U.U.D.M. Project Report 2016:37
Classification of Compact Orientable Surfaces using Morse Theory
Johan Rydholm
Examensarbete i matematik, 15 hp Handledare: Thomas Kragh Examinator: Jörgen Östensson Augusti 2016
Department of Mathematics Uppsala University
Classi cation of Compact Orientable Surfaces using Morse Theory
Johan Rydholm
1 Introduction
Here we classify surfaces up to dieomorphism. The classi cation is done in section Construction of the Genus-g Toruses, as an application of the previously developed Morse theory. The objects which we study, surfaces, are de ned in section Surfaces, together with other de nitions and results which lays the foundation for the rest of the essay. Most of the section Surfaces is taken from chapter 0 in [2], and gives a quick introduction to, among other things, smooth manifolds, dieomorphisms, smooth vector elds and connected sums. The material in section Morse Theory and section Existence of a Good Morse Function uses mainly chapter 1 in [5] and chapters 2,4 and 5 in [4] (but not necessarily only these chapters). In these two sections we rst prove Lemma of Morse, which is probably the single most important result, even though the proof is far from the hardest. We prove the existence of a Morse function, existence of a self-indexing Morse function, and nally the existence of a good Morse function, on any surface; while doing this we also prove the existence of one of our most important tools: a gradient-like vector eld for the Morse function. The results in sections Morse Theory and Existence of a Good Morse Func- tion contains the main resluts and ideas. They have many generalizations and many other applications; to mention one, in [4] Milnor goes on and prove the generalized Poincaré Conjecture1 in dimensions n ≥ 5, using similar results. In this essay, the Classi cation Theorem, great as it may seem, is almost like only a corollary of these two sections.2
1The theorem is: If M is a simply-connected smooth manifold having dimension n ≥ 5 with the (integral) homology of the n-sphere, then M is homeomorphic to the sphere in the n n same dimension S ; if n = 5 or n = 6, then M is even dieomorphic to S . 2A good popular account of the classi cation of surfaces are given in chapter 3 in [7].
2 Contents
Introduction 2
Surfaces 4
Morse Theory 9
Existence of a Good Morse Function 21
Constrution of the Genus-g Toruses 34
3 Surfaces
This section introduces some basics of the theory of manifolds, speci cally we de ne the concept of a surface and a genus-g torus. De nition 1. A smooth manifold of dimension n is a topological space M to- gether with two families of open sets , of n and respectively, {Ui}i∈I , {Vi}i∈I R M and a set of homeomorphisms {φi : Ui → Vi}i∈I - called parametrizations, whose inverses, are called coordinate system or just coordinates - such that [ Vi = M, i∈I and all transition functions −1 φi ◦ φj are smooth; we also require the triad ({Ui}, {Vi}, {φi}) to be maximal in the sense: given ({Ui}, {Vi}, {φi}), if φ : U → V satis es the conditions above together with the triad, then (U, V, φ) is in the triad. Lastly, a smooth manifold is required to be hausdor and second countable. A surface S is a manifold of dimension 2. Moreover, for brevity, we require a surface to be connected, compact and orientable (to be de ned later). In other words, what we will call surfaces are connected, compact and orientable 2-dimensional smooth manifolds. Also, in our de nition the manifolds (and thereforee also the surfaces) are without boundary; sometimes we will nevertheless talk about a boundary, formally this is the set of all points where all neighbourhoods - instead of being homeomor- phic (and by the de nitions below, also dieomorphic) to U open in Rn - are homeomorphic to n n . Sometimes we will have to H = {(x1, . . . , xn) ∈ R |xn ≥ 0} work with surfaces with boundary, when this is the case it will be obvious from context, and it won't cause any special problems. Note also that if the over- lap for two parametrizations is empty, then the transition function is trivially smooth.
De nition 2. A mapping ϕ : M → M 0 between two smooth manifolds M and M 0, of dimension n and m respectively, is said to be smooth at a point p, if for any parametrizations φ : U → V ⊂ M, φ0 : U 0 → V 0 ⊂ M 0 around p and ϕ(p) 0−1 −1 respectively, the map φ ◦ ϕ ◦ φ is smooth at x0 = φ (p). The mapping φ is said to smooth if it is smooth at all points in M.
In particular, a real valued function f : S → R from a surface S is said to be smooth, if for any parametrization φ : U → V ⊂ S, the function f ◦φ is smooth. We can now de ne a notion of equivalence of surfaces, and more generally, equivalence of smooth manifolds, that is when they from our viewpoint are considered the same: the notion of dieomorphic smooth manifolds.
De nition 3. Two smooth manifolds M and M 0 are said to be dieomorphic, when there exists a smooth bijection
ϕ : M −→∼ M 0, such that its inverse ϕ−1 is smooth; such a map is called a dieomorphism.
4 The goal is to classify surfaces up dieomorphism; we will see that the clas- si cation can be done by a class of surfaces to be de ned later, called genus-g toruses. Before that we will de ne tangent spaces and dierentials. De nition 4. A smooth map γ :(−, ) → M from an open interval (−, ) of R to a smooth manifold M, is called a curve in M. Suppose that p ∈ M is a point in the manifold, with γ(0) = p, for some curve γ. Let D denote the set of all real valued functions on M, that are smooth at p. The tangent vector to the curve γ at p is the function γ0 : D → R de ned by d(f ◦ γ) γ0(f) = (t = 0). dt A tangent vector at p is just the tangent vector of some curve. The set of all tangent vectors of M at p is called the tangent space of M at p, and it is denoted TMp.
A result from the theory of smooth manifolds says that TMp is a nite dimensional vector space having the same dimension as M. De nition 5. Let M and M 0 be two smooth manifolds, and let ϕ : M → M 0 be a smooth map. For all p ∈ M and all v ∈ TMp, by a well known result from the theory of smooth manifolds (which is not hard to prove), given any curve γ for which v is the tangent vector of, the linear map
0 0 dϕp : TMp → TMp, dϕp(v) = β (0), where β = ϕ ◦ γ, is independent of the choice of γ. This map is called the dierential of ϕ at p. Next we de ne the notion of a vector eld on a manifold. A special type of vector elds called gradient-like vector elds de ned in the next section will be one of the main tools used. Afterwards we de ne the notion of a smooth manifold being orientable, and give a citerion using möbius bands to determine whether a surface is orientable or not.
The set TM = {(p, v)|p ∈ M, v ∈ TMp} is called the tangent bundle of M. It is well known that the tangent bundle is a smooth manifold of dimension two times the dimension of M. Using this smooth manifold we can give the de nition. De nition 6. A vector eld on a smooth manifold M is a map X : M → TM, such that for each p ∈ M we assaign a vector v ∈ TMp: X : p 7→ (p, v). Since TM is a smooth manifold we have an earlier de nition that guarantees that it makes sense to ask wether X is smooth or not. Hence we can simply de ne a smooth vector eld to be a vector eld X that is smooth by this de nition (de nition 2). De nition 7. We say that a manifold M is orientable if there are parametriza- tions {φi} covering all of M such that for every pair φi : Ui → Vi, φj : Uj → Vj with , the change of coordinates −1 has positive determinant. Vi ∩ Vj 6= ∅ φj ◦ φi For surfaces: A surface S is orientable if and only if it is impossible to embedd a möbius band B in S.3 3 An embedding of B in S is a map e : B → S such that (i) dϕp : TBp → TSp is injective for all p ∈ B, and (ii) ϕ is a homeomorphism from B to ϕ(B).
5 Now we give some facts and terminology about vector elds which will be used in many of the proofs. We begin with trajectorys.
De nition 8. A trajectory in a point p of a smooth vector eld X is map γ :(−, ) → V , from an open interval (−, ) to a neighbourhood V of p such that γ(0) = p and γ0(t) = X(γ(t)). A trajectory of a smooth vector eld X, from a point p to another point q, is a smooth curve γ : (0, 1) → S such that γ(t) → p as t → 0, γ(t) → q as t → 1, and γ0(t) = X(γ(t)). The curve in the de nition above is known to exist and be unique for small enough . Next we mention, a theorem from the theory of dierential equations which will be needed.
Theorem 1. Suppose that X is a smooth vector eld on a manifold M, and let p ∈ M. Then there exists a neighbourhood V ⊂ M of p, an interval (−, ) and a smooth map ϕ :(−, ) × V → M such that the curve
t 7→ ϕ(t, q), t ∈ (−, ), for any q ∈ V , exists and is the unique satisfying
∂ϕ = X(ϕ(t, q)), ϕ(0, q) = q. ∂t The curves t → ϕ(t, q) in the theorem above, for any xed q, whose deriva- tives are the trajectories, are called integral curves of the vector eld. To de ne a genus-g torus, we are going to need the notion of dieomorphisms being isotopic. This is something that also will be used in the sections about Morse functions. We are also going to need connected sums. These two notions allow us to past together surfaces in a unique way.
De nition 9. Suppose that ϕ0 : M → M and ϕ1 : M → M are dieomorphisms on a smooth manifold M. We say that they are smoothly isotopic, if there is a smooth map (a smooth homotopy) F : [0, 1] × M → M such that:
1. F (0, p) = ϕ0(p), for all p ∈ M,
2. F (1, p) = ϕ1(p), for all p ∈ M, and 3. F (t, p): M −→∼ M is a dieomorphism for all t ∈ [0, 1]. ∼ Lemma 1. Any orientation preserving dieomorphism ϕ : S1 −→ S1 is smoothly isotopic to the identity map.
∼ Proof. Let S1 be given by R/2πZ, and let ϕ : R/2πZ −→ R/2πZ be an ori- entation preserving dieomorphism. It is clear that any orientation preseving dieomorphism of R/2πZ onto itself can be given as function on R onto it- self, which on [0, 2π] is a strictly increasing smooth function going from some θ ∈ [0, 2π) to θ + 2π, and which is 2π-periodic. Suppose ϕ goes from θ to θ + 2π. Let the isotopy be de ned by
F (t, x) = tϕ(x) + (1 − t)x.
6 It is clear that F (0, x) = id and F (1, x) = ϕ. We show that it is a dieomor- phism going from tθ to tθ + 2π, for any xed t. Firstly, F is clearly smooth. Moreover, strictly increasing:
0 0 Fx(t, x) = tϕ (x) + (1 − t) > 0, because ϕ0(x) > 0 (and either t > 0 or 1 − t > 0, always t, 1 − t ≥ 0). It starts at tθ, because F (t, 0) = tϕ(0) + (1 − t)0 = tθ, and ends at tθ + 2π, because F (t, 2π) = tϕ(2π) + (1 − t)2π = t(θ + 2π) + (1 − t)2π = tθ + 2π.
De nition 10. Let S1 and S2 be two surfaces. The connected sum of S1 and S2, denoted S1#S2, is obtained in the follwing way. Begin by deleteing an open ball from each surface, denote the balls B1 and B2; then both S1 − B1 and S2 − B2 have a boundary - denoted ∂B1 and ∂B2, respectively - dieomorphic to S1. De ne an orientation preserving dieomorphism ∼ g : ∂B1 −→ ∂B2.
Then S1#S2 is obtaint by taking the disjoint union (S1 −B1)∪g (S2 −B2), where we identify the points on ∂B1 and ∂B2 by the dieomorphism g. The obtained surface S1#S2 is by lemma 1 unique up to dieomorphism. The genus of a surface is the number of holes of a surface. Hence a genus-g torus is like a torus but with g holes, instead of just one. The formal de nition is as follows.
De nition 11. We de ne a genus-g torus, denoted Tg, inductively. A 0-torus is any surface homeomorphic to a sphere. A 1-torus is any surface homeomorphic to the torus; we can take the following one as archetype: 2 3 p 2 2 2 2 T1 = (x, y, z) ∈ R | b − x + y + z = a , where a, b > 0 are constants such that b > a.(a is the radius of the tube, b is the radius from the center of the torus hole to the center of the tube.) Suppose we have de ned , for all . We then de ne as . Tg g ≤ g0 Tg0+1 Tg0 #T1
Figure 1: The connected sum of a 1-torus and a 2-torus, obtaining a 3-torus.
7 We will show, in the last section, that the sphere and the torus both are unique up to dieomorphism, and not only homeomorphism, from which it will follow that all Tg are. The great accomplishment of the classi cation theorem is that it shows that the number of "holes" in a surface is enough to classify them: all surface with g holes are dieomorphic to the genus-g torus, and hence to each other, and so up to dieomorphism the genus-g toruses make up all possible surfaces.
8 Morse Theory
This section introduces some of the central notions of Morse theory. Morse theory is about a class of smooth real valued functions on manifolds (surfaces), which can be seen as a way to de ne and measure the height (possibly negative) of a manifold (surface), such that the changes as the height increases are good (in a sense given later). In the next section we will see how we can deform the measurement - rather then deforming the surface - so that we get a good way to measure height: we show that there exists a good Morse function. One section further we will connect this to the notion of genus. The theory here is only given for surfaces; most results hold also for arbitrary manifolds, only the proofs have to be slightly changed. We begin by de ning the concepts of critical point and critical value, after- wards we de ne the concept of index which allows us to compare dierent types of critical points.
De nition 12. Let S be a surface, and let f : S → R be a real valued function. We say that p ∈ S is a critical point of f if
f 0(p) = 0, where 0 is the induced map of the tangent spaces. In a f : TSp → T Rf(p) parametrization φi, with φi(x0, y0) = p, this means that
∂(f ◦ φ ) ∂(f ◦ φ ) i (x , y ) = i (x , y ) = 0. ∂x 0 0 ∂y 0 0
A critical value x ∈ R of f is a real number that is the image of a critical point. Any points which is not a critical point of f is called a regular point of f, and any real number which is not a critical value of f is called a regular value of f.
De nition 13. Let S be a surface, let p ∈ S be a point, let f : S → R be a real valued function on S, and let φ be a parametrization of S around p with φ(x0, y0) = p. The Hessian of f with respect to φ is de ned as
∂2(f◦φ) ∂2(f◦φ) ! ∂x2 ∂x∂y Hφ(f) = ∂2(f◦φ) ∂2(f◦φ) . ∂x∂y ∂y2
If φ and ψ are two dierent parametrizations of S around p, with φ(x0, y0) = ψ(x0, y0) = p, then the Hessian of f with respect to φ and ψ are related by
T Hψ(f) = J (θ) ◦ Hφ(f) ◦ J(θ), where θ is the change of coordinates from ψ to φ, i.e. θ = φ−1 ◦ ψ, and J(θ) is the Jacobian of θ (J T (θ) is just its transpose). Because of this the following de nition is independent of parametrization.
De nition 14. Let f : S → R be a real valued smooth function on a surface S. We say that a critical point p ∈ S is non-degenerate, if the matrix Hφ(f)(x0, y0) is non singular for any parametrization φ, with φ(x0, y0) = p. Critical points with singular hessian are called degenerate.
9 Later we will show that there for any surface S exists a function f : S → R where all critical points are non-degenerate, with terminology to be de ned: on any surface there exists a Morse function. Hence to nd important imformation of a surface one only needs to consider non-degenerate critical points (this also depends on a lemma roughly saying that we only need to consider regions around critical points). To do so, we use the following de nitions:
De nition 15. Let f : S → R be a smooth real valued function on a surface S. If all critical points of f are non-degenerate, then we say that f is a Morse function on S.
De nition 16. Let f : S → R be a Morse function on a surface S. Let p ∈ S be a critical point. The index of p with respect to f is de ned to be the dimension of the maximal subspace of TSp on which H(f) is negative de nite. Again using the Jacobian, one can show that this de nition is independent of coordinates. Note that the index is 0, 1 or 2, depending on whether p is a local minimum, a sadlepoint or a local maximum respectively. The following theorem, due to Morse, provides the key tool when using the index of a critical point to analyze a surface.
Theorem 2 (Lemma of Morse). Let f : S → R be a real valued smooth function on a surface S. Let p ∈ S be a non-degenerate critical point. Then there exists a parametrization φ : U → V , with φ(x0, y0) = p ∈ V , such that: Index(p) = 0 =⇒ f ◦ φ(x, y) = f(p) + x2 + y2, Index(p) = 1 =⇒ f ◦ φ(x, y) = f(p) + x2 − y2, Index(p) = 2 =⇒ f ◦ φ(x, y) = f(p) − x2 − y2. Proof. Let φ be a parametrization around p. We will construct new coordinates ˜−1 such that ˜−1 2 2. Then (by maybe switching φ = (v1, v2) f ◦ φ = f(p) ± v1 ± v2 place with v1 and v2) it is immidiate from ±2 0 H (f) = φ˜ 0 ±2 that if the index is then 2 2, if the index is then 0 f = f(p) + v1 + v2 1 f = 2 2, and if the index is then 2 2. f(p) + v1 − v2 2 f = f(p) − v1 − v2 −1 Firstly, we can assume (x0, y0) = 0, i.e. φ (p) = 0. Moreover we can assume that f(p) = 0. If we take U small enough we can assume U convex. Then we note that g : U → R de ned by g = f ◦ φ can be written
g(x, y) = xh1(x, y) + yh2(x, y), for some smooth functions h1, h2. Because, by using the fundamental theorem of analysis and the chain rule, we see that
Z 1 Z 1 0 ∂g ∂g g(x, y) = gt(tx, ty)dt = (tx, ty)x + (tx, ty)y dt. 0 0 ∂x ∂y Hence we can let Z 1 ∂g Z 1 ∂g h1(x, y) = (tx, ty)dt, h2(x, y) = (tx, ty)dt. 0 ∂x 0 ∂y
10 Since is a critical point, ∂g ∂g . Therefore, p = 0 h1(0) = ∂x (0) = 0, h2(0) = ∂y (0) = 0 by using the same reasoning as above on f, now on h1, h2, we can write
h1(x, y) = xk11(x, y) + yk12(x, y) h2(x, y) = xk21(x, y) + yk22(x, y), and hence also:
2 2 g(x, y) = x k11 + xy(k12 + k21) + y k12. We can assume that (if , replace with 0 0 k12 = k21 k12 6= k21 k12, k21 k12 = k21 = 1 ). Also note that 2 (k12 + k21) ∂2g ∂2g ! k11(0) k12(0) 1 ∂x2 (0) ∂xy (0) = ∂2g ∂2g , k21(0) k22(0) 2 ∂xy (0) ∂y2 (0) hence the matrix (kij)(0) is non-singular. Now we only need to show that we can pick coordinates so that k12 = k21 = 0, and k11 = ±1, k22 = ±1, in a neighbourhood around 0. We will show this by doing two coordinate changes. The rst coordinate change is p yk12(x, y) u1(x, y) = |k11(x, y)| x + k11(x, y)
u2(x, y) = y.
(It is possible, by a linear change, to assume k11(x, y) 6= 0, maybe adding an extra coordinate change). Since, for the Jacobian J of the mapping (x, y) 7→ (u1, u2) at 0, we have that
∂u1 ∂u1 ! det ∂x ∂y det AB p J(0, 0) = ∂u2 ∂u2 (0) = (0) = A(0) = |k11(0, 0)|, ∂x ∂y 0 1 (A,B)4 and therefore J(0, 0) 6= 0, we can use the inverse function theorem to conclude that (x, y) 7→ (u1, u2) in a small neighbourhood of 0 actually is a coordinate change. Now we see that f expressed in coordinates u1, u2 becomes 2 2 2 2 (x k11 + 2xyk12 + y k22)(u1, u2) = ±u1 + l(u1, u2)u2, where l(u1, u2) is a smooth function. The second coordinate change is just
v1(u1, u2) = u1 p v2(u1, u2) = |l(u1, u2)|u2.
The mapping (u1, u2) 7→ (v1, v2) is even simpler to show is a coordinate change. Finally we get that f expressed in v1, v2 is 2 2 f = ±v1 ± v2. This completes the proof.
4 k k yk k k − k k A = 11 11x x + 12 + p|k | 1 + y 12x 11 12 11x 3/2 11 2 2|k11| k11 k11 k11k11 yk k k12 k11 − k12k11 B = y x + 12 + p|k | 12 + y y y 3/2 11 2 2|k11| k11 k11 k11
11 The following result is now an immediate consequence of the above theorem. As an application on surfaces, by compactness there can only be nitely many non-degenerate critical points. Corollary 1. Non-degenerate critical points of a surface are isolated.
Figure 2: A torus standing on the xy-plane, the height along the z-axis shown by a Morse function f. There are four critical points, p0, p1, p2, p3, having indices 0, 1, 1, 2 respectively. The lines mark the critical values f(pj).
Next we will de ne, and prove the existence of, a gradient-like vector eld for a Morse function f on any surface S. Properties of these vector elds will be used later on.
De nition 17. Let f : S → R be a Morse function on a surface S.A gradient like-vector eld for f, is a smooth vector eld ξ such that 1. ξ(f)(q) > 0 for all regular points q, and 2. given a critical point p of f, there are coordinates φ−1 = (x, y) in a neigh- bourhood U of p, such that f ◦ φ = f(p) ± x2 ± y2 and ξ has coordinates (±x, ±y), where the sign depends on index as above (either both positive if index 0, x positive and y negative if index 1, or both negative if index 2), in all of U. Lemma 2. For every Morse function f on any surface S, there exists a gradient- like vector eld ξ for f. For the proof we will need a fact about partitions of unity. Partitions of unity provides an excellent tool for extending local properties of neighbourhoods to the whole surface.
12 De nition 18. Let S be a surface. A family of open subsets {Ui}i∈I of S is said to be locally nite if every point p ∈ S has a neighbourhood W such that W ∩ Ui 6= ∅ for only a nite number of indices in I. The support of a function f : S → R is the closure of the set of points where f is non zero; we denote the support by supp(f). Let {Vi}i∈I be an open cover of S. We say that a family of smooth functions is a partition of unity subordinate to the cover if: {fi : S → R}i∈I {Vi}
1. For all i ∈ I, fi ≥ 0 and supp(fi) ⊂ Vi.
2. The family {supp(fi)}i∈I is locally nite. 3. P , for all . i∈I fi(p) = 1 p ∈ S Theorem 3. For any surface S and any locally nite cover of neighbourhoods {Vi} of S, there exists a partition of unity subordinate to the cover. We will prove this theorem using a type of functions called bump functions. Bump functions will also be used in later proofs when constructing certain vector elds. De nition 19. A bump function is a smooth nowhere negative real valued function from a manifold (for us will be a surface or some b : M → R≥0 M M subset of R2), having compact support. Here is a proof of existence of a kind of bump functions on surfaces.
Lemma 3. Given any surface S, a parametrization φ : U → V ⊂ S, and any point p ∈ V . There exists a bump function b : S → R on S such that f(S) ⊂ [0, 1], supp(S) ⊂ V and b(q) = 1 in a neighbourhood of p. Proof. The function ( − 1 e x , if x > 0, h(x) = 0, if x ≤ 0 is smooth at all points in R. Let C(s) = {(x, y) ∈ R2||x|, |y| < s} denote the the open square in R2 with side 2s; let C(s) denote its closure. We will begin by constructing a smooth function 2 such that , b0 : R → R b0(x, y) ∈ [0, 1] b0(x, y) = 1 for (x, y) ∈ C(1), and b0(x, y) = 0 for (x, y) 6∈ C(2). Let h(x) h (x) = , 0 h(x) + h(1 − x) so that h0(x) = 1 for x ≥ 1, and h0(x) = 0 for x ≤ 0. Now let
h1(x) = h0(x + 2)h0(2 − x).
Then h1(x) = 1 for |x| ≤ 1, and h1(x) = 0 for |x| ≥ 2. Thus we can let b0 be de ned by
b0(x, y) = h1(x)h1(y). −1 −1 0 Let U = φ (V ) and (x0, y0) = φ (p). Take a square neighbourhood C = −1 {(x, y) ∈ U||x0 − x| < , |y0 − y| < } of φ (p) in U. Let 2 k(x, y) = (x − x , y − y ); 0 0
13 this function maps C0 dieomorphically on C(2). Then the desired bump func- tion is given by ( b ◦ k ◦ ϕ−1(q), if q ∈ V b(q) = 0 0, otherwise.
Proof of theorem 3. For each , let be a bump function so that p ∈ S hp : S → R hp = 1 in a neighbourhood Up ⊂ supp(hp) ⊂ Vi, for some i. Clearly {Up}p∈S is an open cover of S, by compactness we can pick a nite number of the Up's, , that also cover . We have that , for some , U1,...,Un S Ul ⊂ Vil Vil ∈ {Vi} . For all such that , let constantly. For l = 1, . . . , n i Vi 6∈ {Vi1 ,...,Vin } fi = 0 let Vi ∈ {Vi1 ,...,Vin }
hl1 + ··· + hlk fi = Pn , j=1 hj where the comes from the 's such that . It is now hl1 , . . . , hlk Ulj Ulj ⊂ Vi = Vil easy to check that {fi} is a partition of unity subordinate to the cover {Vi}. When we further use bump functions we will sometimes assume they have similar extra properties. We will omit proving that these bump functions actu- ally exists, even though they of course do exists, and the proofs are not hard. Also the proof of lemma 3 contains the crucial ideas to how one does.
Proof of lemma 2. Let p1, . . . , pn be the critical points of f. By theorem 2, for i i i i each pi there is a parametrization φ : U → V such that on U , f ◦ φi(x, y) = 2 2. Let i be neighbourhoods of such that i i. Let f(pi) ± x ± y V0 pi V0 ⊂ V Sn i and Sn i. U0 = i=0 U0 V0 = i=0 V0 Each Sn i is a regular point of , hence, by the implicit function q ∈ S− i=1 V f theorem, we can write f ◦ φq(x, y) = constant + x in Uq, where φq : Uq → Vq is a parametrization. Note that Sn i is compact. Using this and the earlier nding, we S − i=1 V can nd neighbourhoods V1,...,Vk such that 1. Sn i Sk , S − i=1 V ⊂ i=1 Vi
2. V0 ∩ Vj = ∅, for 1 ≤ j ≤ k, and
i 3. there are parametrizations φj : Uj → Vj so that f ◦ φ = constant + xi on Uj, for j = 1, . . . k. Now, for all 1 ≤ i ≤ n, on V i there are vector elds vi with coordinates (±x, ±y), and for each 1 ≤ j ≤ k, on Vj there is the vector eld vj, which in coordinates is ∂/∂x. We can now piece together these vector elds using a partition of unity 1 n 1 n subordinate to the cover V ,...,V ,V1,...,Vk: let {f , . . . , f , f1, . . . , fk} be such a partition of unity, now we can de ne the smooth vector eld by
n k X i i X ξ(q) = f (q)v (q) + fj(q)vj(q). i=1 j=1 It is well de ned because the sum is nite, and smooth because all functions i f , fj, and all locally de ned vector elds are. We now only have to check that ξ(f)(q) > 0 for any regular point q.
14 1 n Let q ∈ S be a regular point. Then at least one of f , . . . , f , f1, . . . , fk is non-zero at q, and all of them are non-negative. Moreover if q is in any of the i sets Vj, then vj(f) = ∂/∂x(x + constant) = 1 > 0; if q is in any of the sets V then vi(f) = (±x, ±y) · (±2x, ±2y) = 2x2 + 2y2 > 0. Thus we obtain a gradient-like vector eld ξ for f on S.
Now let S be a surface and f be a Morse function on S. For a ∈ R, let Sa = {p ∈ S|f(p) ≤ a}, which is a compact submanifold of S, possibly with boundary. We now prove an important lemma about how Sa changes when a increases. It shows that we only need to analyze a surface around the critical points of a Morse function when classifying the surface.
Lemma 4. Let a < b be two real numbers, and suppose f −1[a, b] = {p ∈ S|a ≤ f(p) ≤ b} ⊂ S contains no critical points of f. Then Sa is dieomorphic to Sb. For the proof we need a tool provided by 1-parameter groups of dieomor- phisms. And a lemma about how to generate them, which we state only for surfaces.
De nition 20. A 1-parameter group of dieomorphisms of a manifold M is a smooth map ϕ : R × M → M such that: 1. For each xed t ∈ , the map de ned by M 3 p 7→ ϕ(t, p), denoted ∼ R ϕt : M −→ M, is a dieomorphism. 2. For all , . s, t ∈ R ϕs+t = ϕs ◦ ϕt Given a a 1-parameter group of dieomorphisms ϕ on a smooth manifold M, we de ne a smooth vector eld X on M by
f(ϕ (q)) − f(q) X(f)(q) = lim h , h→0 h where f is any smooth real valued function on M. The vector eld X is said to generate ϕ. Lemma 5. A smooth vector eld X on a surface S, which vanishes outside of a compact subset K of S, generates a unique 1-parameter group of dieomor- phisms ϕ on S. Proof. The rst thing we prove is that ϕ - satisfying the equality above - as- suming it is a 1-parameter groupof dieomorphisms, must be unique, then we go on and show that one exists. Let ϕ be 1-parameter group of dieomorphisms generated by X. For the curve t 7→ ϕt(q) we have that
dϕ (q) f(ϕ (q)) − f(ϕ (q)) t (f) = lim t+h t dt h→0 h f(ϕ (ϕ (q))) − f(ϕ (q)) = lim h t t h→0 h = X(f)(ϕt(q)).
15 Hence, t 7→ ϕt(q) satisfy the dierential equation with initial condition dϕ (q) t = X(ϕ (q)), dt t ϕ0(q) = q, because ϕ0 = id. This dierential equation has a locally unique smooth solution by theorem 1, q ∈ Vq, t ∈ (−q, q), for each q ∈ S which extends uniquely globally because R is connected. This proves the rst part. Cover the compact set K by a nite number of open neighbourhoods V1,...,Vn from the family {Vq}q∈K . Let 0 be the smallest of the numbers i correspond- ing to Vi, 1 ≤ i ≤ n. If we let ϕt(q) = q for q 6∈ K, then the dierential equation above has a unique solution for t ∈ (−0, 0) and all q ∈ S. This function is smooth, and it is obvious that if 0 , then . Hence |s|, |t|, |s + t| < 2 ϕs+t = ϕs ◦ ϕt ∼ each ϕt, t ∈ (−0, 0), is a dieomorphism ϕt : S −→ S. For , we write 0 , , 0 0 . Then, if set |t| ≥ 0 t = n 2 + rt n ∈ Z rt ∈ (− 2 , 2 ) n ≥ 0
ϕt = ϕ 0 ◦ . . . ϕ 0 ◦ ϕr , 2 2 t
0 where ϕ 0 is iterated n times, and if n < 0 do as above but with replaced by 2 2 0 . Then is smooth, well de ned and a dieomorphism for all , and − 2 ϕt t ∈ R clearly satisfy for all . ϕ ϕs+t = ϕs ◦ ϕt s, t ∈ R Proof of lemma 4. Pick a gradient-like vector eld ξ for f. Then, since by as- sumption f −1[a, b] contains no critical points of f, we have that ξ(f)(p) > 0 for all −1 . Multiplying at each point in −1 by 1 we obtain a p ∈ f [a, b] ξ f [a, b] ξ(f) new gradient-like vector eld ξ˜ for f in f −1[a, b] - then ξ˜(f) is constantly equal to 1 in f −1[a, b] - which we can assume, using a bump function, vanishes outside a compact neighbourhood of this set. Hence lemma 5 applies, and so we get a −1 1-parameter group of dieomorphisms ϕ˜ on S. For any ϕ˜t(p), with p ∈ f [a, b] we get that df(ϕ ˜ (p)) t = ξ˜(f)(p) = 1, dt hence f(ϕ ˜t(p)) = f(p) + t, where f(ϕ ˜0(p)) = f(p) comes from ϕ˜0 = id. We now ∼ a b conclude that the dieomorphism ϕ˜b−a : S −→ S sends S onto S , and hence a ∼ b induces, by restriction, a dieomorphism ϕ˜|Sa : S −→ S . For this we only a b a need to show that ϕ˜b−a(S ) = S . To see this, if p ∈ S , then f(ϕ ˜b−a(p)) = b b (b − a) + f(p) ∈ S , because f(p) ≤ a, hence ϕ˜b−a(p) ∈ S ; if on the other hand b, then −1 a, beacuase , hence p ∈ S f(ϕ ˜b−a(p)) = (a − b) + f(p) ∈ S f(p) ≤ b −1 a. ϕ˜b−a(p) ∈ S We conclude this section with the theorem mentioned above ensuring the existence of a function with no degenerate critical points. In the following sections all critical points of a real valued smooth function will therefore be assumed to be non-degenerate.
Theorem 4. For any surface surface S there exists a Morse function f : S → R on S.
What we will do is to embedd the surface in Rm, which always is possible for some large enough 5, and then show that the function , de ned by m Lp : S → R 5See [3].
16 2, where is the euclidean norm in m, contains no degenerate Lp(q) = kp−qk k, k R critical points for almost all p ∈ Rm. (For almost all will here mean the same as all but a set of measure zero). Let N = {(p, v) ∈ S × Rm|v · p = 0}; this is the set of all pairs (p, v) with p ∈ S and v ∈ Rm perpendicular to p. It is easy to show that N is an m-dimensional manifold, sometimes called a normal bundle of S.
De nition 21. Let E : N → Rm be de ned by E(p, v) = p + v, and let q ∈ S. We de ne a focal point of (S, q), to be a point e ∈ Rm such that e = q + v, for some v with (q, v) ∈ N, where the kernel of the jacobian JE(q, v) of E at (q, v) is non-trivial. A focal point of S is a focal point of (S, q), for some q ∈ S.
The goal is to show that a point q ∈ S is a degenerate critical point of Lp, if and only if p is a focal point of (S, q). Then theorem 4 follows from a corollary of the following theorem due to Sard6.
Theorem 5 (Sard's theorem). Let f : U → Rm be a smooth map from an open set n, and are any natural numbers. Let rank U ⊂ R n m C = {x ∈ U| (dfx) < m}. Then the set f(C) ⊂ Rm has measure zero in Rm.
Corollary 2. For almost all x ∈ Rm, x is not a focal point of S. Now we de ne the rst and second fundamental form, they will help us locate the focal points. Let (x, y) = φ−1 be coordinates of an open set V ⊂ S. The inclusion map m determines smooth functions , S → R m u1(x, y), . . . , um(x, y) where we use the notation: 2 m, . u˜ : R ⊃ U → R u˜(x, y) = (u1(x, y), . . . , um(x, y)) And so to distinguish between S and S ⊂ Rm, we write q when we talk about points in the surface and q˜ when we talk about points in the surface embedded in Rm. De nition 22. The rst fundamental form associated with the coordinate sys- tem (x, y) (and with the embedding into Rm) is the symmetric matrix of real valued functions
∂u˜ ∂u˜ ∂u˜ ∂u˜ ! g˜1 g˜2 ∂x · ∂x ∂x · ∂y (gi) = = ∂u˜ ∂u˜ ∂u˜ ∂u˜ . g˜2 g˜3 ∂x · ∂y ∂y · ∂y The second fundamental form is the symmetric matrix of vector valued function
˜ ˜ ˜ l1 l2 (li) = ˜ ˜ l2 l3 de ned as follows. Let the vectors ∂2u˜ , ∂2u˜ and ∂2u˜ , each a˜1 = ∂x2 a˜2 = ∂x∂y a˜3 = ∂x2 be expressed as a sum of a vector tangent to S, and a vector normal to S. We ˜ 7 de ne li as the normal component of ai, i = 1, 2, 3 . 6See [6]. 7 m m We can de ne an embedding of S in R as a smooth map u˜ : S → R , such that (i) the dierential du˜p is injective for all p ∈ S, and (ii) u˜ is a homeomorphism from S to u˜(S). Then m the tangent space of S in R , at any point p, can be de ned as du˜p(TSp). Now it easy to de ne tangential and normal components of vectors.
17 Given a vector n˜ that is orthogonal to S at the point p˜ with kn˜k = 1, the matrix ˜ ˜ n˜ · a˜1 n˜ · a˜2 n˜ · l1 n˜ · l2 = ˜ ˜ n˜ · a˜2 n˜ · a˜3 n˜ · l2 n˜ · l3 is called the second fundamental form in the direction n˜ at the point p˜.
Now we assume that the coordinates are chosen so that (˜gi) at p˜ is the identity matrix. Let n˜ be a unit vector orthogonal to S at p˜. Then we de ne the principal curvatures K1 and K2 at the point p in the normal direction n˜ as ˜ 1 1 the eigenvalues of the matrix (˜n · li); and are called the principal radii K1 K2 ˜ of curvature. If the matrix (˜n · li) is singular, then at least one of K1 and K2 are zero, and hence the corresponding 1 is not de ned.8 Ki Now let lp,˜ n˜ =p ˜ + tn˜ be the line in direction n˜, where n˜ is a normal vector with kn˜k = 1, through the point p˜ ∈ S.
1 Lemma 6. The focal points of (S, q) along ln,˜ q˜ are precisely the points q˜+ n˜. Ki
Proof. Choose m−2 smooth vector elds v1(x, y), . . . , vm−2(x, y), with kvik = 1, along in m, such that at each point, is orthogonal to and each pair of S R vi S vectors vi, vj, i 6= j, are orthogonal to each other. We now introduce coordinates −1 on by φN = (x, y, t1, . . . , tm−2) N
φN (x, y, t1, . . . , tm−2) = (˜u(x, y), t1v1(x, y) + ··· + tm−2vm−2(x, y)).
Then the function E : N → Rm induces a function m−2 E∗ X (x, y, t1, . . . , tm−2) 7→ u˜(x, y) + tivi(x, y). i=1 The partial derivatives of this function are
∗ m−2 ∂E ∂u˜ X ∂vi = + t , ∂x ∂x i ∂x i=1 ∗ m−2 ∂E ∂u˜ X ∂vi = + t , ∂y ∂y i ∂y i=1 ∂E∗ = vi. ∂ti Now consider the following matrix Q, obtained by taking the inner product of the partial derivatives of ∗ with the vectors ∂u˜ ∂u˜ : E ∂x , ∂y , v1, . . . , vm−2
∂E∗ ∂u˜ ∂E∗ ∂u˜ ∂E∗ ∂E∗ ∂x · ∂x ∂x · ∂y ∂x · v1 ... ∂x · vm−2 ∂E∗ ∂u˜ ∂E∗ ∂u˜ ∂E∗ ∂E∗ ∂y · ∂x ∂y · ∂y ∂y · v1 ... ∂y · vm−2 ∂E∗ ∂u˜ ∂E∗ ∂u˜ ∂E∗ ∂E∗ · · · v1 ... · vm−2 Q = ∂t1 ∂x ∂t1 ∂y ∂t1 ∂t1 . . . . . . ...... . . . . ∂E∗ ∂u˜ ∂E∗ ∂u˜ ∂E∗ ∂E∗ · · · v1 ... · vm−2 ∂tm−2 ∂x ∂tm−2 ∂y ∂tm−2 ∂tm−2 8For a more detailed introduction to these concepts see for example [1].
18 To simplify, we can write this matrix as being constituted by four matrices A, B, C, D in the following way:
AB Q = , CD where we after also simplifying gets, ∂u˜ Pm−2 vi ∂u˜ ∂u˜ Pm−2 ∂vi ∂u˜ ∂x + i=1 ti ∂x · ∂x ∂x + i=1 ti ∂x · ∂y A = , ∂u˜ Pm−2 ∂vi ∂u˜ ∂u˜ Pm−2 ∂vi ∂u˜ ∂y + i=1 ti ∂y · ∂x ∂y + i=1 ti ∂y · ∂y m−2 m−2 ! P t ∂vi · v ... P t ∂vi · v B = i=1 i ∂x 1 i=1 i ∂x m−2 , Pm−2 ∂vi Pm−2 ∂vi i=1 ti ∂y · v1 ... i=1 ti ∂y · vm−2 0 0 . . . . C = . . , 0 0
D = Im−2. It is clear that the rank of this matrix is equal to the rank of the jacobian of (because ∂u˜ ∂u˜ are linearly independent), at the corresponding E ∂x , ∂y , v1 . . . , vm−2 points (corresponding, of course, by the coordinates or parametrization). Also we see that the nullity of Q is determined only by the upper left corner matrix A; focusing on this matrix then, we see that by the identities
∂ ∂u˜ ∂v ∂u˜ ∂2u˜ 0 = v · = i + v , ∂x i ∂x ∂x ∂x i ∂x2 ∂ ∂u˜ ∂v ∂u˜ ∂2u˜ 0 = v · = i + v , ∂x i ∂y ∂x ∂y i ∂x∂y ∂ ∂u˜ ∂v ∂u˜ ∂2u˜ 0 = v · = i + v , ∂y i ∂x ∂y ∂x i ∂x∂y ∂ ∂u˜ ∂v ∂u˜ ∂2u˜ 0 = v · = i + v , ∂y i ∂y ∂y ∂y i ∂y2 i = 1, 2, . . . , m − 2, we have that Pm−2 ˜ Pm−2 ˜ g˜1 − i=1 tivi · l1 g˜2 − i=1 tivi · l2 A = Pm−2 ˜ Pm−2 ˜ . g˜2 − i=i tivi · l2 g˜3 − i=1 tivi · l3 Thus we have shown that q˜ + tn˜ is a focal point of (S, q), if and only if A is singular. Now if (gi) is the identity matrix, i.e. g1 = 1, g2 = 0, g3 = 1, then is singular if and only if 1 is an eigenvalue of the matrix ; hence is A t (˜n · li) A singular if and only if t = 1 or t = 1 , and this completes the proof. K1 K2 Proof of theorem 4. Let m be a xed point. For the function m, p ∈ R Lp : S → R which in coordinates is de ned by
2 Lp(˜u(x, y)) = ku˜(x, y) − pk =u ˜(x, y) · u˜(x, y) − 2˜u(x, y) · p + p · p,
19 we have that ∂L ∂u˜ p = 2 (˜u − p), ∂x ∂x ∂L ∂u˜ p = 2 (˜u − p). ∂y ∂y
Thus q˜ =u ˜(x, y) is a critical point of Lp if and only if q˜ − p is orthogonal to S at q˜. Moreover we have
∂2L ∂2u˜ ∂u˜ ∂u˜ p = 2 · (˜u − p) + · , ∂x2 ∂x2 ∂x ∂x ∂2L ∂2u˜ ∂u˜ ∂u˜ p = 2 · (˜u − p) + · , ∂x∂y ∂x∂y ∂x ∂y ∂2L ∂2u˜ ∂u˜ ∂u˜ p = 2 · (˜u − p) + · . ∂y2 ∂y2 ∂y ∂y
Setting p =u ˜ + tn˜ as in lemma 6, we obtain
2 2 ∂ Lp ∂ Lp ! ˜ ˜ ∂x2 ∂x∂y g˜1 − tn˜ · l1 g˜2 − tn˜ · l2 2 2 = 2 . ∂ Lp ∂ Lp ˜ ˜ g˜2 − tn˜ · l2 g˜3 − tn˜ · l3 ∂x∂y ∂y2
Therefore we have that q˜ ∈ S is a degenerate critical point of Lp if and only if p is a focal point of (S, q). Now using corollary 2 we obtain the nal result: For almost all m the function has no degenerate critical points. p ∈ R Lp : S → R
20 Existence of a Good Morse Function
We will here use the properties of Morse functions on surfaces - and the fact that we always can nd a Morse function - to show that we always can nd a good Morse function on a surfaces. We will proceed by rst show that any Morse function can be turned into a self-indexing Morse function, and thereafter prove and use a cancelation theorem to cancel critical points against each other. From this we will in next section show that the surface up to dieomorphism can be determined by the number of critical points having index 1 of the good Morse function, and moreover that this number is the same as two times the genus of the surface.
De nition 23. Let f : S → R be a Morse function. We say that f is self- indexing if, for any critical point p ∈ S, we have
f(p) = index(p).
De nition 24. Let f : S → R be a Morse function. We say that f is a good Morse function if f has only one maximum and one minimum (one critical point with index 2, and one with index 0), and is self-indexing. The goal of this section is the following theorem.
Theorem 6. Every surface S has a good Morse function f : S → R. Proving this is the main step in the classi cation of surfaces. We begin by introducing some new theory. Begin by observing that, by theorem 2, close to a critical point p, on every level set f = f(p) + there is an induced sphere. In local coordinates this is the 1-sphere x2 + y2 = which goes upwards if the index of p is 0 and downwards if the index is 2. If the index instead is 1 we get one 0-sphere which goes upwards,
x2 = y = 0, and one 0-sphere which goes downwards
x = 0 y2 = .
By lemma 4 this extends to an induced sphere on all level sets f = a, as long as we do not pass a critical value on the way from p. Note that if we go upwards from a critical point having index 0, we get an induced sphere having dimension 1; if we go upwards or downwards from a critical point having index 1, we get an induced sphere having dimension 0 (that is, two disjoint points); if we go downwards from a critical point having index 2, we get an induced sphere having dimension 1. Now we rst prove that we can turn any Morse funtion into a self-indexing one.
21 Lemma 7. Let f : S → R be a Morse function. Let p ∈ S be a critical point of f with index 0 (index 2), such that f(p) = a. Then for all b ≤ a (all b ≥ a), there exists a Morse function g : S → R with the same critical points as f having the same indices, with g(p) = b, and such that f and g coincides except on a neighbourhood V of p.
Proof. We prove the case with indexf (p) = 0, the case indexf (p) = 2 is similar. If b = a we are done; so let b < a. Let V be a neibourhood centered at p, such that trough out U = φ−1(V ), we have f ◦ φ−1 = f(p) + x2 + y2; moreover, by possibly making V smaller, we can assume that U = {(x, y) ∈ R2|x2 + y2 < c}, where c is some small real number. Let γ :[a, a + c) → R be a smooth map with the properties: 1. dγ , , dt (t) > 0 t ∈ [a, a + c) 2. γ(a) = b, 3. γ(t) = t, t ∈ (a + c − , a + c), for some small (i.e. for t near a + c).
Now we de ne g : S → R by: ( γ(f(q)), if q ∈ V, g(q) = f(q), if q 6∈ V.
Clearly g is smooth. Also, we have that g(p) = γ(f(p)) = b, and by the chain rule, since dγ , in we have dt (t) > 0 V dg dγ = f 0(q) = 0 =⇒ f 0(q) = 0 =⇒ q = p, dq d(t = f(q)) that is, p is the only critical point of g in V . Moreover,
g(p) = γ f(p) + x2 + y2 , (x, y) ∈ U, and so the hessian of g at p with respect to φ is given by 2γ0(a) 0 , 0 2γ0(a) hence indexg(p) = 0. Finally near a + c, γ(t) = t, hence for (x, y) close to c (i.e. close to the boundary of U),
g(x, y) = γ(f(x, y)) = f(x, y).
What is left to show is that such a γ exists. We proceed similarly to how we constructed bump functions. Let
( − 1 e t , if t > 0, h0(t) = 0, if t ≤ 0, which is a smooth function on all of , with for , and with R h0(t) > 0 t > 0 dh 1 for . Also, for , 0 1 − t . Now let h0(t) = 0 t ≤ 0 t > 0 dt (t) = t2 e > 0
h0(t) h1(t) = . h0(t) + h0(1 − t)
22 Observe that
0 0 0 dh1 h0(t)(h0(t) + h0(1 − t)) − h0(t)(h0(t) − h0(1 − t)) (t) = 2 dt (h0(t) + h0(1 − t)) 0 0 h0(t)h0(1 − t) + h0(t)h0(1 − t) = 2 , (h0(t) + h0(1 − t)) so that dh1(t) for . Also we have that for and dt (t) > 0 t > 0 h1(t) = 1 t ≥ 1 h1(t) = 0 for t ≤ 0. Now we change the interval: let
t − a h (t) = h . 2 1 c −
Now h2 has the same properties as h1 above, only 0 is replaced by a and 1 by a + c − . Now let
γ(t) = th2(t) + (t − (a − b))h2(2a + c − − t), where t ∈ [a, a + c). Clearly condition 2 and 3 holds for this function. For 3, note that dh1(t) is symmetric in the sense that 0 0 . This implies dt h1(t) = h1(1 − t) that 0 0 , hence h2(t) = h2(2a + c − − t) dγ (t) = h (t) + th0 (t) + h (2a + c − − t) − (t − (a − b))h0 (2a + c − − t) dt 2 2 2 2 0 = h2(t) + h2(2a + c − − t) + (a − b)h2(t) > 0, where the inequality comes from , , 0 , and in particular h2(t) ≥ 0 a−b > 0 h2(t) ≥ 0 h2(2a + c − − t) > 0 near and at t = a. If , then dγ in all of might fail. b 6≤ a dt > 0 [a, a + c) Now we will use lemma 7 to "push" all the critical points having index 2 up, and all critical points having index 0 down. We will do this so that for some big A we have index(p) = 2 =⇒ f(p) = A, and for some small B < A we have index(p) = 0 =⇒ f(p) = B, and moreover so that A > max ({f(p)|index(p) = 1}) and B < min ({f(p)|index(p) = 1}). Then f −1[A + , B −], for small > 0, contains only critical points having index 1. By adding some big constant to f we can assume B = 0, and by dividing f by some big constant we can assume A = 2. Now we need 2 lemmas, which will enable us to put all critical points having index 1 on any level set f = a ∈ (0, 2). Let 0 < a < b < 2. In the next lemmas we will assume the reasoning above so that f −1[a, b] contains only critical points ahaving index 1. Also, let ξ be any gradient-like vector eld for f. Lemma 8. Suppose that f −1[a, b] contains only two levels of critical points. That is , 0 0 0 , with and 0 0, for p = {p1, . . . , pn} p = {p1, . . . , pm} f(p) = c f(p ) = c some c, c0 ∈ (a, b), and p ∪ p0 is the set of all critical points in f −1[a, b]. Let Kp be the set of points on trajectories goint from or to any pi ∈ p, and Kp0 be de ned the same way but with p0 instead of p (i.e. points on the induced 0 spheres). Suppose Kp and Kp0 disjoint. Then for any d, d ∈ [a, b] there exists a Morse function g such that:
23 1. ξ is a gradient-like vector eld for g. 2. The critical points of g (in f −1[a, b]) are p∪p0, but with the levels changed to d and d': g(p) = d and g(p0) = d0. 3. g and f agrees near f −1(a) and f −1(b) (and outside of f −1[a, b]), and g = f + constant in neighbourhoods of the points in p and p0.
Note that Kp and Kp0 consists of levels of disjoint spheres having dimension 0.
0 Lemma 9. Let −1 c+c . It is possible to alter in a neighbourhood W = f 2 ξ V of W so that Kp and Kp0 are disjoint, i.e. so that all spheres corresponding to 0 0 are disjoint. p1, . . . , pn, p1, . . . , pm With these lemmas we simply take d = d0 and start just under 1 and start by pushing all levels of critical points greater then 1 down, and all levels smaller then 1 up; the "pushing" is done by repeated use of lemma 8, and lemma 9 guarantees that it is always possible. So proving these two lemmas is enough to prove the rst step towards theorem 6. Corollary 3 (Self-Indexing Theorem). Every surface admits a self-indexing Morse function.
Proof of lemma 8. Note that all trajectories through points outside (Kp ∪ Kp0 ) go from f −1(a) to f −1(b). Let τ : f −1(a) → R be a smooth function which −1 −1 is 0 near Kp ∩ f (a) and 1 near Kp0 ∩ f (a), constructed similarly to the construction of γ in lemma 7 or to the construction of bump functions; this is clearly possible since Kp and Kp0 are disjoint, hence also compact. The function τ extends uniquely to a function τ˜ : f −1[a, b] → R which is constant on each trajectory, and which is 0 near Kp and 1 near Kp0 . De ne the Morse function g : S → R by ( f(q), if q 6∈ f −1[a, b], g(q) = G(f(q), τ˜(q)), if q ∈ f −1[a, b], where G :[a, b] × [0, 1] → [a, b] is any smooth function such that: 1. For all , we have ∂G , and , . x, y ∂x (x, y) > 0 G(a, y) = a G(b, y) = b 2. G(f(p), 0) = G(c, 0) = d and G(f(p0), 1) = G(c0, 1) = d0. 3. for near and and all , ∂G , for in a neigh- G(x, y) = x x 0 1 y ∂x (x, 0) = 1 x bourhood of , and ∂G , for in a neighbourhood of f(p) = c ∂x (x, 1) = 1 x f(p0) = c0. We verify that g satisfy the three conditions. The critical points of g are given by g0(p) = 0, where ∂G ∂G g0(p) = (f(p), τ˜(p))f 0(p) + (f(p), τ˜(p))˜τ 0(p). ∂x ∂y We omit proving that ξ is a gradient-like vector eld also for g; what actually is important about this criterion is that the index of the critical points don't change (and that they are still non-degenerate), but this is straight forward.
24 Since τ˜0(p) = 0, near the critical points of f, all critical points of f are also critical points of g. Also, if q is a regular point of f, then it is a regular point of . To this rst observe that 0 and ∂G , so if 0 g f (q) 6= 0 ∂x 6= 0 τ (q) = 0 we are done. Hence we can assume both f 0(q) 6= 0 and τ 0(q) 6= 0. Note that 0 and 0 are maps from to , and that by de nition of , we have f (q) τ (q) TSq R τ that τ 0(ξ(q)) = 0, while f 0(ξ(q)) 6= 0; therefore we can see f 0(q) and τ 0(q) as two linearly independent vector. So for g0(q) to be equal to zero we must have ∂G ∂G , which is a contradiction. ∂x = ∂y = 0 Near f −1(a) and f −1(b) we have that G(x, y) = x, hence f and g agree near these levels. Clearly g = f + constant near the points in p and p0 (because ∂G for near , ∂G for near 0 and 0 near and ∂x (x, 0) = 1 x c ∂ (x, 1) = 1 x c t = 0 p p0). Proof of lemma 9. Let and (the notation is moti- Sp = Kp ∩ W Sp0 = Kp0 ∩ W vated by the fact that the sets are unions of spheres of critical points having the same level). The proof will be in two steps: First we show that there exists a dieomorphism ∼ , smoothly isotopic to the identity, such that h : W −→ W h(Sp) and are disjoint. Then we use the isotopy to alter on . Sp0 ξ V Note that we can nd small neighbourhoods in of the points Np1 ,...,Npn W in this level on the trajectories going from p1, . . . , pn, respecively, each dieo- morphic to . Let ; this is a set dieomorphic to , R Np = Np1 ∪ · · · ∪ Npn Sp × R where we moreover can assume, letting ∼ be such a dieomor- k : Sp × R −→ Np −1 phism, that k( p × {0}) = p. Also let N0 = Np ∩ p0 , and g = π ◦ k , where S S S |N0 is the natural projection, so that if and only if . π : Sp ×R → R q ∈ N0 x ∈ g(N0) Since is a set consisting of nitely many points in , we can easily g(N0) R pick a point . Now take any isotopy ∗ - using z ∈ R − g(N0) H : [0, 1] × R → R the notation, for xed , ∗ ∗ - such that: t ht (x) = H (t, x) 1. ∗ and ∗ , h0 = id h1(0) = z 2. ∗ for all , and all , where is some large number. ht (x) = x |x| ≥ A t ∈ [0, 1] A Such an isotopy is easily seen possible to construct. Now we we de ne the isotopy H : [0, 1] × W → W : let ( k(p, H(t, x)), if q = k(p, x) ∈ Np ht(q) = q, if q ∈ W − Np.
(Note that it is important that ∗ eventually becomes the identity, for to be ht H a smooth isotopy.) The desired dieomorphism is h = h1. Let c+c0 , and let be so small so that −1 . The integral e = 2 f [e − , e] ⊂ V −1 curves of ξ/ξ(f) determine a family of dieomorphism Φt(q) from f (e − ) to −1 −1 f (t), for 0 ≤ t ≤ , where Φ0 goes from W to itself, and Φ goes from f (e−) onto W . Using this we can de ne a dieomorphism ϕ :[e−]×W −→∼ f −1[e−, e], such that ϕ({e − } × W ) = f −1(e − ) and ϕ({e} × W ) = f −1(e) = W , by
−1 ϕ(t, q) = Φe−t(q).
That it really is a dieomorphism depends on that Φ(t, q) is a 1-parameter group of diemorphisms (see proof of lemma 4).
25 We now de ne a diemorphism H˜ :[e − , e] × W −→∼ [e − , e], by
H˜ (t, q) = (t, ht(q)), where we have changed H slightly so that ht = id, for t near e − , and ht = h, for t near e. Then ξ ξ˜ (q) = (ϕ ◦ H˜ ◦ ϕ−1(q)) 0 ξ(f) de nes a new smooth vector eld. The sought after, altered, vector eld is then given by ( ξ(f)(q)ξ˜ (q), if q ∈ f −1[e − ], ξ˜(q) = 0 ξ(q), elsewhere. It can easily be checked that it is a new gradient-like vector eld for f. We will show that the trajectories going to and from 0 0 are all p1, . . . , pn, p1, . . . , pm disjoint. But this follows from that the new sphere ˜ of at the level is Sp p e h(Sp) and the new sphere ˜ of 0 at the same level is the same as , hence they are Sp0 p Sp0 disjoint. The next step is to cancel out critical points so that, from a self-indexing Morse function, we obtain a new self-indexing Morse function having only 1 maximum point and one minimum point.
Let ξf be a gradient-like vector eld for a Morse function f. Let p0 be a critical point of f having index 0, and let p1 be a critical point of f having index . The sphere induced by going upwards from will be denoted , the 1 p0 Sp0 sphere induced by going downwards from will be denoted . We begin by p1 Sp1 proving the cancelation theorem, then we continue by applying it to reduce the number of critical points having index 0, and nally we reduce the number of critical points having index 2. With theorem 1, we can conclude that, given the gradient-like vector eld ξ on a surface S, for any point p ∈ S there passes a unique trajectory T . In particular if p0 and p1 are critical points having indices 0 and 1 respectively, such that and intersects in one point, then there exists exactly one trajectory Sp0 Sp1 T from p0 to p1. We begin by assuming that f is a Morse function, which is not self-indexing but instead that we have altered f so that there is some small > 0, with ∗ −1 < f(p0) and 1 − > f(p1), so that S = f [, 1 − ] contains only p0 ∗ and p1 as critical points. Then S will be a two dimensional compact smooth −1 manifold with boundary; we denote the boundary components by B0 = f () −1 and B1 = f (1 − ). This is easily seen to be possible in a way, very similar to how we proved the existence of a self-indexing Morse function, using the lemmas 8 and 9.
Theorem 7 (Cancelation Theorem). Let f be the Morse function described above with critical points p0 and p1 having index 0 and 1, respectively, and let be the gradient-like vector eld for above. Suppose that and ξf f Sp0 Sp1 intersects in one point. Then it is possible to alter the gradient-like vector eld
ξf on a small neighbourhood V of the only trajectory T from p0 to p1, producing a, on this neighbourhood, nowhere zero vector eld ξg that is a gradient-like vector eld for a Morse function g, that coincides with ξf outside of V .
26 This means that g is a Morse function with the same critical points as f, having the same indices, except that p0 and p1 are regular points of g; if this is true, then we say that p0 and p1 are cancelable. We begin by proving some lemmas.
Lemma 10. We can alter the gradient-like vector eld ξf in V to a new gradient-like vector eld for f, so that for a smaller neighbourhood VT ⊂ V of , we have that still is in , and there is a coordinate chart −1 2 T T VT φT : VT → R such that −1 −1 φT (p0) = (0, 0), φT (p1) = (0, 1), and −1 φT ∗ ξf (q) = η(x, y) = (x, v(y)), where v is a smooth function of y, such that dv (y) = 1, for y near 0 and 1, dy and 0, if y = 0, 1, v(y) = > 0, if y ∈ (0, 1), < 0, elsewhere. Proof. We can parameterize the closure of the trajectory by a smooth bijective function γ : [0, 1] → T , such that γ(0) = p0 and γ(1) = p1. Put this function on the y-axis in a coorrinate system, so that T corresponds to {0} × [0, 1]. We can nd neighbourhoods U0 and U1 of (0, 0) and (0, 1) respectively, together with parametrizations φ0 : U0 → V0 and φ1 : U1 → V1 onto neighbourhoods V0 and V1 of p0 and p1 respectively, such that ξf = (x, y) in U0 and ξf = (x, 1 − y) in 2 2 2 2 U1 (and f = f(p0) + x + y in U0, f = f(p1) + x − (1 − y) ). By possibly rotating U0 or U1 (or both) we can assume that T in U0 corresponds to φ0(0, y), for y ≥ 0, and T in U1 corresponds to φ1(0, y), for y ≤ 0. Moreover we see that we can make γ coincide with the y-coordinates of φ0(0, y) and φ1(, y) in T ∩ V0 and respectively. Hence we can parametrize along by T ∩ V1 ξf T φ2|T
−1 φ2∗ ξ(γ(y)) = (0, v(y)), for a smooth positive function v, such that v satis es the criterions speci ed in the lemma.
We can take δ > 0 so small so that (δ, 1 − δ) intersects with U0 and U1; and using theorem 1 we can extend around to a samll φ2|T {0} × [δ, 1 − δ] neighbourhood U2 that corrsponds, by a parametrization φ2, to a neigbourhood V2 of the respriction of T to γ[δ, 1−δ], such that φ2 coincide with φ0 on U0 (and V0) and φ1 on U1 (and V1). And we can, by compactness, take V2 so small so that all trajectories from U2 ∩ U0 (V2 ∩ V0) go to U2 ∩ U1 (V2 ∩ V1). Then we can take V = V0 ∪ V1 ∪ V2 as a neighbourhood of T , with cor- responding coordinate neighbourhood U = U0 ∪ U1 ∪ U2 and parametrization φ : U → V . We now de ne the altered vector eld of ξf , which we for now ˜ denote ξ: Begin by take z1 and z2, so that 0 < z1 < z2; preliminary we also require that {(x, y)||x| ≤ y ∈ [0, 1]} ⊂ U, we will make this stronger later. Let η(x, y) = (x, v(y)) be a smooth vector eld on R2. Note that η coincides with −1 in , with −1 in and with −1 on ; i.e. behaves locally φ0∗ (ξf ) U0 φ1∗ (ξf ) U1 φ∗ (ξ) T η
27 like a gradient-like vector eld for f. Now let φ∗(η(x, y)), if q = φ(x, y), |x| ≤ z1, ∗ ˜ (|x|−z1)ξf (q)+(z2−|x|)φ if ξ(q) = z −z , q = φ(x, y), z1 ≤ |x| ≤ z2 2 1 ξf (q), if q = φ(x, y), x ≥ z2.
˜ Hence ξ is a smooth gradient-like vector eld on all of S which coincides with ξf outside of a small neighbourhood V of T , and with coordinates as in the lemma on a smaller neighbourhood VT ⊂ V , where φT is given by the restriction of φ to , . VT φ|VT Now note that, in , ∂f , and ∂f , for some {0} × [δ, 1 − δ] ∂x = 0 ∂y > > 0 positive constant (otherwise ∂f would converge to , by compactness). Also 0 ∂y 0 v(y) > 1 > 0 on [δ, 1 − δ] for some constant, by the same reasoning. Hence letting z2 be small (and we can of course still always take z1 > 0 smaller) we can make sure that
∂f ∂f ∂f ∂f (x, v(y)) · , = x + v(y) > 0. ∂x ∂y ∂x ∂y
Hence ξ˜ is a gradient-like vector eld for f on all of S. ˜ From now on, our gradient-like vector eld ξf will be assumed to be ξ from the previous lemma.
Lemma 11. Given an open neighbourhood V˜ of T one can always nd a smaller neighbourhood V 0 of T with the property that no trajectory leads from V 0 outside of V˜ and back again into V 0. Proof. Towards a contradiction, now suppose that the statement is false. Then we would have a sequence ∞ so that each goes from a point to a point {Ti}i=1 Ti ri si outside V˜ and then to a point ti, such that the sequences ri and ti tends to T as i goes to in nity. S − V˜ is compact, hence si → s 6∈ V˜ as i → ∞. The curve from theorem 1 ϕ(t, s) must go to to B1, otherwise we would get a new trajectory from p0 to p1 (since the curve must go to p0). Then by using the continuous dependence of ϕ(s0, t) on s0 we see that for all s0 in some 0 neighbourhood Vs the trajectory through s goes to B1. The partial trajectories 0 0 Ts0 from s to B1 are compact; hence, in any metric, the least distance d(s ) 0 0 from T to Ts0 depends continuously on s and will be bounded from 0 for all s . Therefore, since , it is impossible that as . The desired ti ∈ Tsi ti → T i → ∞ contradiction.
Now let V˜ be an open neighbourhood such that V˜ ⊂ VT , where VT the neigh- bourhood from lemma 10, and let V 0 be the neighbourhood from the previous 0 lemma - so that T ⊂ V ⊂ V˜ ⊂ V˜ ⊂ VT .
Lemma 12. It is possible to alter the gradient-like vector eld ξf on a compact 0 ∗ subset Vc of V producing a on S nowhere zero vector eld ξg, such that every integral curve of ξg that passes through a point in V˜ was outside V˜ at some time t0 < 0 and will again be outside V˜ at some time t1 > 0.
28 Proof. Start by replacing η(x, y) by η˜(x, y) de ned by
η˜(x, y) = (x, v˜(|x|, y)), where, v˜ = v outside of a compact neighbourhood of φ−1(T ) in U 0 = φ−1(V 0) ∗ and v˜(0, y) < 0. This determines a nowhere zero vector eld ξg on S . In our coordinate system, the integral curves of ξg in VT satisfy dx dy = x, =v ˜(|x|, y). dt dt
Consider the integral curve t 7→ (x(t), y(t)) with initial value (x0, y0) as t in- creases. We show that t 7→ (x(t), y(t)) have to leave U˜ = φ−1(V˜ ). The case where t decreasees is similar. We have that dx , which if we solve the dierential equation becomes dt = x t −1 x(t) = x0e , for some constant x0. If x0 6= 0, then, since U˜ = φ (V˜ ) is compact, hence bounded, (x(t), y(t)) will eventually leave U˜. If instead x0 = 0, then we have dy which is everywhere negative; hence will dt =v ˜(0, y) (x(t), y(t)) leave U˜. In the next we will keep the notations from previous lemmas, in particular
ξg will denote our obtained new vector eld.
∗ Lemma 13. Every trajectory in S of ξg goes from B0 to B1.
0 Proof. By previous lemmas: If an integral curve of ξg is ever in V it eventually 0 gets outside V˜ , and since ξg agrees with ξf outside of Vc ⊂ V it will remain 0 ouside V . Hence it will follow a trajectory of ξ to B1. Likewise one conclude that any trajectory must come from B0.
Lemma 14. The smooth vector eld ξg is a gradient-like vector eld of a Morse function g on S that agrees with f outside of S∗ and inside S∗ in neighbourhoods of B0 and B1.
Proof. The proof will be in two steps: First we show how ξg determines a ∼ ∗ dieomorphism ϕ : [0, 1] × B0 −→ S , such that ϕ({0} × B0) = B0 and ϕ({1} × . Then we construct a Morse function such that B0) = B1 g0 : [0, 1] × B0 → R ∂g , and such that agrees with near and . By the ∂t > 0 g0 f0 = f ◦ ϕ B0 B1 −1 dieormorphism ϕ we will obtain the desired Morse function g = g0 ◦ ϕ from ∗ to with no critical points, which moreover will agree with near and S R f B0 B1. Let ∗ ∗ be the family of integral curves of , from ψ(t, p): R × S → S ξg ∗ theorem 1. On S , ξg is nowhere zero and all curves go from B0 to B1, hence ξg is nowhere tangent to B0 or B1 (that is: nowhere tangent to the boundary of ∗). Let ∗ be the function that for each ∗ assigns the S τ0 : S → R q ∈ S t ∈ R where the integral curve ψ(t, q) reaches B0; and likewise let τ1 send q to the where reaches . Note that and are smooth, which is an t ∈ R ψ(t, q) B1 τ0 τ1 easy consequence of the implicit function theorem. Therefore the projection ∗ π : S → B0, π(p) = ψ(τ0(p), p) is smooth. Now consider the smooth vector eld
τ1(π(p))ξg(p).
29 It's integral curves go from B0 to B1, and they do so in unit time. Hence we could just as well assume that this property were already satis ed by ξg. Now ∼ ∗ the dieomorphism ϕ : [0, 1] × B0 −→ S is given by
ϕ(t, p0) = ψ(t, p0) were ψ is the family of integral curves for ξg, now assumed to have integral curves that go from B0 to B1 in unit time. We can choose δ > 0, such that for all p0 ∈ B0 and all t < δ or t > 1 − δ we have that ∂f 0 > 0. ∂t Let α : [0, 1] → [0, 1] be any smooth function that is equal to 0 in [δ, 1 − δ] and equal to near and . Now we de ne : First let be given by 1 0 1 g0 k : B0 → R
R 1 ∂f0 1 − α(t) (t, p0)dt k(p ) = 0 ∂t . 0 R 1 0 (1 − α(t))dt
By choosing δ small enough, we may assume k(p0) > 0. Now let g0 : [0, 1]×B0 → R be given by Z s ∂f0 g0(s, p0) = α(t) (t, p0) + (1 − α(t))k(p0) dt. 0 ∂t
Since g0 clearly is smooth we only have to verify that it agrees with f0 for s near 0 and 1, and that g0 has no critical points. Using that α(s) = 1, for s near 0 and 1: For these s near 0 we have that Z s ∂f0 g0(s, p0) = (t, p0)dt = f0(s, p0) − f0(0, p0), 0 ∂t where f0(0, p0) = 0, which shows the agreement. For s near 1 we have that Z s ∂f0 g0(s, p0) = α(t) (t, p0) + (1 − α(t))k(p0) dt 0 ∂t Z 1 ∂f0 = α(t) (t, p0) + (1 − α(t))k(p0) dt 0 ∂t Z 1 ∂f0 − α(t) (t, p0) + (1 − α(t))k(p0) dt s ∂t
= 1 − (f0(1, p0) − f0(s, p0)) = f0(s, p0), because f0(1, p0) = 1, and because Z 1 Z 1 ∂f0 α(t) (t, p0)dt + (1 − α(t)) dtk(p0) 0 ∂t 0
1 1 ∂f 1 R R 0 Z ∂f 0 (1 − α(t))dt 1 − 0 α(t) ∂t (t, p0)dt = α(t) 0 (t, p )dt + = 1. ∂t 0 R 1 0 0 (1 − α(t))dt Moreover dg ∂f 0 (s, p ) = α(s) 0 (s, p ) + (1 − α(s))k(p ) > 0. ds 0 ∂s 0 0
30 This completes the proof of Cancelation Theorem. From now on in this section, let f : S → R denote a self indexing Morse funcion, ξ a gradien-like vector eld for f, and let Sa = {p ∈ S|f(p) ≤ a} be the set of all points less then or equal to some "height" a. We will now use Cancelation Theorem together with two lemmas connecting criticial points to attachings of rectangles and discs, and an induction argument, to reduce the number of critical points of a self-indexing Morse function, so that we obtain a Morse function with only one minimum and one maximum.
De nition 25. By an attaching of a rectangle R to Sa we will mean taking the disjoint union a R ∪g S , where we identify points on the boundary of Sa (∂Sa = f −1(a)) with points on two opposite sides of the rectangle by an injective continuous map g : R∗ → ∂Sa. For example R∗ can be given by {x = 0} ∪ {x = 1}, if R = {(x, y) ∈ R2|0 ≤ x ≤ 1, 0 ≤ y ≤ 1}. Likewise, by an attching of a disc D, say D = {(x, y) ∈ R2|x2 + y2 ≤ 1}, we mean taking the disjoint union
a S ∪g D, where we again identify the points on ∂D = S1 by a injective continuous function g : S1 → f −1(a). We will show that the identi cation, when attaching rectangles, can be done by a smooth map g, where we will not use the rectangle R from the de ni- tion above, but a smooth rectangle homeomorphic to R. This implies that in the lemmas below, we will not only get homeomorphisms, but the stronger: dieomorphisms.
Lemma 15. Let 0 < a < 1 and 1 < b < 2. Then Sb is dieomorphic to Sa with a smooth rectangle Ri for each critical point pi of index 1 attached. Proof. For simplicity we will suppose that there only is one critical point having index 1, the general case is similar. Denote this point by p0. 2 2 Let V be a neighbourhood of p0 such that, in local coordinates, f = 1+x −y through out V . Let > 0 be so small so that f −1(1 − ) and f −1(1 + ) both have two components in V . 0 Let V ⊂ V be a smaller neighbourhood of p0. We will now use a smooth function k : S → R on S, such that k is constanty zero on a neighbourhood 0 −1 Vk ⊂ V of p - where Vk is so small so that it does not intersect with f (1 − ) ∪ f −1(1 + ) - is constantly 1 outside of V 0, and increases strictly in between (meaning that in the local polar coordinates , ∂k is strictly greater then (θ, r) ∂r zero). (k is 1 minus a bump function). Then using k and ξ we de ne a new smooth vector eld ξ˜ on S by: ( k(q) ξ(q) , if q ∈ f −1[1 − , 1 + ] ∪ V − {p}, ξ˜(q) = ξ(f)(q) 0, if q = p, and ξ˜vanishes outside a compact neighbourhood of f −1[1−, 1+]∪V . Applying lemma 5, we see that ξ˜generates a 1-parameter group of dieomorphisms ϕ(t, q).
31 −1 Consider ϕ2. We will argue that this diemorphism sends f (1 − ) union the boundary of a subset R of V to f −1(1 + ), and that R is homeomorphic to a rectangle. All p ∈ f −1(1 − ) whose integral curves are always outside of V 0 will be pushed up by unit time (as in lemma 4), and hence be sent to f −1(1+). Those points that go through V 0 will be slowed down, hence they will reach a level set less then 1 + . Now we consider −1, and speci cally what happens when applied to the ϕ2 −1 −1 part of f (1 + ) that is not hit by parts of f (1 − ). By continuity of ϕ2, we can choose V 0 and k so that this is one segment of each of the components −1 of f (1 + ) in V . By the continuity of ϕ2, and by smoothness of k going from to , −1 of the segments will be two curves connecting the two compnents 0 1 ϕ2 of f −1(1 − ). These curves together with the parts of f −1(1 − ) that have integral curves through V 0 enclose a region R be homeomorphic to a rectangle (See picture).
1− 1+ Figure 3: Describes how ϕ2 sends S ∪ R to S .
From now on, whenever we talk about attaching of a rectangle we will mean a smooth attaching of the smooth rectangle from the proof above.
Lemma 16. Let 1 < b < 2, and suppose that S has m maximum points. Then 2 b S = S is dieomorphic to S with m closed discs D1,...,Dm attached. Proof. This follows from that (S − Sb) ∪ f −1(b) is dieomorphic to m disjoint discs.
32 Lemma 17. Suppose that S has has n minima, that is, suppose that there exists n critical points p1, . . . , pn of f having index 0, n ≥ 2. Then there exists critical points r1, . . . , rn−1 of f having index 1 such that that pi and ri are cancelable, for 1 ≤ i ≤ n − 1. Proof. By observing how Sa changes as a increases, we will by reconstructing the surface see that if this is not true, then we could not have a connected surface. Let 0 < a < 1 and 1 < b < 2. First note that Sb must be connected. This is because passing from Sb to the full surface S is done by attaching discs, and attaching discs are done by dieomorphisms of connected boundary components, hence these attchings can't connect disconnected components; only attachments of rectangles can do that. By theorem 2 and lemma 4, Sa is dieomorphic to the union of n disjoint b discs D1,...Dn. As we have noted earlier S must consist of one connected component, i.e. be connected. Take any of Di's, with corresponding critical point pi. Di must be attached by some rectangle, corresponding to a critical point rj - by possibly renamin we can take ri - having index 1 that is also attached to some other disc Dl, l 6= i, with corresponding critical point pl. If not: then Sb could not be connected, since there would be nothing connecting Di to any of the other discs. Moreover, the induced spheres - , going upwards from , and going Spi pi Sri downwards, from ri - must intersect in only one point, because the trajectories from ri must go down to both points pi and pl. Hence pi and ri are cancleable.
Corollary 4. On any surface S, there exists a self-indexing Morse function having only 1 minimum point. The cancelation of the maximum points is just ipping the surface upside down - using the new Morse function 2−f - and then using the previous lemma again. Thus we obtain: Any surface admits a self-indexing Morse function with just one maximum and one minimum, that is, any surface admits a good Morse function.
33 Constrution of the Genus-g Toruses
In this section we let S denota a surface, f : S → R denota a good Morse function on S, and as always, Sa = {p ∈ S|f(p) ≤ a}. From the previous section we know that to classify surfaces, we only need to look at how we can attach rectangles to a disc, and then attach one disc to to the obtained "disc with rectangles" in the end. The rst observation to be made is the following: Since the surface is con- nected and without boundary, there can only be one boundary component of Sb, for 1 < b < 2. We next focus on the rectangles: The attaching of a rectangle can be made in 2 ways: the rst is with no twist, the second with a twist (more twists are homeomorphic to no twist or one twist). However, since S must be orientable there can be no twist, since if there were we would have a homeomorphic copy of a Möbius band inside S, contradicting the assumtion that S is orientable. One way to visualize the attachment of rectangles is to think of it as a procedure, attaching one at a time; but as long as all attachements are to dierent parts of the original disc, even if we see how the boundary components changes as we attach one rectangle at a time, this procedure is equivalent to attach all at the same time at dierent places. We now make the following observation, based on the fact that boundary components here must be closed paths. Lemma 18. 1. If a rectangle is attached to two dierent boundary components, then the number of boundary components reduces by one. 2. If a rectangle is attached to one boundary component, then the numer of boundary components increase by one. Proof. Obvious from picture.
Figure 4: Illustration of the previous lemma.
34 Corollary 5. The number of critical points having index 1 of f must be even. Proof. We start with one boundary component, and we must end with one; hence we must attach an even number of rectangles, and so the number of critical points having index 1 must be even. Now we prove the two starting cases of classi cation. For this we begin by observing that there is only one way to attach a disc. This is a consequence of lemma 1, because the attaching is done by a dieomorphism from S1 onto itself. Theorem 8. Suppose there are no critical points having index 1 of f, then S is dieomorphic to a sphere.
Proof. By lemmas above S must be a dieomorphic to two discs attached to each other smoothly along their boundaries, which is a construction of the sphere. And moreover there is only one way to do this by lemma 1.
If lemma 1 were not true, there might have been two dierent surfaces, not dieomorphic to each other, but both homeomorphic to the sphere (and hence homeomorphic to each other).
Theorem 9. Suppose there are two critical points having index 1, then S is dieomorphic to a torus. Proof. We will think of this as attaching one rectangle at a time. After attaching the rst rectangle we get a cylinder. The cylinder has two boundary components. The second rectangle must be attached to both these boundary components for Sb, 1 < b < 2, to be dieomorphic to a surface with one boundary component. Finally, attaching a disc along the boundary gives us a torus; the hole can be visualized as the region between the second rectangle and the cylinder. The attachments of the two rectangles and the sphere is done in a unique way (up to dieomorphism), and smoothly.
Figure 5: Illustrates the reasoning of the previous proof.
Another way to see it, is to attach one rectangle to each disc, obtaining two cylinders, which obviously forms a torus when attached together.
35 Thus we the torus and the sphere 2 are unique up to dieomorphism, T1 S hence all Tg are, g ≥ 0. Now we use induction to prove the nal theorem of the classi cation.
Theorem 10 (Classi cation of Surfaces). Suppose f has 2g critical points hav- ing index 1. Then S is dieomorphic to the genus-g torus Tg. The proof will be in three main steps: lemma A, lemma B and lemma C. Lemma 19 (A). There is an attachement of two of the rectangles, such that one is inside the other, so that removing this attachment (or doing the attaching) does not change the number of boundary components. Call such attachments for pretszel attachments. Proof. If this wasn't true there could clearly not be only one boundary compo- nent.
Lemma 20 (B). After attaching all rectangles the boundary must be connected. We can now slide the pretzel attachment to the left of some point on the disc - the disc to which the rectngles are attached - or to the left of a previously slided pretzel attachment so that:
1. All rectangles are still attached to disjoint parts of the original disc. 2. The slided pretzel attachment together with the originl disc makes a path along the boundary that does not intersect the boundary of any other rect- angles (i.e. we hve slided the pretzel out of the region of the boundry of the disc were the other rectangles are attached).
Proof. Clearly such slides can be done smoothly on the boundary, hence gives us the same surface up to dieomorphism. The two criterions is easily seen to possible to satisfy, by picture similar to gure 5.
Lemma 21 (C). Using lemma A and B inductively we get a disc with g disjoint pretzel attachments.
Proof. Obvious.
Proof of Classi cation of Surfaces. Attaching a single disc to the disc having g pretzels clearly gives us a genus g-torus.
36 References
[1] Manfredo Perdigão do Carmo, Dierential Geometry of Curves and Surfaces, Upper Saddle River, New Jersey 07458, Prentice-Hall, Inc., 1976. [2] Manfredo Perdigão do Carmo, Riemannian Geometry, Boston, Birkhäuser, 1992.
[3] Ib Madsen, Jørgentornehave, From Calculus to Cohomology, Cambridge, Cambridge University Press, 1997. [4] John Milnor, Lectures on the H-Cobordism Theorem, Princeton, New Jersey, Princeton University Press, 1965. [5] John Milnor, Morse Theory, Annals of Mathematical Studies 51, Princeton, New Jersey, Princeton University Press, 1963. [6] John Milnor, Topology from the Dierentiable Viewpoint, Revised Edition, Princeton, New Jersey, Princeton University Press, 1997. (Orginially pub- lished by The University Press of Virginia, 1965).
[7] Donal O'shea, The Poincaré Conjecture : In Search of the Shape of the Universe, Penguin Books, 2008.
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