Travaux sur les symÃ©tries de Lie des Ã©quations aux ... - DMA - Ens

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284by − 1 Y 2 x 1 2x1 and solving for Xx 2 x: 2(8.91)0 = X 1x 1 x 2 + 1 2 Y x 1 x 1,X 2x 2 x 2 = −(1 + s∗ x 2) Y x 1 x 1 + r Y x 1.This is (8.86) 7 . Differentiating (8.91) 2 with respect to x 1 , taking account of(8.83) 1 , we get (8.86) 16 :(8.92) X 2x 1 x 2 x 2 = r Y x 1 + r Y x 1 x 1.We then replace X 1x 1 from (8.90) in (8.81) 4 :(8.93)0 = X 2x 1 + 2 X 1y + r X 1 + r X 2 + r X 2x 2 + r Y x 1 + r Y y++ r Y x 1 x 1 + s∗ Y x 1 y + s ∗ X 1x 1 x 1.We differentiate this equation with respect to x 2 , knowing Y x 2 = 0:(8.94)0 = Xx 2 1 x + 2 X 1 2 x 2 y + r X 1 + r Xx 1 + r X 2 + r X 22 x + r X 2 2 x 2 x + r Y 2 x 1 + r Y y++ r Y x 1 x 1 + s∗ x 2 Y x 1 y + s ∗ x 2 X 1x 1 x 1 + s∗ X x 1 x 1 x 2.We replace: X 1xfrom (8.89); X 22 x 2 xfrom (8.91) 2 2 ; we differentiate (8.81) 2with respect to x 1 x 1 to replace X 1x 1 x 1 xby r Y 2 x 1 x 1, thanks to (8.83) 1; and wereorganize:(8.95)2 X 1x 2 y +s∗ x Y 2 x 1 y+s ∗ x X 1 2 x 1 x = −X 2 1 x 1 x 2+r X 1 +r X 2 +r X 2x 2+r Y x 1+r Y y+r Y x 1 x 1.We differentiate (8.81) 2 with respect to y and (8.81) 3 with respect to x 1 :(8.96)X 1x 2 y + 1 2 Y x 1 y = 0,Y x 1 y − 2 X 1x 1 x 1 = −X 2x 1 x 2 + r Y x 1 + r Y x 1 x 1.For the three unknowns X 1x 1 x, Y 1 x 1 y, X 1x 2 y, we solve the three equations(8.95), (8.96) 1 , (8.96) 2 , reminding s ∗ x(0) = 0: 2(8.97)X 1x 1 x = r X 1 + r X 2 + r X 21 x + r Y 2 x 1 + r Y y + r Y x 1 x 1 + r X 2x 1 x 2,Y x 1 y = r X 1 + r X 2 + r X 2x + r Y 2 x 1 + r Y y + r Y x 1 x 1 + r X 2x 1 x 2,X 1x 2 y = r X 1 + r X 2 + r X 2x 2 + r Y x 1 + r Y y + r Y x 1 x 1 + r X 2x 1 x 2.We get (8.86) 11 and (8.86) 9 .Thus, we may replace X 1x 1 x 1 and Y x 1 y in (8.81) 4 to get (8.86) 2 :(8.98)X 2x 1 = −2 X 1y + r X 1 + r X 2 + r X 2x 2 + r Y x 1 + r Y y + r Y x 1 x 1 + r X 2x 1 x 2.Next, we differentiate (8.83) 3 with respect to x 1 and we replace: X 1xfrom 1(8.90); X 2xfrom (8.98); Y 1 x 1 y from (8.97) 2 ; X 1x 1 xfrom (8.97) 1 1 ; Y x 1 x 1 x 1from (8.83) 1 ; and we compare with (8.83) 2 ; we differentiate (8.96) 1 with
espect to x 1 and (8.96) 2 with respect to x 1 ; solving, we obtain four newrelations:(8.99)X 1x 1 x 1 x = r X 1 + r X 2 + r X 21 x + r Y 2 x 1 + r Y y + r Y x 1 x 1 + r X 2x 1 x 2,Y x 1 x 1 y = r X 1 + r X 2 + r X 2x + r Y 2 x 1 + r Y y + r Y x 1 x 1 + r X 2x 1 x 2,X 2x 1 x 2 y = r X 1 + r X 2 + r X 2x + r Y 2 x 1 + r Y y + r Y x 1 x 1 + r X 2x 1 x 2,X 2x 1 x 1 x = r X 1 + r X 2 + r X 22 x + r Y 2 x 1 + r Y y + r Y x 1 x 1 + r X 2x 1 x 2.We get (8.86) 19 and (8.86) 14 .Next, in (8.81) 5 , we replace: X 1xfrom (8.90); X 21 xfrom (8.98); Y 1 x 1 yfrom (8.97) 2 ; we get:(8.100)Xy 2 = r X 1 + r X 2 + r Xy 1 + r X 2x + r Y 2 x 1 + r Y y + r Y x 1 x 1++ s ∗ X 2x 1 x + 1 s∗ Y yy + s ∗ X 1x 1 y .We differentiate (8.98) with respect to x 1 and we replace: X 1xfrom (8.90);1X 2xfrom (8.98); Y 1 x 1 y from (8.97) 2 ; Y x 1 x 1 x 1 from (8.83) 1; X 2x 1 x 1 xfrom 2(8.99) 4 ; we get:(8.101)X 2x 1 x 1+2 X 1x 1 y = r X 1 +r X 2 +r Xy 1 +r X 2x 2+r Y x 1+r Y y+r Y x 1 x 1+r X 2x 1 x 2.In (8.82) 2 , we replace: X 1xfrom (8.90); X 11 x 1 xfrom (8.97) 1 1 ; Y x 1 y from(8.97) 2 ; and we reorganize:(8.102)2 Xx 2 2 y −6 X x 1 1 y −X x 2 1 x = −2 Y yy+r X 1 +r X 2 +r X 21 x 2+r Y x 1+r Y y+r Y x 1 x 1+r X x 2 1 x 2.Differentiating (8.81) 3 with respect to y, we replace: Y x 1 y from (8.97) 2 ;Y x 1 x 1 y from (8.99) 2 ; and we reorganize:(8.103)X 2x 2 y −2 X 1x 1 y = −Y yy+r X 1 +r X 2 +r X 2x 2+r Y x 1+r Y y+r Y x 1 x 1+r X 2x 1 x 2.For the three unknowns X 2x 1 x, X 1 1 x 1 y , X 2x 2 y, we then solve the four equations(8.101), (8.102), (8.103), (8.83) 4 (in which we replace: X 1xfrom (8.90);1X 2xfrom (8.98); Y 1 x 1 y from (8.97) 2 ; X 1x 1 xfrom (8.97) 1 1 ):(8.104)X 2x 1 x = r X 1 + r X 2 + r X 11 y + r X 2x + r Y 2 x 1 + r Y y + r Y x 1 x 1 + r X 2x 1 x + r Y yy, 2X 1x 1 y = r X 1 + r X 2 + r Xy 1 + r X 2x + r Y 2 x 1 + r Y y + r Y x 1 x 1 + r X 2x 1 x + r Y yy, 2X 2x 2 y = r X 1 + r X 2 + r Xy 1 + r X 2x + r Y 2 x 1 + r Y y + r Y x 1 x 1 + r X 2x 1 x + r Y yy. 2We get (8.86) 6 and (8.86) 10 . Replacing then X 2x 1 x, X 1 1 x 1 yin (8.100) gives(8.105)Xy 2 = r X 1 +r X 2 +r Xy 1 +r X 2x 2+r Y x 1+r Y y+r Y x 1 x 1+r X 2x 1 x 2+r Y yy.This is (8.86) 5 .285
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Joël M E R K E RÉcole Normale Sup
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§1. INTRODUCTIONSeveral physically
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e a collection of m analytic second
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7where the indices j, l 1 vary in {
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This phenomenon could be explained
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This yields the prolongation of the
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X(x, y) and Y (x, y) such that it m
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2.23. Compatibility conditions for
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This lemma is left to the reader; a
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simplification nor any reordering:(
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aleza, y por otra, las organizacion
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computational level (differential-g
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For instance, in the case m = 2, by
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in (3.11). The second equation that
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Similarly, the second equation take
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order derivatives of X and of the Y
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Lemma 3.32. The system yxx j = F j
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Lemma 3.45. The following quadratic
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often be denoted by the sign “·
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1Now, taking account of the factor
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conditions, totally equivalent to t
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43remaining terms afterwards:(3.71)
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Here, the sign ≡ precisely means:
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Next, replacing plainly (3.64) in (
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49the order of §3.73. We get:(3.89
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51Multiplying by −2 and reorganiz
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⎧Θ l 1y l 2 = −Ll 1l 1 ,l 1 ,y
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+ 1 ∑4 δj l 1Hl k 2L k k,kk− 2
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+ 1 ∑2 δj l 1Hl k 3M l2 ,kk+ 1 4
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In the hardest techical part of thi
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− 1 3+ 1 3− 1 4+ 1 4∑Hl k 1H
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− 1 3∑kL l 1l1 ,k,x Hk l 1+ 1 3
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Here, the sign ≡ means “modulo
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Thirdly, put j := l 2 in (3.108) wi
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69correct. We get:(4.29)0 =?== −
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Next, apply the operator ∑ k Ll 2
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73Writing term by term the substrac
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of the terms of the subgoal (4.29):
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77+ 1 2∑kH k l 1 ,y l 2 Hl 2k15
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79= − X xx Yx j + Y jm∑+++++l 1
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Our first task is to compute the de
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Replacing this expression of A k in
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85have the continuation(5.17) −y
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87and where thirdly (we are nearly
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89obtain:(5.23)III :=m∑m∑l 1 =1
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[GTW1989] GRISSOM, C.; THOMPSON, G.
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93Nonalgebraizable real analytic tu
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and of CR dimension m = n − d ≥
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coordinates z ′ = 2i ln(z/z p ),
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{x ∈ K n : |x| < ρ} for some ρ
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(2) A K-algebraic inversion mapping
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(1) The complex Lie algebra Hol(M,
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Proposition 3.1. Let t ↦→ G i (
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The first relation gives nothing, s
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with |β∗ k | ≥ 1 and integers
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Clearly, the left hand side is an a
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field X 1 ′ := h ∗ (X 1 ). Taki
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which yields after differentiating
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117Im w∆ n(ρ 4 )∆ n(ρ 1 )∆
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[∂ i,ei H ei (t ′ )] t ′ =He
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p = (w p , z p , ζ p , ξ p ) ∈
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Finite nondegeneracy is interesting
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such that |ω j i ∗ ,β∗ i(t,
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for k even, we have Γ k (z (k) ) =
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that for every local holomorphic se
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h(t) = ˜H(t, J l 0µ 0¯h(0)) with
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infinite families of pairwise non b
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functions w(z)). The coefficients R
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Then the Lie criterion states that
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y k ′ = λk k y k and w ′ = µ
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The last statements of Corollaries
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[Sha2000] SHAFIKOV, R.: Analytic co
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Here x = (x 1 , . . ., x n ) ∈ K
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Consequently, for the case κ = 2,
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are devoted to provide a general on
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for l = 1, . . .,m such that the lo
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Lemma 8.1. The following conditions
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155namely X ←→ X + X ←→ X .
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In terms of Sussmann’s approach [
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x κ−1 χ κ−1 + O(|x| κ ) + O
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Observe that these expressions are
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where the first term I involves onl
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I 5 = ∑ i 1I 6 = ∑ i 1 ,i 2I 7
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U 2 U κ−1 , we obtain the five f
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and U 2 U κ−1 , we obtain the fi
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following equations for j = 1, . .
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linear equations:(7.28) ⎧0 = Rx j
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175R j containing at least one part
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177[3] :∑σ∈S κ−1κδ l,....
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the values of the partial derivativ
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also appears some derivatives Q l
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[5] F. ENGEL; LIE, S.: Theorie der
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185Nonrigid sphericalreal analytic
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origin:{ ( ∣ ∣ )AJ 4 1∣∣∣
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I 2 characterizes equivalence to w
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y means of a fundamental, elementar
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2) When M is Levi nondegenerate at
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195holding in C { z ′ , z ′ , w
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Fortunately for our present purpose
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We notice passim that S ≡ Q x (no
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for a certain local K-analytic new
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Conversely, if y xx (x) = F ( x, y(
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205But we may also express the dual
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Then thanks to a straightforward ap
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and we then expand carefully the re
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explicit computation, so let us rew
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213+T ah3Q 2 a Q b∆(aa|b)∆(b|bb
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215Vanishing Hachtroudi curvaturean
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to fix the ideas, it will be assume
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then by replacing the(so obtained)v
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and also in addition the fact that
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and this defines without ambiguity
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the leaves of this local foliation
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Then by differentiating with respec
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∂Here, the coefficients of the2 T
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231for (7.28):( )∂□ ·∂a □
Page 233 and 234: 233Lie symmetriesof partial differe
Page 235 and 236: Differentiating the first equation
Page 237 and 238: Example 1.21. (Continued) With n =
Page 239 and 240: defined by the graphed equations:((
Page 241 and 242: for some two local analytic maps Π
Page 243 and 244: Proof. Let l = l ( x i , y j , yβ(
Page 245 and 246: complementary views on the same obj
Page 247 and 248: Classification problem 3.11. Classi
Page 249 and 250: Definition 4.10. L is an infinitesi
Page 251 and 252: Convention 5.2. The letters R will
Page 253 and 254: Assuming that dimSYM(E 1 ) = 8, tak
Page 255 and 256: Restarting from §4.1, let ϕ a Lie
Page 257 and 258: Here, R j l 1 ,kare universal polyn
Page 259 and 260: Lemma 6.34. Let M be a submanifold
Page 261 and 262: where the expressions D β,β1 are
Page 263 and 264: Then the sum L + L is tangent to M
Page 265 and 266: Lemma 7.11. ([CM1974, BER1999, Me20
Page 267 and 268: where(7.33) ∆ := ∑1kn∂ 2∂x
Page 269 and 270: with r, s being unspecified functio
Page 271 and 272: and moreover, all other, higher ord
Page 273 and 274: To conclude, we replace X xx so obt
Page 275 and 276: Definition 8.43. A real analytic hy
Page 277 and 278: Lemma 8.59. The function b depends
Page 279 and 280: where the letter r denotes an unspe
Page 281 and 282: (8.80) ⎧Y 1,1,1 = Y x 1 x 1 x 1 +
Page 283: (X 1 , X 2 , Y , Xy 1,X 2x, Y 2 x 1
Page 287 and 288: For the two unknowns Xyy 1 and X 2x
Page 289 and 290: Redeveloping the determinant, the v
Page 291 and 292: 291are obtained by equating to zero
Page 293 and 294: Let (z 0 , c 0 ) = (x 0 , y 0 , a 0
Page 295 and 296: we deduce that Γ 5 is submersive (
Page 297 and 298: 11.4. Regularity and jet parametriz
Page 299 and 300: Cramer’s rule, we get(11.13) ⎧
Page 301 and 302: in K[a, z] n+m and in K[x, c] p+m .
Page 303 and 304: of the cancellation properties (11.
Page 305 and 306: 305II: Explicit prolongations of in
Page 307 and 308: Define then the transformed jet ϕ
Page 309 and 310: Let λ ∈ N be an arbitrary intege
Page 311 and 312: 311[Ol1986], [BK1989]):(Y(1.31)⎧
Page 313 and 314: + [Y y 2 − 2 X xy ] y 1 y 2 + [
Page 315 and 316: for some nonnegative integers A, B,
Page 317 and 318: Once the correct theorem is formula
Page 319 and 320: By the classical formula for the de
Page 321 and 322: The induction formulas are⎧(3.4)
Page 323 and 324: We have to compute:( ) ⎛ ⎞∑
Page 325 and 326: Next, we gather the underlined term
Page 327 and 328: + ∑k 1 ,k 2 ,k 3 ,k 4[−δ k 1,k
Page 329 and 330: Correspondingly, we identify the se
Page 331 and 332: must be equal to the number of indi
Page 333 and 334: Y i1 ,i 2 ,i 3 ,i 4written in one o
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335conclusion, we have shown that (
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337Theorem 3.73. For every κ 1 an
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Theorem 3.79. For every integer κ
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+++m∑l 1 ,l 2 =1m∑l 1 ,l 2 ,l 3
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343form:(4.12)κ+1Yκ j = Y jx κ +
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eing related to the number 2 in the
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Appying the chain rule, we may comp
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Secondly:(5.3)Y j i 1 ,i 2= Y jx i
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As explained before the statement o
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g j i 1 ,...,i λand similarly for
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with respect to the variables x i 1
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Assuming F = F(x, y x ) to be indep
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359(III’) ⎧0 = ∑ (δ k 2)j 3H
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Open problem 1.17. For n = 2 establ
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⎧∣ ⎨X 1 xX 1+ y x 1 y x 1 ·1
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eplacing the third column by the se
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⎧ ∣ ∣⎫ ⎨ ∣∣∣∣∣
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e written under the specific form:(
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Indeed, the collection of equations
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3.12. Principal unknowns. As there
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375Sixthly:(3.22) ⎧δ k 1j 1Θ n+
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are a consequence of (I’), (I”)
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379IV: BibliographyREFERENCES[Ar198
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[G1989] GARDNER, R.B.: The method o
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[Me2005a] MERKER, J.: On the local
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