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Important Theorems in Euclidean Topology

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Important Theorems in Euclidean Topology Important Theorems in Euclidean Topology B. Rogers March 2020 1 Introduction The development of topology has been one of the most important milestones in mathematics of the 20th century. Its roots can be traced back to Euler and his 1736 paper on the Seven Bridges of K¨onigsberg [4]. In this essay I aim to present four of the most influential theorems in Euclidean topology [16], namely the Borsuk-Ulam theorem, the Hairy Ball theorem, the Jor- dan Curve theorem, and the Brouwer Fixed Point theorem. Firstly their history will be mentioned, along with a discussion of their statements and applications. Then I shall provide a wider context for the understanding of the relationships between them, including many other theorems within the scope of topological progress of the last century. This essay assumes some familiarity with basic concepts within topology, especially including homotopy theory. 2 The Borsuk-Ulam Theorem The Borsuk-Ulam theorem was first referenced in Lyusternik and Shnirel'man (1930). It was first proved by Borsuk in 1933, who gave Ulam credit for the formulation of the problem in a footnote. [21, p. 25] The Borsuk-Ulam the- orem states that for every continuous map from a 2-sphere into Euclidean 2-space, there exists a pair of antipodal points. [7, p. 133] Antipodal points are points on spheres that are diametrically opposite, so the straight line connecting them passes through the centre of the sphere. More formally: 2 2 Define a continuous map f: S ! R . Then there exist antipodal points w and -w in S2 such that f(w) = f(−w). The proof follows: First we shall show that we need only concern B2, instead of S2. Define a 2 2 map g : S ! R by g(w) = f(w) − f(−w) w 2 S2 1 We want to show that g must vanish at some point of S2. We will only need to use the following property: g(−w) = −g(w) w 2 S2 2 2 Consider the continuous map g : B ! R defined by, p h(x; y) = g(x; y; 1 − x2 − y2)(x; y) 2 B2 According to this mapping, h is the result of flattening the top half of S2 onto B2. Using the aforementioned property we have, h(−z) = −h(z) z 2 S1 The reason we specify z 2 S1 is that this is the unit circle, and so it the set of points in B2 that contain antipodal points of the half-sphere. It is the equator of the half-sphere. It is therefore sufficient to show that any map 2 2 h from B to R satisfying the previous property vanishes at some point of B2. We can construct a proof by contradiction. Suppose that an h does not vanish on B2. We can construct another map: h(z) jh(1)j φ(z) = z 2 B2 jh(z)j h(1) defines a map with our desired properties. More specifically we have defined a map φ : B2 ! S2 that satisfies, φ(−z) = −φ(z) z 2 S1 φ(1) = 1 Consider the path: α(s) = φ(e2πis) 0 ≤ s ≤ 1 We can show that its index is zero.1. We can obtain a contradiction by showing that its index is odd. Define a loop in B2 by, β(s) = e2πis 0 ≤ s ≤ 1 2 2 Then α = φ(β). Since B is a convex subset of R , α is homotopic (with endpoints fixed) to the constant loop in S1 at 1. In other words, α can be shrunk to the point 1 in S1. We can then deduce that the have the same index, 0. 1Note that index is referred to by many authors as degree For more detailed reasoning as to why the index is zero, see the proof of Brouwer Fixed Point theorem 2 Set k : [0,1] ! R such that, φ(e2πis) = e2πik(s) 0 ≤ s ≤ 1 k(0) = 0 Then ind(α) = k(1): Since φ is an odd function: 1 1 exp[2πik(s + )] = − exp[2πik(s)] = exp[2πi(k(s) + ))] 0 ≤ s ≤ 1 2 2 1 For each fixed s 2 [0; 2 ], the number 1 1 k(s + ) − k(s) − 2 2 must be an integer. As the above number depends continuously on s and has a discrete range, it is constant. Set this constant value equal to an integer m, so: 1 1 k(s + ) − k(s) = m + 0 ≤ s ≤ 1 2 2 2 Then, 1 1 ind(α) = k(1) = k(1) − k( ) + k( ) − k(0); 2 2 which gives an odd integer as shown by 1 1 m + + m + = 2m + 1 2 2 Thus we have obtained a contradiction, so h does vanish on B2, and the theorem is proved. One interpretation of this theorem is that if you take a rubber ball, deflate it, crumple it and lay it flat, then there are two points that were antipodal on the surface of the ball that are now lying on top of each other. Another classic interpretation of the theorem is that at any time, there are two antipodal points on the surface of the Earth with equal temperature and, simultaneously, the same pressure. The single dimensional case is easier to prove. We can construct the odd function in an analogous way. As it is both positive and negative in the domain, it follows that the function vanishes in that domain by the Intermediate Value theorem, and the theorem is proved. Interestingly, the Borsuk-Ulam theorem has a combinatorial analog, that is, a combinatorical statement which is equivalent [23]. It is called Tucker's Lemma and this surprising link between topology and combinatorics is part of a field called topological combinatorics. An interesting implication of this theorem is the Ham Sandwich theorem, and its 2 dimension counterpart, the Pancake theorem. The Ham Sandwich theorem states [7, p. 134]: 3 3 Let U, V, and W be 3 bounded connected open subsets of R . 3 Then there is a plane in R that divides each of the sets into two pieces of equal volume. The Pancake theorem states: 2 Let U and V be 2 bounded connected open subsets of R . Then there is a straight line that divides each of U and V in half by area. 3 The Hairy Ball Theorem The Hairy Ball theorem is the statement that there is no continuous field of non-zero tangent vectors on an even-dimensional n-sphere.[22]The the- orem was first stated by Poincar´ein the 19th century and was proved by Brouwer in 1912. When tangent vectors are visualised as hairs, the state- ment becomes `you can't comb a hairy ball without creating a tuft'. In- terestingly, the single-holed torus is the only compact, orientable surface (2 dimensional manifold) that has a non-vanishing tangent vector field![13] In fact, the Hairy Ball theorem can be thought of as a consequence of the remarkable Poincar´e-Hopftheorem. This shows that the sum of the indices of a vector field on a compact, differentiable manifold over all of its (iso- lated) zeros is independent of which vector field is considered.[7, p. 188] Importantly, the number (which must be an integer) is equal to the Euler characteristic of the manifold. The most applicable consequence of this to the Hairy Ball theorem is that if you have a compact, differentiable manifold with a non-vanishing tangent vector field, its Euler characteristic must be zero.[27] Examples of manifolds with zero Euler charactertic are the torus, odd dimension n-spheres, the M¨obiusloop, and Klein bottle. All of these shapes have continuous non-vanishing tangent vector fields,(the latter 2 are non-orientable surfaces, so the torus is the only orientable surface with the desired property) which are fairly easy to imagine (apart from maybe the case of the Klein bottle).[14]. There are many proofs of this famous theorem, which were formulated at various points in time. Now I present a sketch of Milnor's analytic proof of the Hairy Ball theorem[22], as I believe that this is the easiest to follow without a foundation in homotopy theory. Theorem 1: An even-dimensional sphere does not possess any continuously differentiable field of unit tangent vectors. n−1 The sphere S is trivially the set of all vectors u = (u1; :::un) in Euclidean n n space R such that the Euclidean length jjujj = 1. A vector v(u) in R is 4 Figure 1: A failed attempt to comb Figure 2: A successful attempt to a hairy ball [5] comb a torus [6] tangent to Sn−1 at u if the Euclidean inner (dot) product u · v(u) equals zero. We can now show that if n − 1 is odd, it is possible to define a differentiable field of unit tangent vectors on Sn−1, by v(u1; :::; un) = (u2; −u1; :::; un; −un−1) Note that (u2; −u1; :::; un; −un−1)·(u1; :::un) = u2u1 −u1u2 +:::+unun−1 −un−1un = 0 This is why the Hairy Ball theorem is specified only for even-dimensional n-spheres; when n-1 is even, n is odd so in the above equation we would not be left with zero. n Let A be a compact region in R and let x ! v(x) be a continuously differ- entiable vector field which is defined throughout a neighbourhood of A. For each real number t, consider ft(x) = x + tv(x) which is defined for all x in A. Lemma 1: If the parameter t is sufficiently small, then this mapping ft is one-to-one and transforms the region A onto a nearby region ft(A) whose volume can be expressed as a polynomial function of t.
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