Chap. 1. Some Differential Geometric Tools

Transcription

1 Chap. 1. Some Differential Geometric Tools

2 1. Manifold, Diffeomorphism 1.1. The Implicit Function Theorem ϕ : U R n R n p (0 p < n), of class C k (k 1) x 0 U such that ϕ(x 0 ) = 0 rank Dϕ(x) = n p x U Then V = V 1 V 2 U neighborhood of x 0 R n, V 1 R p, V 2 R n p and ψ : V 1 V 2 of class C k such that {x V 1 V 2 ϕ(x) = 0} = {(x 1, x 2 ) V 1 V 2 x 2 = ψ(x 1 )}. Definition: The set {ϕ(x) = 0} is called a differentiable manifold.

3 Examples: (analytic) affine manifold: {x R n Ax b = 0} of dim. p if rank A = n p and b ImA. sphere S 2 of R 3 : {(x, y, z) R 3 (x x C ) 2 + (y y C ) 2 + (z z C ) 2 R 2 = 0} 2-dimensional analytic manifold. R C

4 1.2. Diffeomorphism, homeomorphism ϕ : U R n V R n, of class C k, k 1 is a local diffeomorphism in a neighborhood U(x 0 ) of x 0 U if ϕ is invertible from U(x 0 ) to a nghbrhood V (ϕ(x 0 )) of ϕ(x 0 ) V ϕ 1 is also of class C k. ϕ is a local homeomorphism of class C k if it is of class C k, locally invertible, and if its inverse is continuous. Example: given m 1, ϕ(x) = x m is a diffeomorphism from R + {0} to R + {0} and a homeomorphism from R + to R +.

5 1.3. The Local Inversion Theorem The mapping ϕ is a local diffeomorphism in a neighborhood of x 0 if and only if its tangent linear mapping Dϕ(x 0 ) is one-to-one. Curvilinear coordinates: Consider the change of coordinates y 1 = ϕ 1 (x),..., y n p = ϕ n p (x), y n p+1 = ψ 1 (x),..., y n = ψ p (x), where ψ is chosen s.t. (ϕ, ψ) is a local diffeomorphism. Then the manifold ϕ(x) = 0 is straightened out: y 1 = 0,..., y n p = 0. Example: the manifold defined by ϕ(x 1, x 2, x 3 ) = x x2 2 + x2 3 R 2 = 0 is locally straightened out, in a neighborhood of (1, 0, 0) by the coordinate change y 1 = ϕ(x 1, x 2, x 3 ), y 2 = x 2, y 3 = x 3.

6 2. Tangent bundle, vector fields The tangent space to X at the point x X is the vector space T x X = ker DΦ(x). The tangent bundle TX is the set TX = x X T x X. Remark: a manifold and its tangent space at every point have the same dimension. The tangent space to the sphere x x2 2 + x2 3 R2 = 0 is the 2-dim. vector space spanned by v 1 = (x 2, x 1, 0) and v 2 = (x 3, 0, x 1 ).

7 Vector field, integral curve: A C k (analytic) vector field f on X is a C k (analytic) mapping such that to every x X corresponds the vector f(x) T x X. An integral curve of the vector field f is a local solution of the differential equation ẋ = f(x). Examples: Constant vector field parallel to x 1 : f(x 1,..., x n ) = (1, 0,..., 0). Its integral curves are given by x 1 (t) = x 1 (t 0 ) + t t 0, x 2 (t) = x 2 (t 0 ),..., x n (t) = x n (t 0 ). Linear vector field in n dimensions: f(x) = Ax. curves are x(t) = e A(t t0) x(t 0 ). Its integral f(x 1, x 2 ) = (1, x 1 ). Its integral curves are x 1 (t) = x 1 (t 0 ) + t t 0 and x 2 (t) = 1 2 (t t 0) 2 + x 1 (t 0 )(t t 0 ) + x 2 (t 0 ).

8 Lie derivative: Let h be a C 1 function from R n to R. We call Lie derivative of h with respect to f, denoted by L f h, the derivative of h along the integral curve of f at t = 0. In other words: L f h(x) = d dt h(x t(x)) t=0 = n i=1 f i (x) h x i (x). An arbitrary vector field f may be identified with the first order linear differential operator n L f = f i (x). x i i=1 Example: f(t, x) = (1, v) with v constant (i.e. f = t + v x ), and h(t, x) = x vt. L f h(t, x) = v + v = 0.

9 Image of a vector field, straightening out: The image of the vector field f by the diffeomorphism ϕ, denoted by ϕ f is the vector field ϕ f = (L f ϕ 1 (ϕ 1 (y)),..., L f ϕ n (ϕ 1 (y))) T. Moreover, we say that f is straightened out by diffeomorphism if L f ϕ 1 (ϕ 1 (y)) = 1 et L f ϕ i (ϕ 1 (y)) = 0 i = 2,..., n. x ϕ y x(t) ϕ 1 y(t) t t

10 Examples: f(x) = cos 2 x, x R (f = cos 2 x x ). Integral curves: tan x(t) = t t 0 + tan x(t 0 ). Straightening out diffeomorphism: y = tan x. f(x 1, x 2 ) = x 1 x 2 cos 2 x 2 + cos 2 x x 2 is straightened out by 1 x 2 the diffeomorphism y 1 = tan x 2, y 2 = log x 1 2 1x2 2. f(x) = 1 m x 2 U with x = (x x 1 x 1 x 1, x 2 ) R n R n, 2 U function of x 1 only. The Hamiltonian H = x2 2 2m + U(x 1) is a first integral. If h(x 1, x 2 ) satisfies h H h H = 1, then f is straightened x 1 x 2 x 2 x 1 out by the diffeomorphism (x 1, x 2 ) (H, h).

11 First integral: A first integral of f is a function γ satisfying L f γ = 0. In other words, γ is constant along the integral curves of f. Given a vector field f of class C k (k arbitrary integer), let x 0 X be a transient point (f(x 0 ) 0). There exists a local coordinate system (ξ 1,..., ξ n ) of class C k at x 0 for which f is straightened out: L f ξ 1 = 0,..., L f ξ n 1 = 0 and L f ξ n = 1. Thus f admits a set of n 1 independent first integrals. Example: Back to f(x 1, x 2 ) = x 1 x 2 cos 2 x 2 + cos 2 x x 2. The 1 x 2 function y 2 = log x x2 2 is a first integral.

12 Lie bracket: L f L g h L g L f h = n i=1 n j=1 ( ) g f i f j g i x j h. j x j x i The Lie bracket of the vector fields f and g is the vector field defined by: L [f,g] = L f L g L g L f. In local coordinates: [f, g] = n i=1 Example: f = x 1, g = n j=1 ( ) g f i f j g i x j. j x j x i x 2 + x 1 x 3. [f, g] = x 3.

13 Properties: skew symmetry : [f, g] = [g, f]; [αf, βg] = αβ[f, g] + (αl f β)g (βl g α)f, for every pair (α, β) of C functions; Jacobi identity: [f 1, [f 2, f 3 ]] + [f 2, [f 3, f 1 ]] + [f 3, [f 1, f 2 ]] = 0. Also: Given a diffeomorphism ϕ from X to Y. Let f 1 and f 2 be arbitrary vector fields of X and ϕ f 1 and ϕ f 2 their images in Y. We have [ϕ f 1, ϕ f 2 ] = ϕ [f 1, f 2 ].

14 Interpretation: exp-εf(expεg(expεf(x))) exp-εg(exp-εf(expεg(expεf(x)))) g f expεg(expεf(x)) ε 2 [f,g](x) x expεf(x) exp( ɛg) exp( ɛf) exp ɛg exp ɛf(x) = x + ɛ 2 [f, g](x) + O(ɛ 3 ).

15 3. Systems of first-order PDE s L g1 y = 0. L gk y = 0 where g 1,..., g k are regular vector fields such that rank (g 1 (x),..., g k (x)) = k x V. Denote by D = span {g 1,..., g k }. D is called a distribution. We say that D is involutive if for every pair (f 1, f 2 ) D D, we have [f 1, f 2 ] D. Frobenius theorem: There exists a local diffeomorphism ϕ that straightens out g 1,..., g k simultaneously, namely such that, if we note ξ = ϕ(x), we have, in ϕ(v ), ϕ D(ξ) = span {ϕ g 1 (ξ),..., ϕ g k (ξ)} = span { ξ,..., 1 ξk } if and only if D is involutive.

16 Setting z = y ϕ 1, we get L g y = 0 for all g D or z = 0,..., z = 0 ξ 1 ξ k thus z(ξ 1,..., ξ k, ξ) = constant for all ξ = (ξ k+1,..., ξ n )such that ξ ϕ(v ). Thus the solution y is deduced from z by y = z ϕ. Remark: The distribution spanned by f = and g = + x x 1 x 1 2 x 3 is not involutive since [f, g] =. Thus, the only solution to x 3 L g1 y = L g2 y = 0 is y = constant. Consider now the distribution spanned by f and g 1 = clearly involutive. Then y is an arbitrary function of x 3. x 2, which is

17 4. Cotangent bundle, differential forms By Frobenius theorem, T x X = span { x,..., 1 x1 } for a suitable coordinate system. Denote by TxX its dual vector space with dual basis {dx 1,..., dx n }, defined by < x, dx i j >= δ i,j, i, j = 1,..., n. T X = TxX is called the cotangent bundle. x X If ϕ is differentiable from X to R, its differential is defined by n ϕ dϕ = dx x i. i=1 i More generally, a differential 1-form is a differentiable mapping n x ω(x) TxX, i.e. ω(x) = ω i (x)dx i. It is of class C k if all i=1 ω i s are C k. The set of 1-forms is denoted by Λ 1 (X).

18 ω Λ 1 (X) is said exact if there exists a differentiable function ϕ such that ω = dϕ. Exactness necessary condition: ω i = ω j i, j. x j x i We define the (interior) product of a 1-form ω and a vector field f by n < ω, f >= ω i f i. i=1 If ω = dϕ, < ω, f >=< dϕ, f >= L f ϕ. Example: h(t, x) = x vt. Thus dh = vdt + dx and, with f = t + v x, we have L fh =< f, dh >= v + v = 0.

19 Example: Find y such that L g1 y = 0,..., L gk y = 0 is equivalent to find ω exact such that < g 1, ω >= 0,..., < g k, ω >= 0 or: find ω vanishing on G = span {g 1,..., g k } T X and exact. 2-forms: Λ 1 (X) Λ 1 (X) = Λ 2 (X): skew symmetric tensor product of 2 copies of Λ 1 (X). Its basis is the collection of all pairs dx i dx j for i j, with dx i dx j = dx j dx i i, j. A 2-form is thus of the form ω = ω i,j dx i dx j. 1 i<j n Exterior derivative of 1-forms: For an arbitrary C k 1-form ω, its exterior derivative dω is the 2-form given by dω = ω i dx x i dx j = ( ωi ω ) j dx i,j j x 1 i<j n j x i dx j. i

20 Thus a 1-form ω is exact if dω = 0. Poincaré s lemma: On a retractable open subset of X, ω is exact if and only if dω = 0. Image of a form: Given ω T Y and ϕ local diffeomorphism from X to Y, the backward (or induced) image of ω by ϕ is the 1- form in T X, noted ϕ ω, and defined by < ϕ ω, v >=< ω, ϕ v > for every v T X. Clearly, ϕ ω = (ω i ϕ) ϕ i dx x j. i,j j Extension to 2-forms: ϕ (ω θ) = ϕ ω ϕ θ. In particular: d(ϕ ω) = ϕ (dω).

21 Lie derivative of a form: The Lie derivative L f ω of ω along f T X is defined by < L f ω, g >= L f < ω, g > < ω, [f, g] > for every vector field g T X. n ( Equivalently: L f ω = L f ω i + < ω, f ) > dx x i. i i=1 Example: dl f ω = 0. again, h = x vt, ω = dh, f = t + v x. We have

22 5. Pfaffian systems, integrability Given r independent 1-forms ω 1,..., ω r in Λ 1 (X), an integral manifold Y of the Pfaffian system ω 1 = 0,..., ω r = 0 is a submanifold of X such that there exists a differentiable mapping ϕ from Y to X such that ϕ ω 1,..., ϕ ω r identically vanish on Y. The Pfaffian system {ω 1,..., ω r } is algebraically equivalent to the system {ω 1,..., ω r } of degree 1 if any ω i can be expressed as a linear combination of the ω j s: ω i = r j=1 ω j θ j with suitably chosen forms θ j s. The algebraic closure of the Pfaffian system is the system ω 1 = 0, dω 1 = 0,..., ω r = 0, dω r = 0. The Pfaffian system is closed if it is equivalent to its closure.

23 An integral manifold of a Pfaffian system is also an integral manifold of its closure. A system of r independent forms is completely integrable if there exist r differentiable independent functions y 1,..., y r on an open U X such that the system is algebraically equivalent on U to dy 1 = 0,..., dy r = 0. In other words, the system admits r locally independent first integrals. Frobenius Theorem (dual): A necessary and sufficient condition for a Pfaffian system of rank r to be completely integrable on U X is that it is closed in U. In this case, there exists an open dense U 1 U such that through every point of U 1 passes an integral manifold of dimension n r. Moreover the distribution corresponding to the tangent space of any integral manifold is involutive.

24 Interpretation: If the Pfaffian system ω 1 = 0,..., ω r = 0 is closed, let Y be an integral manifold. There exists ϕ : Y X smooth such that, identically on Y, ϕ ω 1 = 0,..., ϕ ω r = 0. If v T Y, < ϕ ω i, v >=< ω i, ϕ v >= 0, i = 1,..., r. Thus Ω = span {ω 1,..., ω r } identically vanishes on ϕ T Y whose dimension is n r. The involutivity of T Y implies that ϕ T Y = span { y,..., 1 y } n r which yields that Ω = span {dy n r+1,..., dy n }. In other words, the integral manifold Y is given by the r independent first integrals dy n r+1 = 0,..., dy n = 0, or y n r+1 = C 1,..., y n = C r. Thus, if T Y = span {g 1,..., g n r }, we have solved the system of PDE s: L gi y j = 0, i = 1,..., n r, j = n r + 1,..., n.