INVERSE FUNCTION THEOREM and SURFACES in Rn Let F ∈ C K(U;Rn)

INVERSE FUNCTION THEOREM and SURFACES IN Rn

Let f ∈ Ck(U; Rn), with U ⊂ Rn open. Assume df(a) ∈ GL(Rn), where a ∈ U. The Inverse Function Theorem says there is an open neighborhood n n V ⊂ U of a in R so that f|V : V → R is a homeomorphism onto its image W = f(V ), an open subset of Rn; the inverse g = f −1 : W → V is a Ck 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 −1 (that is, A is invertible, and |v|X ≤ C|w|Y , where C = ||A ||) we compute successive approximations, starting from an arbitrary a ∈ X and solving the sequence of linear problems with given ‘right-hand side’:

Ax1 = y − φ(a), Ax2 = y − φ(x1), Ax3 = y − φ(x2), ···

Convergence of (xn) is established by setting this up as a ﬁxed point problem, for F (x) = A−1[y − φ(x)]:

x1 = F (a), x2 = F (x1), ··· 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) =y, ¯ y, y¯ ∈ Y :

|x − x¯| = |A−1[y − y¯] − A−1[φ(x) − φ(¯x)]| ≤ |A−1|(|y − y¯| + Lip(φ)|x − x¯|),

1 (1 − λ)|x − x¯| ≤ |A−1||y − y¯|, 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.) n Proof. Let x ∈ X. For the sequence of iterates xn = F (x), the contrac- m m tion property easily implies d(xm+1, xm) ≤ λ d(x1, x) := λ δ, and then if n > m ≥ N:

λN d(x , x ) ≤ d(x , x ) + . . . d(x , x ) ≤ (λn−1 + . . . λm)δ ≤ δ, n m n n−1 m+1 m 1 − λ so (xn) is a Cauchy sequence, and xn → p ∈ X. Then p is a ﬁxed point:

d(F (p), p) ≤ d(F (p),F (xn)) + d(xn+1, p) ≤ λd(xn, p) + d(xn+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 −1 we estimate, for x ∈ B¯r(a) (using a = A [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 ∈ Bs(b) (open ball in (1−λ)r −1 Y ), where s ≤ |A−1| (with λ = |A |Lip(φ) < 1 as before). We summarize 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 → Rn, where U ⊂ Rn is an open set, suppose f is differentiable at a ∈ U, with df(a) ∈ GL(Rn) (that is, df(a) is an invertible linear map.) Then we have, for a function r : U → Rn: 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 → Rn (U ⊂ Rn open) be strongly differentiable at a ∈ U, and assume df(a) ∈ GL(Rn) (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 Rn. (ii) The inverse map g = f −1 : W → V is strongly differentiable at b = f(a); (iii) If f ∈ C1(U; Rn), then V can be chosen so that g = f −1 is differentiable in W ; (iv) If f ∈ Ck(U; Rn) (with k ≥ 1) then g ∈ Ck(W ; Rn). Proof. (i): Since f is strongly differentiable at a, we may find an open ball V = Br(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 diﬀeren- tiability of the inverse (of a homeomorphism). Note that the inverse of the

3 homeomorphism f(x) = x3 of the real line is not differentiable at x = 0. This can only happen since f 0(0) = 0. Differentiability of the inverse. Let f : U → V be a homeomorphism, where U, V are open in Rn, with inverse g : V → U. If f is differentiable at a ∈ U and df(a) ∈ GL(Rn), 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 ∈ Rn with |w| < dist(b, ∂V )) by:

g(b + w) = g(b) + [df(a)]−1[w] + s(w), we need to show limw→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:

s(w) = v − [df(a)]−1[df(a)[v] + r(v)] = −[df(a)]−1[r(v)], so: |s(w)| |r(v)| |v| ≤ |df(a)−1| , |w| |v| |f(a + v) − f(a)| and the claim follows from limv→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 Lips- chitz 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|−|v−u| ⇒ |v−u| ≤ (|df(a)|−)−1|w−z| := M|w−z|.

|s(w)−s(z)| ≤ |df(a)−1||r(v)−r(u)| ≤ |df(a)−1||v−u| ≤ M|df(a)−1||w−z|

4 Statement (ii) in the IFT follows directly from this.

Proof (iii) in the IFT: If f ∈ C1(U, Rn), since df : U → L(Rn) is continuous, df(a) ∈ GL(Rn) and GL(Rn) is open in L(Rn), we have df(x) ∈ n GL(R ) for x in a neighborhood V1 ⊂ V of a. Then the lemma on differentiability of the inverse guarantees g = f −1 is differentiable at each point of W = f(V1). Before proving (iv) we must take a short detour. Lemma 1. The inversion map f(X) = X−1 is smooth from GL(Rn) to L(Rn) ∼ M(n)(n × n matrices.) Proof. First, f is continuous at any X ∈ GL(Rn), since it is differentiable, with df(X)[V ] = −X−1VX−1. We know the directional derivative is −1 −1 ∂V f(X) = −X VX . Given V ∈ M(n), let BV : M(n) × M(n) → M(n) be the bilinear map BV (X,Y ) = XVY . For fixed V ∈ M(n), the directional n derivative map ∂V f : GL(R ) → M(n) is the composition:

−1 −1 ∂V f = −B◦(f, f), where (f, f): GL(n) → M(n)×M(n) maps X 7→ (X ,X ).

n 1 Thus ∂V f is continuous in GL(R ) for all V ∈ M(n), so f is C . Then 1 (f, f) is C , and since BV (being bilinear) is smooth, it follows the compo- 1 2 sition ∂V f is C (for each V ∈ M(n)), so f is C . Repeating this argument (or by induction) we see that f is a Ck map, for each k ≥ 1. Corollary 1 (of lemma 1). Let f : U → V be a homeomorphism between open subsets U, V of Rn, of class Ck (k ≥ 1). If g = f −1 is diﬀerentiable in V , then g is of class Ck. Proof. It follows from the Chain Rule that, if g is diﬀerentiable in V , we have df(x) ∈ GL(Rn) for all x ∈ U and dg(y) = [df(g(y))]−1. Thus dg : V → L(Rn) may be written as the composition:

V → U → GL(Rn) → L(Rn) dg = Inv ◦ df ◦ g, where Inv : GL(Rn) → L(Rn) is the inversion map (which, according to the Lemma, is smooth.) Thus if f is Ck (or df is Ck−1), it follows dg is Ck−1 (or g is Ck.) Clearly this Corollary implies part (iv) of the IFT. Corollary 2 (of the proof of the IFT). (Diﬀerentiable perturbation of the identity.) Let U ⊂ Rn be open and convex, φ : U → Rn of class C1, with ||dφ(x)|| ≤ λ < 1 for all x ∈ U. Then f : U → Rn given by f(x) = x + φ(x)

5 is a diffeomorphism from U onto its image f(U), an open subset of Rn. If U = Rn, then f(U) = Rn. If f is Ck (k ≥ 1), then g = f −1 is Ck. 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) = In + 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(Rn). (That is, state a precise theorem and prove it.)

Implicit Function Theorem. Consider f : U → Rn diﬀerentiable, where U ⊂ Rm 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) ∈ Rn. 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 Rp to Rn.

To make this precise, let K = Ker(df(a)) be the nullspace of df(a), a p-dimensional subspace of Rm, and let E ⊂ Rm be a complementary (n- dimensional) subspace, so that Rm = K ⊕ E is a direct sum decomposition n and the restriction df(a)|E is a linear isomorphism from E to R . We may m m regard R as a cartesian product: R = K × E, and write x = (x1, x2) with respect to this product decomposition. Theorem. Suppose f is strongly diﬀerentiable at a ∈ U, with df(a) ∈ L(Rm,Rn) surjective (and m > n). There exists an open neighborhood m V ⊂ U of a = (a1, a2) in R , an open neighborhood W1 of a1 in K and a diﬀerentiable map h : W1 → E so that h(a1) = a2 and M ∩V coincides with the graph of h:

M ∩ V := {x ∈ V ; f(x) = b} = {(x1, h(x1)); x1 ∈ W1}.

If f is a Ck map, then so is h. If f is C1 in V , the diﬀerential of h at a point x ∈ W1 is:

−1 dh(x) = −[d2f(x, h(x))] ◦ d1f(x, h(x)) ∈ L(K; E),

n n where d1f(x, h(x)) ∈ L(K; R ); d2f(x, h(x)) ∈ Iso(E; R ).

6 Proof. Consider the map f¯ : U → K × Rn:

f¯(x) = (π(x), f(x)), f¯(a) = (a1, b), where π : Rm = K × E → K is projection onto the factor K. Then f¯ is strongly diﬀerentiable at a, with derivative:

df¯(a)[v] = (π[v], df(a)[v]) ∈ K × Rn, v ∈ K × E ∼ Rm.

This diﬀerential is an isomorphism, since df¯(a)[v] = 0 implies π[v] = 0, so v = (0, v2) ∈ {0} × E, hence df(a)[v] = 0 implies v = 0 (since the restriction of df(a) to {0} × E is an isomorphism to Rn.) Thus the Inverse Function Theorem implies the existence of neighbor- m n hoods V of a in R , W = W1 × W2 of (a1, b) in K × R , so that f¯ : V → W is a diﬀeomorphism, with inverse g : W → V of class Ck if f is Ck.

For y1 ∈ W1, z ∈ W2, g(y1, z) = (g1(y1, z), g2(y1, z)) is mapped by f¯ to (g1(y1, z), f(g(y1, z)), and since this must equal (y1, z), it follows that g1(y1, z) = y1, for all y1 ∈ W1. Deﬁne:

h : W1 → E, h(y1) = g2(y1, b).

Clearly h(a1) = g2(a1, b) = a2, since g(a1, b) = (a1, a2) (given f¯(a1, a2) = k k (a1, f(a2)) = (a1, b).) Since g ∈ C (W, K × E), we have h ∈ C (W1,E). And we see that, for x = (x1, x2) ∈ V ⊂ K × E:

f(x) = b ⇔ f¯(x) = (x1, b) ⇔ g(x1, b) = x ⇔ g2(x1, b) = x2 ⇔ h(x1) = x2.

To compute the diﬀerential of h at x1 ∈ W1, note that, with G : W1 → m R ,G(x1) = (x1, h(x1)):

n 0 = d(f ◦ G)(x1) = d1f(x1, h(x1)) + d2f(x1, h(x1)) ◦ dh(x1) ∈ L(K,R ),

n n where d1f(x) ∈ L(K,R ), d2f(x) ∈ L(E,R ) are partial derivatives (with respect to the splitting Rm = K × E), with the second one an isomorphism n for x ∈ V (taking a smaller V if needed), since d2f(a) ∈ Iso(E,R ) and f 1 is C ; and dh(x1) ∈ L(K,E). Solving for dh(x1), we ﬁnd:

−1 dh(x1) = −[d2f(x1, h(x1))] ◦ d1f(x1, h(x1)) ∈ L(K,E).

Remark. In particular dh(a1) = 0, since d1f(a) = df(a)|K×{0} = 0. Deﬁnition. For f : U → Rn of class C1 (U ⊂ Rm open, m ≥ n), a point b ∈ Rn is a regular value of f if df(x) ∈ L(Rm,Rn) is surjective, for each x in the level set Mb = {x ∈ U; f(x) = b}.

7 Submersions are open maps. A C1 map f : U → Rn (U ⊂ Rm open, m > n is a submersion if df(x) ∈ L(Rm,Rn) 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 n f|V : V → R is an open map. (ii) If f is a C1 submersion from U to Rn, 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 × Rn

(where Rm = K ×E is deﬁned by choosing a complement E to the kernel K of df(a)) is a diﬀeomorphism from V to a neighborhood W1 × W2 of (a1, b) n in K × R (b = f(a), a = (a1, a2) with a1 ∈ K, a2 ∈ E.) On V , f = π2 ◦ f¯, n where π2 : W1 × W2 → W2 is projection onto the second factor W2 ⊂ R . ¯ Since π2 and f are open, this shows f|V is open.

(ii) From part (i), for each point x ∈ U there is a neighborhood Vx ⊂ U so that f|Vx is an open map. If A ⊂ U is open in U, we have A = ∪x∈U Vx ∩A, n so f(A) = ∪x∈U f(Vx ∩ A), where Wx = f(Vx ∩ A) is open in R . Hence n f(A) = ∪x∈U Wx is open in R . Remark. In particular, under the conditions of part(i), the point f(a) is in the interior of f(V ). Images of Ck parametrizations. A Ck map (k ≥ 1) f : U → Rn (U ⊂ Rn open, n > m) is an immersion if df(x) ∈ L(Rm,Rn) is injective, for each x ∈ U. f is a Ck parametrization of its image M = f(U) ⊂ Rn if it is an injective immersion and deﬁnes a homeomorphism from U to M, where M ⊂ Rn has the induced topology. n k Proposition 3. Any image M ⊂ R of a C parametrization φ : W0 → m M = f(W0)(W0 ⊂ R 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 Ck map.)

Proof. Let p ∈ M, p = φ(x0), x0 ∈ W0. Consider the subspace Tp = m m dφ(x0)[R ], and a complementary subspace Np ∈ R , which deﬁnes a direct m m sum decomposition R = Tp ⊕ Np; let π : R → Tp be the associated

8 projection. (Note dim(Tp) = m.)The composition π ◦ φ : W0 → Tp has diﬀerential at x0 given by:

d(π ◦ φ)(x0) = dπ(p) ◦ dφ(x0).

m This is a linear isomorphism from R to Tp, since π is the identity on Tp m and dφ(x0) is an isomorphism from R to Tp (φ is an immersion.)

By the Inverse Function Theorem, we may ﬁnd neighborhoods V0 ⊂ W0 of x0 and Dp ⊂ Tp of π(p) so that π ◦ φ is a diﬀeomorphism F : V0 → Dp (of class Ck since φ is Ck.)

Let U = φ(V0) ⊂ M. This is an open neighborhood of p in M, since it −1 −1 coincides with π (Dp) ∩ M. Consider g : Dp → U given by g = φ|V0 ◦ F . Clearly g is a homeomorphism (and of class Ck, as the composition of Ck m maps). And π ◦ g = idDp shows that, with respect to the splitting R = k Tp ⊕ Np, g has the form g(x) = (x, h(x)), for a C map h : Dp → Np. Thus g is a Ck graph parametrization of the open set U ⊂ M. Exercise 2. Let h : U → Rn be a Ck map (k ≥ 1), U ⊂ Rm open. The graph of h is the subset of Rm × Rn:

G = {(x, y) ∈ U × Rn; y = h(x)}

(i) Show that φ : U → Rm × Rn, φ(x) = (x, h(x)) is a Ck 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 × Rn to U. (ii) Show there exists F : U × Rn → Rn of class Ck, with surjective diﬀerential at each point, so that G = F −1(0) (the level set of 0 ∈ Rn.)

m-dimensional surfaces in Rn. Deﬁnition. A non-empty subset M ⊂ Rn is an m-dimensional surface of class Ck (also known as ‘m-dimensional submanifold of Rn, of class Ck’) if each p ∈ M admits a neighborhood U ⊂ M (in the topology induced from k m E) which is the image φ(U0) of a C parametrization φ : U0 → E (U0 ⊂ R open.)

Remark. It can be shown that there is no loss of generality in taking U0 connected in this deﬁnition; in fact we could assume U0 is homeomorphic to the open ball in Rm.

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 Ck submersion F : W → Rp (where p = dim(E) − m) we have U = F −1(0).

(ii) For each p ∈ M we may ﬁnd a direct sum decomposition E = Tp⊕Np, k U1 ⊂ Tp open, a C map h : U1 → Np and a neighborhood W of p in E so that U = M ∩ W = {(x, h(x)) : x ∈ U1}. 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 Rn. For instance, no compact surface in Rn can be covered by a single parametrization (in particular, no compact surface can be globally a graph). There are also 2-surfaces in R3 (and compact 2-surfaces in R4) which are not globally level sets for a regular value (although this is always true for a simply-connected 2-surface in R3.) The Implicit Function Theorem can be stated as saying that, if b ∈ Rn n m is a regular value for f : U → R (U ⊂ R open, m > n), then Mb is locally a graph, hence is a surface in Rn (of dimension m − n). Example 1. Consider the subset of R2: 1 M = {(x, sin ); x > 0} ∪ {(x, y); x = 0, y ∈ (−1, 1)}. x This is not a one-dimensional submanifold of R2, 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 R2 (so we see that the disjoint union of two submanifolds of the same dimension is not always a submanifold of Rn.) Example 2. Consider the subset of R2 made up of a circle and one of its tangent lines:

M = {(x, y) ∈ R2; y = 0 or x2 + (y − 1)2 = 1}.

This is not a submanifold of R2. Locally near any point, a submanifold must be expressible as a graph, with respect to some direct-sum splitting of Rn.

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 ⊂ Rn be a Ck m-surface (k ≥ 1, 1 ≤ m < n.) The tangent space to M at p is the set of all velocity vectors at p of C1 curves contained in M. Precisely:

n 0 n TpM = {v ∈ R ; v = α (0), where α : I → R , α(t) ∈ M∀t, α(0) = p}. (Here I ⊂ R is any fixed open interval containing 0 and α is assumed to be C1.) 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), φ0(s) > 0, φ(0) = 0. Then if φ0(0) = 1, we have (α ◦ φ)0(0) = α0(0). A more serious problem is that it is not at all clear that this is a vector n space! (At least it is easy to show it is a cone in R : v ∈ TpM, λ ∈ R ⇒ λv ∈ TpM.) We need the following proposition: Proposition 4. Let M be a C1 m-surface in Rn, let p ∈ M. n (i) If M is given near p by a parametrization φ : U0 → R , φ(x0) = p, φ(U0) = U ⊂ M, then:

m n m TpM = dφ(x0)[R ] = {v ∈ R ; v = dφ(x0)[w] for some w ∈ R .} (This is clearly a subspace of Rn, of dimension m). (ii) If M is given near p by a graph parametrization (there is a neighborhood W of p in Rn = E ⊕ F and a map h : V → F , V ⊂ E open) so that M ∩ W = {(x, h(x)); x ∈ V } and p = (x0, h(x0)), x0 ∈ V , then:

TpM = {v ⊕ dh(x0)[v], v ∈ E.} (iii) If M is given near p as the level set M ∩ W = {x ∈ W ; F (x) = b}, for some C1 submersion F : W → Rn−m,W a neighborhood of p in Rn, b = F (p), then: TpM = Ker(dF (p)), a subspace of Rn. Remark: indeterminacy. The space in (i) might depend on φ (replac- ing φ by φ ◦ ψ, where ψ is a diffeomorphism of V0 fixing x0 gives another n parametrization with domain V0). Similarly changing the splitting of R 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 Rn 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 : U0 → U ⊂ M is a local parametrization of a neighborhood of p ∈ M with φ(x0) = p and F : U0 → U0 is a diﬀeomorphism of U0 ﬁxing x0, then (evidently) ψ = φ ◦ F is again a parametrization of U, with ψ(x0) = p. m m m Then dψ(x0)(R ) = dφ(x0)(R ), since dF (x0) is an isomorphism of R .

Exercise 3. (i) Show that if φ : U0 → U ⊂ M, ψ : U0 → U ⊂ M are m parametrizations (U0 ⊂ R open) satisfying φ(x0) = ψ(x0) = p ∈ M, then m m dφ(x0)(R ) = dψ(x0)(R ). Hint: use the proof of proposition 5 at the end of this section to find a diffeomorphism F of U0 fixing x0 so that ψ = φ ◦ F , then the observation in the previous paragraph. (ii) Show that if p ∈ Rn is a regular value of f : U → Rn (U ⊂ Rm open, m > n) and φ : Rn → Rn 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 m equal (call them Mp) and Ker(df(x)) = Ker(dg(x)) (as subspaces of R ), for each x ∈ Mp. 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 → Rn is 1 −1 1 a C curve with image in U ⊂ M, then γ = φ ◦ α : I → U0 is a C curve. 0 m 0 Let v0 = γ (0) ∈ R . Then α (0) = dφ(x0)[v0], 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 ⊂ Rn be an m-dimensional surface of class Ck, f : I ⊂ Rn be a Ck map (I ⊂ Rp open). If f(I) ⊂ U, where U ⊂ M is k m the image of a C parametrization φ : U0 → U (U0 ⊂ R open), then the composition φ−1 ◦ f : I → Rm is of class Ck. Proof. It is enough to show φ−1 ◦ f is of class Ck in a neighborhood of an arbitrary point t0 ∈ I. This would follow easily if we could show that, in some neighborhood V = W ∩ M ⊂ U of p = f(t0) = φ(x0), the inverse

12 −1 k m φ : V → U0 extends to a C map g : W → R . n m To see this, consider the splitting R = E ⊕ F , where E = dφ(x0)(R ) n and F is any complementary subspace of R . Let {v1, . . . , vn−m} be a basis of F , and extend φ to Φ : U × Rn−m → Rn, as follows:

n−m X n−m Φ(x, y) = φ(x) + yivi, (x, y) ∈ U × R . i=1

n−m The diﬀerential of Φ at (x0, 0) ∈ U × R is:

X m n−m dΦ(x0, 0)[(v, w)] = dφ(x0)[v] + wivi ∈ E ⊕ F, (v, w) ∈ R × R .

It is clear that this is an isomorphism of Rn. By the Inverse Function m n−m Theorem, there are neighborhoods V0 ⊂ U0 of x0 in R , W0 of 0 in R n k and W of p in R so that Φ is a C diffeomorphism from V0 × W0 to W . −1 Let g = Φ : W → V0 × W0. Since Φ(x, 0) = φ(x) ∈ W ∩ M := V ⊂ U for −1 x ∈ V0, we see that the restriction of g to V coincides with (φ )|V . −1 −1 Observing that in the neighborhood f (V ) ⊂ I of t0 we have φ ◦ f = g ◦ f and that the latter composition is Ck concludes the proof of the proposition. Problems from [Fleming]. 1. (7, p.147). Let g : U → Rn be a C1 map (U ⊂ Rn open), with dg(x) ∈ n 2 GL(R ) for all x ∈ U. Suppose y0 6∈ g(U), and let ψ(x) = |g(x)−y0| . Show that dψ(x) 6= 0, for any x ∈ U. (This means: for any x ∈ U, we may find v ∈ Rn so that dψ(x)[v] 6= 0). Hint: Use fact that if A ∈ GL(Rn), we have AT [w] 6= 0 if w 6= 0, so v = AT w satisfies hw, A[v]i= 6 0. Here the superscript T denotes ‘transpose’.

2. (8, p.147). Let g : Rn → Rn be a C1 map. Suppose there exists c > 0 so that |g(x) − g(y)| ≥ c|x − y|, for all x, y ∈ Rn. Show that: (i) g is injective; (ii) dg(x) ∈ GL(Rn) for all x ∈ Rn; (iii) g(Rn) = Rn. Hint: For (ii), assume dg(x)[v] = 0 for some 0 6= v ∈ Rn, and use the definition of differential to contradict the condition given. n For (iii): By contradiction, if y0 6∈ g(R ), consider the function ψ(x) = 2 |g(x) − y0| . By the previous problem, this function has no critical points. n On the other hand, ψ attains its infimum (call it M) over R : let (xn) be

13 n 2 a sequence in R so that ψ(xn) = |g(xn) − y0| → M. Then (g(xn)) is a bounded sequence. Show this implies (xn) is bounded, and therefore has a convergent subsequence. Conclude. n 3. (11. p. 160). Let C = (cij) ∈ GL(R ) be a symmetric matrix, and consider M = {x ∈ Rn; hx, Cxi = 1} (for the usual inner product in Rn.) (i) Show that M is a compact submanifold of Rn, of dimension n − 1. Hint: M = {x ∈ Rn|F (x) = 1} for a submersion F : Rn → R. Don’t forget to show M is compact!

(ii) Show that the equation of the tangent space Tx0 M, x0 ∈ M, is:

n Tx0 M = {x ∈ R ; hx, Cx0i = 0}.