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1 September 12, 2014 1 September 12, 2014 Definition 1.1. f : U ⊂ Rn ! Rm is smooth if it has continuous partial derivatives of all orders. Need U to be open in order to define derivative. In order to define for arbitrary subsets, need local extension: Definition 1.2. f : X ⊂ Rn ! Rm is smooth if at each point x 2 X, there is an open set x 2 U ⊂ Rn and a smooth map F : U ! Rm such that F jU\X = f, i.e. f has a local extension in a neighborhood of each point. Recall open subsets of X in subspace topology are precisely sets of the form U \ X, so smoothness is a \local" property { f : X ! Rn is smooth if it smooth in a neighborhood of each point of x (in the subspace topology). Definition 1.3. f : X ⊂ Rn ! Y ⊂ Rm is a diffeomorphism if f and f −1 are smooth (and f is a bijection). examples: circles = knots 6= triangle, etc. Definition 1.4. A subset X ⊂ RN is a(smooth) k-dimensional manifold if it islocally diffeomorphic to Rk, that is every x 2 X has a neighborhood (open set) x 2 V ⊂ X that is diffeomorphic to some open set U ⊂ Rk. A diffeomorphism φ : U ! V is a parametrization of (the neighborhood) V . The inverse diffeomorphism φ−1 : V ! U is called a coordinate system on V . −1 −1 When φ is written in coordinates φ = (x1; : : : ; xk), the smooth func- tions x1; : : : ; xk are called coordinate functions. We say that dim X = k is the dimension of X. We can think of U and V as identified by phi { given coordinates (x1; : : : ; xk) on U, this gives us a point on V by using the parametrization φ, and given a point v 2 V , we can obtain the coordinates \of v" by taking (x1(v); : : : ; xk(v)). Example: we can show that S1 = f(x; y) 2 R2 : x2 + y2 = 1g is a 1-dimensional manifold. 1 We can parametrizep the set of (x; y) 2 S on the upper semicircle, i.e. 2 y > 0. φ1(x) = (x; 1 − x ) takes the open interval (−1; 1) to the upper −1 semicircle with a smooth map. The inverse φ1 (x; y) = x is smooth. For the 1 p 2 1 lower semicircle, we take φ2(x) = (x; − 1 − x ). This parametrizes all of S p 2 except for two points (±1; 0). To parametrize, we can take φ3 = ( 1 − y ; y) p 2 1 and φ4 = (− 1 − y ; y). These four parametrizations cover S and, for each point (at least) one of these gives a local parametrization. Proposition 1.5. Given X ⊂ RN and Y ⊂ RM manifolds, X × Y is a submanifold of RN+M . Also, dim X × Y = dim X + dim Y . Proof. Let k = dim X and l = dim Y . Suppose (x; y) 2 X × Y . Then, there exists and open set W ⊂ Rk and a local parametrization φ : W ! φ(W ) ⊂ X, as well as an open subset U ⊂ Rl along with a local parametrization : U ! Y . Define φ × : W × U ⊂ Rk+l ! X × Y by φ × (w; u) = (φ(w); (u)). One can check that φ × is a local parametrization of X × Y in a neighborhood of (x; y). Definition 1.6. If X and Z are both manifolds in RN and Z ⊂ X, then Z is a submanifold of X. 2 September 15, 2014 Suppose f is a smooth map from (an open set of) Rn to Rm and x is in the domain. Then, for any vector h 2 Rn, the derivative of f in the direction of h at x is f(x + th) − f(x) dfx(h) = lim : t!0 t n m We can define dfx : R R by h 7! dfx(h). This map is linear and has matrix form 2 @f1 ::: @f1 3 @x1 @xn 6 . 7 4 . 5 @fm ::: @fm @x1 @xn Proposition 2.1 (Chain Rule). d(g ◦ f)x = dgf(x)dfx: A linear map can be thought of as its best linear approximation. We can use derivatives to identify the best linear approximation a manifold X at x. Let φ : U ! X be a local parametrization around x, and assume φ(0) = x for convenience. Then, the best linear approximation is u 7! φ(0) + dφ0(u) = x + dφ0(u): 2 Definition 2.2. The tangent space of X at x, Tx(X) is the image of the map k N dφ0 : R ! R . N Tx(X) is a vector subspace of R , and the translation x + Tx(X) is the \best linear approximation" of X at x. For this to be a real definition, need to make sure it is well-defined. Let : V ! X be another parametrization. Then, by shrinking U and V , can assume that φ(U) = (V ). The map h : −1 ◦φ : U ! V is a diffeomorphism (composition). Then, φ = ◦ h, so by chain rule dφ0 = d 0 ◦ dh0. Hence, the image of dφ0 is contained in the image of d 0. The reverse also holds similarly, so they must be identical. Proposition 2.3. dim Tx(X) = dim X. Proof. Let φ : U ! V be smooth, and let Φ0 : W ! U be a smooth map −1 0 0 that extends φ . Then Φ ◦ φ = id, so dΦx ◦ dφ = id. This implies that dφ must be an isomorphism, i.e. dim = k. 3 September 17, 2014 1 For an examplep of a tangent space, we can take the parametrization of S , φ : x 7! (x; 1 − x2). Then 1 dφ = p x : 1−x2 1 pFor each point x 2 R, this gives a vector that is tangent to S at (x; 1 − x2). We can do a similar example for S2, with a parametrization φ :(x; y) 7! (x; y; p1 − x2 − y2). Then, 2 1 0 3 dφ = 6 0 1 7 : 4p x p y 5 1−x2−y2 1−x2−y2 The column space of the matrix gives a 2-dimensional subspace of R3 which is tangent to S2 at φ(x; y). Now want to take derivative of f : X ! Y . 3 We take parametrizations φ : U ⊂ Rk ! RN and : V ⊂ Rl ! RM . WLOG take φ(0) = x; (0) = y. We have commutative diagram: X ! (f) Y U ! (h = −1 ◦ f ◦ φ) V −1 So we can define dfx = d 0 ◦ dh0 ◦ dφ0 . We claim this is independent of parametrization. Indeed, if Ψ and Φ are two different parametrizations of X and Y , and H = Ψ−1 ◦ f ◦ Φ, then h = −1 ◦ Ψ ◦ Ψ−1 ◦ f ◦ Φ ◦ Φ−1 ◦ φ = −1 ◦ Ψ ◦ H ◦ Φ−1 ◦ φ. Then, −1 −1 −1 −1 −1 d 0 ◦ dh0 ◦ dφ0 = d ◦ d ◦ dΨ ◦ dH ◦ dΦ ◦ φ ◦ dφ = dΦ ◦ dH ◦ dΦ : A similar argument shows: Proposition 3.1 (Chain Rule). d(g ◦ f)x = dgf(x) ◦ dfx. Proof. Let φ : U ! X, : V ! Y and ν : W ! Z. Then h = −1 ◦ f ◦ φ : U ! V and j = ν−1 ◦ g ◦ : V ! W , and j ◦ h : U ! W . −1 By definition, d(g ◦ f)x = dν0 ◦ d(j ◦ h)0 ◦ dφ0 . N By the chain rule in R , d(j ◦ h) = (dj)0 ◦ (dh)0, and also (dj)0 ◦ (dh)0 = −1 (dj)0 ◦ d 0 ◦ d ◦ (dh)0. Substituting yields the desired equality. 4 September 19, 2014 We want to use the tangent space to study smooth manifolds. Definition 4.1. A function f : X ! Y is a local diffeomorphism if for every x 2 X, there exists a neighborhood x 2 U that maps diffeomorphically to a neighborhood f(U) of y = f(x). In order to be a local diffeomorphism, note that dfx : Tx(X) ! Ty(Y ) must be an isomorphism. This follows from a chain rule argument if f : U ! V is a diffeomorphism. There is a version of the inverse function theorem for smooth manifolds: Theorem 4.2 (Inverse Function Theorem). Suppose that f : X ! Y is a smooth map whose derivative dfx at the point x is an isomorphism (of vector spaces). Then f is a local diffeomorphism at x. As an example, f : R ! S1 : t 7! (cos t; sin t) is a local diffeomorphism. 4 5 September 22, 2014 A local diffeomorphism says that locally, X and Y \look the same". We can make this more precise by noting that we can find local coordinates around x and y so that f(x1; : : : ; xk) = (x1; : : : ; xk). In particular, if φ : U ! X is a parametrization around x 2 X, and dfx is an isomorphism, then the composition : f ◦ φ : U ! Y is a parametrization around y = f(x) 2 Y (here it may be necessary to shrink U to a smaller set so that f is a local diffeomorphism). Definition 5.1. Two maps f : X ! Y and f 0 : X0 ! Y 0 are equivalent if there exist diffeomorphisms α : X0 ! X and β : Y 0 ! Y so that f ◦α = β ◦f 0 (i.e. diagram commutes). f is locally equivalent to f 0 at x if there is a local diffeomorphism at x and a local diffeomorphism at y = f(x) that makes the diagram commute. The inverse function theorem can be interpreted to say that if dfx is an isomorphism, then f is locally isomorphic to the identity. If dim X < dim Y , then IFT cannot apply, since the matrix dfx is not square. The closest thing we can hope for is dfx is injective. Definition 5.2. f : X ! Y is an immersion at x if dfx : Tx(X) ! Ty(Y ) is injective.
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