Inductive and Recursive Moving Frames for Lie Pseudo-Groups

Transcription

1 Inductive and Recursive Moving Frames for Lie Pseudo-Groups Francis Valiquette (Joint work with Peter) Symmetries of Differential Equations: Frames, Invariants and Applications Department of Mathematics and Statistics May 19, 01 Francis Valiquette Recursive Moving Frames 05/19/01 1 /

2 Introduction Take the Best of the Two Methods Equivariant method Cartan s method d[λ(ω)] = λ[dω + v ( ) (ω)] dθ = π θ + T (θ θ) Recursive Moving Frames Notation: z = (x, y, u), with u = u(x, y). z (n) = (x, y, u (n) ) = n th order jet. equality modulo contact forms. Francis Valiquette Recursive Moving Frames 05/19/01 /

3 Lie Pseudo-Groups Lie Pseudo-Groups Example A Lie pseudo-group G is the generalization of a (local) Lie group action: Z = ϕ(z). It is characterized by its determining system: F (n ) (z, Z (n ) ) = 0. n = order of G. Let f D(R) and g C (R), then X = f (x), Y = e(x, y) = y f (x) + g(x), U = u + e x f x is a Lie pseudo-group G D(R 3 ) with determining system (n = 1) X y = X u = 0, Y u = 0, Y y = X x, U = u + Y x X x. Francis Valiquette Recursive Moving Frames 05/19/01 3 /

4 Lie Pseudo-Groups Prolonged Action G induces an action on J n z (n) g (n) z (n) : J 0 : X = f (x), Y = e(x, y) = y f (x) + g(x), U = u + e x f x, J 1 : J : J 3 : U X = u x f x U XX = u xx f x U XY = u xy f x + e xx e x u y f x f xxe x fx 3, U Y = u y f x + e xxx e xx u y e x u xy f xx u x f 3 x + e x u yy + 3e x f xx u y 4e xx f xx 3e x f xxx fx 4 f xx U XYY = u xyy fx f xxx f xx u y e x u yy f 3 x f xx u yy + e x u yyy f 4 x fx 4 + f xx, fx + 8 e xfxx, fx 5, U YY = u yy fx,, U YYY = u xyy fx 3, Francis Valiquette Recursive Moving Frames 05/19/01 4 /

5 Moving Frames Moving Frames Once the prolonged action becomes free choose K n J n. Solve the normalization equations g (n) z (n) K n to obtain ρ (n) (z (n) ): ρ (n) (z (n) ) ρ (n) (z (n) ) z (n) K n z (n) Substitute ρ (n) (z (n) ) into g (n) z (n) ρ (n) (z (n) ) z (n) = n th order normalized invariants. Francis Valiquette Recursive Moving Frames 05/19/01 5 /

6 Recursive Moving Frame Stategy Moving frames: Prolong to J Implicit differentiation unwieldy expressions!!! U XX = uxx f x normalize exxx exx uy ex uxy fxx ux + fx 3 Recursive moving frames: Prolong to J k normalize prolong to J k+l normalize + e x uyy + 3ex fxx uy 4exx fxx 3ex fxxx fx 4 Use the Maurer Cartan structure equations and the recurrence relations. + 8 ex f xx fx 5. Francis Valiquette Recursive Moving Frames 05/19/01 6 /

7 Lie Algebra and Maurer Cartan Forms g: g = { v = 3 a = 1 ζ a (z) z a v tangent to the G-orbits }. F (n ) (z, Z (n ) ) = 0 linearization at 1 (n ) L (n ) (z, ζ (n ) ) = 0. v g v is a solution of L (n ) (z, ζ (n ) ) = 0. g : Let µ a B = M C forms of D(R3 ) µ a B G? Define the lift map: λ(ζ a B ) = µa B, λ(za ) = Z a = g z a. The M C forms µ a B G satisfy the linear constraints L (n) (Z, µ (n) ) = λ[l (n) (z, ζ (n) )] = 0, n n. Francis Valiquette Recursive Moving Frames 05/19/01 7 /

8 Maurer Cartan Equations of G ( dµ a C = C = (A,B) ( ) C µ a A,b µ b A B) L ( ) (Z,µ ( ) ) = 0 X = f (x), Y = e(x, y) = y f (x) + g(x), U = u + e x : f x Let v = ξ x + η y + φ u : ξ y = ξ u = 0, η u = 0, ξ x = η y, φ = η x, φ u = 0, µ x Y = µx U = 0, µy U = 0, µx X = µy Y, µu = µ y X, µu U = 0. Basis of g = {µ x X k = µ X k, µ y X k = ν X k : k 0}: dµ = µ X µ, dν = ν X µ + µ X ν, dµ X = µ XX µ, dν X = ν XX µ + µ XX ν, dµ XX = µ XXX µ + µ XX µ X, dν XX = ν XXX µ + µ XXX ν + λ Francis Valiquette Recursive Moving Frames 05/19/01 8 /

9 Recurrence Relations Lift of submanifold jets = prolonged action: λ(x) = X, λ(y) = Y, λ(u x i y j ) = U X i Y j. Lift of horizontal forms: λ(dx) d H X = ω x, λ(dy) d H Y = ω y. Lift of vector field jets: λ(ζb a ) = µa B. Recurrence relations: dx ω x + µ, dy ω y + ν, du X i Y j U X i+1 Y j ωx + U X i Y j+1 ωy + λ(φ i,j ), where v ( ) = v + φ i,j g ( ). u i,j i+j 1 Francis Valiquette Recursive Moving Frames 05/19/01 9 /

10 Example For X = f (x), Y = e(x, y) = y f (x) + g(x), U = u + e x f x : dx = ω x + µ, dy = ω y + ν, du U X ω x + U Y ω y + ν X, du X U XX ω x + U XY ω y + ν XX U X µ X U Y ν X, du Y U XY ω x + U YY ω y + µ XX U Y µ X, du XX U XXX ω x + U XXY ω y + ν XXX U X µ XX U Y ν XX U XX µ X U XY ν X, du XY U XXY ω x + U XYY ω y + µ XXX U Y µ XX U XY µ X U YY ν X, du YY U XYY ω x + U YYY ω y U YY µ X,. Francis Valiquette Recursive Moving Frames 05/19/01 10 /

11 Recursive Algorithm Step -1: Compute ω x = d H X = f x dx, ω y = d H Y = e x dx + f x dy. Step 0: Order 0 normalizations: K 0 = {x = y = u = 0} 0 = X = f, 0 = Y = e, 0 = U = u + e x f x e x = u f x. Then ω x = f x dx, ω y = d H Y = f x (dy u dx). Order 0 recurrence relations: 0 = dx = ω x + µ, 0 = dy = ω y + ν, 0 = du U X ω x + U Y ω y + ν X, µ = ω x, ν = ω y, ν X (U X ω x + U Y ω y ). Francis Valiquette Recursive Moving Frames 05/19/01 11 /

12 µ = ω x = f x dx, ν = ω y = f x (dy u dx), Order 0 M C structure equations: ν X (U X ω x + U Y ω y ). dµ = µ X µ, dν = ν X µ + µ X ν, dω x µ X ω x, dω y ν X ω x + µ X ω y. Hence, µ X df x f x ( U Y u ) y ω x, ν X (U X ω x + U Y ω y ). f x U X =?, U Y =?, U X, U Y pseudo-group prolonged coefficients (r (1) 0 = r 1 r 0 = ). Francis Valiquette Recursive Moving Frames 05/19/01 1 /

13 Step 1: Order 1 recurrence relations: Recursive Moving Frames du X U XX ω x + U XY ω y + ν XX U X µ X U Y ν X, du Y U XY ω x + U YY ω y + µ XX U Y µ X, Cross-section: K 1 = K 0 {u x = u y = 0}: µ X df x f x Order 1 recurrence relations: + u y f x ω x, ν X 0. µ XX (U XY ω x + U YY ω y ), ν XX (U XX ω x + U XY ω y ). Order 1 M C structure equations: dµ X = µ XX µ, dν X = ν XX µ + µ XX ν, with r (1) 1 =. Hence, µ XX U XY ω x u yy f x dµ X µ XX ω x, dν X ν XX ω x µ XX ω y, ω y, ν XX = U XX ω x U XY ω y. Francis Valiquette Recursive Moving Frames 05/19/01 13 /

14 Step : Order recurrence relations: du XX U XXX ω x + U XXY ω y + ν XXX U XX µ X U XY ν X, du XY U XXY ω x + U XYY ω y + µ XXX U XY µ X U YY ν X, du YY U XYY ω x + U YYY ω y U YY µ X. Cross-section (u yy > 0): K = K 1 {u xx = u xy = 0, u yy = 1}: µ XX ω y, ν XX 0. Order recurrence relations: µ XXX (U XXY ω x + U XYY ω y ), ν XXX (U XXX ω x + U XXY ω y ), µ X 1 (U XYY ω x + U YYY ω y ). Next, 1 = U YY = u yy f x f x = u yy. Francis Valiquette Recursive Moving Frames 05/19/01 14 /

15 Coordinate Expressions Invariant horizontal coframe: ω x = u xx dx, ω y = u yy (dy u dx). Invariant total derivative operators: D X = 1 uyy (D x + u D y ), D Y = 1 uyy D y. Normalized invariants: U XYY ω x + U YYY ω y µ X u u yyy + u xyy + u y u yy u 3/ yy ω x + u yyy uyy 3/ ω y. Francis Valiquette Recursive Moving Frames 05/19/01 15 /

16 Step 3, Step 4,... Step k 3: The M C forms of order k are known up to r (1) of order k. k 1 = lifted invariants Normalize the r (1) k 1 = unknown lifted invariants of order k to 0. Order k recurrence relations symbolic expressions for the order k + 1 normalized M C forms. Order k M C structure equations expressions for the order k + 1 normalized M C forms with degree of indeterminancy r (1) k =. Francis Valiquette Recursive Moving Frames 05/19/01 16 /

17 Final Results Cross-section: K = {x = y = 0, u yy = 1, u x k = u yx k = 0 : k 0}. Normalized M C forms: µ = ω x, ν = ω y, µ X U XYY ω x + U YYY µ X k+1 U Y X k 1 ωy, ν X k 0, k 1. Recursive construction of the normalized M C forms: ω y, dµ = µ X µ, dω x = U YYY ωy ω x, dµ X = µ XX µ, dµ X = µ XX ω x, dµ XX = µ XXX µ + µ XX µ X, dµ XX µ XXX ω x + U XYY.. µ XX ω x, Francis Valiquette Recursive Moving Frames 05/19/01 17 /

18 dµ X = µ XX ω x : µ XX = D X dµ X = D X (µ X ) d(d X µ X ) = D X (µ X ) du XYY. dµ XX µ XXX ω x + U XYY µ XX ω x : ( µ XXX = D X dµ XX U ) XYY µ XX ω x = D X (µ XX ) + U XYY µ XX. In general: µ X k+1 = D X (µ X k ) + k 1 U XYY µ X k, k. Conclusion: The normalized M C forms of order are recursively constructed from µ X U XYY ω x + U YYY Francis Valiquette Recursive Moving Frames 05/19/01 18 / ω y.

19 Fundamental Basis Theorem Maurer Cartan Invariants Definition Let µ a B G i=1 I a B;i ωi, with #B = k I a B;i = k th order M C invariants. µ XX = D X (µ X ) du XYY & µ X k+1 = D X (µ X k ) + k 1 U XYY µ X k, k : D X (µ X k ) D X (M C inv.) + M C inv. k 1 U XYY µ X k M C inv. correction terms Consequence 1: The M C invariants of order can be expressed in terms of the M C invariants of order 1 and their invariant derivatives. Francis Valiquette Recursive Moving Frames 05/19/01 19 /

20 Fundamental Basis Theorem Fundamental Basis Theorem Consequence : U XYY and U YYY generate the algebra of differential invariants (Peter & Juha Gröbner basis). Proof: Write U X i+1 Y j = D X (U X i Y j ) + Nx i,j, U X i Y j+1 = D Y (U X i Y j ) + Ny i,j and observe that N k i,j contains normalized invariants of order i + j, M C invariants U XYY, U YYY and their invariant derivatives. Then, J 1 : U X = D X (U) + N x, U Y = D Y (U) + N y, J : U XX = D X (U X ) + N x X, U XY = D X (U Y ) + N x Y, U YY = D Y (U Y ) + N y Y, Francis Valiquette Recursive Moving Frames 05/19/01 0 /

21 Fundamental Basis Theorem J 3 :... Completeness of (X, Y, U ( ) ) X = Y = U = 0 Theorem Let G be a Lie pseudo-group of order n ( 1). Then the M C inv. of order n, and the normalized inv. of order 0, form a generating set of the algebra of differential invariants. Francis Valiquette Recursive Moving Frames 05/19/01 1 /

22 Final Remarks Final Remarks Why is the recursive method interesting? Avoids computing the prolonged action. Avoids computing the phantom invariants U X k+1, U YX k, k 0. Avoids computing ρ (n). Only computes the essential quantities: ω x, ω y, D X, D Y, M C invariants. Can also compute invariant contact forms invariant variational bicomplex. Method applies to finite-dimensional Lie group actions G. For G = G, the Fundamental Basis Thm = Evelyne s result (007). Francis Valiquette Recursive Moving Frames 05/19/01 /