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

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78operator is given by D = ∂∂x + ∑ ml 1 =1 yl 1compute first(5.1) ⎧DDX = D ⎪⎨⎪⎩[X x += X xx + 2m∑l 1 =1m∑l 1 =1y l 1x · X y l 1]y l 1x · X xy l 1 +m∑x ·l 1 =1 l 2 =1∂+ ∑ m∂y l 1 l 1 =1 yl 1 xx ·m∑y l 1x y l 2x · X y l 1y l 2 +∂∂y l 1 x, wem∑l 1 =1y l 1xx · X y l 1 ,and(5.2) ⎧DDY ⎪⎨j = D⎪⎩[Y jx += Y jxx + 2m∑l 1 =1m∑l 1 =1y l 1x · Y jy l 1]y l 1x · Y jxy l 1 + m∑m∑l 1 =1 l 2 =1y l 1x y l 2x · Y jy l 1y l 2 + m∑l 1 =1y l 1xx · Y jy l 1 .Now, we can <strong>de</strong>velope the equation 0 = −DY j · DDX + DX · DDY j ,which yields(5.3)0 = −+[[Y jx ++X x ++m∑l 1 =1m∑l 1 =1 l 2 =1m∑l 1 =1m∑y l 1x · Y jy l 1m∑]·[X xx + 2y l 1x y l 2x · X y l 1y l 2 +y l 1x · X y l 1]·m∑l 1 =1 l 2 =1[m∑l 1 =1m∑l 1 =1Y jxx + 2 m∑l 1 =1y l 1x y l 2x · Y jy l 1y l 2 + m∑l 1 =1y l 1x · X xy l 1 +y l 1xx · X y l 1]+y l 1x · Y jxy l 1 +y l 1xx · Y jy l 1]=
79= − X xx Yx j + Y jm∑+++++l 1 =1m∑y l 1x ·m∑l 1 =1 l 2 =1m∑m∑l 1 =1 l 2 =1 l 3 =1m∑l 1 =1m∑y l 1xx ·m∑l 1 =1 l 2 =1xx X x +[−2 X xy l 1 Yx j + 2 Y j X xy l 1 x−]−X xx Y j + Y jy l 1 xx X y l 1 +[y l 1x y l 2x · −X y l 1y l 2 Yx j + Y jy l 1y X l 2 x−m∑]−2 X xy l 2 Y j + 2 Y j X y l 1 xy l 2 y l 1 +[y l 1x y l 2x y l 3x · −X y l 2y l 3 Y j + Y jy l 1 y l 2y X l 3 y l 1[−X y l 1 Y jx + Y jy l 1 X x]+[]y l 1xx y l 2x · −X y l 1 Y j + Y j X y l 2 y l 1 y l 2 .]+The goal is to show that after solving these m equations for j = 1, . . .,mwith respect to the yxx l , l = 1, . . .,m, one obtains the expression (3.33) ofLemma 3.32, or equivalently, using the ∆ notation instead of the squarenotation, one obtains(5.4)⎧0 = yxx j · ∆ ( x|y 1 | · · · |y m) + ∆ ( x|y 1 | · · · | j xx| · · · |y m) +m∑+ y l 1x · [2∆ ( x|y 1 | · · · | j xy l 1| · · · |y m) −⎪⎨⎪⎩++l 1 =1m∑m∑l 1 =1 l 2 =1m∑m∑l 1 =1 l 2 =1 l 3 =1−δ j l 1∆ ( xx|y 1 | · · · |y m)] +y l 1x y l 2x · [∆ ( x|y 1 | · · · | j y l 1y l 2| · · · |y m) −m∑−2 δ j l 1∆ ( xy l 2|y 1 | · · · |y m)] +y l 1x y l 2x y l 3x[−δjl 1∆ ( y l 2y l 3|y 1 | · · · |y m)] .Unfortunately, the equations (5.3) are not solved with respect to the yxx,jbecause in its last line, we notice that the y l 2 xx are mixed with the y l 1 x . Consequently,we have to solve a linear system of m equations with the unknowns
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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
Page 27 and 28: in (3.11). The second equation that
Page 29 and 30: Similarly, the second equation take
Page 31 and 32: order derivatives of X and of the Y
Page 33 and 34: Lemma 3.32. The system yxx j = F j
Page 35 and 36: Lemma 3.45. The following quadratic
Page 37 and 38: often be denoted by the sign “·
Page 39 and 40: 1Now, taking account of the factor
Page 41 and 42: conditions, totally equivalent to t
Page 43 and 44: 43remaining terms afterwards:(3.71)
Page 45 and 46: Here, the sign ≡ precisely means:
Page 47 and 48: Next, replacing plainly (3.64) in (
Page 49 and 50: 49the order of §3.73. We get:(3.89
Page 51 and 52: 51Multiplying by −2 and reorganiz
Page 53 and 54: ⎧Θ l 1y l 2 = −Ll 1l 1 ,l 1 ,y
Page 55 and 56: + 1 ∑4 δj l 1Hl k 2L k k,kk− 2
Page 57 and 58: + 1 ∑2 δj l 1Hl k 3M l2 ,kk+ 1 4
Page 59 and 60: In the hardest techical part of thi
Page 61 and 62: − 1 3+ 1 3− 1 4+ 1 4∑Hl k 1H
Page 63 and 64: − 1 3∑kL l 1l1 ,k,x Hk l 1+ 1 3
Page 65 and 66: Here, the sign ≡ means “modulo
Page 67 and 68: Thirdly, put j := l 2 in (3.108) wi
Page 69 and 70: 69correct. We get:(4.29)0 =?== −
Page 71 and 72: Next, apply the operator ∑ k Ll 2
Page 73 and 74: 73Writing term by term the substrac
Page 75 and 76: of the terms of the subgoal (4.29):
Page 77: 77+ 1 2∑kH k l 1 ,y l 2 Hl 2k15
Page 81 and 82: Our first task is to compute the de
Page 83 and 84: Replacing this expression of A k in
Page 85 and 86: 85have the continuation(5.17) −y
Page 87 and 88: 87and where thirdly (we are nearly
Page 89 and 90: 89obtain:(5.23)III :=m∑m∑l 1 =1
Page 91 and 92: [GTW1989] GRISSOM, C.; THOMPSON, G.
Page 93 and 94: 93Nonalgebraizable real analytic tu
Page 95 and 96: and of CR dimension m = n − d ≥
Page 97 and 98: coordinates z ′ = 2i ln(z/z p ),
Page 99 and 100: {x ∈ K n : |x| < ρ} for some ρ
Page 101 and 102: (2) A K-algebraic inversion mapping
Page 103 and 104: (1) The complex Lie algebra Hol(M,
Page 105 and 106: Proposition 3.1. Let t ↦→ G i (
Page 107 and 108: The first relation gives nothing, s
Page 109 and 110: with |β∗ k | ≥ 1 and integers
Page 111 and 112: Clearly, the left hand side is an a
Page 113 and 114: field X 1 ′ := h ∗ (X 1 ). Taki
Page 115 and 116: which yields after differentiating
Page 117 and 118: 117Im w∆ n(ρ 4 )∆ n(ρ 1 )∆
Page 119 and 120: [∂ i,ei H ei (t ′ )] t ′ =He
Page 121 and 122: p = (w p , z p , ζ p , ξ p ) ∈
Page 123 and 124: Finite nondegeneracy is interesting
Page 125 and 126: such that |ω j i ∗ ,β∗ i(t,
Page 127 and 128: 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 □
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233Lie symmetriesof partial differe
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Differentiating the first equation
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Example 1.21. (Continued) With n =
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defined by the graphed equations:((
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for some two local analytic maps Π
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Proof. Let l = l ( x i , y j , yβ(
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complementary views on the same obj
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Classification problem 3.11. Classi
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Definition 4.10. L is an infinitesi
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Convention 5.2. The letters R will
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Assuming that dimSYM(E 1 ) = 8, tak
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Restarting from §4.1, let ϕ a Lie
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Here, R j l 1 ,kare universal polyn
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Lemma 6.34. Let M be a submanifold
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where the expressions D β,β1 are
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Then the sum L + L is tangent to M
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Lemma 7.11. ([CM1974, BER1999, Me20
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where(7.33) ∆ := ∑1kn∂ 2∂x
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with r, s being unspecified functio
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and moreover, all other, higher ord
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To conclude, we replace X xx so obt
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Definition 8.43. A real analytic hy
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Lemma 8.59. The function b depends
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where the letter r denotes an unspe
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(8.80) ⎧Y 1,1,1 = Y x 1 x 1 x 1 +
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(X 1 , X 2 , Y , Xy 1,X 2x, Y 2 x 1
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espect to x 1 and (8.96) 2 with res
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For the two unknowns Xyy 1 and X 2x
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Redeveloping the determinant, the v
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291are obtained by equating to zero
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Let (z 0 , c 0 ) = (x 0 , y 0 , a 0
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we deduce that Γ 5 is submersive (
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11.4. Regularity and jet parametriz
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Cramer’s rule, we get(11.13) ⎧
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in K[a, z] n+m and in K[x, c] p+m .
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of the cancellation properties (11.
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305II: Explicit prolongations of in
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Define then the transformed jet ϕ
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Let λ ∈ N be an arbitrary intege
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311[Ol1986], [BK1989]):(Y(1.31)⎧
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+ [Y y 2 − 2 X xy ] y 1 y 2 + [
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for some nonnegative integers A, B,
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Once the correct theorem is formula
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By the classical formula for the de
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The induction formulas are⎧(3.4)
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We have to compute:( ) ⎛ ⎞∑
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Next, we gather the underlined term
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+ ∑k 1 ,k 2 ,k 3 ,k 4[−δ k 1,k
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Correspondingly, we identify the se
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must be equal to the number of indi
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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
Page 383 and 384:
[Me2005a] MERKER, J.: On the local
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