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

114exist algebraic functions B ′ k (¯z′ ) such that(3.39) B ′ k (¯z′ ) ≡∑n−1l=1∂ zl ϕ(0, ¯f ′ (¯z ′ )) ∂ z ′kf ′l (0) ≡ ε k ¯f ′ k (¯z′ ).We have proved that the components f k ′(z′ ) are all algebraic.Finally, coming back to the relation (3.36), we want to prove that thederivatives ∂ zl ϕ(f ′ (z ′ ), ¯f ′ (¯z ′ )) are all algebraic. However, the first termof (3.36) is not algebraic in general. Fortunately, using the fact thatΘ ′= ¯w ′ + O(2), we see that there exists a unique algebraic solution¯w ′ = Λ ′ (z ′ , ¯z ′ ) of the implicit equation Θ ′ (z ′ , ¯z ′ , ¯w ′ ) = 0, namely satisfyingΘ ′ (z ′ , ¯z ′ , Λ ′ (z ′ , ¯z ′ )) ≡ 0. Then by replacing ¯w ′ by Λ ′ in (3.36), we getthat there exist algebraic functions C k ′(z′ , ¯z ′ ) such that(3.40) C ′ k(z ′ , ¯z ′ ) ≡∑n−1l=1∂ zl ϕ(f ′ (z ′ ), ¯f ′ (¯z ′ )) ∂ z ′kf ′l(z ′ ).Since f ′ is tangent to the identity map, we can solve by Cramer’srule this linear system for the derivatives ∂ zl ϕ, which yields that the∂ zl ϕ(f ′ (z ′ ), ¯f ′ (¯z ′ )) are all algebraic. Since f ′ (z ′ ) is also algebraic, weobtain in sum that the derivatives ∂ zl ϕ(z, ¯z) are all algebraic. In conclusion,taking Theorem 2.1 for granted, the proof of Theorem 1.5 is complete.3.4. Proof of Theorem 1.4. Let M : v = ϕ(z¯z) in C 2 with Hol(M) generatedby ∂ w and iz∂ z . Without loss of generality, we can assume thatϕ(r) = r + O(r 2 ). Let M ′ be an algebraic equivalent of M. Let t = h ′ (t ′ ),or z = f ′ (z ′ , w ′ ), w = g ′ (z ′ , w ′ ) be a local holomorphic equivalence satisfyingh ′ (M ′ ) = M. Let t ′ = h(t) be its inverse. Then Hol(M ′ ) istwo-dimensional and generated by h ∗ (∂ w ) and h ∗ (iz∂ z ). First of all, usingthe algebraicity of the CR automorphism group of M ′ and proceedingas in the proof of Proposition 3.1, we can prove that there exist two generatorsof Hol(M ′ ) of the form X 1 ′ = b ′ (w ′ ) ∂ w ′ and X 2 ′ = a ′ (z ′ ) ∂ z ′ whereb ′ and a ′ are algebraic and satisfy b ′ (0) = 1 and a ′ (z ′ ) = iz ′ + O(z ′2 ).Furthermore, we may assume that h ′ is tangent to the identity map and thath ′ ∗ (b′ (w ′ ) ∂ w ′) = ∂ w and h ′ ∗ (a′ (z ′ ) ∂ z ′) = iz∂ z . As in (3.24), it follows thatb ′ (w ′ ) g w ′ ′(w′ ) ≡ 1 and a ′ (z ′ ) f z ′ ′(z′ ) ≡ if ′ (z ′ ). Let w ′ = Θ ′ (z ′ , ¯z ′ , ¯w ′ ) be thecomplex algebraic equation of M ′ . Then we get the following power seriesidentity:(3.41) g ′ (Θ ′ (z ′ , ¯z ′ , ¯w ′ )) − ḡ ′ ( ¯w ′ ) ≡ 2i ϕ(f ′ (z ′ ) ¯f ′ (¯z ′ )),