MATH 18.095 Recall That a Group Is a Set G Together with an Operation

MATH 18.095 Recall That a Group Is a Set G Together with an Operation

MATH 18.095 SQUARE DANCE PROBLEM SET TARA HOLM 1. GROUPS Recall that a group is a set G together with an operation · that satisfies • Closure: g · h is again an element of G (g · h 2 G); • Associativity: g · (h · k) = (g · h) · k; • Identity: there is an element e such that e · g = g = g · e; and • Inverses: for each g 2 G there is an h so that g · h = e = h · g. EXERCISES ON GROUPS Problem 1. Show that the set of integers Z with operation addition is a group. Problem 2. Show that the set of integers Z with operation multiplication is NOT a group. 1 2 Problem 3. Show that the circle S (thought of as angles in R ) with operation angle addition is a group. 2. TOPOLOGICAL SPACES A bijection is an exact pairing between the elements of two sets X and Y.A homeomorphism between two topological spaces X and Y is a continuous bijection f : X Y such that f-1 : Y X is also continuous. We say two spaces X and Y are homeomorphic if there is a homeomorphism f : X Y. ! ! π π Example 1. The function f(x) = tan(x) from the open interval X = - 2 ; 2 to the real numbers ! -1 Y = R is a homeomorphism. Indeed, tan(x) is a continuous function, and it’s inverse f (x) = tan-1(x) is also continuous. Two spaces X and Y are homotopy equivalent to one another if they can be transformed into one another by bending, squishing, stretching and expanding operations. 2 2 2 Example 2. A solid disk D = f (x; y) 2 R j x + y ≤ 1 g is homotopy equivalent to a point. Each point in the unit disk can be squished towards the origin along the ray connecting it to the origin. 2 1 The punctured plane R n f(0; 0)g is homotopy equivalent to the unit circle S . The homotopy equivalence squishes each ray to the point on the unit circle that it intersects. A covering space of a topological space X is a space Xe together with a surjective local homeo- morphism p : Xe X: Example 3. Any space X with p : X X the identity map p(x) = x is a covering space of itself. 1 2 For any natural number n, the circle S (thought! of as angles in R ) is an n-fold cover of itself, 1 1 where the map p : S S multiplies! an angle by n. ! MATH 18.095 SQUARE DANCE PROBLEM SET TARA HOLM EXERCISES ON TOPOLOGICAL SPACES Problem 4. Consider each of the capital letters of the alphabet in a sans serif font, as written here: ABCDEFGHIJKLMNOPQRSTUVWXYZ . Group the letters into subsets of those that are homeomorphic to one another. To get you started, we note that X and Y are NOT homeomorphic. This is because if you remove the central point, the X gets separated into four separate pieces; however, there is no point in Y whose removal splits the Y into four pieces. Problem 5. Consider each of the capital letters of the alphabet in a sans serif font, as written here: ABCDEFGHIJKLMNOPQRSTUVWXYZ . Group the letters into subsets of those that are homotopy equivalent to one another. To get you started, we note that X and Y ARE homotopy equivalent. This is because each letter can be squished to a point. The letter O is NOT homotopy equivalent to those because it has a hole and so cannot be squished to a point. Problem 6. Show that the torus is a 2-fold cover of the Klein bottle. To get you started, you will want to think about how to build the torus and Klein bottle by gluing the edges of a rectangle together, as indicated in the figures below. Then try to figure out how to decompose the rectangle that makes the torus into two Klein bottles. (a) (b) FIGURE 1. As (a) indicates, we first identify the top and the bottom edge to make a cylinder. We then identify the two ends, with the same orientation, to make a torus (b). (a) (b) FIGURE 2. In (a), we again identify the top and the bottom edge to make a cylinder. We then identify the two ends, the other way around, to make a Klein bottle (b). In three dimensions, the Klein bottle must pass through itself in order identify the two ends correctly. In four dimensions, the Klein bottle need not intersect itself..

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