SMSTC (2017/18) Geometry and Topology 2.

SMSTC (2017/18) Geometry and Topology 2 Lecture 1: Differentiable Functions and Manifolds in R n Lecturer: Diletta Martinelli (Notes by Bernd Schroers) a wwwsmstcacuk 11 General remarks In this lecture we review properties of differentiable functions R n R m and explain how they can be used to define parametrised surfaces and (sub)manifolds in R n The two key results are the inverse mapping theorem and the implicit function theorem We will prove both and illustrate them with examples The proof of the inverse mapping theorem requires the contraction mapping theorem in metric spaces, which is an important tool in many branches of analysis We therefore prove it here as well One of messages of this lecture is that differentiating a function f : R n R m is more than computing partial derivatives Differentiation of a function at a point p means finding the best linear approximation at p The linear approximation itself is a linear map f (p) : R n R m (or, as you shall see in later lectures, more precisely a linear map of tangent spaces T p R n T f(p) R m ) The linear map can be represented by a matrix of partial derivatives but it is the properties of the linear map which determine the properties of f sufficiently close to p The material in this lecture takes up an unfortunate position in the university curriculum in the UK and the USA: it is usually omitted from undergraduate courses because it is considered too abstract and difficult, and then, if at all, taught in the context of differentiable manifolds at the graduate level Here we steer an intermediate course, formulating and proving the inverse mapping and implicit function theorems for functions on R n R m There are relatively few books that cover the material at the same level, but [1] and [2] are notable exceptions 12 Differentiable maps R n R m Notation: We denote vectors in R n by ordinary Roman letters x, y, and write x for the norm (usually the Euclidean norm) We denote the components of x R n by x 1,, x n We use column or row notation for vectors based on notational convenience The idea of differentiation is that of approximation by a linear function: Definition 11 Let U be an open set in R n Then f : U R m is called differentiable in a U if there is a linear map L : R n R m and a function r : U R m which is continuous in a and vanishes at a so that for all x U a DilettaMartinelli@edacuk f(x) = f(a) + L(x a) + r(x) x a (11) 1 1

SMST C: Geometry and Topology 2 1 2 y f y=f(a)+f (a)(x a) a x Figure 11: Differentiation as approximation by a linear map The linear map L is uniquely determined by this condition (check it!) and is called the derivative of f at a, written f (a) or df a While all terms in the equation (11) are well-defined, the definition does not state which vector spaces the linear maps L acts on and takes its value in The full answer to this question is that the map L is really a map of tangent spaces L : T a R n T f(a) R m A careful definition of tangent spaces would take us to far afield at the moment, but in the current context we can think of T a R n simply as the vector space obtained from the affine space R n by identifying the point a as the origin, ie, as the space of vectors beginning at a In order to determine if a function is differentiable we need to have a candidate linear map L and then we need to check if satisfies r(x) = f(x) f(a) L(x a) x a The candidate linear map we use is given by the Jacobian matrix (12) lim r(x) = 0 (13) x a (a) f m (a) x n (a) f m x n (a) Warning:: The function f may not be differentiable in a point a even if the Jacobian matrix exists there - see the Exercises for this lecture! If the Jacobian matrix exists and if (12) satisfies the requirement (13) then we write (a) f (a) = f n (a) x n (a) f n x n (a)

SMST C: Geometry and Topology 2 1 3 Example 11 With x(r, θ) = (r cos θ, r sin θ), find x where it is defined Compare r with r One can show that continuous partial differentiability is a sufficient condition for differentiability: Lemma 11 If all first partial derivatives of f : U R m exist and are continuous in a, then f is differentiable in a and f equals the Jacobian matrix at a Most of the results for differentiation of functions f : R R hold for maps f : R n R m but need to be suitably interpreted For real numbers α, β and differentiable functions f, g : R n R m we have (αf + βg) = αf + βg If we have a product on R n (for example the scalar product, or the vector product in the case of n = 3), the product or Leibniz rule hold, but both sides now need to be interpreted as linear maps: (fg) (a) = f (a)g(a) + f(a)g (a) In the chain rule for f : R n R m and g : R k R n the right hand side has to be read as matrix multiplication: (f g) (a) = f (g(a))g (a) Finally, we have Lemma 12 If a function f : U R n is differentiable in a U with invertible derivative f (a) and has an inverse f 1 which is continuous in b = f(a), then the inverse is also differentiable in b, with derivative ( f 1 ) (b) = ( f (a) ) 1, where the right hand side is the matrix inverse Example 12 Note that the existence and non-vanishing of f at a point is not sufficient for local invertibility You are asked to explore this in one of the exercises for this lecture by considering the function { 0 if x = 0 f(x) = x + x 2 cos 1 x if x 0 The function f is differentiable in x = 0 and f (0) 0 but that there is no neighbourhood U of 0 where f is invertible Recall the mean value theorem in R: Theorem 11 Suppose the function f : [a, b] R is continuous and differentiable in (a, b) Then there is a c (a, b) so that f f(b) f(a) (c) = b a The theorem holds in a modified form for real valued functions on R n : Theorem 12 If f : U R is differentiable and the line ab joining a, b U lies in U, then there is a point c on that line so that f(b) f(a) = f (c)(b a)

SMST C: Geometry and Topology 2 1 4 For a function f : U R m one can still apply the mean value theorem to each component f j of f, getting m points c j on the line ab where f j (b) f j (a) = f j (c j)(b a) If one has a bound B on the matrix norm of f (c) which is uniform on the line ab then one can deduce a useful inequality: sup c ab f (c) B, f(b) f(a) B b a (14) 13 Banach contraction mapping theorem As preparation for the proof of the inverse mapping theorem we prove a very general and powerful result of maps in metric spaces Definition 12 A metric on a set X is a mapping d : X X R + 0 with d(x, y) = d(y, x) symmetry (15) d(x, y) = 0 x = y definiteness (16) d(x, y) d(x, z) + d(z, y) triangle inequality (17) for all x, y, z X Definition 13 A metric space (X, d) is complete if every Cauchy sequence has a limit in (X, d) Example 13 R n with metric d(x, y) = x y T(x 0 ) x2 x1 x0 x Figure 12: The sequence (x n ) used in the contraction mapping theorem Definition 14 Let (X, d) be a metric space A mapping T : X X is said to be a contraction mapping if there is a k (0, 1) so that for all x, y X d(t x, T y) kd(x, y) The constant k is called a Lipschitz constant Note that contraction mappings are continuous

SMST C: Geometry and Topology 2 1 5 Theorem 13 (Banach contraction mapping theorem) Let (X, d) be a complete metric space and T : X X a contraction mapping with Lipschitz constant k Then 1 For any x 0 X, the sequence (x n ) defined via converges x n = T (x n 1 ), n N 2 All sequences (x n ) defined in this way have the same limit x (independent of x 0 ) and x is the unique solution of x = T (x) It is called the fixed point of T Proof 1 Recalling we have the inequality m l=1 k l = 1 km+1 1 k 1 1 k d(x n+m, x n ) d(x n+m, x n+m 1 ) + + d(x n+1, x n ) (k n+m 1 + k n )d(x 1, x 0 ) kn 1 k d(x 1, x 0 ) Hence the sequences (x n ) are Cauchy sequences for any x 0 X The completeness of X then guarantees the existence and uniques of the limit for each of these sequences 2 Fix a x 0 and denote the limit of (x n ) by x Then d(x, T (x )) d(x, x n ) + d(x n, T (x )) d(x, x n ) + kd(x n 1, x ) 0 for n Hence T (x ) = x and thus x is a fixed point of T Let x be another fixed point Then d(x, x) = d(t (x ), T ( x)) kd(x, x) (1 k)d(x, x) 0 Since (1 k) is positive, we deduce that d(x, x) = 0 and hence x = x The uniqueness of the fixed point implies that all sequences (x n ) have the same limit 14 The inverse mapping theorem Theorem 14 Let U R n be open and f : U R n continuously differentiable Assume that for a U, the linear map f (a) : R n R n is invertible, ie, det f (a) 0 Then there exists a neighbourhood V of a such that 1 f V is injective, 2 f(v ) is open, 3 The inverse function g of f V is continuously differentiable Proof Assume without loss of generality that a = 0, f(a) = 0 and f (0) = id This can always be achieved by multiplying f from the left with the matrix f (0) 1

SMST C: Geometry and Topology 2 1 6 We define the map φ y : U R n via φ y (x) = x f(x) + y, and note that every fixed point of φ y is an inverse image of y contraction mapping theorem, and need to check its conditions We would like to apply the First we need to show that φ y maps a complete metric space to itself Thus we are looking for a closed ball B(0, 2r) of radius 2r around the origin so that for every y B(0, r), the map φ y maps B(0, 2r) to itself and is contracting Using f (0) = id and the continuity of f we conclude that there is a r > 0 so that for all x B(0, 2r) id f (x) 1 2 The inequality (14) deduced from the mean value theorem applied to φ 0 (x) = x f(x) then gives Thus, by the triangle inequality so that φ y : B(0, 2r) B(0, 2r) as claimed x f(x) 1 2 x φ y (x) x f(x) + y 2r, Next we need to establish that φ y is a contraction mapping Using again the inequality (14) for φ y and x 1, x 2 B(0, 2r) we deduce φ y (x 2 ) φ y (x 1 ) 1 2 x 1 x 2, so that φ y is indeed a contraction and has a unique fixed point x, ie, for each y B(0, r) there is a unique x B(0, 2r) so that f(x) = y We thus know that f maps the preimage of B(0, r) bijectively to B(0, r) By the continuity of f this preimage is an open set; we denote it by V To show, finally, that the inverse map g of f V is continuously differentiable we only need to show that it is continuous, by Lemma 12 To show continuity we observe implies φ 0 (x 1 ) = x 1 f(x 1 ), φ 0 (x 2 ) = x 2 f(x 2 ), x 1, x 2 B(0, 2r) or, by the same inequality as above, so that, with y i = f(x i ), i = 1, 2, x 2 x 1 = φ 0 (x 2 ) φ 0 (x 1 ) + f(x 2 ) f(x 1 ) x 2 x 1 1 2 x 2 x 1 + f(x 2 ) f(x 1 ) g(y 2 ) g(y 1 ) 2 y 2 y 1, which implies continuity of g and completes the proof

SMST C: Geometry and Topology 2 1 7 15 The implicit function theorem We are interested in systems of equations of the form f 1 (x 1,, x m, u 1,, u p ) = 0 f m (x 1,, x m, u 1,, u p ) = 0, (18) and would like to know if we can express x 1,, x m as functions of u 1,, u p We write D for an open subset of R m R p, and combine the functions f 1,, f m into one function f : D R m We write f (1) for the derivative of f with respect to the first m variables x 1,, x m, so f (1) is represented by the Jacobian matrix f m Theorem 15 (Implicit function theorem) Assume that D R m R p is open and that f : D R m is continuously differentiable Assume further that x m f m x m f(a, b) = 0 and det f (1) (a, b) 0 Then there exists a neighbourhood U of b and a uniquely determined continuously differentiable map ϕ : U R m so that, for all u U, f(ϕ(u), u) = 0 Proof Define the function F : D R m R p via F (x, u) = (f(x, u), u) One checks that F is continuously differentiable, with Jacobian and ( f F = (1) f (2) ) 0 id R p detf = detf (1) The map F satisfies the assumptions of the inverse mapping theorem, so for each (a, b) D there is a neighbourhood W so that F W has a continuously differentiable inverse G : F (W ) W The equation F (x, u) = (y, v) is then equivalent to G(y, v) = (x, u) with x = g(y, v), u = v,

SMST C: Geometry and Topology 2 1 8 where g : R m R p R m is continuously differentiable We claim that ϕ(v) = g(0, v) is the required map Indeed, we have F G =id, so or, with y = 0 and u = v, f(g(y, v), v) = y f(ϕ(u), u) = 0 This is the only solution of f(x, u) = 0, since g(f(x, u), u) = x with f(x, u) = 0 implies g(0, u) = ϕ(u) = x The derivative of the map ϕ can be computed directly from f(ϕ(u), u) = 0 Using the chain rule, one finds ϕ (u) = f (1) (ϕ(u), u) 1 f (2) (ϕ(u), u) 16 Submanifolds of R n We begin by considering parametrised p-dimensional surfaces in R n (p < n): Definition 15 Let U R p be an open set and σ : U R n a continuously differentiable map If σ has rank p (ie, if it is injective), we call σ a parametrised p-surface in R n Definition 16 A parametrisation σ : V R n is equivalent to the parametrisation σ : U R n if there is a diffeomorphism φ : V U so that σ = σ φ Example 14 The stereographic parametrisation of the 2-sphere σ(u 1, u 2 ) = 1 1 + u 2 1 + (2u 1, 2u 2, 1 u 2 u2 1 u 2 2) (19) 2 (projection from the South pole) can be traded for σ(x 1, x 2 ) = (x 1, x 2, 1 x 2 1 x2 2 ) (110) on the upper hemisphere - see Fig 13 Definition 17 Let g : R n R n p be continuously differentiable and assume that g has maximal rank n p everywhere Then we call the set M := {x g(x) = 0} of zeros of g a p-dimensional submanifold of R n (assuming that M is not the empty set)

SMST C: Geometry and Topology 2 1 9 S 2 p= σ (u) u σ Figure 13: Parametrising the the sphere Note that all level sets {x f(x) = c} for a constant c R n p are included in this definition: we simply take g = f c Example 15 We have already encountered the 2-sphere of radius r: S 2 = {x R 3 x 2 1 + x 2 2 + x 2 3 = r 2 } Lemma 13 Every p-dimensional submanifold of R n can locally be described by a parametrised p-surface in R n Proof This is a direct consequence of the implicit function theorem 17 Sard s theorem In defining a submanifold as a level set of a function f : R n R n p we need to check the maximal rank requirement on the derivative f Sard s theorem tells us that if we pick a value c in the image set of a suitably smooth f : R n R n p at random then the level set {x f(x) = c} will almost certainly be a submanifold To formulate the theorem we need some terminology Definition 18 For a differentiable function f : U R n R m with m < n, a point a U is called a critical point if the rank of f (a) is smaller than the maximal rank m The set of critical points is called the critical set of f Theorem 16 (Sard s Theorem) Let f : R n R n p be a k-times differentiable function, where k max{p + 1, 1} Write X for the critical set of f Then f(x) has Lebesgue measure 0 in R n p

SMST C: Geometry and Topology 2 1 10 Exercises 1H 1H1 (a) Consider the function f : R 2 R 2 given by 0 if (x 1, x 2 ) = (0, 0) f(x 1, x 2 ) = x 1 x 2 if (x 1, x 2 ) (0, 0) x 2 1 +x 2 2 Show that both partial derivatives exist at (x 1, x 2 ) = (0, 0) but that the function is not differentiable there (b) Prove Lemma 12 (c) (i) Suppose a function f : U R n R m is such that all partial derivatives exist and are continuous in U Show that f is then continuously differentiable in U (ii) Let (a, b) be an open interval and x 0 (a, b) Suppose a continuous function f : (a, b) R is differentiable in (a, b) \ {x 0 }, and the limit lim x x 0 f (x) exists Show that the function is then differentiable in x 0 and that f (x 0 ) = lim x x 0 f (x), so that the function is in fact continuously differentiable at x 0 (iii) Consider now a continuous function U R n R m and a point x 0 U Suppose that all the partial derivatives exist and are continuous in U \ {x 0 }, and that, moreover, the limits of all partial derivatives exist as x x 0 Show that the function is then continuously differentiable in all of U (iv) Use (iii) to show that the function f : R 2 R 2 given by ( ) x2 x f(x) = 1 r, r = x x 1 x 2 r 2 1 + x2 2, is continuously differentiable in all of R 2, and compute its derivative at the origin (d) Consider the function f(x) = { 0 if x = 0 x + x 2 cos 1 x if x 0 Show that f is differentiable in x = 0 and that f (0) 0 but there is no neighbourhood U of 0 where f is invertible Why does the inverse mapping theorem not apply? (e) Suppose that the cubic equation t 3 y 1 t 2 + y 2 t y 3 = 0 has three distinct real zeros when (y 1, y 2, y 3 ) = (b 1, b 2, b 3 ) R 3 Deduce that the equation has three distinct real zeros for (y 1, y 2, y 3 ) in a sufficiently small neighbourhood of (b 1, b 2, b 3 ) Hint: Write the coefficients of a cubic polynomial as a function of the zeros and use the inverse mapping theorem (f) What is bigger, e π or π e? Justify your answer with a full proof

SMST C: Geometry and Topology 2 1 11 References [1] Michael Spivak, Calculus on Manifolds, W A Benjamin, New York 1965 [2] Tom M Apostol, Mathematical Analysis, Addison-Wesley, 1974