The Topological Degree and Applications Tyson Liddell Supervisor: Dr J´er^omeDroniou Monash University February 21, 2015 1 1 Abstract In many areas of mathematics and the applied sciences, numerous problems boil down to solving equations that are often nonlinear. The Brouwer degree is a power tool used in analysis and topology that, under certain conditions, can guarantee solutions to nonlinear equations exist. This report will present a construction of the Brouwer degree and list a few of its properties. Some applications of the Brouwer degree are then presented; in particular, a new result that pertains to the orientation preserving structure of derivatives of continuous injective maps. 2 Introduction 2.1 Topological Degree n n Consider some open, bounded subset Ω ⊂ R and any continuous map f : Ω ! R . In many prob- lems in this general setting we are often posed with a question that reduces to knowing something n about the equation f(x) = y for a given y 2 R . Specifically, often it would be very useful to know immediately whether or not such an equation has any solutions x 2 Ω. For exactly this purpose we introduce a tool called the topological degree which, for the sake of brevity, from this point forward we may refer to as the degree. As we shall see, the degree can provide us with knowledge of existence of solutions to the previous equation. Of course, it may not be immediately obvious as to why this is useful. Consider the following problems. Given a square matrix with non-negative entries, does it have n a real eigenvalue? Does a continuous map f : A ⊂ R ! A, where A is compact and convex, have a fixed point? Are certain continuous map surjective? As we will show, all of these classical questions can be answered with relative ease using the degree and the fact that it gives a definite answer to the existence of solutions to the earlier equation. As a further example, the reader may also recall the Jordan curve theorem, paraphrased by the intuitively obvious statement that a closed curve in 2 R that does not intersect itself partitions the plane into two parts, an `inside' and `outside'. This is surprisingly difficult to show by elementary means; however, using the degree it can be proved it with little effort. The origin of the degree lies in algebraic topology, however we will use principally analytic arguments with some elementary topology. 2.2 Outline We will begin this report by providing further motivation and intuition for the degree by touching on the one-dimensional case. Following this we construct the topological degree for Euclidean space, known as the Brouwer degree, from which we will then demonstrate its use by solving some of the earlier mentioned problems. Up until this point we draw strongly from the work in [Dei85]. We then present an original result pertaining to the orientation preserving structure of derivatives of continuous injective maps. 2.3 Notation A full notation reference is provided in Appendix A, however we take this opportunity to mention some of what will be covered. Unless otherwise explicitly stated, Ω will denote an open and bounded n subset of R . The set of all continuous (or k-times differentiable with continuous derivatives up to 2 n order k respectively) functions f :Ω ! R will be denoted by C(Ω) = C0(Ω) (or Ck(Ω) respectively). 1 k Similarly, the notation C (Ω) will be used to denote the set \k2NC (Ω). For differentiable functions k on compact sets we define for any k 2 N [ f0g the set of all restricted functions C (Ω) = ffj Ω : n n f 2 Ck(R )g. We will use j · j to denote the Euclidean norm and for any f 2 C(S) where S ⊂ R is compact we use the notation jfj0 = maxfjf(x)j : x 2 Sg. The open and closed balls of radius r centered at x will be denoted by Br(x) and Br(x) respectively. The notation `id' will be used n n n for the identity map id : R ! R which may also be restricted to a domain of a subset of R n n dependent on the context. For any y 2 R and A ⊂ R the distance of y to A will be denoted %(y; A) = inffjy − aj : a 2 Ag. For a function f : A ! B and some y 2 B, the inverse image f −1(fyg) = fx 2 A : f(a) = yg will be denoted by the obvious shorthand f −1(y). n A function f 2 Ω ! R is differentiable at a 2 Ω if and only if there exists a linear map n n n n n f 0(a): R ! R and remainder r : R ! R such that for all h 2 R satisfying a + h 2 Ω we have f(a + h) = f(a) + f 0(a)h + r(h) 0 with limh!0 jr(h)j=jhj = 0. In this case f (a) is unique and we denote the Jacobian at a by 0 −1 n Jf (a) = det(f (a)). Furthermore, if f (y) = fx1; : : : ; xmg, we call y 2 R a regular value of f if Jf (xi) 6= 0 for i = 1; 2; : : : ; m. However, if Jf (xi) = 0 for some i, then y is called singular value −1 and the corresponding xi a critical point of f. This definition implies that if f (y) = ; then y is a regular value of f. 3 The degree in one dimension Let us motivate the basic idea of the topological degree with the following simple example. 0:2 0:4 0:6 0:8 1 Figure 1: Graph of an arbitrary function satisfying the assumptions of Proposition 3.1 . Proposition 3.1. Let f : [0; 1] ! R be a continuously differentiable function with f(0) 6= 0, f(1) 6= 0 and let A = f −1(0) be the set of zeroes of f. If for all x 2 A we have f 0(x) 6= 0, that is 0 is a regular value of f, then A is a finite set and 1 X (sgn f(1) − sgn f(0)) = sgn f 0(x) (1) 2 x2A 3 where sgn : R ! {−1; 0; 1g denotes the sign function defined by 8 −1; x < 0 <> sgn(x) = 0; x = 0 : :>1; x > 0 Proof. To show that A is finite, let us assume for a contradiction that A is an infinite set. Those familiar with basic analysis will immediately recognise that [0; 1] is a compact set and as such A ⊂ [0; 1] must have an accumulation point γ 2 [0; 1]. That is there exists a sequence (xn) in A nfγg such that limn!1 xn = γ. As f(xn) = 0 for all n 2 N, the continuity of f implies that 0 f(γ) = f(limn!1 xn) = limn!1 f(xn) = 0 and so γ 2 A. Furthermore, by assumption f (γ) 6= 0 and so, by the Inverse Function Theorem (B.4), there exists a neighbourhood Uγ of γ such that the restriction fjUγ : Uγ ! f(Uγ ) is a diffeomorphism. Therefore, as fjUγ is injective we must have f(x) 6= 0 for all x 2 Uγ nfγg. However, this is a contradiction as γ is an accumulation point of A and thus at least one xn belongs to Uγ n(γ). Therefore, A is a finite set as stated. We now show that (1) holds. If A = ; then we must have sgn f(α) = sgn f(β) 2 {−1; 1g for all α; β 2 [0; 1], otherwise, by the intermediate value theorem, we would have A non-empty. Therefore, clearly (1) is true using the empty sum convention. Now for the case where A is non-empty, let us order the elements of A = fa1; : : : ; ang so that a1 < a2 < : : : < an. By the continuity of f, for each disjoint interval I 2 f[0; a1), (a1; a2);:::; (an; 1]g there exists cI 2 {−1; 1g such that for all x 2 I, sgn f(x) = cI by the same argument we used for the case of the empty set. Furthermore, 0 0 by differentiability, f(ai + h) = f(ai) + hf (ai) + r(h) where for small h, jhf (ai)j > jr(h)j because 0 0 f (ai) 6= 0. Therefore, f(ai + h) = hf (ai) + r(h) and for small h, by the previous inequality, 0 0 sgn f(ai ± h) = sgn[±hf (ai)] = ± sgn[hf (ai)]. In other words, sgn f(ai + h) = − sgn f(ai − h). What we have shown is the obvious fact that the sign of f must alternate at each zero and remain 0 constant everywhere else as our intuition tells us. It is also obvious that sgn f (ai) indicates how the sign of f changes across the zero ai. Using this fact, if we let α1 = 0; α2 2 (a1; a2); α3 2 (a2; a3); : : : ; αn 2 (an−1; an); αn+1 = 1 and apply telescopic summation we obtain n n X X 0 sgn f(1) − sgn f(0) = (sgn f(αi+1) − sgn f(αi)) = 2 sgn f (ai) i=1 i=1 and rearranging gives the result at (1). This completes the proof. On first glance Proposition 3.1 may seem obvious and its proof a waste of mathematical machin- ery. However, we will shortly generalise this idea to continuous functions on arbitrary Euclidean n n spaces (R ! R ) where we are not so well served by our intuition alone.