On the classification of certain curves up to projective tranformations Mehdi Nadjafikhah Abstract The purpose of this paper is to classify the curves in the form y 3 = c 3 x 3 +c 2 x 2 + c 1x + c 0, with c 3 0, up to projective transformations, and then show that, in the regular case, the necessary and sufficient condition for the two curves y 3 = c 3x 3 + c 2x 2 + c 1x + c 0 and y 3 = c 3x 3 + c 2x 2 + c 1x + c 0 with c 3 c 3 0, are equivalece to a projective transformation is that 27c 0 c 2 3 9c 1 c 3 c 2 + 2 c 2 3 3c 1 c 3 c 2 2 3 = 27 c 0 c 2 3 9 c 1 c 3 c 2 + 2 c 2 3 3 c 1 c 3 c 2 2 3. Furthermore, several special cases are considered. M.S.C. 2000: 53C10. Key words: Lie transformation theory, equivalence of submanifolds, symmetry, differential equations. 1. Introduction Let R 2 have the standard structure with the identity chart x, y and x := x and y := y are the standard vector fields on it. Let 1.1 C: y 3 = c 3 x 3 + c 2 x 2 + c 1 x + c 0 have the induced differentiable structure from R 2. Let 1.2 P 2, R = GL3, R/{λI 3 λ 0} be the Lie group of real projective transformations in the plan; that is, the transformation in the form a11 x + a 12 y + a 13 x, y, a 21x + a 22 y + a 23 1.3 a 31 x + a 32 y + a 33 a 31 x + a 32 y + a 33 such that a ij R and det[a ij ] 0. The LP 2, R Lie algebra of this Lie group is spanned by { span R x, y, x x, y x, y y, x 2 x + xy y, xy x + y 2 } 1.4 y Differential Geometry - Dynamical Systems, Vol.6, 2004, pp. 14-22. c Balkan Society of Geometers, Geometry Balkan Press 2004.
On the classification of certain curves 15 over R. P 2, R as a Lie group acts on R 2 and so, on M the set of all sub-manifolds in the form 1.1 of R 2. The two curves C 1 and C 2 in the form 1.1 are said to be equivalence if there exists an element T of P 2, R as 1.3 such that T C 1 = C 2. This relation partition M into disjoint cossets. The main idea of this paper is to show that there exists a bijective relation between M/P2, R and R := R { }. Then, it is shown that Main Theorem. a If c 3 0 and 3c 1 c 3 c 2 2 0, then the necessary and sufficient condition for the two curves C and C are congruent up to a projective transformation is that c 3 0, 3 c 1 c 3 c 2 2 0 and 27c 0 c 2 3 9c 1 c 3 c 2 + 2 c 2 3 3c 1 c 3 c 2 2 3 = 27 c 0 c 2 3 9 c 1 c 3 c 2 + 2 c 2 3 3 c 1 c 3 c 2 2 3. b If c 3 0 and 3c 1 c 3 c 2 2 = 0, then the necessary and sufficient condition for the two curves C and C are congruent up to a projective transformation is that c 3 0 and 3 c 1 c 3 c 2 2 = 0. c If c 3 = 0 and c 2 0, then the necessary and sufficient condition for the two curves C and C are congruent up to a projective transformation is that c 3 = 0 and c 2 0. d If c 3 = c 2 = 0 and c 1 0, then the necessary and sufficient condition for the two curves C and C are congruent up to a projective transformation is that c 3 = c 2 = 0 and c 1 0. e If c 3 = c 2 = c 1 = 0 and c 0 0, then the necessary and sufficient condition for the two curves C and C are congruent up to a projective transformation is that c 3 = c 2 = c 1 = 0 and c 0 0. f If c 3 = c 2 = c 1 = c 0 = 0, then the necessary and sufficient condition for the two curves C and C are congruent up to a projective transformation is that c 3 = c 2 = c 1 = c 0 = 0. We done this by studying the structure of symmetries of the equations of an special differential equation which M is it s solution. 2. Forming the problem In this section, we convert the given problem, which in fact, is a classification problem of special submanifolds of the Cartesian plane R 2 up to projective transformations, to the problem of classifying the solutions of a special differential equation. By calculating the fourth order derivative of the equation 1.1, we find that 2.1 E : 6y 2 y + 3yy 2 + 4yy y 3 + y 2 y 4 = 0
16 Mehdi Nadjafikhah which is an ordinary differential equation. Solving this equation, it turn out that d 3 dx y 3 = 0, and then, y 3 is a third order polynomial in x, and belong to M. Therefore, M is just the set of solution of the equation E. Hence, the problem of classifying 3 the integral curves of the equation 1.1 up to projective transformation, is equivalent to the problem of classifying the curves in the form 1.1 up to projective transformation. That is, the problem of classifying the P 2, R invariant solutions of the equation E. 3. Symmetry group of the equation E In order to find the symmetries of the differential equation E, we use the method, which is described in the page 104 of [4]. Let G be the symmetry group of the equation E, and G := LG be it s Lie algebra. Assume X := ξx, y x + ηx, y y be an arbitrary element of G, and prolong it to the G 4, which is acting on the E J 4 R 2. We find that: 360 y 2 ξ yyyy + 8yξ yyy + 12ξ yy y 5 +24 12η yy 24ξ xy + 8yη yyy 24ξ xyy + y 2 η yyyy 4y 2 ξ xyyy y 4 12 5y 2 ξ yyy + 6yξ yy + 6ξ y y + 6ξ xx 12η xy +12ξ xxy η xyyy 12yη xyy y 2 + 3y 2 ξ xxyy y 3 4 54yξ y + y 2 ξ yy y 3 + 38ξ x 6η y + 16yξ xy 3.1 6yη yy y 2 η yyy + 4y 2 ξ xyy y 6η xx + 12yη xxy 4yξ xxx 12y 2 ξ xxy + 3y 2 η xxyy y 2 + 5y 2 ξ y y 4 + 42η 8yξ x + 4η y + y 2 η yy 4y 2 ξ xy y 3 +30yyξ yy + 4ξ y y 2 + 64η x + 8yη xy 6yξ xx 3y 2 ξ xxy + 2y 2 η xyy y + y8η xxx + 4yη xxy yξ xxxx + y2η + yη y 4yξ x y 4 + 2y4η x 3yξ xx + 2yη xx 5yξ y y y 3 + 64yη y 8yξ x + 2η + y 2 η yy 4y 2 ξ xy y 2 +2y6η xx + 3yη xxy 2ξ xxx y + y 2 η xxxx = 0 Taking in to account that y, y, y 3 and y 4 are independent in the fourth order jet space J 4 R 2 of R 2, we find a system of partial differential equations from the equation 3.2, such as y 3.2 3.3 3.4 3.5 y 2 ξ yyyy + 8yξ yyy + 12ξ yy = 0 12η yy 24ξ xy + 8yη yyy 24ξ xyy y 2 + η yyyy 4y 2 ξ xyyy = 0 6ξ xx 12η xy + 12ξ xxy 12yη xyy 2y 2 η xyyy + 3y 2 ξ xxyy = 0 y 2 ξ yyy + 6yξ yy + 6ξ y = 0
On the classification of certain curves 17 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 6η xx + 12yη xxy 4yξ xxx 12y 2 ξ xxy + 3y 2 η xxyy = 0 8ξ x 6η y + 16yξ xy 6yη yy y 2 η yyy + 4y 2 ξ xyy = 0 4ξ y + yξ yy = 0 8η xxx + 4yη xxy yξ xxxx = 0 4η x + 8yη xy 6yξ xx 3y 2 ξ xxy + 2y 2 η xyy = 0 yξ yy + 4ξ y = 0 2η 8yξ x + 4η y + y 2 η yy 4y 2 ξ xy = 0 ξ y = 0 2η + yη y 4yξ x = 0 4η x 3yξ xx + 2yη xx = 0 6η xx + 3yη xxy 2ξ xxx = 0 4yη y 8yξ x + 2η + y 2 η yy 4y 2 ξ xy = 0 η xxxx = 0 By solving the equation 3.13 we conclude that ξ = fx is a function of x. Then, the equations 3.5, 3.8 and 3.11 are automatically satisfied, and provide no advantages. Now, by solving the equation 3.14 we find that there exists a function g such that η = 4 3 yf x + 1 y gx. Using the equation 3.4, it turn out that f x = 0, or 2 fx = Ax + B for arbitrary numbers A and B. Further, by the equation 3.18 it is conclude that g 4 x = 0, or gx = Cx 3 + Dx 2 + Ex + F for the arbitrary constants C, D, E and F. Furthermore, the other equations 3.2 to 3.18 are satisfied, and the this system are consistent. Therefore, The most general vector field X G is in the form X = a 1 x 2 + a 2 x + a 3 x + a 4 + a 1 xy + a 5x 3 + a 6 x 2 + a 7 x + a 8 y 2 y for the arbitrary constants a i, i = 1,, 8. In this manner, we have Theorem 1. The Lie algebra of symmetry group of the equation E is an 8 dimensional R Lie algebra which is spanned by the vector fields X 1 := x X 2 := x x X 3 := y y X 4 := x 2 x + xy y X 5 := 1 y 2 y X 6 := x y 2 y X 7 := x2 y 2 y X 8 := x3 y 2 y
18 Mehdi Nadjafikhah and have the commutator table X 1 X 2 X 3 X 4 X 1 0 X 1 0 2X 2 + X 3 X 2 X 1 0 0 X 4 X 3 0 0 0 0 X 4 2X 2 X 3 X 4 0 0 X 5 0 0 3X 5 3X 6 X 6 X 5 X 6 3X 6 2X 7 X 7 2X 6 2X 7 3X 7 X 8 X 8 3X 7 3X 8 3X 8 0 X 5 X 6 X 7 X 8 3.19 X 1 0 X 5 2X 6 3X 7 X 2 0 X 6 2X 7 3X 8 X 3 3X 5 3X 6 3X 7 3X 8 X 4 3X 6 2X 7 X 8 0 X 5 0 0 0 0 X 6 0 0 0 0 X 7 0 0 0 0 X 8 0 0 0 0 { } Theorem 2. Let A := span R X 1, X 2, X 3 and { B := span R X 4, X 5, X 6, X 7, X 8 }, then a A is a Lie subalgebra of G, and at the same time the Lie algebra of LP 2, R. b The connected Lie subgroup A of P 2, R corresponding to A, is consist of the transformation in the form 3.20 T A α1,α 2,α 3 : x, y α 1 + α 2 x, α 3 y where α 1, α 2 and α 3 are arbitrary constants with α 2 α 3 0. c B is an ideal of G which is the semi-product of A and B: G = A s B. d G LP 2, R = A. e The connected Lie subgroup B of P 2, R corresponding to B, is consist of the transformation in the form x 3.21 T B α4,α 5,α 6,α 7,α 8 : x, y x +, 1 xα 4 y + 3 y 1 xα 3 + 3tα 5 x 3 + α 6 x 2 + α 7 x + α 8 4 where α 4, α 8 arbitrary constants, and α 4 is small enough.
On the classification of certain curves 19 Proof: By the table 3.19, we have [A, A] A, [G, B] B and G = A B as vector spaces. Therefore, a and c are satisfied. In order to proving b, it is enough to find the integral curves of the each vector fields X 1 := x, X2 := x x and X 3 := ỹ ỹ on the manifold { x, ỹ, x, ỹ R}, which are initialized in the point x, y. We find out that these are x, y x + t, y, x, y e t x, y and x, y x, e t y, respectively. Because the Lie bracket of the each X i s with the generators 1.4 is not belongs to LP 2, R, d is also satisfied. For e, we proceed as b and find the integral curves corresponding to X 4, X 5, X 6, X 7 and X 8, respectively. Therefore, we have x x, y 1 xt, y x, y x, 3 y 1 xt 3 + 3t x, y x, 3 y 3 + 3tx x, y x, 3 y 3 + 3tx 2 x, y x, 3 y 3 + 3tx 3 and proof is complete. 4. The action on the M In this section we describe the action of G, and then the Lie group P 2, R on the manifold R 2, and then prolong it to an action on M. First, we need a chart ϕ : M R 4 on M such as ϕ{y 3 = c 3 x 3 + c 2 x 2 + c 1 x + c 0 } = c 3, c 2, c 1, c 0. Then, we describe the action of G on the manifold M. We have Theorem 3. If ϕc = c 3, c 2, c 1, c 0 and α 1,, α 8 are arbitrary numbers and α 4 be small as enough, then T A α1,0,0c ϕ c 3, 3α1c 2 3 + 2α 1 c 2 + c 1, 3α 1 c 3 + c 2, α1c 3 3 + α1c 2 2 + α 1 c 1 + c 0 T A 0,α2,0C ϕ α2c 3 3, α2c 2 2, α 2 c 1, c 0 T A 0,0,α3 C ϕ α3 1 3, α3 1 2, α3 1 1, c 0 T B α4,0,0,0,0c ϕ 6c 3 6α 4 c 2 + 6α4c 2 1 6α4c 3 0, 2c 2 4α 4 c 1 + 6α 4 c 0, 3α 4 c 0 + c 1, c 0 4.1 T B 0,α5,0,0,0C ϕ 6c 3, 2c 2, c 1, c 0 2α 5 T B 0,0,α6,0,0C ϕ 6c 3, 2c 2, c 1 3α 6, c 0 T B 0,0,0,α7,0C ϕ 6c 3, 2c 2 6α 7, c 1, c 0 T B 0,0,0,0,α8 C ϕ 6c 3 18α 8, 2c 2, c 1, c 0 and G is generated by the 8 elements 4.2. transitive. Further, The action of G on M is
20 Mehdi Nadjafikhah Proof: By applying the formulas 3.20 and 3.22, we find out the first fact. In order to show the second fact, we consider that, if we set α 1 = 0, α 2 = 0, α 3 = 0, α 4 = 0, α 5 = c 0 /2, α 5 = c 1 /3, α 5 = c 2 /6 and α 5 = c 3 /18 in the equations 4.2 and assuming 4.2 T = T B 0,0,0,0,α8 T B 0,0,0,α7,0 T B 0,0,α6,0,0 T B 0,α5,0,0,0 T B α4,0,0,0,0 T A 0,0,α3 T A 0,α2,0 T A α1,0,0 then, we find that T {y 3 = 0} = {y 3 = c 3 x 3 + c 2 x 2 + c 1 x + c 0 }, and the proof is complete. 5. Classification In this section, with a view to the above facts, we find the invariants of action A on M, to classifying the curves in the form 1.1. For this, we calculate the Killing vector fields of the action of A on R 2, by the infinitesimal method. Theorem 4. By the above assumptions, if à and B are the set of Killing vector fields corresponding to A and B respectively, and ci := c i, then à = span R { X 1, X 2, X 3 }, B = span R { X 4, X 5, X 6, X 7, X 8 }, and X 1 = 3c 3 c2 + 2c 2 c1 + c 1 c0 X 2 = 3c 3 c3 + 2c 2 c2 + c 1 c1 X 3 = c 3 c3 + c 2 c2 + c 1 c1 c 0 c0 X 4 = c 2 c3 + 2c 1 c2 + 3c 0 c1 X5 = c0 X6 = c1 X7 = c2 X8 = c3 Proof: In accordance with the definition of the such vector fields page 56 of [3], if X be the corresponding Killing vector field to X, then X = d dt F X {y 3 = c 3 x 3 + c 2 x 2 + c 1 x + c 0 }. Where, F X is the flow of X. We proceed the computations by t=0 the parts b and e of Theorem 2. The next step is to find the invariants of A on M. Theorem 5. 5.1 The action of A on M has only one functionally independent invariant: I := 27c 0c 2 3 9c 1 c 3 c 2 + 2 c 2 3 3c 1 c 3 c 2 2 3 Proof: If F c 3, c 2, c 1, c 0 be the an arbitrary invariant of the action A on M, then by the page 62 of [3] X i F = 0, for i = 1, 2, 3. In this manner we have a system of partial differential equations: 3c 3 F c2 + 2c 2 F c1 + c 1 F c0 = 0 3c 3 F c3 + 2c 2 F c2 + c 1 F c1 = 0 c 3 F c3 + c 2 F c2 + c 1 F c1 c 0 F c0 = 0 Solving this system, leads that F must be a function of I in 5.1.
On the classification of certain curves 21 Theorem 6. If c 3 0 and 3c 1 c 3 c 2 2 0, then the necessary and sufficient condition for the two curves C and C are congruent up to a projective transformation is that c 3 0, 3 c 1 c 3 c 2 2 0 and 5.2 27c 0 c 2 3 9c 1 c 3 c 2 + 2 c 2 3 3c 1 c 3 c 2 2 3 = 27 c 0 c 2 3 9 c 1 c 3 c 2 + 2 c 2 3 3 c 1 c 3 c 2 2 3. Proof: In view that the necessary condition for C 1, C 2 M be equivalent with respect to the action of A that is, have a same A orbit, is that IC 1 = IC 2, the necessity part is proved. For the sufficiency part, we find a transformation T α,β,γ A such that T α,β,γ {y 3 = 3 i=0 c ix i } = {y 3 = 3 i=0 c ix i }. Then, it is necessary that 5.3 5.4 5.5 5.6 γ 3 c 0 = c 3 α 3 + c 2 α 2 + c 1 α + c 0 γ 3 c 1 = 3c 3 α 2 β + 2c 2 αβ + c 1 β γ 3 c 2 = 3c 3 αβ + c 2 β 2 γ 3 c 3 = c 3 β 3 By the 5.3 we have γ = β 3 c3 c3. Then, by the 5.5 we conclude that α = c3 c2β c2 c3 3c 3 c 3, and by the 5.4, we have β = c 3 3c 1 c 3 c 2 2 c 3 3 c 1 c 3 c 2 2. If in the case 3 c 1 c 3 c 2 2 = 0, then by the equations 5.3 to 5.6, it is necessary that 3c 1 c 3 c 2 2 = 0, and two denominators are zero. Now, by letting this values in the 5.3, we conclude the equation 5.2. We further consider the case c 3 = 0, and prove the following Theorem 7. a If c 3 = 0 and c 2 0, then the necessary and sufficient condition for the two curves C and C are congruent up to a projective transformation is that c 3 = 0 and c 2 0. b If c 3 = c 2 = 0 and c 1 0, then the necessary and sufficient condition for the two curves C and C are congruent up to a projective transformation is that c 3 = c 2 = 0 and c 1 0. c If c 3 = c 2 = c 1 = 0 and c 0 0, then the necessary and sufficient condition for the two curves C and C are congruent up to a projective transformation is that c 3 = c 2 = c 1 = 0 and c 0 0. d If c 3 = c 2 = c 1 = c 0 = 0, then the necessary and sufficient condition for the two curves C and C are congruent up to a projective transformation is that c 3 = c 2 = c 1 = c 0 = 0. Proof: In each case, we find a transformation T α,β,γ A such that T α,β,γ {y 3 = 3 i=0 c ix i } = {y 3 = 3 i=0 c ix i }. In the case a we assume that 5.7 γ = 2/3 c2 α, b = c 2 c 1 c 1 c 2 α, α = c 2 c 2 2c 2 c 2 c 2 4c 0 c 2 c 2 1 4 c 0 c 2 c 2 1
22 Mehdi Nadjafikhah and the condition 4c 0 c 2 c 2 1 = 0 is equivalence with 4 c 0 c 2 c 2 1. In the case b we assume that 5.8 γ = c 3 3, b = c 0 c 2 2 c 0 c 1, α = c 2 1 c 1 and the condition c 1 = 0 is equivalent with c 1. In the case c, the two curves are y = 1 and, in the case d, the two curves are y = 0. Now the main theorem is a conclusion of the Theorems 6 and 7. References [1] E. Cartan, La Theorie des Groups Finis et Continus et la Geometrie differentielle Traitees par la Methode du Repere Mobile, Cahiers Scientifiques, Vol. 18, Gauthier- Villars, Paris, 1937. [2] M. Fels, and P. J. Olver, Moving coframes. I : A practical algorithm, Acta Appl. Math., Math. 51 1998 161-213. [3] P. J. Olver, Equivalence, Invariants, and Symmetry, Cambridge University Press, Cambridge University Press, 1995. [4] P. J. Olver, Application of Lie Groups to Differential Equations,GTM vol. 107, Springer Verlag, New Yourk, 1993. Author s address: Mehdi Nadjafikhah Iran University of Science and Technology, Faculty of Mathematics, Department of Pure Mathematics, Narmak-16, Tehran, Iran E-mail: m nadjafikhah@iust.ac.ir