advertisement

Problem 1: Prove that the function f : R → R defined by f (0) = 0 and f (x) = exp(−1/x2 ) for x 6= 0 is infinitely differentiable everywhere. Solution: First prove by induction on k that f (k) (x) = pk ( x1 )f (x) for all x 6= 0, where pk (t) is a polynomial in t. (In fact, p0 (t) = 1 and pk+1 (t) = −t2 p0k (t) + 2t3 pk (t).) It then also follows by induction on k that f (k) is differentiable at x = 0 (the only worrisome case) with f (k+1) (0) = 0; indeed, if this holds for k, then h1 (f (k+1) (h) − f (k+1) (0)) = h1 pk+1 ( h1 )f (h) → 0 as h → 0 because exp grows faster than any polynomial, proving the claim for k + 1 as well. Problem 2: Define M = {(x, y, eiθ ) ∈ R2 × S 1 : x sin 2θ = y cos 2θ } as a topological subspace of R2 × S 1 with its standard topology. Construct a differentiable atlas on M consisting of charts (U, h) and (V, k), as follows: define H : R × (−π, π) → M by H(x, θ) = (x, x tan 2θ , eiθ ) and let U = im(H) and h = H −1 ; similarly, define K : R × (0, 2π) → M by K(y, θ) = (y cot 2θ , y, eiθ ) and let V = im(K) and k = K −1 . Finally, define f : M → R2 by f (x, y, eiθ ) = (x, y). For any given q ∈ R2 , determine whether or not q is a regular value of f , and find f −1 (q). Solution: Fix q = (x, y) ∈ R2 . Then a point p ∈ R2 × S 1 lies in f −1 (q) iff p = (x, y, eiθ ) with x sin 2θ = y cos 2θ . If x 6= 0, then tan 2θ = xy , which determines θ uniquely modulo 2πZ, so that eiθ is uniquely determined. A similar reasoning holds if y 6= 0. Thus, in these two cases, f −1 (q) consists of precisely one point. This leaves the case (x, y) = (0, 0), but then any θ is allowed, so that f −1 (q) = S 1 . Counting dimensions, the regular value theorem tells us that q = (0, 0) is not a regular value of f . For q = (x, y) with x 6= 0, the unique p ∈ M with f (p) = q lies in U , so in order to see that q is a regular value, it suffices to check that det Jach(p) (f ◦ H) 6= 0. But (f ◦ H)(x, θ) = (x, x tan 2θ ), so this determinant equals x2 (sec 2θ )2 6= 0. Likewise for y 6= 0. Problem 3 (Exercise 2.3 in the book): Let M be an n-dimensional manifold with maximal atlas Amax M , let f : M → R be a differentiable function, and let p ∈ M . Prove or disprove that n −1 the map Amax M,p → R given by (U, h) 7→ gradh(p) (f ◦ h ) defines an element of Tp M . Solution: For M = R consider the two charts (U, h), (V, k) with U = V = R, h(x) = x, and k(x) = tx, where t 6= 0 is fixed. Then gradh(x) (f ◦ h−1 ) = f 0 (x), gradk(x) (f ◦ k −1 ) = 1t f 0 (x), and Jach(x) (k ◦ h−1 ) = t, so the required property fails unless t = ±1 or f 0 (x) = 0. Problem 4 (Question (3), p. 45): Let M be a manifold, let N be a submanifold of M , and let p ∈ N . Given a derivation µ of EN,p , define a derivation Φ(µ) of EM,p via Φ(µ)(f ) = µ(f |N ) for all f ∈ EM,p . Prove that Φ is injective and that im(Φ) is the space of all derivations ν of EM,p such that ν(f ) = 0 for all f ∈ EM,p with f |N = 0. Solution: The key is that there exists a linear and multiplicative map EN,p → EM,p , g 7→ g ext , with g ext |N = g for all g ∈ EN,p . (If dim M = m, dim N = n, and if h is a submanifold chart for N , one can define g ext (h−1 (x1 , . . . , xm )) = g(h−1 (x1 , . . . , xn )). There are many possible choices for h, hence many different ways of constructing such an “extension operator” g 7→ g ext .) Given this, we begin by observing that µ(g) = Φ(µ)(g ext ) for all g ∈ EN,p , so that Φ(µ) = 0 ⇒ µ = 0, proving that Φ is injective. Next, ν = Φ(µ) implies that ν(f ) = µ(f |N ) = 0 whenever f |N = 0. Vice versa, if ν is any derivation of EM,p with f |N = 0 ⇒ ν(f ) = 0, then ν(f ) = ν((f |N )ext ) for all f ∈ EM,p (because (f − (f |N )ext )|N = 0), so that ν = Φ(µ) with µ defined by µ(g) = ν(g ext ) for all g ∈ EN,p . Notice that µ is indeed a derivation of EN,p because ν is a derivation of EM,p and because the map g 7→ g ext is linear and multiplicative.