INVERSE THEOREM and SURFACES IN Rn

Let f ∈ Ck(U; Rn), with U ⊂ Rn open (k ≥ 1). Assume df(a) ∈ GL(Rn), where a ∈ U. The says there is an open n n neighborhood V ⊂ U of a in R so that f|V : V → R is a onto its W = f(V ), an open 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 : f(x) = y. The idea is that if f is close (in some sense) to a linear 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 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 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 , 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 A will satisfy a Lipschitz condition only in some open 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 . 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). Exercise 1. Let f : R → R have the form f(x) = ax + g(x), where a > 0 and g is Lipschitz on R, with constant M satisfying M < a. (i) Show that f is injective. (ii) Show that f is surjective. Remark use only the definitions, quoting no theorems other than the fixed point theorem for contractions in part (ii).

A property of C1 maps. Let f : U → Rm be C1, U ⊂ Rn open. Given a ∈ U, we have the remainder:

ra(x) = f(x) − f(a) − df(a)[x − a], x ∈ U.

n m Then dra(x) = df(x) − df(a) is continuous from U to L(R ,R ), and van- ishes at x = a. We conclude that, for any  > 0, there exists an r > 0 so that ||dra(x)|| ≤  if x ∈ Br(a). By the Mean Value Inequality:

||ra(x) − ra(y)|| ≤ ||x − y||, for x, y ∈ Br(a).

In other words: ra is a Lipschitz map with arbitrarily small Lipschitz con- stant, in a sufficiently small neighborhood of a. Equivalently:

r (x) − r (y) lim a a = 0. x→a,y→a ||x − y||

Note that this is equivalent to saying that, for all x, y ∈ Br(a)(setting ra(x, y) = ra(x) − ra(y)): r( x, y) f(y) − f(x) = df(a)[y − x] + r(x, y), lim a = 0. x→a,y→a ||x − y||

Remark: Note that only continuity of df at a is used. Conversely, if this condition holds at a ∈ U, then df is continuous at a. To see this, combining: f(x) − f(y) = df(a)[x − y] + ra(y, x), f(y) − f(x) = df(x)[y − x] + rx(y),

3 we find: (df(x) − df(a))[y − x] = −[ra(y, x) + rx(y)]. Given  > 0, we find ρ > 0 and for each x ∈ U, 0 < ηx < ρ so that:

||ra(x, y)|| ≤ ||y−x|| for x, y ∈ B2ρ(a); ||y−x|| ≤ ηx ⇒ ||rx(y)|| ≤ ||y−x||.

Let x ∈ U satisfy ||x−a|| ≤ ρ, and let u ∈ Rn be a unit vector. Then letting y = x + ηxu we have: x, y ∈ B2ρ(a) and ||y − x|| ≤ ηx, so

n ηx||(df(x)−df(a))[u]|| ≤ ||ra(x, y)||+||rx(y)|| ≤ 2ηx, ∀u ∈ R with ||u|| = 1.

Thus, for the operator norm in L(Rn; Rm):

||df(x) − df(a)|| ≤ 2, establishing continuity of df at a.

Turning to the case m = n, let f ∈ C1(U; Rn), with df(a) ∈ GL(Rn) (that is, df(a) is an invertible linear map) for all a ∈ U. Then: f(x) = f(a)+df(a)[x−a]+ra(x) = df(a)[x]+φ(x), φ(x) = f(a)−df(a)[a]+ra(x), so φ is Lipschitz in U if and only if ra is (with the same Lipschitz constant.) Thus φ has arbitrarily small Lipschitz constant in a sufficiently small ball at a, assuming f ∈ C1. This motivates the hypothesis in the statement of the Inverse Mapping Theorem.

Inverse Mapping Theorem. Let f : U → Rn (U ⊂ Rn open) be of class C1 in U. Let a ∈ U, and assume df(a) ∈ GL(Rn) (that is, df(a) is an .) 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 differentiable in W , and dg(y) = [df(g(y)]−1, for all y ∈ W ; (iii) If f ∈ Ck(U; Rn) (with k ≥ 1) then g ∈ Ck(W ; Rn). 1 Proof. (i): Since f is C , 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 per- turbation of the isomorphism df(a). Conclusion (i) follows from Proposition 1.

4 Prior to proving (ii), we consider a general result on differentiability of the inverse (of a homeomorphism). Note that the inverse of the homeomor- phism 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 homeomor- phism, where U, V are open in Rn, with inverse g : V → U. If f is differ- entiable at a ∈ U and df(a) ∈ GL(Rn), then g is differentiable at b = f(a), with dg(b) = [df(a)]−1. 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 that |f(a + v) − f(a)|/|v| is bounded below, for |v| sufficiently small, if df(a) is an isomorphism: |f(a + v) − f(a)| v |r (v)| c ≥ |df(a)[ ]| − a ≥ 0 , |v| |v| |v| 2 where |df(a)u| ≥ c0 > 0, if |u| = 1. Proof (ii) in the IMT: If f ∈ C1(U, Rn), since df : U → L(Rn) is continuous, df(a) ∈ GL(Rn) and GL(Rn) is open in L(Rn), taking the neighborhood V ⊂ U of a from part (i) small enough we guarantee df(x) ∈ GL(Rn) for x ∈ V . Then the lemma on differentiability of the inverse (just proved) guarantees g = f −1 isdifferentiable in W = f(V ), and dg is as claimed. Before proving (iii) we must take another short detour. Lemma 1. The inversion map f(X) = X−1 is smooth from GL(Rn) to L(Rn) ∼ M(n)(n × n matrices.)

5 Proof. First, f is continuous at any X ∈ GL(Rn), since it is differen- tiable, with df(X)[V ] = −X−1VX−1. We know the directional 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 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 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) and g is Ck−1, it follows dg is Ck−1 (or g is Ck.) This proves g is Ck inductively. Clearly this Corollary implies part (iii) of the IMT. 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) 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 2. Generalize Corollary 2 to maps of the form f(x) = Ax + φ(x), where A ∈ GL(Rn) and φ ∈ C1(U; Rn). (That is, state a precise theorem and prove it.)

6 Theorem (for maps). 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 of f through a: M = {x ∈ U; f(x) = b}, where b = f(a) ∈ Rn. The Implicit Function Theorem (for maps) says that, locally in a neigh- borhood 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 df is continuous at a ∈ U, with df(a) ∈ L(Rm,Rn) surjective (and m > n). There exists an open neighborhood V ⊂ U of m 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)) ∈ K × E; 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 ).

Proof. (i) 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 in U, with differential continuous at a, given by: 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, and then df(a)[v] = 0 implies v = 0 (since the restriction of df(a) to {0} × E is an isomorphism to Rn.)

7 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.

(ii) 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) = (a1, f(a2)) = (a1, b).) k k 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.

(iii)To compute the differential of h at x1 ∈ W1, note that, with G : W1 → m R ,G(x1) = (x1, h(x1)), from f ◦ G ≡ b in W1 we have:

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 (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), concluding the proof. 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}. Informally, the theorem says that the preimage of a regular value is locally a graph. 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 f : X → Y 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.

8 Proposition 2. Let f : U → Rn be C1 (U ⊂ Rn open, m > n). (i) If df(a) ∈ L(Rm; Rn) is surjective, then there exists a neighborhood n V of a so that the restriction 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¯, where π2 : W1 × W2 → W2 is projection onto the n ¯ 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 ). Exercise 3. (i) If m < n, the set of T ∈ L(Rm; Rn) so that Ker(T ) = {0} is open in L(Rm; Rn) (with respect to the operator norm). Hint: If Ker(T ) = {0}, inf{||T v||; ||v|| = 1} = c0 > 0 (why?) So if ||S − T || < c0/2... (ii) If m > n, the set of T ∈ L(Rm; Rn) so that Ran(T ) = Rn is open in L(Rm; Rn) (with respect to the operator norm). n Hint: Ran(T ) = R if, and only if, we can find n unit vectors v1, . . . , vn m in R , so that T v1, . . . , T vn are linearly independent. If this is the case, Pn inf{||c1T v1 + ... + cnT vn||; i=1 |ci| = 1} = m0 > 0 (why?) Now assume ||S − T || < m0/2n, and show that c1Sv1 + ... + cnSvn = 0 (where we may Pn assume i=1 |ci| = 1–why?) is impossible unless all ci = 0. Images of Ck parametrizations. A Ck map (k ≥ 1) f : U → Rn (U ⊂ Rm 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. Recall that this last condition is equivalent to: for any V ⊂ U open, f(V ) = M ∩ W , for some W ⊂ Rn open (that is, f is an open map).

9 n k Proposition 3. Any image M = φ(W0) ⊂ R of a C parametrization n m k φ : W0 → R (W0 ⊂ R open, m < n) is locally a graph (of a C map defined on an open subset of an m-dimensional , with values in an n − m-dimensional one.) That is, each point p ∈ M admits a neighborhood W ⊂ Rn (open in Rn), so that M ∩ W is the graph of a Ck map.)

Proof. Let p ∈ M, p = φ(x0), x0 ∈ W0. Consider the subspace Tp = m n n dφ(x0)[R ] ⊂ R , and a complementary subspace Np ⊂ R , which defines a n n direct sum decomposition R = Tp ⊕ Np; let π : R → Tp be the associated projection. (Note dim(Tp) = m, since dφ(x0) is injective.)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. Note that (by the homeomorphism condition in −1 the definition of parametrization) U = W ∩ M, where W ⊂ π (Dp) is open n −1 in R . Consider g : Dp → U given by g = φ|V0 ◦ F . Since g is injective, surjective, continuous and open, it is a homeomorphism (and of class Ck, as k −1 the composition of C maps). And π◦g = (π◦φ|V0 )◦F = idDp shows that, m with respect to the splitting R = Tp ⊕ Np, g has the form g(x) = (x, h(x)), k k for a C map h : Dp → Np. Thus g is a C graph parametrization (with domain Dp) of the set U = W ∩ M. This concludes the proof of Proposition 3. Exercise 4. 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)}

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. (m < n) 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

10 each p ∈ M admits a neighborhood U ⊂ M (in the topology induced from n k m R ) 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. 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 ⊂ Rn open) so that for some Ck submersion F : W → Rp (where p = n − m) we have U = F −1(0). (ii) For each p ∈ M we may find a direct sum decomposition Rn = k Tp ⊕ Np, U1 ⊂ Tp open, a C map h : U1 → Np and a neighborhood W of p n in R 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.)

11 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. 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 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 neigh- borhood 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)),

12 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. 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 5. (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 con- tained in the set of velocity vectors of curves on M through p is clear. To prove that any velocity vector of a 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,

13 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 −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;

14 (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 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 submanifold of Rn, of dimension n − 1. Hint: M = {x ∈ Rn|F (x) = 1} for a submersion F : Rn → R.

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

n Tx0 M = {x ∈ R ; hx, Cx0i = 0}.

15