lemma 3.2. Let, 2 G +. Then for all x, y T S M x 2, y 2 =, 2 Mx, y Proof. The first components of the map q, p, 2 q, p is the adjoint of the inverse of the derivative of the second component, and the second component of the map x, y x 2, y 2 is the adjoint of the inverse of the derivative of the first. Also, the first component of M is half the inverse of the derivative of the stereographic projection. Therefore it suffices to show that, for all x, y T S, the second component of M x 2, y 2 is equal to the second component of, 2 Mx, y. 2 Observe that on the energy hypersurface K + 0 = {q, p q = p 2 + } Aq, p = `2 < p, q >; p 2 + q 2 < p, q > p = `2 q < p, q >; 2 < p, q > p and q Bp = ` p 2 p + ; 2p p + = ` q p 2 ; 2 q p = `2 q p 2 ; 2 q p. Recall that A and B are the building blocks of the Ligon Schaaf map [0], see [4] II.3.4.

8 8 KNÖRRER AND TRUBOWITZ Set a = 2 + 2, b = Then the second component of, 2 Mx, y is a + x x + b b + x x + a = a + x + b x b + x + a x = 2 + x 2 x = + x 2 x 2 which is the second component of M x 2, y 2. Since M, ı = ı, 0, it follows that ϕ +, 2 = M 2, ı 2 for all, 2 G + If, as before, one interprets ϕ + as the Kustaanheimo Stiefel map, then this formula is the relation between the Kustaanheimo Stiefel regularization and the Moser regularization described in [8]. 4. Three dimensions, positive energy The quaternions can be viewed as the subalgebra of the algebra Mat2 2, C of complex 2 2 matrices. In fact, Mat2 2, C = H R C = H H where the multiplication on H H is given by, β, β = ββ, β + β In this description, the group SL2, C corresponds to G = {, β H H 2 β 2 =, β + β = 0} For, β G and q, p H pure H pure we set, β q, p = βp + q βp +, p ββp + Theorem 4.. Let, β G. Then for all q, p H pure H pure. i, β q, p H pure H pure. ii, β, β q, p =, β, β q, p for all, β G. iii K, β q, p = K q, p. Furthermore the map q, p, β q, p is symplectic with respect to the symplectic form Re dq dp. The proof is similar to the case of negative energy. We consider the map ϕ : {, β G β 0} K 0, β β ı β, β = lim, β ı t t 2, tı ϕ maps the flow t, β cosh t, ı sinh t on G to the regularized Kepler flow. In the case of positive energy, one restricts the Kustaanheimo Stiefel map to I = {z, w I z 2 w 2 = }. Under the identification of C 2 C 2 with H H this set corresponds to Ĩ = {u, v I u 2 v 2 = }. Ĩ can parametrized by G through the map, β β,. The composition of this isomorphism with the Kustaanheimo Stiefel map KS coincides with ϕ.

9 EXPLICIT SYMMETRIES OF THE KEPLER HAMILTONIAN 9 Let H 3 + = {x 0, x, x 2, x 3 R 4 x 2 0 x2 x2 2 x2 3 =, x 0 > 0} be the upper sheet of the three dimensional hyperboloid. Further denote by T H 3 + = {x, y x H 3 +, y R4 x 0 y 0 x y x 2 y 2 x 3 y 3 = 0, y 2 0 y2 y2 2 y2 3 = } the unit tangent bundle of H 3 +. The Belbruno map is {x, y T H+ 3 x N} R 3 R 3 x0 x, y 2 y, y 2, y 3 + y 0 2 x, x 2, x 3, x, x 2, x 3 x 0 The second component is the stereographic projection from the three dimensional hyperboloid. Here, we identify R 4 with R H pure by mapping x 0, x, x 2, x 3 to x 0, x ı + x 2 j + x 3 k. Under this identification 3 the unit tangent bundle of the hyperboloid corresponds to the set T H of pairs x 0, x, y 0, y R H pure 2 for which x 0, x x 0, x =, 0, y 0, y y 0, y =, 0 x 0, x y 0, y y 0, y x 0, x = 0, 0 In this language the Belbruno map is B : { x 0, x, y 0, y T H x 0, x, 0} H pure H pure x0, x, y 0, y x0 y + y x, x x 0 The group G acts on R H pure by, β, x0, x, β x 0, x, β Lemma 4.2. Let, β G. Then for all x 0, x, y 0, y T H B, β x 0, x, β,, β y 0, y, β =, β B x 0, x, y 0, y The proof is again similar to the case of negative energy. Also, for, β G ϕ, β = B, β, 0, β,, β 0, ı, β See also [4]. 5. Three dimensions, zero energy In this situation the symmetry group is { G 0 = Hunit, c H, c + c = 0} c with the multiplication c = c c + c 3 The isomorphism between H H = H R C and Mat2 2, C, mentioned above, maps R H pure to the space of Hermitian matrices.

10 0 For KNÖRRER AND TRUBOWITZ G c 0 and q, p H pure H pure set whenever defined q, p = cp + q cp + c, pcp + Theorem 5.. Let G c 0. Then for all q, p H pure H pure. i q, p H c pure H pure. [ [ ] ii c q, p = c] c q, p for all c iii K 0 q, p = K c 0 q, p. Furthermore the map q, p c q, p is symplectic with respect to the symplectic form Re dq dp. The proof is similar to the case of negative energy. Again, we consider the map c G 0. ϕ 0 : {, c G 0 c 0} K 0 0 c c ı c, c = lim ı t, tı t c ϕ 0 maps the flow t = on G c ıt c t ı 0 to the regularized Kepler flow. Also observe that ϕ 0 is the composition of the maps G 0 H pure H pure ; c c, ı and H pure \ {0} H pure H pure H pure ; x, y xyx, x This gives the regularization described in [2], Theorem 3, or [2]. References [] H.Bacry, H.Ruegg, J.Souriau: Dynamical groups and spherical potentials in classical mechanics. Commun. Math. Phys. 3, [2] J.Belbruno: Two Body motion under the inverse square central force and equivalent geodesic flows. Celestial Mechanics 5, [3] B.Cordani: The Kepler Problem. Birkhäuser Verlag 2003 [4] R.Cushman, L.Bates: Global Aspects of Classical Integrable Systems. Birkhäuser Verlag 997 [5] V.Guillemin, Shl.Sternberg: Variations on a Theme of Kepler. AMS Colloquium Publications 42, 990 [6] G.Györgyi: Kepler s equation, Fock variables, Bacry s generators and Dirac brackets. Nuov. Cim. 53 A, [7] M.Koecher, R.Remmert: Hamilton s Quaternions. In: H.-D. Ebbinghaus et al.: Numbers. Springer Verlag 99, pp [8] M.Kummer: On the regularization of the Kepler problem. Commun. Math. Phys. 84, [9] P.Kustaanheimo, E.Stiefel: Perturbation theory of Kepler motion based on spinor regularization. J. Reine Angew. Math. 28,

11 EXPLICIT SYMMETRIES OF THE KEPLER HAMILTONIAN [0] T.Ligon, M. Schaaf: On the global symmetry of the classical Kepler Problem. Reports on Mathematical Physics 9, [] J.Moser: Regularization of Kepler s problem and the averaging method on a manifold. Comm. Pure Appl. Math. 23, [2] Y.Osipov: The Kepler problem and geodesic flows in spaces of constant curvature. Celestial Mechanics 6, [3] H.Rogers: Symmetry transformations of the classical Kepler problem. J. Math.Phys 4, [4] M.Sani: Regularisierungen des Kepler schen Problems für den Fall von positiver Energie in der Formulierung mit Quaternionen. Bachelor Thesis, Zürich [5] J.Sourlier: Symmetries of the Kepler Problem. Bachelor Thesis, Zürich [6] M.Vivarelli: The KS transformation in hypercomplex form. Celestial Mechanics 29, [7] J.Waldvogel: Quaternions and the perturbed Kepler problem. Celestial Mech. Dynam. Astronom. 95, Department of Mathematics, ETH Zürich, CH 8092 Zürich, Switzerland address: knoerrer@math.ethz.ch, trub@math.ethz.ch