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

29410.16. Covering property. We now formulate a central concept.Definition 10.17. The pair of foliations (F v , F p ) is covering at the originif there exists an integer k such that the generic rank of Γ k is (maximalpossible) equal to dim K M . Since for a 1 = 0, the dual (k + 1)-th chainΓ ∗ k+1 identifies with the k-th chain Γ k, the same property holds for the dualchains.Example 10.18. With n = 1, m = 2 and p = 1 the submanifold defined byy 1 = b 1 and y 2 = b 2 + xa is twin solvable, but its pair of foliations is notcovering at the origin. Then SYM(M ) is infinite-dimensional, since fora = a(y 1 ) an arbitrary function, it contains a(y 1 ) ∂ + a(b 1 ) ∂ .∂y 1 ∂b 1Because we aim only to study finite-dimensional Lie symmetry groups ofpartial differential equations, in the remainder of this Part I, we will constantlyassume the covering property to hold.By Lemma 10.15, there exist two well defined integers µ and µ ∗ suchthat e 3 , e 4 , . . .,e µ+1 > 0, but e µ+l = 0 for all l 2 and similarly,e ∗ 3 , e∗ 4 , . . .,e∗ µ ∗ +1 > 0, but e∗ µ ∗ +l = 0 for all l 2. Since the pair of foliationsis covering, we have the two dimension equalities{ n + p + e3 + · · · + e µ+1 = dim K M = n + m + p,(10.19)p + n + e ∗ 3 + · · · + e∗ µ ∗ +1 = dim K M = n + m + p.By definition, the ranges of Γ µ+1 and of Γ ∗ µ ∗ +1 cover (at least; more is true,see: Theorem 10.28) an open subset of M . Also, it is elementary to verifythe four inequalitiesµ 1 + m, µ ∗ 1 + m,(10.20)µ µ ∗ + 1, µ ∗ µ + 1.In fact, since Γ k+1 with x 1 = 0 identifies with Γ ∗ k , the second line follows.Definition 10.21. The type of the covering pair of foliations (F v , F p ) is thepair of integers(10.22) (µ, µ ∗ ), with max(µ, µ ∗ ) 1 + m.Example 10.23. (Continued) We write down the explicit expressions of Γ 4and of Γ 5 :(10.24)⎧⎪⎨Γ 4 (x 1 , a 1 , x 2 , a 2 ; 0) = ( x 1 + x 2 , x 2 a 1 , a 1 + a 2 , −x 1 a 1 − x 1 a 2 − x 2 a 2 , ) ,Γ 5 (x 1 , a 1 , x 2 , a 2 , x 3 ; 0) = ( x 1 + x 2 + x 3 , x 2 a 1 + x 3 a 1 + x 3 a 2 , a 1 + a 2⎪⎩)− x 1 a 1 − x 1 a 2 − x 2 a 2 .Here, dimM = 3. By computing its Jacobian matrix, Γ 5 is of rank 3 atevery point (x 1 , a 1 , 0, −a 1 , −x 1 ) ∈ K 5 with a 1 ≠ 0. Since (obviously)(10.25) Γ 5(x1 , a 1 , 0, −a 1 , −x 1)= 0 ∈ M,