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

138and T 0 M χ ′ = ′ {v′ = 0}, we have c = 0. Next, in the equation(7.20) ⎧⎪⎨d(y 2 + y 6 + y 9 + y 10 χ(y)) ≡ [ay + b(y 2 + y 6 + y 9 + y 10 χ(y))] 2 ++ [ay + b(y 2 + y 6 + y 9 + y 10 χ(y))] 6 + [ay + b(y 2 + y 6 + y 9 + y 10 χ(y))] 9 +⎪⎩+ [ay + b(y 2 + y 6 + y 9 + y 10 χ(y))] 10 χ ′ (ay + b(y 2 + y 6 + y 9 + y 10 χ(y))),we firstly see that b = 0, and then from(7.21) d(y 2 + y 6 + y 9 + y 10 χ(y)) ≡ a 2 y 2 + a 6 y 6 + a 9 y 9 + a 10 y 10 χ ′ (ay),we see that a = d = 1. In other words, h = Id, whence y ′χ ′ (y ′ ) ≡ χ(y). This proves Lemma 7.1.= y andIn the remainder of §7, we shall exhibit other classes of hypersurfaceswith a control on their CR automorphism group. Since the computations aregenerally similar, we shall summarize them.7.4. Some classes of strong tube hypersurfaces in C n . GeneralizingLemma 7.1, we may state:Lemma 7.2. The real analytic hypersurfaces M χ1 ,...,χ n−1⊂ C n of equation∑n−1(7.22) v =k=1[ε k y 2 k + y 6 k + y 9 ky 1 · · ·y k−1 + y n+8kχ k (y 1 , . . ., y n−1 )],where ε k = ±1, are pairwise not biholomorphically equivalent strong tubes.Proof. The associated system of partial differential equations is of the form(7.23) {∂2zk z kw = − iε k − (15i/2 4 ) (∂ zk w) 4 + O((∂ z1 w) 7 ) + · · · + O((∂ zn−1 w) 7 ),∂ 2 z k1 z k2w = 0, for k 1 ≠ k 2 .Using the formulas (7.8) and inspecting the coefficients of the monomials inthe Wl 1 up to order five in the (n −1) equations extracted from the set of Lieequations(7.24) {R2k,k + (15i/2 2 )(Wk 2 ) Rk 1 + O((W1 1 ) 6 ) + · · · + O((Wn−1) 1 6 ) = 0,R 2 k 1 ,k 2= 0, for k 1 ≠ k 2 ,we get ∂ zl R ≡ ∂ w R ≡ ∂ zl Q k ≡ ∂ w Q k ≡ 0 for l, k = 1, . . .,n − 1. Thus,M χ1 ,...,χ n−1is a strong tube.Next, reasoning as in the end of the proof of Lemma 7.1, we see firstthat an equivalence between M χ1 ,...,χ n−1and M ′ χ ′ 1 ,...,χ′ n−1must be of the formz k ′ = ∑ n−1l=1 λl k z l, w ′ = µ w, where λ l k , 1 ≤ l, k ≤ n − 1 and µ are real.Inspecting the terms of degree 9, 10, . . ., n + 7, we get λ l k = 0 if k ≠ l, i.e.