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

with |β∗ k | ≥ 1 and integers j∗, 1 . . .,j∗ m with 1 ≤ j∗ k ≤ d such that the realmapping( )∂|β∗ 1| ϕ j 1(3.19) ψ(y) :=∗(y), . . ., ∂|βm ∗ | ϕ j m ∗(y)=: y ′ ∈ R m∂y β1 ∗ ∂y βm ∗is of rank m at the origin in R m .Proof. We follow the definition of finite nondegeneracy given in §1.2. Letr j (t, ¯t) := v j − ϕ j (y) = 0 be the defining equations of M. Let L k :=∂¯zk + ∑ dj=1 ϕ j,¯z k∂ ¯wj , k = 1, . . .,m, be a basis of (1, 0)-vector fields tangentto∑M. We write the first order terms in the Taylor series of ϕ j (y) as ϕ j (y) =nl=1 λ j,l y l + O(|y| 2 ). Then the holomorphic gradient of r j is given by(3.20) ⎧∇ ⎪⎨ t (r j ) = (∂ z1 r j , . . .,∂ zm r j , ∂ w1 r j , . . .,∂ wd r j )= i2 −1 (∂ y1 ϕ j , . . .,∂ ym ϕ j , 0, . . .,0, −1, 0, . . ., 0)⎪⎩= i2 −1 (λ j,1 , · · · , λ j,m , 0, . . ., 0, −1, 0, . . ., 0), at the origin.On the other hand, since for β = (β 1 , . . .,β m ) ∈ N m with |β| ≥ 1 theorder |β| derivation L β := L β 11 · · ·Lβm m acts on functions of y as the operator(2i) −|β| ∂y β , we can compute(3.21) {βL (∇t (r j )) = (L β ∂ z1 ϕ j , . . ., L β ∂ zm ϕ j , 0, . . .,0, . . ., 0)= i −|β|+1 2 −|β|−1 (∂ β y ∂ y 1ϕ j , . . .,∂ β y ∂ y mϕ j , 0, . . ., 0, . . .,0).By inspecting the expressions (3.20) and (3.21), we see thatSpan{(L β (∇ t (r j )))(0) : β ∈ N m , j = 1, . . .,d} = C n if and onlyif Span{(∂ β y ∂ y1 ϕ j (0), . . ., ∂ β y ∂ ym ϕ j (0)) : β ∈ N m , |β| ≥ 1, j =1, . . ., d} = R m . This last condition is clearly equivalent to the one statedin Lemma 3.2.We can prove now that the inverse mapping ψ ′ (y ′ ) of the mappingψ(y) defined by (1.1) (or (3.19)) has algebraic first order derivatives.By Step 1, there exists a biholomorphic transformation Φ mapping thestrong tube M onto the algebraic pseudotube M ′ with the property thatΦ ∗ (∂ ti ) = c ′ i(t ′ i) ∂ t ′∑ i. Writing Φ(t) = (h 1 (t), . . .,h n (t)), we have Φ ∗ (∂ ti ) =nl=1 h l,t i(t) ∂ t ′l= c ′ i (t′ i ) ∂ t ′ , so h ii(t) depends only on t i which yieldsΦ(t) = (h 1 (t 1 ), . . .,h n (t n )). We shall use the convenient notation t ′ i =h i (t i ) and t i = h ′ i(t ′ i) for the inverse h ′ i := h −1i , i = 1, . . ., n. If accordingly,Φ ′ (t ′ ) = t denotes the inverse of Φ(t) = t ′ , we have Φ ′ ∗ (c′ i (t′ i ) ∂ t ′) = ic ′ i (t′ i ) h′ i,t(t ′ ′ i ) ∂ t i= ∂ ti , which shows that c ′ i (t′ i ) h′ ii,t i(t ′ ′ i ) ≡ 1. Since c′ i (0) =1, we see that h ′ i (t′ i ) = ∫ t ′ i0 1/[c′ i (σ)] dσ is the complex primitive of an algebraicfunction. This observation will be important.109