INVERSE FUNCTION THEOREM and SURFACES IN R n

INVERSE FUNCTION THEOREM and SURFACES IN R n Let f C k (U; R n ), with U R n open. Assume df(a) GL(R n ), where a U. The Inverse Function Theorem says there is an open neighborhood V U of a in R n so that f V : V R n is a homeomorphism onto its image W = f(v ), an open subset of R n ; the inverse g = f 1 : W V is a C k map. The inverse function theorem as an existence theorem. Given f : X Y (X, Y Banach spaces) and y Y, we seek to solve the nonlinear equation: f(x) = y. The idea is that if f is close (in some sense) to a linear operator A L(X, Y ) for which the problem is uniquely solvable, and (for a given b Y ) we happen to know a solution a X to f(x) = b, then for values y close to b the equation should have a unique solution (close to a). For example, suppose f has the form f(x) = Ax + φ(x), where A L(X, Y ) is invertible (with bounded inverse) and φ : X Y is Lipschitz, with a small Lipschitz constant Lip(φ). Then to solve Ax + φ(x) = y assuming the linear problem Av = w can be solved uniquely for any w Y (that is, A is invertible, and v X C w Y, where C = A 1 ) we compute successive approximations, starting from an arbitrary a X and solving the sequence of linear problems with given right-hand side : Ax 1 = y φ(a), Ax 2 = y φ(x 1 ), Ax 3 = y φ(x 2 ), Convergence of (x n ) is established by setting this up as a fixed point problem, for F (x) = A 1 [y φ(x)]: x 1 = F (a), x 2 = F (x 1 ), Convergence to a fixed point is guaranteed provided F is a contraction of X: F (x) F ( x) λ x x, where 0 < λ < 1. Since F (x) F ( x) = A 1 [φ(x) φ( x)], it suffices to require Lip(φ) < 1/ A 1. Then the estimate (for two solutions of f(x) = y, f( x) = ȳ, y, ȳ Y : x x = A 1 [y ȳ] A 1 [φ(x) φ( x)] A 1 ( y ȳ + Lip(φ) x x ), 1

(1 λ) x x A 1 y ȳ, λ = A 1 Lip(φ) < 1 shows that the inverse map g(y) = x is Lipschitz, with constant Lip(g) A 1 /(1 λ). We recall here the standard fixed point theorem for contractions. Contractions have unique fixed points. Let (X, d) be a complete metric space, F : X X a λ-contraction, where 0 < λ < 1: d(f (x), F (y)) λd(x, y), x, y X. Then F has a unique fixed point p X, which is globally attracting (F n (x) p, x X.) Proof. Let x X. For the sequence of iterates x n = F n (x), the contraction property easily implies d(x m+1, x m ) λ m d(x 1, x) := λ m δ, and then if n > m N: d(x n, x m ) d(x n, x n 1 ) +... d(x m+1, x m ) (λ n 1 +... λ m )δ λn 1 λ δ, so (x n ) is a Cauchy sequence, and x n p X. Then p is a fixed point: d(f (p), p) d(f (p), F (x n )) + d(x n+1, p) λd(x n, p) + d(x n+1, p) 0. It follows easily from the contraction property that F can only have one fixed point. Typically the perturbation φ of the invertible linear map A will satisfy a Lipschitz condition only in some open set U X. Then given a point b Y for which the problem has a unique solution a U (that is, f(a) = b) and a nearby point y Y, we seek solutions x B r (a), (closed ball), where r > 0 is small enough that this ball is contained in U. The successive approximations scheme still works, provided we guarantee this ball is invariant under F. So we estimate, for x B r (a) (using a = A 1 [b φ(a)]): F (x) a = A 1 [y φ(x)] a = A 1 [y b] A 1 [φ(x) φ(a)] A 1 ( y b + Lip(φ) x a ) A 1 (s + Lip(φ)r), and this is bounded above by r provided we pick y B s (b) (open ball in Y ), where s (1 λ)r (with λ = A 1 Lip(φ) < 1 as before). We summarize A 1 as follows: 2

Proposition 1. Perturbations of invertible linear maps are homeomorphisms. Let f : U Y (U X open) have the form f(x) = Ax + φ(x), where A L(X, Y ) is boundedly invertible and φ is Lipschitz with Lip(φ) < A 1 1. Then V = f(u) is open in Y, and f : U V is a homeomorphism with Lipschitz inverse. If U = X, f is a homeomorphism onto Y (with Lipschitz inverse). Turning to the case of f : U R n, where U R n is an open set, suppose f is differentiable at a U, with df(a) GL(R n ) (that is, df(a) is an invertible linear map.) Then we have, for a function r : U R n : f(x) = f(a)+df(a)[x a]+r(x) = df(a)[x]+φ(x), φ(x) = f(a) df(a)[a]+r(x), so φ is Lipschitz in U if and only if r is (with the same Lipschitz constant.) Recall the condition r is Lipschitz with s small constant, in a sufficiently small ball with center a corresponds exactly to strong differentiability at a. This motivates the hypothesis in the statement of the Inverse Function Theorem. Inverse Function Theorem. Let f : U R n (U R n open) be strongly differentiable at a U, and assume df(a) GL(R n ) (that is, df(a) is an isomorphism.) Then there exists a neighborhood V U of a so that: (i) The restriction f V is a bi-lipschitz homeomorphism onto its image W = f(v ), and W is open in R n. (ii) The inverse map g = f 1 : W V is strongly differentiable at b = f(a); (iii) If f C 1 (U; R n ), then V can be chosen so that g = f 1 is differentiable in W ; (iv) If f C k (U; R n ) (with k 1) then g C k (W ; R n ). Proof. (i): Since f is strongly differentiable at a, we may find an open ball V = B r (a) so that for x, x in V : f(x) = f(a)+df(a)[x a]+r(x), where r(x) r( x) λ x x with λ df(a) 1 < 1. In other words (since f(a) df(a)[a] is a constant) in this ball V, f is a perturbation of the isomorphism df(a). Conclusion (i) follows from the proposition. Prior to proving (ii) and (iii), we consider a general result on differentiability of the inverse (of a homeomorphism). Note that the inverse of the 3

homeomorphism f(x) = x 3 of the real line is not differentiable at x = 0. This can only happen since f (0) = 0. Differentiability of the inverse. Let f : U V be a homeomorphism, where U, V are open in R n, with inverse g : V U. If f is differentiable at a U and df(a) GL(R n ), then g is differentiable at b = f(a), with dg(b) = [df(a)] 1. If f is strongly differentiable at a, then g is strongly differentiable at b. Proof. Defining s(w) (for w R n with w < dist(b, V )) by: g(b + w) = g(b) + [df(a)] 1 [w] + s(w), we need to show lim w 0 s(w)/ w = 0. Let v = g(b+w) g(b), and compute: df(a)[v]+r(v) = f(a+v) f(a) = f(g(b)+g(b+w) g(b)) b = b+w b = w. Thus: so: s(w) = v [df(a)] 1 [df(a)[v] + r(v)] = [df(a)] 1 [r(v)], s(w) w df(a) 1 r(v) v v f(a + v) f(a), and the claim follows from lim v 0 r(v)/ v = 0, the fact that v 0 iff w 0 (since f and g are both continuous), and the fact (proved earlier) that f(a + v) f(a) / v is bounded below, for v sufficiently small (since df(a) is an isomorphism.) Turning to strong differentiability, if we set v = g(b+w) g(b) (as before) and u = g(b + z) g(b), it follows as above that we have the relation, for the first-order Taylor remainders (of f at a and of g at b): s(w) s(z) = [df(a)] 1 [r(v) r(u)], w z = df(a)[v u] + r(v) r(u). Recall f is strongly differentiable at a iff the remainder r satisfies a Lipschitz condition with arbitrarily small constant, in a sufficiently small ball with center a. As the following estimates show, this implies s has the same property, in a sufficiently small ball with center b; thus g is strongly differentiable at b. w z df(a) v u ɛ v u v u ( df(a) ɛ) 1 w z := M w z. s(w) s(z) df(a) 1 r(v) r(u) ɛ df(a) 1 v u ɛm df(a) 1 w z 4

Statement (ii) in the IFT follows directly from this. Proof (iii) in the IFT: If f C 1 (U, R n ), since df : U L(R n ) is continuous, df(a) GL(R n ) and GL(R n ) is open in L(R n ), we have df(x) GL(R n ) for x in a neighborhood V 1 V of a. Then the lemma on differentiability of the inverse guarantees g = f 1 is differentiable at each point of W = f(v 1 ). Before proving (iv) we must take a short detour. Lemma 1. The inversion map f(x) = X 1 is smooth from GL(R n ) to L(R n ) M(n) (n n matrices.) Proof. First, f is continuous at any X GL(R n ), since it is differentiable, with df(x)[v ] = X 1 V X 1. We know the directional derivative is V f(x) = X 1 V X 1. Given V M(n), let B V : M(n) M(n) M(n) be the bilinear map B V (X, Y ) = XV Y. For fixed V M(n), the directional derivative map V f : GL(R n ) M(n) is the composition: V f = B (f, f), where (f, f) : GL(n) M(n) M(n) maps X (X 1, X 1 ). Thus V f is continuous in GL(R n ) for all V M(n), so f is C 1. Then (f, f) is C 1, and since B V (being bilinear) is smooth, it follows the composition V f is C 1 (for each V M(n)), so f is C 2. Repeating this argument (or by induction) we see that f is a C k map, for each k 1. Corollary 1 (of lemma 1). Let f : U V be a homeomorphism between open subsets U, V of R n, of class C k (k 1). If g = f 1 is differentiable in V, then g is of class C k. Proof. It follows from the Chain Rule that, if g is differentiable in V, we have df(x) GL(R n ) for all x U and dg(y) = [df(g(y))] 1. Thus dg : V L(R n ) may be written as the composition: V U GL(R n ) L(R n ) dg = Inv df g, where Inv : GL(R n ) L(R n ) is the inversion map (which, according to the Lemma, is smooth.) Thus if f is C k (or df is C k 1 ), it follows dg is C k 1 (or g is C k.) Clearly this Corollary implies part (iv) of the IFT. Corollary 2 (of the proof of the IFT). (Differentiable perturbation of the identity.) Let U R n be open and convex, φ : U R n of class C 1, with dφ(x) λ < 1 for all x U. Then f : U R n given by f(x) = x + φ(x) 5

is a diffeomorphism from U onto its image f(u), an open subset of R n. If U = R n, then f(u) = R n. If f is C k (k 1), then g = f 1 is C k. Proof. Since U is convex, it follows φ is a λ-contraction. Thus f is a homeomorphism onto an open set f(u). Since df(x) = I n + dφ(x) and dφ(x) < 1, we know df(x) is an isomorphism, for each x U. Thus g = f 1 is differentiable in f(u). (By definition, this says f is a diffeomorphism.) Exercise 1. Generalize this to f(x) = Ax + φ(x), where A GL(R n ). (That is, state a precise theorem and prove it.) Implicit Function Theorem. Consider f : U R n differentiable, where U R m is open and m > n: m = n + p. Suppose we have a point a U where df(a) is surjective. We would like to use the Inverse Function Theorem to say something about the level set of f through a: M = {x U; f(x) = b}, where b = f(a) R n. The Implicit Function Theorem says that, locally in a neighborhood of a, M coincides with the graph of a map h from an open subset of R p to R n. To make this precise, let K = Ker(df(a)) be the nullspace of df(a), a p-dimensional subspace of R m, and let E R m be a complementary (ndimensional) subspace, so that R m = K E is a direct sum decomposition and the restriction df(a) E is a linear isomorphism from E to R n. We may regard R m as a cartesian product: R m = K E, and write x = (x 1, x 2 ) with respect to this product decomposition. Theorem. Suppose f is strongly differentiable at a U, with df(a) L(R m, R n ) surjective (and m > n). There exists an open neighborhood V U of a = (a 1, a 2 ) in R m, an open neighborhood W 1 of a 1 in K and a differentiable map h : W 1 E so that h(a 1 ) = a 2 and M V coincides with the graph of h: M V := {x V ; f(x) = b} = {(x 1, h(x 1 )); x 1 W 1 }. If f is a C k map, then so is h. If f is C 1 in V, the differential of h at a point x W 1 is: dh(x) = [d 2 f(x, h(x))] 1 d 1 f(x, h(x)) L(K; E), where d 1 f(x, h(x)) L(K; R n ); d 2 f(x, h(x)) Iso(E; R n ). 6

Proof. Consider the map f : U K R n : f(x) = (π(x), f(x)), f(a) = (a1, b), where π : R m = K E K is projection onto the factor K. Then f is strongly differentiable at a, with derivative: d f(a)[v] = (π[v], df(a)[v]) K R n, v K E R m. This differential is an isomorphism, since d f(a)[v] = 0 implies π[v] = 0, so v = (0, v 2 ) {0} E, hence df(a)[v] = 0 implies v = 0 (since the restriction of df(a) to {0} E is an isomorphism to R n.) Thus the Inverse Function Theorem implies the existence of neighborhoods V of a in R m, W = W 1 W 2 of (a 1, b) in K R n, so that f : V W is a diffeomorphism, with inverse g : W V of class C k if f is C k. For y 1 W 1, z W 2, g(y 1, z) = (g 1 (y 1, z), g 2 (y 1, z)) is mapped by f to (g 1 (y 1, z), f(g(y 1, z)), and since this must equal (y 1, z), it follows that g 1 (y 1, z) = y 1, for all y 1 W 1. Define: h : W 1 E, h(y 1 ) = g 2 (y 1, b). Clearly h(a 1 ) = g 2 (a 1, b) = a 2, since g(a 1, b) = (a 1, a 2 ) (given f(a 1, a 2 ) = (a 1, f(a 2 )) = (a 1, b).) Since g C k (W, K E), we have h C k (W 1, E). And we see that, for x = (x 1, x 2 ) V K E: f(x) = b f(x) = (x 1, b) g(x 1, b) = x g 2 (x 1, b) = x 2 h(x 1 ) = x 2. To compute the differential of h at x 1 W 1, note that, with G : W 1 R m, G(x 1 ) = (x 1, h(x 1 )): 0 = d(f G)(x 1 ) = d 1 f(x 1, h(x 1 )) + d 2 f(x 1, h(x 1 )) dh(x 1 ) L(K, R n ), where d 1 f(x) L(K, R n ), d 2 f(x) L(E, R n ) are partial derivatives (with respect to the splitting R m = K E), with the second one an isomorphism for x V (taking a smaller V if needed), since d 2 f(a) Iso(E, R n ) and f is C 1 ; and dh(x 1 ) L(K, E). Solving for dh(x 1 ), we find: dh(x 1 ) = [d 2 f(x 1, h(x 1 ))] 1 d 1 f(x 1, h(x 1 )) L(K, E). Remark. In particular dh(a 1 ) = 0, since d 1 f(a) = df(a) K {0} = 0. Definition. For f : U R n of class C 1 (U R m open, m n), a point b R n is a regular value of f if df(x) L(R m, R n ) is surjective, for each x in the level set M b = {x U; f(x) = b}. 7

Submersions are open maps. A C 1 map f : U R n (U R m open, m > n is a submersion if df(x) L(R m, R n ) is surjective, for each x U. Recall a map between topological spaces is open if it maps open subsets of X to open subsets of Y. Homeomorphisms are clearly open maps, and the composition of open maps is open. Proposition 2. (i) If f is strongly differentiable at a U, and df(a) is surjective, then there exists a neighborhood V of a so that the restriction f V : V R n is an open map. (ii) If f is a C 1 submersion from U to R n, then f is an open map. Proof. (i) In the proof of the Implicit function Theorem, we found a neighborhood V of a so that the map f(x) = (π(x), f(x)) K R n (where R m = K E is defined by choosing a complement E to the kernel K of df(a)) is a diffeomorphism from V to a neighborhood W 1 W 2 of (a 1, b) in K R n (b = f(a), a = (a 1, a 2 ) with a 1 K, a 2 E.) On V, f = π 2 f, where π 2 : W 1 W 2 W 2 is projection onto the second factor W 2 R n. Since π 2 and f are open, this shows f V is open. (ii) From part (i), for each point x U there is a neighborhood V x U so that f Vx is an open map. If A U is open in U, we have A = x U V x A, so f(a) = x U f(v x A), where W x = f(v x A) is open in R n. Hence f(a) = x U W x is open in R n. Remark. In particular, under the conditions of part(i), the point f(a) is in the interior of f(v ). Images of C k parametrizations. A C k map (k 1) f : U R n (U R n open, n > m) is an immersion if df(x) L(R m, R n ) is injective, for each x U. f is a C k parametrization of its image M = f(u) R n if it is an injective immersion and defines a homeomorphism from U to M, where M R n has the induced topology. Proposition 3. Any image M R n of a C k parametrization φ : W 0 M = f(w 0 ) (W 0 R m open, m < n) is locally a graph. (That is, each point p M admits a neighborhood V M in which M admits a parametrization as the graph of a C k map.) Proof. Let p M, p = φ(x 0 ), x 0 W 0. Consider the subspace T p = dφ(x 0 )[R m ], and a complementary subspace N p R m, which defines a direct sum decomposition R m = T p N p ; let π : R m T p be the associated 8

projection. (Note dim(t p ) = m.)the composition π φ : W 0 T p has differential at x 0 given by: d(π φ)(x 0 ) = dπ(p) dφ(x 0 ). This is a linear isomorphism from R m to T p, since π is the identity on T p and dφ(x 0 ) is an isomorphism from R m to T p (φ is an immersion.) By the Inverse Function Theorem, we may find neighborhoods V 0 W 0 of x 0 and D p T p of π(p) so that π φ is a diffeomorphism F : V 0 D p (of class C k since φ is C k.) Let U = φ(v 0 ) M. This is an open neighborhood of p in M, since it coincides with π 1 (D p ) M. Consider g : D p U given by g = φ V0 F 1. Clearly g is a homeomorphism (and of class C k, as the composition of C k maps). And π g = id Dp shows that, with respect to the splitting R m = T p N p, g has the form g(x) = (x, h(x)), for a C k map h : D p N p. Thus g is a C k graph parametrization of the open set U M. Exercise 2. Let h : U R n be a C k map (k 1), U R m open. The graph of h is the subset of R m R n : G = {(x, y) U R n ; y = h(x)} (i) Show that φ : U R m R n, φ(x) = (x, h(x)) is a C k parametrization of G. Hint: to show φ 1 is continuous on G, show that it is Lipschitz (by estimating from below φ(x) φ(y) ), or that it is the restriction to G of the projection from U R n to U. (ii) Show there exists F : U R n R n of class C k, with surjective differential at each point, so that G = F 1 (0) (the level set of 0 R n.) m-dimensional surfaces in R n. Definition. A non-empty subset M R n is an m-dimensional surface of class C k (also known as m-dimensional submanifold of R n, of class C k ) if each p M admits a neighborhood U M (in the topology induced from E) which is the image φ(u 0 ) of a C k parametrization φ : U 0 E (U 0 R m open.) Remark. It can be shown that there is no loss of generality in taking U 0 connected in this definition; in fact we could assume U 0 is homeomorphic to the open ball in R m. 9

It follows from the work in the previous section that this definition is equivalent to either of the following two: (i) Each p M admits a neighborhood U M (of the form U = W M, W E open) so that for some C k submersion F : W R p (where p = dim(e) m) we have U = F 1 (0). (ii) For each p M we may find a direct sum decomposition E = T p N p, U 1 T p open, a C k map h : U 1 N p and a neighborhood W of p in E so that U = M W = {(x, h(x)) : x U 1 }. Colloquially, we speak of M being given by local parametrizations, M being locally a regular level set or M being locally a graph. Remark. In the global sense, the three definitions describe different classes of subsets of R n. For instance, no compact surface in R n can be covered by a single parametrization (in particular, no compact surface can be globally a graph). There are also 2-surfaces in R 3 (and compact 2-surfaces in R 4 ) which are not globally level sets for a regular value (although this is always true for a simply-connected 2-surface in R 3.) The Implicit Function Theorem can be stated as saying that, if b R n is a regular value for f : U R n (U R m open, m > n), then M b is locally a graph, hence is a surface in R n (of dimension m n). Example 1. Consider the subset of R 2 : M = {(x, sin 1 ); x > 0} {(x, y); x = 0, y ( 1, 1)}. x This is not a one-dimensional submanifold of R 2, since it can be shown any submanifold is locally connected, and this set isn t: any point on the vertical line segment has a neighborhood in M (in the induced topology) with infinitely many connected components. If we remove the vertical line segment, the resulting set is a submanifold of R 2 (so we see that the disjoint union of two submanifolds of the same dimension is not always a submanifold of R n.) Example 2. Consider the subset of R 2 made up of a circle and one of its tangent lines: M = {(x, y) R 2 ; y = 0 or x 2 + (y 1) 2 = 1}. This is not a submanifold of R 2. Locally near any point, a submanifold must be expressible as a graph, with respect to some direct-sum splitting of R n. 10

Thus it cannot have any branching points (as the origin is in this case). If we remove the origin, the resulting set is a submanifold. Tangent spaces. Let M R n be a C k m-surface (k 1, 1 m < n.) The tangent space to M at p is the set of all velocity vectors at p of C 1 curves contained in M. Precisely: T p M = {v R n ; v = α (0), where α : I R n, α(t) M t, α(0) = p}. (Here I R is any fixed open interval containing 0 and α is assumed to be C 1.) Indeterminacy. Clearly many curves α yield the same v. For instance, let φ : I I be a strictly increasing differentiable diffeomorphism of I fixing 0: t = φ(s), φ (s) > 0, φ(0) = 0. Then if φ (0) = 1, we have (α φ) (0) = α (0). A more serious problem is that it is not at all clear that this is a vector space! (At least it is easy to show it is a cone in R n : v T p M, λ R λv T p M.) We need the following proposition: Proposition 4. Let M be a C 1 m-surface in R n, let p M. (i) If M is given near p by a parametrization φ : U 0 R n, φ(x 0 ) = p, φ(u 0 ) = U M, then: T p M = dφ(x 0 )[R m ] = {v R n ; v = dφ(x 0 )[w] for some w R m.} (This is clearly a subspace of R n, of dimension m). (ii) If M is given near p by a graph parametrization (there is a neighborhood W of p in R n = E F and a map h : V F, V E open) so that M W = {(x, h(x)); x V } and p = (x 0, h(x 0 )), x 0 V, then: T p M = {v dh(x 0 )[v], v E.} (iii) If M is given near p as the level set M W = {x W ; F (x) = b}, for some C 1 submersion F : W R n m,w a neighborhood of p in R n, b = F (p), then: T p M = Ker(dF (p)), a subspace of R n. Remark: indeterminacy. The space in (i) might depend on φ (replacing φ by φ ψ, where ψ is a diffeomorphism of V 0 fixing x 0 gives another parametrization with domain V 0 ). Similarly changing the splitting of R n used in (ii) changes the map h, so we might get a different subspace. And the submersion F in (iii) is not uniquely defined, either. 11

The fact that these subspaces of R n coincide is a consequence of the proposition, which shows they all coincide with the set given in the geometric definition (as the set of velocity vectors of curves through p.) But the following can be shown directly. If φ : U 0 U M is a local parametrization of a neighborhood of p M with φ(x 0 ) = p and F : U 0 U 0 is a diffeomorphism of U 0 fixing x 0, then (evidently) ψ = φ F is again a parametrization of U, with ψ(x 0 ) = p. Then dψ(x 0 )(R m ) = dφ(x 0 )(R m ), since df (x 0 ) is an isomorphism of R m. Exercise 3. (i) Show that if φ : U 0 U M, ψ : U 0 U M are parametrizations (U 0 R m open) satisfying φ(x 0 ) = ψ(x 0 ) = p M, then dφ(x 0 )(R m ) = dψ(x 0 )(R m ). Hint: use the proof of proposition 5 at the end of this section to find a diffeomorphism F of U 0 fixing x 0 so that ψ = φ F, then the observation in the previous paragraph. (ii) Show that if p R n is a regular value of f : U R n (U R m open, m > n) and φ : R n R n is a diffeomorphism fixing p (φ(p) = p), then p is a regular value for g = φ f, the level sets {x f(x) = p} and {x g(x) = p} are equal (call them M p ) and Ker(df(x)) = Ker(dg(x)) (as subspaces of R m ), for each x M p. Proof (of proposition 4). That the vector spaces in (i) and (ii) are contained in the set of velocity vectors of curves on M through p is clear. To prove that any velocity vector of a curve on M through p is in the vector space in (i) we need proposition 5 below. This shows that if α : I R n is a C 1 curve with image in U M, then γ = φ 1 α : I U 0 is a C 1 curve. Let v 0 = γ (0) R m. Then α (0) = dφ(x 0 )[v 0 ], as we wanted to show. Note that (ii) is a special case of (i). Having verified (i), we know the set of velocity vectors of curves on M through p is a vector space of dimension m. Under the hypothesis of part (iii), any such velocity vector is clearly contained in Ker(dF (p)); since both are m-dimensional vector spaces, they must coincide. Proposition 5. Let M R n be an m-dimensional surface of class C k, f : I R n be a C k map (I R p open). If f(i) U, where U M is the image of a C k parametrization φ : U 0 U (U 0 R m open), then the composition φ 1 f : I R m is of class C k. Proof. It is enough to show φ 1 f is of class C k in a neighborhood of an arbitrary point t 0 I. This would follow easily if we could show that, in some neighborhood V = W M U of p = f(t 0 ) = φ(x 0 ), the inverse 12

φ 1 : V U 0 extends to a C k map g : W R m. To see this, consider the splitting R n = E F, where E = dφ(x 0 )(R m ) and F is any complementary subspace of R n. Let {v 1,..., v n m } be a basis of F, and extend φ to Φ : U R n m R n, as follows: n m Φ(x, y) = φ(x) + y i v i, (x, y) U R n m. i=1 The differential of Φ at (x 0, 0) U R n m is: dφ(x 0, 0)[(v, w)] = dφ(x 0 )[v] + w i v i E F, (v, w) R m R n m. It is clear that this is an isomorphism of R n. By the Inverse Function Theorem, there are neighborhoods V 0 U 0 of x 0 in R m, W 0 of 0 in R n m and W of p in R n so that Φ is a C k diffeomorphism from V 0 W 0 to W. Let g = Φ 1 : W V 0 W 0. Since Φ(x, 0) = φ(x) W M := V U for x V 0, we see that the restriction of g to V coincides with (φ 1 ) V. Observing that in the neighborhood f 1 (V ) I of t 0 we have φ 1 f = g f and that the latter composition is C k concludes the proof of the proposition. Problems from [Fleming]. 1. (7, p.147). Let g : U R n be a C 1 map (U R n open), with dg(x) GL(R n ) for all x U. Suppose y 0 g(u), and let ψ(x) = g(x) y 0 2. Show that dψ(x) 0, for any x U. (This means: for any x U, we may find v R n so that dψ(x)[v] 0). Hint: Use fact that if A GL(R n ), we have A T [w] 0 if w 0, so v = A T w satisfies w, A[v] 0. Here the superscript T denotes transpose. 2. (8, p.147). Let g : R n R n be a C 1 map. Suppose there exists c > 0 so that g(x) g(y) c x y, for all x, y R n. Show that: (i) g is injective; (ii) dg(x) GL(R n ) for all x R n ; (iii) g(r n ) = R n. Hint: For (ii), assume dg(x)[v] = 0 for some 0 v R n, and use the definition of differential to contradict the condition given. For (iii): By contradiction, if y 0 g(r n ), consider the function ψ(x) = g(x) y 0 2. By the previous problem, this function has no critical points. On the other hand, ψ attains its infimum (call it M) over R n : let (x n ) be 13

a sequence in R n so that ψ(x n ) = g(x n ) y 0 2 M. Then (g(x n )) is a bounded sequence. Show this implies (x n ) is bounded, and therefore has a convergent subsequence. Conclude. 3. (11. p. 160). Let C = (c ij ) GL(R n ) be a symmetric matrix, and consider M = {x R n ; x, Cx = 1} (for the usual inner product in R n.) (i) Show that M is a compact submanifold of R n, of dimension n 1. Hint: M = {x R n F (x) = 1} for a submersion F : R n R. Don t forget to show M is compact! (ii) Show that the equation of the tangent space T x0 M, x 0 M, is: T x0 M = {x R n ; x, Cx 0 = 0}. 14