id = w n = w n (m+1)/2

Transcription

1 Samuel Lee Algebraic Topology Homework #4 March 11, 2016 Problem ( 1.2: #1). Show that the free product G H of nontrivial groups G and H has trivial center, and that the only elements of G H of finite order are the conjugates of finite order elements of G and H. Proof. Suppose that w = w 1... w m Z(G H). Then w commutes with every element of G H and in particular must commute with any g G. So wg = gw implies that every letter w i in w is an element of G since there are no relations between G and H. Similarly w must commute with arbitrary h H, so each w i must be an element of H. This cannot happen unless w is the empty word, so the center of G H is trivial. Let w = w 1... w m G H be of finite order n so that w n = id. If m is even, then w 1 and w m belong to different groups and then w n cannot be the identity since there are no possible reductions in w n = w 1... w m w 1... w m... w 1... w m. (1) Thus n is odd and we can then notice that, since the w i s alternate group membership, w i and w m+1 i are both elements in either G or H for i = 1,..., m. In order for w n = id, (??) must have significant cancellation between adjacent words. Specifically, since w is reduced, w m w 1 = id, w m 1 w 2 = id,..., w (m+1)/2+1 w (m 1)/2 = id. Then (??) becomes id = w n = w n (m+1)/2 which implies that w (m+1)/2 is of finite order n. The relationship w m+1 i w i = id for i = 1,..., (m 1)/2 implies that w was a conjugate of w (m+1)/2 (conjugated by the element w 1,..., w (n 1)/2. Problem ( 1.2: #2). Let X R n be the union of convex open sets X 1,..., X m such that X i X j X k for all i, j, k. Show that X is simply connected. Proof. When n = 1, X is just a convex open set, which is simply connected. So suppose that the result is true for all integers less than m and X = X 1 X m. By assumption, π 1 (X 1 X m 1 ) = {0}. Moreover, X 1 X m 1 is open and (X 1 X m 1 ) X m = (X 1 X m ) (X 2 X m ) (X m 1 X m ) (2) is nonempty since in the hypothesis that X i X j X k we can take i = j so that the intersection of any two X i is nonempty. In order to apply Van Kampen s theorem, we must show that (??) is path connected. Each X i X m is path connected, seeing that it is the intersection of two convex sets and therefore convex hence path connected. So let x i X i X m and x j X j X m. Again applying the condition that the intersection of any three X k are nonempty, there must be an x ij X i X j X m. Then a path from x i to x j can be found composing the line from x i to x ij in X i X m with the line from x ij to x j in X j X m. Now we are in a position to apply Van Kampens theorem. We have that π 1 (X m ) = {0} and again, by assumption, π 1 (X 1 X m 1 ) = {0}. So π 1 (X) = π 1 (X 1 X m 1 ) π 1 (X m )/N = {0} {0}/N = {0}, regardless of N, meaning that X is simply connected.

2 Problem ( 1.2: #4). Let X R 3 be the union of n lines through the origin. Compute π 1 (R 3 X). Solution: The family f t (x) = x(1 t) + t x x is a deformation retract of X onto the sphere S 2 with 2n punctures. This is in turn homotopy equivalent to a disk with 2n 1 punctures, by recognizing one of the holes as the boundary (or recalling that S 2 \ {x} D 2 ). If we fix a basepoint x 0 in this space, and take a loop at x 0 around each of the 2n 1 punctures, the space is deformation retractable onto these 2n 1 embmedded copies of S 1, as demonstrated below. The resulting space Y is thus a wedge sum of 2n 1 circles therefore π 1 (Y ) = π 1 (S 1 ) π 1 (S 1 ) = Z Z (2n 1 copies). It kind of looks like this : Problem ( 1.2: #7). Let X be the quotient space of S 2 obtained by identifying the north and south poles to a single point. Put a cell complex structure on X and use this to compute π 1 (X). Solution: We can form X by gluing the boundary of a disk into a loop that goes once longitudinally around the torus. This can be described via cell complexes as follows. To form the torus, we can glue all endpoints of two one cells e 1 1, e 1 2 to one 0-cell. If a is a loop with basepoint x = e 0 going once around e 1 1 and b is a loop with basepoint x going once around e 1 2, then we glue a 2-cell e 2 1 along abab to form the torus T 2. To obtain X from this cell structure, one can glue another 2-cell e 2 2 via the map taking it s boundary to a. Then by proposition 1.26 (using Van Kampen s theorem on cell complexes), π 1 (X) = π 1(T 2 ) N where N = [a] = Z. The fundamental group of the torus T 2 is Z Z = a b, hence π 1 (X) = a b a = Z.

3 Problem ( 1.2: #8). Compute the fundamental group of the space X obtained from two tori S 1 S 1 by identifying a circle S 1 {x 0 } in one torus with the corresponding circle S 1 {X 0 } in the other torus. Solution: Let X 1 and X 2 denote the two tori. The fundamental group of each is π 1 (X i ) = a i, b i [a i b i ] (3) where the elements a i and b i correspond to a meridian and longitudinal loop around X i, respectively. Then X is the space X = X 1 X 2 /[a 1 ] [a 2 ]. Let A i be an open neighborhood of X i in X(this is X i with flaps ). The intersection A 1 A 2 is the product S 1 B where B is an space that looks like an open X (i.e is a product of S 1 with the flaps ) and which deformation retracts onto S 1 ; so π 1 (A 1 A 2 has fundamental group Z. Define, according to notation in Hatcher, i αβ : π 1 (A α A β ) π 1 (A α ) to be the homomorphism induced by the inclusion A α A β A α and j α : π 1 (A α ) π 1 (X) to be the homomorphism induced by inclusion A α X. Now since A 1 A 2 is open and path connected, we have π 1 (X) = π 1(A 1 ) π 1 (A 2 ) N (4) where N is the normal subgroup generated by i 1,2 (c)i 2,1 (c) 1 with c = Z π 1 (A 1 A 2 ). Well i 1,2 a 1 and i 2,1 (c) = b 1 2 (sorry for the somewhat abusive notation here, the b 1 2 refers to the inverse in π 1 (A 2 ) of the element corresponding to the path b i ). Thus N = a 1 a 1 2 so (??) becomes π 1 (X) = a1, b 1 [a 1 b 1 ] a 2, b 2 [a 2 b 2 ] a1 a 1 2 = a1, b 1, a 2, b 2 [a 1 b 1 ], [a 2, b 2 ] a1 a 1 2 = a 1, b 1, a 2, b [a1 2 b 1 ], [a 2, b 2 ], a 1 = a 2 Problem ( 1.2: #14). Consider the quotient space of a cuve I 3 obtained by identifying each square face with the opposite square face via the right hand screw motion consisting of a translation by one unit in the other direction perpendicular to the face combined with a one quarter twist of the face about its center point. Show this quotient space X is a cell complex with two 0-cells, four 1-cells, three 2-cells and one 3-cell. Using this structure, show that π 1 (X) is the quaternion group {±1, ±i, ±j, ±k} of order eight. Solution: A presentation for the quaternion group is Q = 1, i, j, k i 2 = j 2 = k 2 = ijk = 1, ( 1) 2 = 1. (5) By labeling edges of one face as a, b, c and d and then applying the described identification ( pushing through the cube and rotating 1/4 twist clockwise, a repreesntation of X is labeled in the diagram. Points labeled 1 are realized to be identified and similarly so are points labeled 2. Using 1 = x 0 as a basepoint define A = ac B = ab D = cb.

4 Now since each face of the square is convex, each loop can be homotopied across the face of a square. For example, on the front face of the diagram, ab ' dc. Also, four paths in a, b, c, d or their reverses which enclose a face gives a trivial element, for example abcd. With this in mind: A2 = a c a c 'caca 'abdc 'abab=b 2 'cadb 'dcab 'cbcb 'dbdb ' D2 'dbca and ABD = a c a b c b ' a b d d a b ' a b a b = B 2.

5 B 4 = (B 2 ) 2 a b a b a b a b a d c b a d c b c b c d c d a d c a d d b a a d c a b d c c d d id. In summary, A 2 = B 2 = D 2 = ABD and (B 2 ) 2 = (ABD) 2 = id. Identifying A i, B j, D k, ABD 1 gives us a presentation identical to that of the quaternions (??): π 1 (X) = A, B, D, ABD A 2 = B 2 = D 2 = ABD, (ABD) 2 = id. Problem ( 1.2: #16). Show that the fundamental group of the surface X of infinite genus is free on an infinite number of generators. I claim that X deformation retracts onto a (countably) infinite wedge of circles. To start take a subset X 0 of X, obtained by cutting along two circles between a genus. X 0 is a twice punctured Figure 1: X k with loops R and G around the punctures. torus which deformation retracts onto three circles. To see this it is helpful to consider the square model of the torus as in Figure??., remove a point and then blow it up to the boundary of the square. Then we have two connected bands (a fattened S 1 S 1 ) from which another point can be removed and then deformation retracted to three circles. Hence π 1 (X 0 ) = a 0, b 0, c 0 where a 0, b 0 and c 0 are single loops around each circle. Similarly, π 1 (X k ) = a k, b k, c k for each of the other sections. It is important to recognize that one circle, say a k, in S 1 S 1 S 1 corresponds to a loop, which I will also call a k, around one puncture of the torus (G in the illustrations) while a loop around the other puncture corresponds to b k c k in the wedge of circles (labeled R in Figure?? and??). Now let A k denote an open neighborhood of X k in X. The intersection of any two adjacent A k s is a cylinder, having fundamental group π 1 (A k A k+1 ) = b k c k = a k+1 = Z. The relation between the wedge of the circles (S 1 S 1 S 1 ) k and (S 1 S 1 S 1 ) k+1 is then that X k X k+1 deformation retracts onto (S 1 S 1 S 1 ) k (S 1 S 1 S 1 ) k+1 ak b k+1 c k+1

6 Figure 2: Square model of Xk and what the loops R and G look like in the punctured band view. Once the other puncture is fattened up a little, we can deformation retract onto S 1 S 1 S 1. which is a wedge of four circles, namely ak identified with bk+1 ck+1, bk and ck. In essence as k increases, we add the circle ak+1 to the wedge sum and as k decreases, ak is added to the wedge sum of the spaces corresponding to lower k. Continuing in this fashion we see that X deformation retracts onto a countable wedge of circles. For each copy of S 1, there is a loop going once around. Hence _ π1 (Sk1 ) = Z Z Z = ai i Z, π1 (X) = k=1 the free group on infinitely many generators. This is what the gluing of the deformation retracts of each Xk looks like (note that for each Xk, the xk s can be slid so that it is actually a wedge sum (in other words the three circles should meet at the same point xk ).: