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

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

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h(t) = ˜H(t, J l 0µ 0¯h(0)) with respect to t that(6.26) J l 0µ 0h(0) = ([∂tα ˜H(t, J l 0µ 0¯h(0))]t=0 ) |α|≤l0 µ 0.Consequently, if K is the holomorphic map <strong>de</strong>fined by(6.27) K(J l 0µ 0) := ([∂ α t˜H(t, J l 0µ 0] t=0 ) |α|≤l0 µ 0),we get the equality J l 0µ 0= K(J l 0µ 0) for every Jl 0 µ 0∈ Ẽ, which provesthat Ẽ is totally real. It follows that E is totally real, which completes theproof.Lemma 6.6. The submanifold E is naturally equipped with a local realalgebraic <strong>Lie</strong> group structure in a neighborhood of J κ 0Id .Proof. In<strong>de</strong>ed, let us parametrize E by a real algebraic mapping(6.28) R c 0∋ (e 1 , . . .,e c0 ) ↦−→ j κ0 (e) ∈ C Nn,κ 0 ,where c 0 is the dimension of E. Here, to avoid excessive formal complexity,we shall avoid to mention all the polydiscs of variation of the variab<strong>les</strong>. Fore ∈ E, we shall use the notation(6.29) H(t; e) := H(t, j κ0 (e)).Let e ∈ E and e ′ ∈ E, set J κ 0:= j κ0 (e) and ′ J κ 0:= j κ0 (e ′ ). Then we can<strong>de</strong>fine the <strong>Lie</strong> group multiplication µ J by(6.30) µ J ( ′ J κ 0, J κ 0) := ([∂ α t (H(H(t, Jκ 0), ′ J κ 0))] t=0 ) |α|≤κ0 .Accordingly, in terms of the coordinates (e 1 , . . .,e c0 ) on E, the <strong>Lie</strong> groupmultiplication µ is <strong>de</strong>fined by(6.31) µ(e, e ′ ) := (j κ0 ) −1 (µ J (j κ0 (e ′ ), j κ0 (e))) ∈ R c 0It follows from the algebraicity of the mappings H and j κ0 that the mappingsµ J and µ are algebraic.We must check the associativity of µ, namely µ(µ(e, e ′ ), e ′′ ) =µ(e, µ(e ′ , e ′′ )). So we set h(t) := H(t, j κ0 (e)), h ′ (t) := H(t, j κ0 (e ′ ))and h ′′ (t) := H(t, j κ0 (e ′′ )). By the <strong>de</strong>finition (6.30), we haveµ J (j κ0 (e), j κ0 (e ′ )) = J κ 0(h ◦ h ′ )(0). Applying then Theorem 6.4, weget H(t, J κ 0(h ◦ h ′ )(0)) ≡ (h ◦ h ′ )(t). Consequently, using again (6.30)and the associativity of the composition of mappings, we may compute(6.32) ⎧µ J (µ J (j κ0 (e), j κ0 (e ′ )), j κ0 (e ′′ )) = µ J (J κ 0((h ◦ h ′ )(0), j κ0 (e ′′ ))⎪⎨⎪⎩= J κ 0((h ◦ h ′ ) ◦ h ′′ )(0)= J κ 0(h ◦ (h ′ ◦ h ′′ ))(0)= µ J (j κ0 (e), J κ 0(h ′ ◦ h ′′ )(0))= µ J (j κ0 (e), µ J (j κ0 (e), j κ0 (e ′ ))),131

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