Travaux sur les symétries de Lie des équations aux ... - DMA - Ens

360use the fact that every cubic polynomial equation of the form(1.13)0 ≡ A ++n∑k 1 =1n∑B k1 · y x k 1 +n∑n∑k 1 =1 k 2 =1 k 2 =1n∑n∑k 1 =1 k 2 =1C k1 ,k 2· y x k 1 y x k 2 +D k1 ,k 2 ,k 3· y x k 1 y x k 2 y x k 3is equivalent to the annihilation of the following symmetric sums of its coefficients:(1.14) ⎧0 = A,⎪⎨ 0 = B k1 ,0 = C k1 ,k 2+ C k2 ,k 1,⎪⎩0 = D k1 ,k 2 ,k 3+ D k3 ,k 1 ,k 2+ D k2 ,k 3 ,k 1+ D k2 ,k 1 ,k 3+ D k3 ,k 2 ,k 1+ D k1 ,k 3 ,k 2.for all k 1 , k 2 , k 3 = 1, . . .,n.In conclusion, the functions G j1 ,j 2, H k 1j 1 ,j 2, L k 1j 1and M k 1in the statementof Theorem 1.7 are far from being arbitrary: they satisfy the complicatedsystem of first order partial differential equations (I’), (II’), (III’) and (IV’)above.Our proof of Theorem 1.7 is similar to the one provided in [Me2004], inthe case of systems of second order ordinary differential equations, so thatmost steps of the proof will be summarized.In the end of this paper, we will delineate a complicated system of secondorder partial differential equations satisfied by G j1 ,j 2, H k 1j 1 ,j 2, L k 1j 1andM k 1which is the exact analog of the system described in the abstract. Themain technical part of the proof of Theorem 1.7 will be to establish thatthis second order system is a consequence, by linear combinations and bydifferentiations, of the first order system (I’), (II’), (III’) and (IV’).Open question 1.15. Are Theorems 1.1 and 1.7 true under weaker smoothnessassumptions, namely with a C 2 or a W 1,∞locright-hand side ?We refer to [Ma2003] for inspiration and appropriate tools.Open question 1.16. Deduce from Theorem 1.7 an explicit necessary andsufficient condition for the associated submanifold of solutions y = b +Π(x i , a k , b) to be locally equivalent to Y = B + X 1 A 1 + · · · + X n A n .As an application, this would characterize local sphericity of a Levi nondegeneratehypersurface M ⊂ C n+1 with n 2.Generalizing the Lie-Tresse classification would be a great achievement.