Degree of circle maps and Sard s theorem.

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1 February 24, Degree of circle maps and Sard s theorem. We are moving toward a structure theory for maps certain maps of the interval and circle. There are certain fundamental notions that we need to introduce. Degree of Circle maps Consider the map φ(t) = e 2πit from R onto S 1. We call this the standard covering map from R onto S 1. The map φ is C, locally one-to-one, and, if z = e 2πiα, then the pre-image φ 1 (x) is the countable set of points α ± 2πn where n Z. For α R, and 0 < ɛ < π, let Uα,ɛ be the ball of radius ɛ centered at α in R. The map φα,ɛ = φ Uα,ɛ is a diffeomorphism from Uα,ɛ onto an open neigborhood of e 2πiα in S 1. We call the map φα,ɛ a local parametrizaion of S 1 at z = e 2πiα. We call the inverse map ψ z,ɛ = φ 1 α,ɛ a local coordinate map on S 1 centered at z. Letting V z,ɛ = φ(uα,ɛ), we call the pair (ψ z,ɛ, V z,ɛ) a local chart on S 1 centered at z. Let f : S 1 S 1 be a C 1 self-map. Let z = e 2πiα, f(z) = e 2πiβ, and ɛ > 0 be small. We call the map φ 1 β,ɛ fφ α,ɛ = ψ f(z),ɛfψz,ɛ 1 a local representative of f at z. When it is clear from the context what we are referring to, we leave out the α and/or ɛ and simply refer to the local representative of f at z as ψ f(z) fψz 1. Let z = e 2πiα be a point in S 1 such that Df(z) 0, and let f = ψ f(z) fψz 1 be a local representative at z. If D( f)(ψ(z)) > 0, we say that f preserves orientation at z. If D( f)(ψ(z)) < 0, we say that f reverses orientation at z. It can be shown that this is independent of the choice of local coordinate maps at z. Recall, that we have defined Df(z) 0 to mean that, some (or any) local coordinate representative f = ψ f(z) fψz 1 has a non-zero derivative at ψ(z). A regular value of f is a point y S 1 such that for each x f 1 y, we have Df(x) = 0. That is, there are no critical points of f in the set of pre-images of y. Note that if y is not in the image of f, then it is a regular value. We wish to show that any C 1 map from an interval I into R has many regular values. To do this, we recall the notion of Lebesgue measure. Recall that, given an open set U R, we can write U as a finite or coutable union of disjoint intervals U = i A U i. We define the Lebesgue

2 February 24, measure of an open interval (a, b) to be its length b a. We write this as m((a, b)). We allow intervals of infinite length in this consideration. Then, we define the Lebesgue measure of U to be the number m(u) = i A m(u i ). The outer measure of any subset E R is defined to be the number m (E) = inf U E,U open m(u). Given a subset E R, denote by E c the complement of E. A set E is called measurable if for any subset F we have m (F ) = m (F E) + m (F E c ). We set m(e) = m (E) for measurable sets. Facts. 1. The set E of measurable sets forms a σ algebra. That is, a countable union of measurable sets is again measurable, and the complement of any measurable set is measurable. 2. Open (and closed) sets are measurable. 3. If E 1, E 2,... is a sequence of disjoint measurable sets, then m( E i ) = i m(e i ). This property is called countable additivity. 4. Every subset of a set of measure zero is measurable and has measure zero. 5. For every x R, m({x}) = One can prove that non-measurable sets exist, but they are hard to construct. So, we assume that all sets we consider are measurable.

3 February 24, For a subset E S 1, we say that E is measurable if, for any coordinate chart (U, φ) we have that φ(u E) is measurable in R. In this case, this simply means that if pick a point z S 1 and think of S 1 \ {z} as an interval, then E \ {z} is measurable in this interval. For N an interval, the circle, or the reals R, we say that a subset E N has full measure if its complement N \ E has measure zero. The next result can be used to show that C 1 maps have many regular values. Theorem 0.1 (Sard s Theorem). Let f : R R be a C 1 map. Then, the set of regular values has full measure in R. Proof. Let C be the set of critical points; i.e., x C iff Df(x) = 0. For each integet i, let C i = C [i, i + 1]. Note that C is a closed subset of R, so each set C i is compact. Also, so is f(c i ). Since f(c) = i f(c i ), it suffices to prove that m(f(c i )) = 0 for every i. Let ɛ > 0. Since x Df(x) is uniformly continuous on [i 1, i+1], we can choose δ > 0 such that if x, y [i 1, i+1] and x y < δ, then Df(x) Df(y) < ɛ. In particular, if x C i and, we and y x < δ, then Df(y) < ɛ. Let B δ (x) denote the interval of length 2δ centered at x. Then, m(f(b δ (x))) < ɛ m(b δ (x). Let n be a positive integer with 1 < δ, and, for j = 0, 1,..., n 1, let n U j = [i + j, i + j+1 ]. n n For each j, we have m(f(u j Ci )) < 2ɛ. n Now, we have n such intervals, so, m(f(c i )) j m(f(u j Ci )) < n 2ɛ n < 2ɛ. Since ɛ was arbitrary, this proves the theorem. QED. Remark. For an open subset U R n and a C r map f : U R p, one defines a regular value y to be a point y with is either not in the image of f or is such that for each x f 1 (y) the derivative Df(x) has maximal rank. Using Lebesgue measure in R p, the general version of Sard s theorem says that if r > max(0, n p), then the set of regular values has full measure. For n = p, one only needs r = 1, and the proof is similar to the one we gave for n = 1. One makes use of the fact that, for an n dimensional cube

4 February 24, Q, the measure of f(q) is the integral of the absolute value of the Jacobian determinant det(df(x)) as x runs through Q. Now, we return to one dimensional maps f : N N. If y is a regular value of f, and x f 1 y define { +1 if f preserves orientation at x sign(x) = 1 if f reverses orientation at x We define the degree of the C 1 map f : S 1 S 1 as follows. Let y be a regular value of f. Set deg(f) = x f 1 y sign(x). y. This is an integer, and turns out to be independent of the regular value Examples. 1. If f is not surjective, then deg(f) = Let d be a non-zero integer. The map f d (z) = z d has degree d. There are some important properties of the degree of a circle map. Let f : X X and g : X X be two self-maps of a topological space X. We say that f is homotopic to g if there is a continuous map F : X [0, 1] X such that, for each x X, we have F (x, 0) = f(x) and F (x, 1) = g(x). Properties. 1. deg(f g) = deg(f)deg(g). 2. If f, g are homotopic C 1 self-maps of S 1, then they have the same degree. 3. Conversely, if f and g are C 1 self-maps of the circle with the same degree, then they are homotopic. 4. If f : S 1 S 1 has degree d, then there is a map F : R R such that (a) F (x + 1) = F (x) + d for all x, and (b) πf = fπ.

5 February 24, Any two such maps F and G are such that there is an integer n such that F (x) = G(x) + n for all x. Any such map F is called a lift of f (to R). We usually fix the choice of a lift F by requiring that F (0) [0, 1). Note that the last property gives a simple geometric description of maps of C r maps of the circle of degree d. For instance, suppose d > 1. Then, one takes a C r map F from the unit interval I into R such that F (1) = F (0) + d and the derivatives of F at 0 and 1 are equal. One extends F to points x [j, j + 1] for j 0 by setting F (x) = F (x j) + jd. One can check that this gives a well-defined map from R to R. The graph of the map F over [j, j + 1] is a translate of that over [0, 1] up or down by jd units according to whether j > 0 or j < 0. We leave the analogous interpretations for d 0 to the reader. Exercises. 1. Let d be an integer. A map F : R R such that F (0) = 0 is the lift of a map f : S 1 S 2 of degree d (via the projection map x e 2πx ) if and only if F (x) = d x + G(x) where G(x + 1) = G(x) for all x. That is, G is a periodic function of period From the preceding exercise, the map F (x) = d x + A sin(2πx) is the lift of a map of the circle of degree d. Show that if 5πA > d, then F has negative Schwarzian derivative. Remark. The degree of maps can be defined for continuous maps of orientable manifolds (e.g. the n sphere S n ). Properties 1 and 2 above hold for arbitrary manifolds. Property 3 above holds for S n.