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

for k even, we have Γ k (z (k) ) = L zk(Γ k−1 (z (k−1) )), the following two propertieshold:127(6.14){J l h(Γ k (z (k) )) = J l h(Γ k−1 (z (k−1) )), if k is even;J l¯h(Γk (z (k) )) = J l¯h(Γk−1 (z (k−1) )),if k is odd.Let now κ 0 := l 0 (µ 0 + 1) be the product of the Levi type with the Segretype of M plus 1 and consider the open subset of the κ 0 -order jet spaceC Nn,κ 0 defined by the inequality |Jκ 0− J κ 0Id | < ε. Let z (k) ∈ ∆ mk as in §5.6above. Since the maps Γ k are holomorphic and satisfy Γ k (0) = 0, we maychoose δ > 0 sufficiently small in order that the following two conditionsare satisfied for every k ≤ µ 0 and for and for every |z (k) | < δ:(6.15) |Γ k (z (k) )| < ρ 3 and |J κ 0h(Γ k (z (k) )) − J κ 0Id | < ε.This choice of δ is convenient to make several susbtitutions by means offormulas (6.12). The formulas (6.16) that we will obtain below stronglydiffer from the previous formulas (6.12), because they depend on the jet ofh at the origin only.Lemma 6.3. Shrinking ε if necessary, for every integer k ≤ µ 0 + 1 andfor every integer l ≥ 0, there exists a complex algebraic mapping Π l,k withvalues in C N n,l defined for |t|, |τ| < ρ3 and for |J kl 0− J kl 0Id| < ε, whichis relatively polynomial with respect to the higher order jets Ji α with |α| ≥kl 0 +1, i = 1, . . .,n, and which depends only on the defining functions ξ j −Θ j (ζ, t) of M , such that the following two families of conjugate identitiesare satisfied(6.16){J l h(Γ k (z (k) )) = Π l,k (Γ k (z (k) ), J kl 0+l¯h(0)), if k is odd;J l¯h(Γk (z (k) )) = Π l,k (Γ k (z (k) ), J kl 0+l¯h(0)),if k is even.Proof. For k = 1, replacing (t, τ) by Γ 1 (z (1) ) in the first relation (6.12) andusing the second property (6.14), we get(6.17){J l h(Γ 1 (z (1) )) = Π l (Γ 1 (z (1) ), J l 0+l¯h(Γ1 (z (1) ))) == Π l (Γ 1 (z (1) ), J l 0+l¯h(0)),so the lemma holds true for k = 1 if we simply choose Π l,1 := Π l . Byinduction, suppose that the lemma holds true for k ≤ µ 0 . To fix the ideas,let us assume that this k is even (the odd case is completely similar). Thenreplacing the arguments (t, τ) in the first relation (6.12) by Γ k+1 (z (k+1) ),using again the second property (6.14), and using the induction assumption,namely using the conjugate of the second relation (6.16) with l replaced by