Note: all spaces are assumed to be path connected and locally path connected.

Transcription

1 Homework 2 Note: all spaces are assumed to be path connected and locally path connected. 1: Let X be the figure 8 space. Precisely define a space X and a map p : X X which is the universal cover. Check in detail that the Deck group is isomorphic to F 2, the free group on two letters. (For example, we may define X R 2, or even X (R 2 ).) Given g F 2 we write g = w 1 w 2... w k where each w i is equal to either a, b, a 1, or b 1, and w i w 1 i+1, thus g determines its factorization into w i. Define e(g) = (x(g), y(g)) R 2 by x(g) = w i=a σ σ3 i, and y(g) = w i=b σ σ3 i, where σ = ±1. For example, e(a 1 ) = ( 1 3, 0), e(bab 1 ) = ( 1 9, 8 27 ). Notice that e : F 2 R 2 is an injective function. Also notice that e(g 1 g 2 ) = e(g 1 )+3 k e(g 2 ), where g 1 is a word of length k, as long as the final letter of g 1 is not the inverse of the first letter of g 2. Given t [0, 1) and gw k+1 F 2, we define e(gw k+1, t) = e(g)+t3 k e(w k+1 ). Thus t e(gw k+1, t) is the straight line path between e(g) and e(gw k+1 ). Again notice that e is an injective function. We define X to be the image of e : F 2 [0, 1) R 2, topologized as a subset of R 2. To every point q = e(w 1... w k+1, t) X we associate a path γ q : [0, 1] X by γ q (s) = { e(w 1,..., w j+1, j (s 1)) if s [1 2 j, 1 2 j 1 ) for some j = 0,... k 1 e(w 1,..., w k+1, t(1 + 2 k (s 1))) if s [1 2 k, 1]. Then γ q (0) = 0, γ q (1) = q, and H(q, s) = γ q (s) defines a homotopy from the identity map to a constant, showing that X is contractible and therefore simply connected. Define p(q) = i wk+1 (e 2πit ), where i wk+1 is the inclusion of S 1 into X via one of the two factors deending on whether w k+1 = a ±1 or w k+1 = b ±1, additionally composing with the reflection if w k+1 is an inverse that is i w 1 (z) = i wk+1 (z). It is obvious from the construction that p : X X k+1 is a covering space. We define two homeomorphisms f a and f b D( X). Let X a X be the set of q X which can be presented as q = e(a 1 g, t) for some t [0, 1) and g F 2 so that the first letter of g is not a. Notice then that X \ X a is the connected component of X \ {e(a 1 )} containing 0. Let Due February 12th 1

2 2 f a (q) = { 1 3 q + ( 1 3, 0) if q X \ X a 3(q ) if q X a. Notice that f a is a homeomorphism, and that f a (e(g)) = e(ag) for all g F 2 (whether or not the first letter of g is a 1!). We define f b similarly. Then for g = w 1... w k, define f g = f wk... f w1, where f w 1 = fw 1 i i by definition. This defines a group homomorphism F 2 tod( X), and it is injective since f g (0) = e(g). To see that this map is surjective, notice that the set of points q X which are not locally homeomorphic to an interval, but instead to an X, are exactly the points presented as e(g). Thus for any f D( X), f(0) = e(g) for some g, and therefore (fg 1 f)(0) = 0, which shows that fg 1 f is the identity map since it is a Deck transformation fixing a point. b: Let a and b represent the two generators of F 2 corresponding to the two loops of X. Let H be the free cyclic group generated by a. Describe the covering space p H : X H X which represents this subgroup. Do the same thing for the free cyclic group H 2 generated by a 2. (A drawing suffices, but make sure to make clear what the covering maps are, not only the spaces). c: Do that same for the normalizers N and N 2 of H and H 2, respectively. Prove that the Deck group of these covering spaces are respectively isomorphic to F 2 /N and F 2 /N 2.

3 3 I meant to ask for N and N 2 being the normal closure, which is turns out is different from the normalizer. The problem won t be graded but for the curious here s a solution. F 2 /N is the quotient group of F 2 when we additionally may replace a = 1 in any word; obviously this is isomorphic to {b k } = Z. The Z action in the picture is by vertical translation. In F 2 /N 2 we may only replace a 2 = 1, thus a general element of the group can be written as g = b i1 ab i2 a... ab i k. We can represent b as the Deck transformation which is again just a vertical shift. The Deck transformation a switches the vertical line at the far left with the vertical line on the other side of the a loop. As in problem (a) we see that these are group isomorphisms because every point in p 1 (x 0 ) is the image of x 0 for some Deck transformation. d: Let X 3 be the triple figure 8, described as a subset of R 2 as {x = sin(θ), y = cos(3θ)}. Notice that X 3 is invariant under 180 degree rotation, the quotient is homeomorphic to X, and this quotient is a covering map. Show that π 1 (X 3 ) is isomorphic to the free group on three letters as follows. First, find three loops in X 3 that you suspect are the generators (keep track of basepoints). Show that the group they generate is a free group by computing their image in π 1 (X). Finally, show that this group is the entire image. There are two points in p 1 (x 0 ) in X 3 (those which are locally homeomorphic to X), we choose the one on the left. Let A be the loop in X 3 which travels around the leftmost loop, let B be the loop traveling around the center loop, and let C be the loop which travels along the bottom branch of the center loop, then goes around the rightmost loop, then goes back to the basepoint by traveling along the lower branch of the center loop in reverse. It is clear then that p (A) = a, p (B) = b 2, and p (C) = bab 1. Let G F 2 be the group generated by p (A), p (B), and p (C), we want to show that G is free. That is, if w = w 1,... w k is any word in A, B, C and their inverses, we

4 4 need to show that p (w) 1 unless w = 1. Since obviously the group generated by p (A) and p (B) is free, any nontrivial word with p (w) = 1 must have at least one C ±1, by conjugation and inversion we can assume that w 1 = C. But then it is impossible to ever cancel the first b in p (w). Indeed, there will always be an odd power of b between the first a and the first instance of p (A), since no product from p (C) and p (B) will ever be a word with exactly one odd power of b. Therefore p (A ±1 ) can never be used to cancel the first a coming from p (w 1 ). Of course nothing from p (B) can cancel this a either. Finally for another letter p (C ± 1) to cancel this initial a, we would have to have no b s between, contradicting the fact w 2 C 1. To see that π 1 (X 3 ) is generated by A, B, and C, notice G F 2 is normal: conjugation by a preserves G since a G and conjugation by b sends p (A), p (B), and p (C) to p (C), p (B), and p (B)p (A)p (B 1 ), respectively. The quotient, which is obtained by letting a = b 2 = 1, is Z 2. Thus G is index 2. p (π 1 (X 3 )) is also index 2, since p 1 (x 0 ) has two points. Since G p (π 1 (X 3 )) and they have the same (finite) index they are equal. 2a: Let be a locally simply connected compact metric space. X is compact if and only if π 1 (X) is finite. Show that If π 1 (X) is infinite, then p 1 (x 0 ) X is an infinite set with no limit points, so it is not compact. Conversely, if A X is an infinite set, then p(a) has a limit point a X. Choose a neighborhood U X containing a which is evenly covered by p, and a slightly smaller open set V whose closure is contained in U. Then there are infinitely many points in p(a) V, so there are infinitely many points in A p 1 (V ). Writing p 1 (V ) is a finite union of sets V i X, there must be at least one V i containing infinitely many points of A. Then p(a V i ) also has a limit point, which is inside V. Since p : V i V is a homeomorphism (since V is contained in U), we see that A has a limit point in V i. b: A finite generating set of a group G is a finite list of elements g 1,..., g k G so that every element g G can be written as product of the g i and their inverses: g = g ±1 i 1 g ±1 i 2... g ±1 i N. Given a finite generating set {g i } k i=1 of G, define for every g G the nonnegative integer g = min{n; g = g ±1 i 1... g ±1 i N }, the number of generators required to write G. For g, h G define ρ(g, h) = gh 1. Show that ρ is a metric on G. Symmetry is obvious, as in non-degeneracy. The triangle inequality follows since obviously gh g + h for any g, h G. c: Let M and N be metric spaces. A map f : M N is called a quasiisometry if there is a constant C > 1, so that for every x, y M, 1 C ρ M (x, y) C ρ N (f(x), f(y)) Cρ M (x, y) + C,

5 5 and furthermore every point z N is distance no more than C away from the image of f. Show that, given two finite generating sets of G, the two resulting metrics are quasi-isometric, under the identity map. If A and B are two generating sets of G, then every element of A can be written with no more than a generators from B, and every element of B can be written with no more than b generators from A. This implies g A a g B and g B b g A. Let C be the maximum of a and b. d: Let X be a compact metric space, which is locally simply connected. Show that π 1 (X) is finitely generated. Show that we can define a metric on the universal cover X so that the covering map p : X X is a local isometry. Then, show that X is quasi-isometric to π 1 (X), with the metric defined above. Choose a finite open cover by simply connected sets {U i }. Choose a point x ij U i U j whenever this set is non-empty. We choose paths β ij connecting the basepoint x 0 to x ij, and α ijk connecting x ij to x jk, contained entirely in U j (so α ijk is uniquely defined up to homotopy). Given any triple i, j, k, let γ ijk = β ij α ijk β 1 jk. We claim that these loops [γ ijk] generate π 1 (X). To see this, let γ be a loop. We can subdivide γ into arcs γ [ta,t a+1] contained in some U i, which presents [γ] = [β i1i 2 α i1i 2i 3 α i2i 3i 4... α in 2 i N 1 i N β 1 i N 1 i N ]. We can then insert a β ij β 1 ij between all α curves to express this as a product of γ ijk. Choose a radius r 0 > 0 so that the balls B r0 (x) are simply connected and evenly covered at every point x X. Given two points x 0 and x 1 in X, consider finite sequences a 1,..., a k X so that a 1 = x 0 and a k = x 1, so that each pair a i, a i+1 are contained in the same component of p 1 (B r0 (x)) for some x. Then define ρ( x 0, x 1 ) = inf{ i ρ(p(a i), p(a i+1 ))}, where ρ is the metric on X and the infimum is over all finite sequences as described. Clearly ρ is locally isometric to ρ via p, using the triangle inequality. We identify π 1 (X) with the set p 1 (x 0 ), and we claim that this inclusion is a quasi-isometry. First, notice that the new metric on X, ρ (x 1, x 2 ) = ρ(p 1 (x 1 ), p 1 (x 2 )) is equal to ρ whenever x 1 and x 2 are distance less that r 0 apart, and therefore it gives the same topology on X (the distance between sets is the infimal distance between points). ρ( x, p 1 (x 0 )) = ρ (p( x), x 0 ) since the Deck group acts by isometries: the distance between a point in p 1 p( x) and any other point y X is the same as the distance between x and the corresponding Deck translate of y. The metric ρ is bounded, since X is compact, therefore every point in X is a bounded distance away from p 1 (x 0 ). That is, the inclusion π 1 (X) X is a quasi-surjection. For any g π 1 (X) we write g p 1 (x 0 ) X the corresponding point, so 1 = x 0. Since both the group metric on π 1 (X) and the ρ norm on p 1 (x 0 ) are invariant under the group action (acting by left multiplication on π 1 (X), and via Deck transformations on p 1 (x 0 )), to prove that π 1 (X) is quasi-isometric to

6 6 X it suffices to show that 1 C g ρ( g, x 0) C g for all g π 1 (X). Let R > 0 be the diameter of X, and let {g j } π 1 (X) be the set of elements so that ρ( g j, x 0 ) < 2R+r 0, then {g j } is a finite set, which we will prove generates π 1 (X). Given g p 1 (0), let a 1,... a k be a sequence so that ρ(a i, a i+1 ) < r 0 and ρ( g, x 0 ) sum i ρ(a i, a i+1 ) < ρ( g, x 0 ) + ε. If ρ(a i, a i+2 ) < r 0 then by omitting a i+1 the sequence better approximates ρ( g, x 0 ), we may therefore assume this never happens. Thus by the triangle inequality we have that for every i either ρ(a i, a i+1 ) or ρ(a i+1, a i+2 ) must be greater than 1 2 r 0, so 1 2 r 0k < ρ( g, x 0 ) + ε. For each a i, let g i be the (possibly non-unique) element of p 1 (x 0 ) closest to a i, so ρ( g i, a i ) < R. Therefore by the triangle inequality ρ( g i+1, g i ) < 2R + r 0, so g i+1 g 1 i is in our generating set {g j } (again using that Deck transformations are isometries). Thus g k = a k = g is a product of elements in the generating set, and no more than k of them. In particular g k 2 r 0 ( ρ( g, x 0 ) + ε). Conversely, let g = g j1... g jk, and let g i = g j1... g ji. By the triangle inequality ρ( g, x 0 ) i ρ( g i+1, g i ), and ρ( g i+1, g i ) = ρ( g ji+1, x 0 ) < 2R+r 0. Thus ρ( g, x 0 ) < (2R + r 0 ) g. e: Let X be a closed n-manifold with a flat Riemannian metric. It is a basic fact from Riemannian geometry that the universal covering space of X is R n, where the covering map p : R n X is a local isometry (i.e. Dp preserves the metric on each tangent fiber). Let g 1,..., g k be a finite generating set of π 1 (X). Let B(N) π 1 (X) consist of those elements g satisfying g N. Show that there is a constant C > 0, so that for any N > 0 B(N) has no more than CN n elements. Let B n (R) R n be the ball of radius R centered around x 0. Let r 0 be a radius so that any two balls of radius r 0 centered at distinct points of p 1 (x 0 ) are disjoint. Then if B n (R) contains k elements of p 1 (x 0 ), it follows that B n (R+r 0 ) contains k balls of radius r 0, and therefore A(R + r 0 ) n kar n 0, where Ar n is the volume of the ball of radius r. That is, k ( R r 0 + 1) n. Since π 1 (X) is quasi-isometric to R n, it follows that B n ( N C 0 ) p 1 (x 0 ) B(N) B n (C 0 N) p 1 (x 0 ) for some constant C 0. So if B(N) has k elements of π 1 (X), then B n (C 0 N) p 1 (x 0 ) does as well, implying that k ( NC0 r 0 + 1) n CN n. f: Show that the manifold X = (S 1 S 2 )#(S 1 S 2 ) does not admit a flat Riemannian metric. Here # denotes the connect sum of manifolds. Since π 1 (X Y ) = π 1 (X) π 1 (Y ), we see that π 1 (S 1 S 2 ) = Z. Also π 1 (S 1 S 2 \ D 3 ) = Z since the generator S 1 {point} doesn t pass through D 3 if we choose D 3 to be small. By definition (S 1 S 2 )#(S 1 S 2 ) = (S 1 S 2 \ D 3 ) D 3 (S 1 S 2 \ D 3 ) so by Van-Kampen s theorem we have that π 1 ((S 1 S 2 )#(S 1 S 2 )) = F 2. Taking the standard generating set a, b of F 2, we see that B(N) consists of all of the group elements we can write with N or fewer letters. The number of such is exponential in N (greater than 3 N ), so the previous result implies we cannot have a flat metric.