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

112y ′ in terms of ỹ ′ as y ′ k = C ′ k (ỹ′ ). Finally, we get∂ψ k′ (3.32) (ỹ ′∂yl′ 1 , . . .,ỹ′ m ) = B′ k,l (iC ′ (ỹ ′ ), −iC ′ (ỹ ′ )),where the right hand sides are algebraic; this shows that the partial derivatives∂ y ′ lψ k ′ are algebraic functions of ỹ′ .To obtain the equivalent formulation of Theorem 1.1, we observe the following.Lemma 3.3. For every k, l = 1, . . .,m, the functions ∂ y ′kψ ′ l (y′ ) arealgebraic functions of y ′ if and only if for every k 1 , k 2 = 1, . . .,m,the second derivative ∂ 2 y k1 y k2(y) is an algebraic function of ψ(y) =(∂ y1 ϕ(y), . . ., ∂ ym ϕ(y)).Proof. Differentiating the identities y k ≡ ψ k ′ (ψ(y)), k = 1, . . ., m, withrespect to y l , we getm∑m∑(3.33) δk l ≡ ∂ y ′jψ k(ψ(y)) ′ ∂ yj ψ j (y) ≡ ∂ y ′jψ k(y ′ ′ ) ∂y 2 my lϕ(y).j=1Applying Cramer’s rule, we see that there exist universal rational functionsR k,l such that{ ∂2yk y lϕ(y) ≡ R k,l ({∂ y ′k2ψ k ′ 1(y ′ )} 1≤k1 ,k 2 ≤m})(3.34)j=1≡ R k,l ({∂ y ′k2ψ ′ k 1(∂ y1 ϕ(y), . . ., ∂ ym ϕ(y))} 1≤k1 ,k 2 ≤m}).This implies the equivalence of Lemma 3.3.In conclusion, taking Theorem 2.1 for granted, the proof of Theorem 1.1is now complete.3.3. Proof of Theorem 1.5. Let M : v = ϕ(z, ¯z) be a rigid Levi nondegeneratehypersurface in C n passing through the origin. We may assume thatv = ∑ n−1k=1 ε k |z k | 2 +ϕ 3 (z, ¯z), where ε k = ±1 and we may write ϕ 3 (z, ¯z) =∑ n−1k=1 [¯z k ϕ 3 k (z) + z k ¯ϕ 3 k (¯z)] + ϕ4 (z, ¯z), with ϕ 4 (0, ¯z) ≡ ϕ 4 z k(0, ¯z) ≡ 0 andϕ 3 k = O(2). After making the change of coordinates z′ k := z k + ε k ϕ 3 k (z),w ′ := w, we come to the simple equation v ′ = ∑ n−1k=1 ε k |z k ′ |2 + χ ′ (z ′ , ¯z ′ ),where χ ′ (0, ¯z ′ ) ≡ χ z ′k(0, ¯z ′ ) ≡ 0, considered in Theorem 1.5.Assume that M is strongly rigid, locally algebraizable and let M ′ be analgebraic equivalent of M. Let t ′ = h(t) be such an equivalence, or in ourprevious notation z ′ = f(z, w) and w ′ = g(z, w). We note z = f ′ (z ′ , w ′ )and w = g ′ (z ′ , w ′ ) the inverse equivalence. Since M is strongly rigid,namely Hol(M) is generated by the single vector field X 1 := ∂ w , it followsthat Hol(M ′ ) is also one-dimensional, generated by the single vector