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

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136where each term Φ l1 ,...,l n−1is a certain linear partial differential expressioninvolving the derivatives of Q 1 , . . .,Q n−1 , R up to order two with coefficientsbeing holomorphic functions of (z, w). The determination of a systemof generators X 1 , . . .,X c of Sym(E M ) is obtained by solving the infinitecollection of these linear partial differential equations Φ l1 ,...,l n−1= 0(cf. [Ol1986], [Su2001a,b], [GM2001a,b,c]). We shall apply this generalprocedure to provide different families of nonalgebraizable real analytic hypersurfacesin C n .7.3. Hypersurfaces in C 2 with control of their CR automorphism group.The goal of this paragraph is to construct some classes of strong tubes,namely tubes having the smallest possible CR automorphism group. Westart with the case n = 2 and study afterwards the case n ≥ 3 in the nextsubparagraph. Let M χ be the strong tube hypersurface in C 2 defined by theequation(7.11) M χ : v = ϕ(y) := y 2 + y 6 + y 9 + y 10 χ(y).where χ is a real analytic function defined in a neighborhood of the originin R.Lemma 7.1. The hypersurfaces M χ are pairwise not biholomorphicallyequivalent strong tubes.Proof. To check that M χ is a strong tube, it suffices to show that every hypersurfaceof the form v = y 2 + y 6 + O(y 9 ) is a strong tube (the term y 9will be used afterwards). Writing v = (w − ¯w)/2i and y = (z − ¯z)/2i,considering w as a function of z and ¯w, ¯z as constants, the differentiation ofw with respect to z in (7.11) yields:(7.12) ∂ z w = 2y + 6y 5 + O(y 8 ).The implicit function theorem yields:(7.13) y = (1/2)∂ z w − (3/2 5 )(∂ z w) 5 + O((∂ z w) 8 ).One further differentiation of equation (7.12) with respect to z gives:(7.14) ∂ 2 zz w = −i − (15i) y4 + O(y 7 ).Replacing y in this equation by its value obtained in (7.13), we obtain thefollowing second order ordinary equation E M satisfied by ∂ z w and ∂ 2 zzw:(7.15) ∂ 2 zzw = −i − (15i/2 4 )(∂ z w) 4 + O((∂ z w) 7 ).In the four dimensional jet space J1,1 2 (C) equipped with the coordinates(z, w, W 1 , W 2 ) the equation of the corresponding complex hypersurface∆ M is of course:(7.16) W 2 = −i − (15i/2 4 )(W 1 ) 4 + O((W 1 ) 7 ).
Then the Lie criterion states that a holomorphic vector field X = Q ∂ z +R ∂ w belongs to Sym(E M ) if and only if its second prolongation X (2) =Q ∂ z + R ∂ w + R 1 ∂ W 1 + R 2 ∂ W 2 is tangent to ∆ M , where the coefficientsR 1 and R 2 are given by the formulas (7.8) specified for n = 2, namely:(7.17) ⎧⎪⎨R 1 = ∂ z R + [∂ w R − ∂ z Q] W 1 − ∂ w Q (W 1 ) 2 .R 2 = ∂zz 2 ⎪⎩R + [2∂2 zw R − ∂2 zz Q] W 1 + [∂ww 2 R − 2∂2 zw Q] (W 1 ) 2 − ∂ww 2 Q (W 1 ) 3 ++ [∂ w R − 2∂ z Q] W 2 − 3∂ w Q W 1 W 2 .The tangency condition yields the following equation which is satisfied on∆ M , i.e. after replacing W 2 by its value given by (7.16):(7.18) R 2 + (15i/2 2 )R 1 (W 1 ) 3 + O((W 1 ) 6 ) = 0.By expanding equation (7.18) in powers of W 1 up to order five, we obtainthe following system of six linear partial differential equations which mustbe satisfied by the derivatives of Q and R up to order two:⎧(e 0 ) : ∂zz 2 R − i(∂ wR − 2∂ z Q) ≡ 0.(e 1 ) : 2∂ 2 zw R − ∂2 zz Q ≡ 0.137(7.19)⎪⎨(e 2 ) : ∂ 2 ww R − 2∂2 zw Q ≡ 0.(e 3 ) : −∂ 2 ww Q + 15i2 2 ∂ z R ≡ 0.⎪⎩(e 4 ) : − 15i2 4 (∂ w R − 2∂ z Q) + 15i2 2 (∂ w R − ∂ z Q) ≡ 0.(e 5 ) : − 15i2 4 (−3∂ w Q) − 15i2 2 (∂ w Q) ≡ 0.It follows from the equation (e 5 ) that ∂ w Q ≡ 0 which implies ∂ 2 ww Q ≡ 0.Then by equation (e 3 ) we obtain ∂ z R ≡ 0, implying ∂ 2 zzR ≡ 0. From equation(e 0 ) we get ∂ w R ≡ 2∂ z Q and, from equation (e 4 ), we get ∂ w R ≡ ∂ z Q.Consequently ∂ z R ≡ ∂ w R ≡ ∂ z Q ≡ ∂ w Q ≡ 0. Since the two vector fields∂ z and ∂ w evidently belong to Sym(E M ), it follows that dim C Sym(E M ) =2. Finally, this implies that dim R Aut CR (M) = 2 and that Aut CR (M) isgenerated by ∂ w + ∂ ¯w and ∂ z + ∂¯z .Next, let χ(y) and χ ′ (y ′ ) be two real analytic functions, and assume thatM χ and M ′ χ ′ are biholomorphically equivalent. Let t′ = h(t) be such anequivalence. Reasoning as in §4 and taking into account that both are strongtubes, we see that h ∗ (∂ z ) and h ∗ (∂ w ) must be linear combinations of ∂ z ′ and∂ w ′ with real coefficients. It follows that h must be linear, of the form z ′ =az +bw, w ′ = cz +dw, where a, b, c and d are real. Since T 0 M χ = {v = 0}
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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
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) ) =
Page 129 and 130: that for every local holomorphic se
Page 131 and 132: h(t) = ˜H(t, J l 0µ 0¯h(0)) with
Page 133 and 134: infinite families of pairwise non b
Page 135: functions w(z)). The coefficients R
Page 139 and 140: y k ′ = λk k y k and w ′ = µ
Page 141 and 142: The last statements of Corollaries
Page 143 and 144: [Sha2000] SHAFIKOV, R.: Analytic co
Page 145 and 146: Here x = (x 1 , . . ., x n ) ∈ K
Page 147 and 148: Consequently, for the case κ = 2,
Page 149 and 150: are devoted to provide a general on
Page 151 and 152: for l = 1, . . .,m such that the lo
Page 153 and 154: Lemma 8.1. The following conditions
Page 155 and 156: 155namely X ←→ X + X ←→ X .
Page 157 and 158: In terms of Sussmann’s approach [
Page 159 and 160: x κ−1 χ κ−1 + O(|x| κ ) + O
Page 161 and 162: Observe that these expressions are
Page 163 and 164: where the first term I involves onl
Page 165 and 166: I 5 = ∑ i 1I 6 = ∑ i 1 ,i 2I 7
Page 167 and 168: U 2 U κ−1 , we obtain the five f
Page 169 and 170: and U 2 U κ−1 , we obtain the fi
Page 171 and 172: following equations for j = 1, . .
Page 173 and 174: linear equations:(7.28) ⎧0 = Rx j
Page 175 and 176: 175R j containing at least one part
Page 177 and 178: 177[3] :∑σ∈S κ−1κδ l,....
Page 179 and 180: the values of the partial derivativ
Page 181 and 182: also appears some derivatives Q l
Page 183 and 184: [5] F. ENGEL; LIE, S.: Theorie der
Page 185 and 186: 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
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[Me2005a] MERKER, J.: On the local
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