1 What is a Manifold?

Transcription

1 Three-Dimensional Geometry and Topology Alvin Jin These are the notes that accompany an independent study on three-dimensional geometry and topology that closely follows Thurston s volume 1 book of the same title. 1 What is a Manifold? Every day objects we see that are examples of two-manifolds, or surfaces are doughnuts, balls, and pretzels among other things. However, three-manifolds are a bit harder to visualize. We will imagine them as alternate universes. We realize manifolds as 1. the solution space of some set of conditions 2. the parameter space for some family of mathematical objects and more other methods. So, how can we recognize the identity of a manifold? 1.1 Polygons and Surfaces The simplest and most symmetric surface is a sphere, followed by a torus. Geometrically, it has symmetry as a surface of revolution in space. However, it also has hidden symmetry in that we can topologically glue together the sides of a square to get a torus. b >> a > > a 1 >> b 1 Figure 1: A torus Problem 1: One-Point Compactification The one-point compactification R n of R n is the topological space obtained by adding a point to R n whose neighborhoods are of the form (R n /B) for all bounded sets B. 1

2 a. Check that the one-point compactification of R n is homeomorphic to the sphere S n. Idea: Consider the canonical stereographic projection and take the inverse of it. We also note that this is equivalent to showing that R n is homeomorphic to S n /{p} where p is the north pole (0,..., 0, 1). Proof. As stated above, we can just show R n is homeomorphic to S n /{p}. Recall that for the stereographic projection of the sphere onto the plane, we took R 2 as a hyperplane R 2 {0} and drew straight lines through the north pole and another point on the sphere to get our projection. We extend this idea to R n : let R n {0} be the hyperplane in R n+1. Consider the north pole p = (0,..., 1) S n. Then consider the stereographic projection f : S n /{p} R n defined by f(x) = 1 1 x n+1 (x x n+1 p) This brings any point on x S n /{p} with the (n + 1) coordinate x n+1 to a unique point on the hyperplane. It has an inverse defined by f 1 (y) = f 1 : R n S n /{p} 1 y (2y + ( y 2 1)p) This brings any point on the plane to S n {p}. which is a continuous function because the denominator never vanishes. b. Consider an ordinary torus in S 3 = R 3, and show that the interchange of curves a and b in Figure 1 can be achieved by moving the torus in S 3 (hold until Section 2.7). Idea (?) We want to be able to reverse the roles of the longitudinal and latitudinal curves. We know that this can be done when we puncture the torus and turn it inside out (doing the puncturing at ). c. Show that this cannot be done in R 3. Idea: We can t create such a puncture in R 3 just by moving the torus. We can also view the torus as a hexagon with opposite sides associated with one another. This gives a different symmetry for us to think about. These two descriptions (square and hexagon) of the torus are related to the tilings of the Euclidean plane. Consider an infinite collection of identical squares such as that in figure 1 or an infinite collection of identical 2

3 hexagons as described before. If we start with a single polygon and then add more polygons layer by layer, identifying edges of the new ones with the same corresponding edges for the old ones (ie a a, b b, c c) then each new tile fits in only one way. So, we can tile the Euclidean plane with squares or hexagons. We can consider the wallpaper groups that cover the Euclidean plane. The tilings show us that the plane is a covering space for the torus. We have that the covering map for the square tiling is the map that identifies corresponding points in each square. Since we know that the plane is simply connected, we have a universal cover of the torus. Then here, we have a single group of homeomorphisms of the plane (that take any point into another point that has the same image on the torus under the covering map). For the square tiling, we have that the covering transformations are the translations that preserve vertices. So, the torus is the quotient space of the plane by the action of the translation covering group Note that covering transformations are Euclidean isometries so we can give the torus a Euclidean structure. We do this by taking x T 2 and taking a neighborhood U of x small enough so that the inverse image of U in the plane is made of connected components homeomorphic to U under the covering map p. By shrinking U, we can guarantee that the distance between two components is greater than the diameter of these components. 1.2 Hyperbolic Surfaces We can also make a polygon for a genus-two surface. The word is aba 1 b 1 cdc 1 d 1. A <<< <<<< << > > << <<< <<<< Figure 2: Genus-2 Torus question we have is if the Euclidean plane can be tiled by these octogons. The answer: no. Why? The interior angle of a regular octogan is 135. But, we need 8 of these octogons to fit around a vertex which requires This tiling doesn t work in Euclidean space, but what if we could choose regular octogons with 45 angles such that 8 would fit around a vertex. To work in a space where we aren t bounded by the parallel axiom of Euclid, we consider hyperbolic geometry. H n can be represented in different ways, which will be discussed later. The hyperbolic plane H 2 is homeomorphic to R 2, and the Poincare disk model maps it onto the open unit disk D in the Euclidean plane. Hyperbolic straight lines appear as arcs of 3

4 circles orthogonal to the boundary D of D and every arc orthogonal to D is a hyperbolic straight line. However, we know that there exists one special case: any diameter of the disk Figure 3: Poincare disk model straight lines. is a limit of circles orthogonal to D and it is also a hyperbolic straight line. Along the diameters, reflections are our normal Euclidean reflections. However, for all other lines, we have inversions that generalize reflections. Definition If C is a circle in the Euclidean plane, the inversion i C in C is the unique map from the complement of the center of C into itself that fixes every point of C, exchanges the interior and exterior of C and takes circles orthogonal to C to themselves. Example. Let C = x 2 + y 2 = 1 be the circle centered at the origin with radius one. Then we look at points x 1 = (0, 1), x 2 2 = ( 3, 0), x 3 = (0, 1), x 4 = (2, 3) and their new coordinates under i C : ( x 1 = 0, 1 ) x 1 = (0, 2) 2 x 2 = ( 3, 0) x 2 = ( 13 ), 0 We now show that inversions are well-defined. x 3 = (0, 1) x 3 = (0, 1) ( 1 x 4 = (2, 3) x 4 = 2, 1 ) Problem 2: Inversions a. Show the following standard result from Euclidean plane geometry: If A is a point outside a circle C and l is a line through A intersecting C at P and P, the product AP AP is independent of l and is equal to AT 2, where AT is a ray tangent to C at T. This product is the power of A with respect to C. To solve this, we use the property of similar triangles. Consider the line segments P T and P T. Without loss of generality, let d(a, P ) < d(a, P ). Then we note that 4

5 m T P A = 1 2 mpt. Similarly, m P T A = 1 2 mpt. Then m T P A = m P T A. So, the angle-angle theorem tells us that P T A T P A. So, we have which is what we wanted. AP AT = AT AP b. Use this to show that the definition of an inversion makes sense. We want to show that inversion does not depend on choice of a point. By the first part, we know that for A outside a circle C and a line l through A intersecting C at P and P, AP AP is independent of l and equal to AT 2. So, consider a point p outside our circle C with center c and the image of p under inversion p. Then cp cp = r 2. But since cp > r, cp < r so that the exterior and interior of circles are switched independent of the choice of p. For any point on the circle C, say r, we have cr cr = r 2 = cr = r, so that the circle is mapped to itself. Lastly, to show that orthogonal circles are mapped to themselves, consider the circle of inversion C with radius rm centered at O and an orthogonal circle C o with radius R, centered at A. Then for an intersection point T of these two circles, OT = r and OA = R. Now, take a line going through P and P on the orthogonal circle. Then OP OP = r 2 and so that the image of P is on the orthogonal circle. c. Prove that if C has center O and radius r, the image P = i C (P ) is the point on the ray OP such that OP OP = r 2. (Similar to above) Proposition: Properties of inversions. If C is a circle in the Euclidean plane, i C is conformal, that is, it preserves angles. Also, i C takes circles not containing the center of C to circles, circles containing the center to lines, lines not containing the center to circles containing the center, and lines containing the center to themselves. Proof. Take two vectors at x / C so that we can construct a circle tangent to each vector and orthogonal to the circle of inversion. These circles are preserved by inversion by definition and meet at the same angle at their other point of intersection. So, we have conformality everywhere but on C. To show conformality on C, we can use continuity. To show circles and lines tangent to C are taken to circles and lines (disrespectively), we let circle mean circle or line. For any point x C, note that the plane is filled by a family F O of circles orthogonal to C at x and by a family F T or circles angent to C at x. Any circle from F T meets any circle from F O perpendicularly at both their intersection points. So, F T forms the set of orthogonal trajectories of F O. F O is preserved by i C by definition 5

6 so since inversion is conformal, the family of orthogonal trajectories of F O is preserved, and any circle tangent to C is taken to another circle tangent to C. Any circle other than a line Figure 4: Family of circles. through the center of C can be expanded or shrunk to a circle tangent to C by a homothety centered at the center of C. The circle s image under inversion shrinks or expands by the same factor. Since the image of the tangent circle is a circle, so is the image of the original circle. Lines through the center go to themselves also (by the above exercise). Example. Mechanical linkages. Figure 5: Mechanical Linkage Another example is a the inverted chess board, where we invert the board along a circle centered at (0, 0). 6

7 Example. Inverted chess board. Figure 6: Inverted chess board where we take inversion around a circle centered around the origin. If a hyperbolic line appears in the Poincare model as a circle orthogonal to D, the hyperbolic reflection in this line appears as the Euclidean inversion in this circle. For arcs, we just see the restriction to D of the Euclidean inversion in the circle. By the above proposition, inversions map D into itself pointwise and preserves hyperbolic lines. In fact, this inversion is an isometry in a certain sense, as we shall soon define distance in hyperbolic space. We will do this by describing the Riemannian metric of the hyperbolic plane (i.e. assigning an inner product for the tangent space at that point). This will be done in a later section. Now, we consider two orthogonal hyperbolic lines L and M, seen in the Poincare model as Euclidean arcs of circles orthogonal to D, and another Euclidean circle C that intersects D in the same two points as L. Figure 7: A hypercircle. 7

8 We know that C is orthogonal to M so that by definititon of inversion, the hyperbolic reflection in M leaves L and C the same. So, since we want reflection to preserve distance in the hyperbolic sense, corresponding points of C on both sides of M must be equally distant from L. So, we vary M among the lines orthogonal to L and see that all points of C are equally distant from L so C is an equidistant curve. In fact, all of the segments in the region bounded by C and L have the same hyperbolic length. If we apply a hyperbolic reflection to the whole space, the angle α at the tips of the region do not change since inversions preserve angles. The width of the region stays the same since we want reflections to preserve length. So, l is a function of α and has a finite derivative at α = 0 since the Euclidean length l E of any particular transversal segment is roughly proportional to α for small α. Euclidean and hyperbolic lengths should be proportional to the first order. In fact, for two points a and b on the circle C, we define the distance function to be ( d h (a, b) = ln de (a, l 0 )d E (b, l 1 ) d E (a, l 1 )d E (b, l 0 )) where d h is the hyperbolic distance, d E is the Euclidean distance, and l 0, l 1 D such that d E (a, l 0 ) d E (a, l 1 ) and d E (b, l 1 ) d E (b, l 0 ). The derivative dl at α = 0 is taken to be 1. dα By this construction, we can define the Riemannian metric. To find the length of a tangent vector v at x, draw the line L orthogonal to v through x and the equidistance circle C. The length of v for small v is roughly the hyperbolic distance between C and L, which is about the Euclidean angle between C and L where they meet. Now, we consider the angle α t of the region built on tv with t approaching zero so that the length of v is dαt at t = 0. dt But, we also know that two vectors at x that have the same Euclidean length also have the the same hyperbolic length, which follows from the existence of a hyperbolic reflection fixing x that takes one vector to the other along with the fact that the derivative of an inversion at a point on the inversion circle is an orthogonal map. Since Euclidean and hyperbolic vector lengths at a point are proportional, we know that the inner products are proportional. In particular, we have that the Poincare model is conformal since the Euclidean and hyperbolic angles are equal. Now, we tile the hyperbolic plane using regular octagons. By the conformality of the Poincare model, we need to tile the plane using angles of 45. Now, we construct the model by placing a regular octagon centered at the origin. Since the edges of the octogon in the hyperbolic model will bend a little bit, for small enough side lengths s of the octogon, the interior angles are close to 135. By moving the vertices of the octogon toward and away from the origin, we can get really large angles and really small angles so that the intermediate value theorem tells us that there is some octagon with interior angles that are exactly 135. So, the existence of the octagons octagon with angles π/4 is guaranteed and we can tile the plane with these octagons. 8

9 Figure 8: Regular octagons tiling the plane. The octagons look different depending on where we are in the model (they shrink quickly as we approach the boundary of D). We can construct these small octagons through use of hyperbolic isometries (e.g. reflections). By this tiling, we know that the genus-two surface can be given a hyperbolic structure, a geometry where the surface looks locally like the hyperbolic plane. 1.3 The Totality of Surfaces Gluing edges of polygonal regions in two dimensional space in a manner such that we get a simplicial complex gives us a two-dimensional manifold. We first suppose that the total number of boundary edges is even for k oriented polygonal regions. We give the edges orientations induced by the orientations of the regions. To glue the edges of the polygonal regions together, we pair the edges such that for each pair, we have an identification with either an orientation-preserving or orientation-reversing homeomorphism. Definition An orientation for a finite-dimensional vector space is an equivalence class of ordered bases that can be taken to another by linear transformations with positive determinant. A linear transformation between oriented vector spaces is orientation-preserving or orientation-reversing depending on whether its determinant is positive or negative. Definition A manifold M is orientable if the tangent space to M at all points can be oriented consistently. This means that M can be covered by coordinate patches such that the derivative map of any coordinate map at any x M is an orientation-preserving map between T x M and R n with its standard orientation. M is oriented if such a choice of orientations have been made. A local diffeomorphism between oriented manifolds is orientation-preserving or orientation-reversing according to what its derivative is at each point. 9

10 Examples. Any reflection in R n is orientation reversing. Rotations and translations, on the other hand, are orientation preserving. Let us consider the rotation matrix [ ] cos θ sin θ R(θ) = sin θ cos θ such that points (x, y) are taken to (x, y ) where The determinant of our rotation matrix is so that rotation is orientation preserving. x = x cos θ y sin θ y = x sin θ + y cos θ cos 2 θ + sin 2 θ = 1 For reflections, consider the reflection across the y-axis so that a point (x, y) is taken to ( x, y). The matrix is [ ] 1 0 M = 0 1 so that the determinant is 1 so we have an orientation-reversing map Problem 3: Homeomorphisms of Intervals Prove that two homeomorphisms of an interval to itself are isotopic if and only if they both preserve orientation, or both reverse orientation. Note that an isotopy between two maps f, g : X Y is a homotopy F : X [0, 1] Y between f and g such that every stage F t : X Y, for t [0, 1], is a homeomorphism onto its image. Proof. Idea: We use the fact that the homeomorphisms must be monotone increasing or decreasing to preserve orientation. ( ) Let f, without loss of generality, be an orientation-preserving map. Consider the canonical straight-line homotopy F (s, t) = (1 t)f + tg for t [0, 1] which is an isotopy. ( ) Let f and g be two isotopic homeomorphisms of an interval I, to itself. Then there exists a homotopy F : I [0, 1] I such that F (s, 0) = f and F (s, 1) = g and for t [0, 1], F t : I I is a homeomorphism onto I. Then f and g must be monotone. In particular, this monotonicity preserves or reverses orientability. 10

11 It s hard to visualize what surface is obtained when we glue a many-sided polygon in a givern pattern. To remedy this problem, we use a numerical invariant, the Euler number of a surface that lets us recognize surfaces quickly. In fact, we know that all compact surfaces can be classified by their orientation and Euler number. We define the Euler number χ(s) of a surface S to be χ(s) = V (S) E(S) + F (S) where V (S), E(S), F (S) are the number of vertices, edges, and faces of the surface. For example, the torus when viewed as a quotient space (ie in the form of a square with sides identified as in the beginning of these notes) has one face, two edges, and one vertex so χ(t 2 ) = = 0 The sphere can be divided into four triangles to form a tetrahedron with six edges and four vertices so χ(s 2 ) = = 2 When we cut a surface into polygons and their edges and vertices, we create a cell division. A cell is a subset C X, where X is any Hausdorff space, homeomorphic to an open disk of some dimension, with the condition that the homeomorphism can be extended to a continuous map from the closed disk into X, called the cell map. A face is a two-cell, an edge is a one-cell, and a vertex is a zero-cell. A cell division of X is a partition of X into cells, in such a way that the boundary of any n-cell is contained in the union of all cells of dimension less than n. If X is a differentiable manifold, then we assume that our cell divisions are differentiable. That is, for each cell C, the cell map can be realized as a differentiable map from a convex polyhedron onto the closure of C, having maximal rank everywhere. In particular, when we think of differentiability at a point, we require that the map be defined on a neighborhood of the point in R n. So, if X R n is not an open set, we say that a map f : X R m is differentiable if it is the restriction of a differentiable map on an open neighborhood of X (ie all partial derivatives are continuous). Intuitively, we can think of differentiable cell divisions as corresponding to cutting a surface into polygons. There are some exceptions - such as the sphere. We express the sphere as a union of a vertex and a face. The Euler number of a space X having a finite cell division is defined as the sum of the numbers of even-dimensional cells, minus the sum of the numbers of odd-dimensional cells. We know also that the Euler number is independent of cell division. Theorem: Invariance of Domain. If a subset A R n is homeomoprhic to an open subset of R n, A is open. An m-dimensional manifold cannot have a subset homeomorphic to R n for n > m. By the invariance of domain, a surface cannot have cells of dimension greater than two (it is a two-manifold). To gain some intuition for the Euler number invariance for cell divisions, consider the division of a one-cell. If we divide an edge by placing a vertex in the middle of the edge, we add one new edge and one new vertex. So, the Euler number remains the same. 11

12 For a two-cell, we have the same concept. We divide the face by a new edge between two of its vertices to get one two cell and one one-cell so that the Euler number remains the same. We could show invariance of the Euler number by using these divisions of edges and faces along with the inverse division to transform any finite cell division to another. But a more elegant method is looking at vector fields on the surface. We first consider an example: the sphere S 2. We can give the sphere a cell division that is realized as a convex polyhedron in E 3. Arrange the polyhedron so that no edge is horizontal (with respect to each axis) so that there is exactly one uppermost vertex U and one lowermost vertex L. We put at positive unit charge at each vertex and a negative unit charge at the center of each edge while a positive unit charge in the middle of each face. The idea is to show that the charges cancel except for those at L and U. So, to do this, we displace the vertex and edge charges into a neighboring face and then group together all the charges in each face (we move each charge horizontally and counterclockwise as viewed from above) so that all the charges in the faces cancel and we are left with the net charge from an open interval along the polyhedron s boundary, which consists of edges and vertices which alternate. Since the first and last parts of this interval are edges, we have a surplus of one negative charge so that the total charge in each face is zero. A picture of this is to see that we only have one vertex and two edges connected by that vertex, along with the face that determines the net charge on the face when the charges move along a vector field. So, we just have L and U left, which gives us +2 as a charge. We can now generalize this idea to any differentiable surface with a differentiable triangulation. So, our cell division has faces as triangles in such a way that the cell map for any face is an embedding when taking each side of the model triangle onto an edge of the cell division, and the cell map for the edge is compatible with the cell map for the face. In particular, they differ by an affine map between domains. Proposition: Nonvanishing vector fields. If a differentiably triangulated closed surface admits a nowhere zero tangent vector field, its Euler number is zero. Proof. Suppose that the vector field is everywhere transverse to the triangulation as in our construction of the cell division of the sphere. We will use the same argument to subdivide the triangulation to make the vector field nearly constant within each triangular face. At each vertex, we put a positive unit charge and at the midpoint of each edge, we put a negative unit charge. In the center of the face of the triangle, we put a positive unit charge. In face, for each face in some coordinate chart, the direction of the field changes at most by ɛ and the direction of the edges changes by at most ɛ. So, given that transverse triangulation, we apply the same idea of moving the vertex and edge charges in the direction of the vector field. If the vertex s charge moves into a face, then the charges for the two adjacent edges go into the face. So, in each face, either one edge charge gets pushed in or two edge charges and one vertex get pushed in. Three or zero charges cannot occur since the field would not be constant. So, in both cases, all the charges cancel out and we have a total charge of zero so our Euler number is zero. We finish the proof by showing that triangulations can be changed so as to become transverse to the field and so the field and edge directions are 12

13 nearly constant within each face. To do this, we first cover the surface with a finite number of ccoordinate patches. Then, we draw equally spaced lines parallel to each edge and subdivide the triangle finely so that the star of each vertex v lies in a single coordinate patch and that the direction of each edge and of the field in the star of v changes by no more than ɛ in these coordinates. Next, we consider mapping the sets of directions of edges and of the field as intervals on the circle of length bounded by ɛ. We can make the intervals of directions of the edges avoid the interval of directions of the field by moving v in the appropriate direction and extending the movement to each edge incident on v by a Euclidean similarity (ie a transformation that multiplies all distances by a constant factor) that keeps the other endpoints of the edge fixed. We extend this transformation to all vertices simultaneously. We partition the vertices into three colors: red, green, and blue. In particular, no two vertices of the same color are joined by an edge. We first adjust all red vertices, then all green so that all edges are transversal. The torus T 2 has nowhere zero vector fields. We can consider a uniform field on E 2 and quotient by Z 2. So, we have that the Euler number is zero. For other surfaces, most of them do not admit a nowhere zero vector field. However, we can find a vector field that is zero at isolated points. By the above proof, we note that charges cancel in regions away from the zeros of such a field. So, now we need to study the behavior near the zeros of the field. So, let X be a vector field on a surface with an isolated zero at a point z. As in the proof of the theorem, we construct a small polygon containing z in its interior, having edges transverse to X. Again, we place a positive unit charge on each vertex and a negative unit charge on each edge with a positive unit charge in the interior of the polygon. We flow the charges off the boundary of the polygon by using X. We define the index of X at z, denoted i(x, z), is the sum of the charges in the interior of the polygon after the operation of the flow. Figure 9: Vector field with index 1. 13

14 Proposition: Index independence. If X is a vector field with an isolated zero z the index of X is independent of the polygon enclosing z; it only depends on the restriction of X to an arbitrarily small neighborhood of z. Proof. Given one polygon containing z with edges transverse to X, we show that a smaller polygon with the same properties gives the same index. We subdivide the annulus between two polygons into triangles, and move the vertices to make the edges transverse to X. The idea of the triangulation of the annulus can be summarized as follows: we triangulate the annulus in the plane by considering a star-shaped set D, which can be triangulated with a point v as the vertex of D. So, given two polygons that are star-shaped with respect to v such that one is contained in the interior of the other, we triangulate the annulus between their boundaries. Now, we suppose that we have a polygon which is star-shaped with respect to the isolated zero z of X. The boundary of the polygon can be made transverse to X by moving vertices in the radial direction so that the polygon after the transformation is still star-shaped. So, as stated above, we divide the annulus between two polygons into triangles and move the vertices to make the edges transverse to X. The field causes some charge to enter the annulus across the outer boundary and some charge to leave across the inner boundary so that the net flux is zero. Example: Index of a Vector Field Consider the vector field defined by F = ( y, x). There is a singularity at the origin. We parameterize a loop by γ = (r cos t, r sin t). We have the winding number of the loop Γ = F γ is the index of our field. In particular, we just need to evaluate, for u = r sin t and v = r cos t: (F, 0) = 1 udv vdu 2π u 2 + v 2 = 1 2π = 1 2π = 1 F γ 2π 0 2π 0 udv vdu u 2 + v 2 ( r sin t)( r sin t) (r cos t)( r cos t) dt r 2 So, if we move once around the origin in the counterclockwise direction, our vector field goes 360 as well and our vector field has index 1. We see that the simplest vector fields with isolated zeros are linear vector fields in the plane. The origin is a zero of any linear vector field and it is isolated if and only if the linear map has non-zero determinant. We also note that if a vector field with isolated zeros is homotoped in a way such that the points where the field vanishes do not change, the indices at the zeros must remain constant since they are integers. So, two linear vector fields whose 14

15 determinants have the same sign have the same index, since they are homotopic through linear vector fields of non-zero determinant. Proposition: Poincare index theorem. If S is a smooth surface and X is a vector field on S with isolated zeros, the Euler number of S is χ(s) = i(x, z) z zeros Consequently, the Euler number of a surface does not depend on the cell division used to compute it - it is a topological invariant. Proof. We start with a finite cell divisiion of S and replace it with a differentiable traingulation. We subdivide and move the triangulation as necessary to make all the zeros lie inside faces with no more than one zero to a face. We then enclose each zero with a polygon contained in the face, where the polygon is transverse to the field (the edges are transverse). We triangulate the annulus by taking away the polygon from the face and then make the rest of the triangulation transverse as in the proof of the proposition about nonvanishing vector fields. Each polygon s contribution to the Euler number is the index of the vector field at the corresponding zero. Each triangle s contribution outside the polygon is zero. So, the formula follows. The Euler number does not depend on cell division since every closed surface admits a vector field with isolated zeros. It turns out that we can completely determine a surface by its Euler number and whether or not it is orientable. We know that orientable sufraces S can be obtained by a gluing of the form a 1 b 1 a 1 1 b a g b g a 1 b 1 and nonorientable surfaces where g is the genus of the surface. a 1 a 1 a 2 a 2...a g a g In a previous section, we constructed a hyperbolic structure for the oriented genus-two surface. We can make an analagous construction for any surface of negative Euler number by gluing the sides of a 2n-gon, with n 3 in such a way that all vertices are identified to a single vertex. The angle sum of the Euclidean polygon is greater than 2π so that there exists a symmetric 2n-gon of the appropriate size in hyperbolic space whose sides glue up to form a smooth hyperbolic structure on the surface. If a surface has Euler number zero, then it can be obtained by gluing sides of a square so that all vertices are identified together so we get a Euclidean structure. The surface is of the form aba 1 b 1 : a torus, or aba 1 b: a Klein bottle. S 2 and its quotient space RP 2 (positive Euler number) have spherical structures, which we call elliptic. The connected sum of two connected n-manifolds M 1 and M 2 is a manifold M 1 #M 2 obtained by removing copies of the n-disk D n from M 1 and M 2 and gluing the two resulting boundary spheres together. 15 g g

16 1.3.2 Problem 4: Connected sum Show that if S 3 = S 1 #S 2 are surfaces, χ(s 3 ) = χ(s 1 ) + χ(s 2 ) 2. What happens for manifolds of other dimensions? We triangulate the cylinder that connects the two surfaces as follows: We identify b 1, b 2, b 3 b 1 b 2 b 3 >> >> >> a > > a >>> >>> >>> c 1 c 2 c 3 Figure 10: A triangulation of a cylinder. with edges in S 1 and c 1, c 2, c 3 with edges in S 2. All vertices of our triangulation are identified with vertices of S 1 and S 2. We remove a face from both S 1 and S 2 when attaching the cylinder. So, if the number of vertices, edges, and faces of S 1 and S 2 are V 1, E 1, F 1 and V 2, E 2, F 2 then the Euler number of the connected sum is V 1 + V 2 E 1 E F 1 + F = χ(s 1 ) + χ(s 2 ) 2 This holds true for manifolds of any finite dimension. We can make our definition of connected sum more precise by choosing diffeomorphic embeddings φ 1 : D n M 1 and φ 2 : D n M 2 of the closed n-disk, remove the two images of D n from the union M 1 M 2 and identify the boundaries φ 1 ( D n ) and φ 2 ( D n ) by the map φ 2 φ 1 1. The choice of φ 1 and φ 2 does not matter much since there is essentially only one way to embed a disk in a connected manifold up to orientation. In particular, if we change φ 1 by an isotopy, the topology doesn t change because any isotopy between embeddings of D n into an n-manifold M can be extended to an isotopy of the whole manifold. Now, we associate the frame, or the ordered basis of our space, at φ(0) given by the image under Dφ(0) of the canonical basis vectors of R n with an embedding φ : D n M. The embeddings are isotopic if and only if their associated frames can be continuously deformed into tone another. So, there are two isotopy classes of diffeomorphic embeddings D n M if M is orientable and only one if M is non-orientable. If an orientation is chosen for our manifold, the two classes are determined by whether the embedding preserves or reverses orientation. If two manifolds are oriented, we form the connected sum so that the resulting manifold 16

17 has an orientation which agrees with the original orientations away from the disks which are removed. In particular, only one of φ 1 and φ 2 must be orientation-preserving. If one of the manifolds is non-orientable, the result of the connected sum does not depend on choice of orientation for the gluing map. However, when two manifolds are orientable but not oriented, the two possible choices of sign for the gluing map might yield different results. However, this does not happen for surfaces because every closed orientable surface admits an orientation-reversing homeomorphism and that the connected sum operation is a commutative semigroup. We note that S 2 acts as an identity element for # and that the torus T 2 and the projective plane RP 2 generate the commutative semigroup of homeomorphism classes of surfaces under # and finally, T 2 #RP 2 = RP 2 #RP 2 #RP Some Three-Manifolds Now, we go through some examples of three-manifolds Example 1. The three-torus We start with a simple example of a three-manifold. We obtain the three-torus by starting with a cube and gluing each face of the cube to the opposite parallel face by parallel translation (think of the two-torus and the gluing is analagous). To visualize the three-torus, we imagine a cube as a room in which we are standing in. We identify the ceiling and ground, and opposite walls. So, if we look ahead, we will see our back and if we look down, we will see the top of our heads. There are six images of ourselves in the standard directions in neighboring rooms but there are also rooms that neighbor diagonally. We would see an infinite repeating array of images of ourselves in the rooms in which the lines of sight continue indefinitely Example 2. The three-sphere We consider the unit sphere S 3 in R 4 given by x x x x 2 4 = 1 It is hard to visualize this but we can imagine S 3 from inside of S 3. To help motivate what S 3 might be like, we start with a person in S 2. Light rays curve around to follow the surface. We imagine person A at the north pole and person B creeping away. As B moves away, its image as seen by A at first grows smaller, although not quite as fast as it would in the plane. Once B reaches the equator, B s image grows larger with continued progress until at the south pole, the image of B fills up the entire background of the field of vision of A in every direction. The same phenomenon takes place in the three-sphere. As a C person passes the halfway mark (1/4 of the circumference), of the sphere in the point of view of a stationary 17

18 observer D, the image of the C gets slowly larger. So, when the antipodal point is reached, the entire field of vision will be taken up. At any other point than the antipode, person D can see person C is two directions (ie the major axis and the minor axis of the sphere defined by geodesics). Person D will see himself in addition to person D for any position of D except the antipodal point Example 3. Elliptic space The elliptic space is just the sphere with antipodal points identified. Topologically, the n- dimensional elliptic space is just the projective n-space RP n. It is the quotient of the sphere by a group of isometries (the identity and antipodal map) so it inerhits a geometry, just as the torus inherits a geometry from the Euclidean plane. We can use elliptic space to realize geometric ideas of the sphere more easily. For example, any two points in elliptic space determine a unique line. In elliptic space, the maximum distance is π/2 whereas in the 3-sphere, the maximum distance is π relative to a stationary observer. The apparent size is a monotone decreasing function of distance. Also note that an observer can see himself filling up the background. The image in front of the observer will be an upside down magnified back of the observer Example 4. Poincare dodecahedral space We obtain this space from gluing the opposite sides of a dodecahedron. The corners of pentagons making up opposite faces of a dodecahedron are out of phase so that the gluing is just like that of the torus. A dodecahedron has 30 edges, which are glued in ten groups of three. For the 20 vertices, we consider the spherical triangles obtained by intersecting tiny spheres about vertices with the dodecahedron. Four triangles come together at each vertex so we get a tetrahedron. So, we have five groups of four vertices. A question that remains is: can we make the geometry of this space locally Euclidean by some particular gluing pattern? We would need the angles between the faces around an edge to add up to 2π. Assuming regular pentagons, we need each angle to be 2π/3. However, the dihedral angles of a dodecahedron are about , which is slightly less than what we want. To remedy this, we use the geometry of the three-sphere. The dihedral angles of a polyhedron in the three-sphere are larger than those of a Euclidean polyhedron. When the distance from the center to the vertices if π/2, the dodecahedron is geometrically a two-sphere with angles π. On the other hand, a small dodecahedron will look like a Euclidean dodecahedron. So, since we can continuously increase the size of the dodecahedron, we can find one with angles exactly 2π/3 radians. We can unroll the Poincare dodecahedron space to obtain a tiling of S 3. So, the dodecahedral space is a quotient of S 3 by a group of isometries. In fact, we can tile the three sphere with 120 dodecahedra. 18

19 Figure 11: A dodecahedron. We obtain the space by gluing opposite faces together so that edges are glued in triples (with 1/10 of a clockwise revolution so the vertices match up) Example 5. Seifert-Weber dodecahedral space If we glue the opposite faces after 3/10 of a clockwise revolution instead of 1/10, edges are identified in six groups of five. All twenty vertiecs are glued together and instead of a tetrahedron, we get a regular icosahedron around vertices. This manifold is called the Seifert-Weber dodecahedral space. We need 72 c irc angles to do a gluing geometrically so we use the three-dimensional hyperbolic space H 3, which can be mapped into the interior of a three-dimensional ball, just as in two dimensions. This description of the ball is the Poincare ball model of hyperbolic space. Planes are setors of spheres orthogonal to the boundary of the ball. The angle between two hyperbolic planes is the same as the angle between the two spheres. Since the spheres intersect the boundary of the ball S 2 perpendicularly, the angle between hyperbolic planes is also the angle between their circles of intersection with the sphere. The angles of a regular dodecahedron in hyperbolic space are smaller than they are in euclidean space. In the limiting case, as the dodecahedron became large, its vertices would almost touch S. 2 The limit is the ideal dodecahedron with vertices on S. 2 We note that the ideal dodecahedron has π/3 radian dihedral angles by looking at the placement of three circles on the boundary of the ball. By a similar logic as before, we can smoothly shrink or expand a doddecahedron so that the angles vary from 60 to , so that there exists a dodecahedron with angles exactly 72. We glue this dodecahedron to make a geometric form of the Seifert-Weber dodecahedral space, which corresponds to a tiling of H 3 by these dodecahedra with 72 angles Example 6. Lens spaces We consider a ball with its surface divided into two hemispheres along the equator. If we glue one hemispherical surface to the other with no twist, we get S 3. The construction of S 3 is analogous to dividing the boundary of a disk into two intervals and gluing one to the other to match each endpoint with itself to form S 2. If the hemispheres are glued with a q/p clockwise revolution, where q and p are relatively prime integers, then a each point along 19

20 the equator is identified to p 1 other points. The resulting identification speed is the lens space. To construct a geometric model, we need the angle between the upper surface and lower surface to be 2π/p. We do this in S 3. Any great circle in S 3 has a whole family of great two-spheres passing through it. Choose two that meet at the angle 2π/p so that when the two faces are glued together, neighborhoods of p points on the rim of the lens are identified and fit exactly, which corresponds to a tiling of S 3 by p lenses. 20