‘}V4f31_] RETURNING MATERIALS: P1ace in book drop to LJBRARJES remove this checkout from ‘J-llflfll-IL. your record. fifl§§_w111 be charged if book is returned after the date stamped beWOw. _ \ fem _.__,_.—-'—A 2 2p — ACTIONS ON THE 2-DIMENSIONAL AND THE SOLID KLEIN BOTTLES BY Fawaz Mohammad Abudiak A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1984 ABSTRACT Z 2p - ACTIONS ON THE 2-DIMENSIONAL AND THE SOLID KLEIN BOTTLES BY Fawaz Mohammad Abudiak In this thesis PL homeomorphisms of periods p and 2p are classified on both the 2—dimensional Klein bottle K2 and the solid Klein bottle K, where p is an odd prime number. It is shown that up to weak equivalence there is only one class of homeomorphisms of period p on K2 and only three equivalence classes of homeomorphisms of period 2p on K2, distinguished by the fixed point . th sets of their p powers. Also, free cyclic actions of odd period are class- ified on K as well as cyclic actions of period 2p. In the first case it is shown that, up to weak equival- ence, only one such action exists, while in the second case there are three such homeomorphisms, distinguished by the fixed point sets of their pth power. Finally, semi-free action on K are classified for any finite period n. It is shown that these exist only for n equals two and for all odd values of n such that F(hn) = a. TO MY PARENTS ii ACKNOWLEDGMENTS I wish to express my sincere gratitude to Professor Kyung Whan Kwun, my thesis advisor, for suggesting the problem and for his continuous support, patience, and encouragement during the completion of the problem. iii TABLE OF CONTENTS INTRODUCTION . . . . . . . . . . . . . . . . . . . . CHAPTER 0. PRELIMENARIES AND DEFINITIONS. . . . . . CHAPTER 1. 252p - ACTIONS ON THE Z-DIM KLEIN BOTTLE SECTION 1.1. Zip - ACTIONS ON THE 2-DIM KLEIN BOTTLE. . . . . . . . . . . . . . SECTION 1.2. 222p - ACTIONS ON THE 2-DIM KLEIN BOTTLE. . . . . . . . . . . . . . CHAPTER 2. Zfizp - ACTIONS ON THE SOLID KLEIN BOTTLE K . . . . . . . . . . . . . . FR . . SECTION 2 1 EE 212k+1 ACTIONS ON K SECTION 2.2. fl: - ACTIONS ON K . . . . . . ZP SECTION 2.3. SEMI-FREE ACTIONS ON K . . . . . BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . iv Page INTRODUCTION Let X and X' be topological spaces. Two homeomorphisms h and g on X and X' respectively are said to be weakly equivalent (written h mm 9) if there exists a homeomorphism t :X + X' such that t-lgt = hl for some positive integer i # 1. All maps and spaces considered in this thesis are in the piecewise linear category. In this thesis we classify piecewise linear homeo- morphisms of periods p and 2p, p an odd prime, on both the 2-dim and the solid Klein bottles. Chapter 1 deals with the homeomorphisms of periods p and 2p on the 2—dim Klein bottle K2, p odd prime. Proposition 1 of section 1.1 asserts that up to weak equivalence there is a: unique homeomorphism of period p on K2. Theorem 1 of section 1.2 gives all the homeomorphisms of period 2p on K2 up to weak equival- ence, p odd prime. In fact there are three such homeomorphisms hi distinguished by the fixed point sets of hip, i = 1, 2, 3. Chapter 2 is divided into three sections. Section 2.1 deals with the free homeomorphisms of odd period on the solid Klein bottle K. Proposition 3 of this section states that there is only one such homeomorphism up to weak equivalence. Section 2.2 provides a complete classification of homeomorphisms of period 2p on K, p odd prime. The main theorem of this section is the classification Theorem 1 which asserts that up to weak equivalence there are only three such homeomorphisms hi, distinguished by F(hip), i = 1, 2, 3. Finally section 2.3 deals with the semi-free periodic actions on K. These actions are given in Theorem 1 of that section. CHAPTER 0 PRELIMINARIES AND DEFINITIONS Throughout this thesis, we work in the PL (piece- wise linear) category and all spaces and maps will be piecewise linear. . . . . . n A homeomorphism h : X + X is periodic if h = identity for n >1 in Z5. If n = 2, h is said to be an involution. Let h be a periodic map of a space X. The cyclic group generated by h will be denoted by <11>. If h is periodic on X, then the orbit space of h is the quotient space obtained by identifying x with hl(x) for all i and all x in X. The orbit space of h will be denoted by X/<11>. The identification map ph : X + X/‘

is called the orbit map. When there is no confusion ph will be simply written as p. Two actions of < h> and < f> on X are said to be weaklygequivalent if there is a homeomorphism t of -l> = < f> and tht-1 = f1 for some X such that § x/ i'e pf: = tph. With a special consideration of the fixed point, sets «of h and f this implies that t f t is a linear transformation (automorphism) of X with respect to h i-e it belongs to the group of deck transformations A(X, h) with the connection with the proposition on page 5, this implies weak -1 equivalence. If tht = f, then h and f are said to be equivalent. The set {xexlh(x) x} of the fixed points of h will be denoted by F(h) or Fix(h). Concerning the action we assume that (l) for every hi 6
  • , F(hi) is a subcomplex of X; (2) the natural cell structure of the orbit space X/ and the orbit map p : X + X/‘ are simplicial and (3) p maps each simplex homeomorphically. According to [16} these conditions are not restrictive, from the PL point of view. The following proposition concerning the free action of periodic homeomorphisms proves to be very useful in proving weak equivalence, where h : X + X acts freely or is free if F(h) = a = F(hl) for all i for which h1# identity. Proposition. Let (X, < h>) be a free periodic action in which h is of finite period, and X is a connected manifold. Let X = X/ be the orbit space and p : X-+X be the orbit map. Let xe X, E = p(x). Then X is a connected manifold, (X, p) a regular cover, and 3:__ w1(§.§)/p,2=K. Proof Let B = K/. By [4] we have x(k) + (p—l)x(F(h)) = px(B), where X indicates the Euler characteristic. Since x(K) = O, (p-l)x(F(h)) = px(B). Since p is odd, dim (F(h)) # 1. So F(h) is either O-dim or g. Assume that dim (F(h)) = O and that F(h) consists of k points. Assume k >0. Then (p—l)k = pX(B) and this implies that k = 3%; x (B) - k>-o yields that (p-1)|X(B) and since B is a surface, x(B)152, so p S 3. But p is an odd prime so p = 3. Hence x(B) = 2 and then B 2 52, the 2-sphere. But this is impossible since K is nonorientable. 80 k must be 0 contradicting our assumption k >0. So F(h) = fl. Moreover, x(B) = 0. Since F(h) = fl, BB = fl and since B is nonorientable B N K. Proof of Proposition 1. It is clear that F(hl) = fl and K/ 8K.

    and act freely on K. Let qi :K‘tK/‘, i = l, 2, be the orbit maps. Let t be any homeomorphism: K/

    '*K/

    . Since F(hi) = fl, g1 and q2 are p-covering projections of K by K. But “1(K) has a unique normal subgroup 10 of index p, so t can be lifted to a homeomorphism + K such that the diagram below is commutative rfl x E K :7 4;; K q1 q2 N! J, t K/ ; K/ i.e q2t - tql. By the commutativity of the diagram _ h _ q1 q2 q2 q1 1 t t \r Ks >K 4 . --l ‘— . --l — . We obtain q1 = qlt hzt. That is t h2t is a non- trivial covering transformation on K with respect to ql. But the group of covering transformations A(K,q1) ; (see Chapter 0). So E-qszf = hll for some 1 S i < p. This shows that h'rwhl and completes the proof of proposition 1. 11 Section 1.2. ZZZP-ACTIONS ON THE 2-DIMENSIONAL KLEIN BOTTLE. Theorem 1. If h : K + K is a PL homeomorphism of period 2p, where p is an odd prime, then h is weakly equivalent to one of the following 29- periodic homeomorphisms depending on whether F(hp) is fl, 3 S1 1 . l . or 2 S U S respectively h ([er) =[323 1 r h ([rz]) =15in 2 r h ( [r z ]) = [:w r z] 3 n' where w e l/p. The proof of theorem 1 occupies the rest of this section. Since h is of period 2p, h2 is of period p on K, so by lemma 2 of Section 1.1, F(hz) = fl. The map h induces an involution h- on K/

    , uniquely determined by h, such that h-q = qh where q : K + K/

    is the orbit map. p is an involution on K. By [12] (also see Now h [8]) F(hp) is one of the following sets: (5, two points, one 2-sided nonseparating simple closed curve and two points, one 2-sided separating simple closed curve and finally two 12 one-sided simple closed curves. In other words F(hp) is homeomorphic to one of the following sets: g, 50, SltJSO, Sl, SlIJSl. So we consider the following two cases , P _ Case 1. F(h ) - fl P Case 2: F(h ) ¢ a Case 1. F(hp) = fl. In this case h acts freely on K. We prove the following: Proposition 2. If h : K—*K is a homeomorphism of period 2p on K, p odd prime, and if h acts freely, then h is where h is as in theorem 1. weakly equivalent to hl' 1 Proof 2 Let h — h2. Let Mi - K/ and q1 .1<-*Mi: <;'.: Mi-+Mi/, i = l, 2, be the orbit maps, where h 1 and h 2 are the induced involutions by hl and h2 on M1 and M2 respectively. Note that F(h-i) = qi(F(hip)), hence F(h'i) = d, i = 1, 2. As in lemma 2 Section 1.1, K/ 2 K. By lemma 2, Mi 2 K, so M./.‘::K/ 2: K. Note also K/ 2: K/ . 1 l l l l . = '. . : + . < '.>. h . is a Let 91 q lql K Ml/ h 1 Note t at q1 V p- covering projection for i = l, 2 and q i is a 2— covering projection for i l, 2. Since F(hip) = fl = F(hz) = F(h), acts freely on K, so gi is 13 a 2p- covering projection for i = l, 2. Let t : K/ + K/

    be any homeomorphism. By [14] up to equivalence there is only one subgroup of indes 2 of "1(K) corresponding to the double cover of K by K. So M1 and M2 are equivalent, hence 3 t : M1 2 M2 such that tq'l = q' t ”H 7: w t— Ml 7‘ 2 ' ' q l q 2 K/ t 4, K/

    Now since ql and q are p- covering projections 2 and fll(K) has a unique normal subgroup of index p, t can be lifted to t : K E K such that qu = q2:. ' = ' = o = = I Hence tq lq1 q 2qzt i.e tgl gzt. As in the proof of proposition 1, section 1.1, :11 :h-1 is a 2 covering transformation on K. Note that 91 and 92 are 2p- covering transformations and the group of covering transformations of K with respect to 91, A(K, gl);; (see Chapter 0). So we obtain that -l i :h2: = hl for some 1 S i < 2p. 14 Hence h2 = h Nw hl. This finishes the proof of proposition 2 and case 1. Case 2. F(hp) # fl. By [12] (or [8]) F(hp) is homeomor- phic to one of the following sets: SO, SOIJSl, Sl, 51:331. If F(hP)I~SO = {x, y}, then since h(F(hp)) = F(hp) and F(h) = o, h(x) = y and h(y) = x. But this implies that hzx = x and hzy = y contradicting 2 p .v 0 the fact that F(h ) = g. So F(h ) cannot be ~S . If F(hp) z 810 SO, then since F(hp) is . . O O l l invariant under h, we get h(S ) = S and h(S ) z S . As above this contradicts F(hz) = fl. Hence F(hp) O 1 cannot be NS 6's . Now we are left with the two possibilities: F(hp) N S1 or S1 0 51. Lemma 3. If F(hp)” s1 or 510 31, then K/ is homeomorphic to Mobius band or an annulus respectively. Proof Let B = K/. Since hp is an involution on K, we have by [4] that x(K) + x(F(hp)) = 2x(B). But F(hp) is one-dim, so x(F(hp)) = 0. And since X(K)==O, x(B) = 0. So B is homeomorphic to one of the spaces: 2 . . K, T == the 2-dim torus, Mobius band M, an annulus A. 15 Now we claim that 3(B) = q(F), where q :K + B is the orbit map and F = F(hp). If xe‘K - F, then 2 x has a neighborhood U 3 Hi and q(U )z U . q(U ) x x x x is a neighborhood of q(x) e B and since q(Ux) NUx a*JR 2, q(x) eInt(B). If x eF, then x has a neighborhood 2 2 Ux mm . F n Ux separates Ux into two components ‘3 IR+ and q(Ux) z 1122+ and is a neighborhood of q(x) e B, i.e q(x)€ BB. This proves the claim. So 3B # a, 2 hence B cannot be iaK or T , for these have no boundary. If F #5 , then B has one boundary component If Fivsllel, then 3B has two boundary com- ponents and B 3 A. U Now, let h2 and h3 be as in theorem 1. We prove the following lemma. Lemma 4. Let h : K-rK be a periodic homeomorphism of period 2p, p odd prime, and assume that F(hp)iUSl. Then h ~w hz. Proof h induces h- :K/-*K/ of period p and h acts freely on K/. Similarly h2 induces h.2: K/-*K/ of period p and h-2 acts freely 16 on K/. By lemma 3 above K/s3M, a Mobius band and k/~M. Let F = F(hp) and F2 = F(th). Let K' = K/ , K2 = K/ and q : K + K', ql: K + K2 be the orbit spaces and orbit maps respectively. Also let q.1: K2 + K2/ zK/

    , q': K+K'/ 23K/ be the orbit maps of K2 and K'. ql(F2) = stssl and q(F) = Jissl are the boundaries of K2 and K' respectively. q'1(J2) and q'(J) are the boundaries of K/

    and K/ respectively. Since h- and h; are free of period p on M, an arguement like the one used in section 1.1 using the Euler characteristic yields K2/iaM ~ K'/. Let t : q'(J)-*qtl(J ) be a 2 homeomorphism. Extend t to a homeomorphism on all of K/ still call it t. "H K > K q ‘[ ql K' E > K2 q' J<11 K/ t is K/ 17 Since q' and q3_ are p- covering projections and 1r1(M) gm has an unique normal subgroup of index p, we can lift t to a homeomorphhml E': K' + K2 such that q3.t = tq' and E(J) = J2. Now q1 :K - F2 + K2 - J2 and q : K - F-+K' - J are double covering projections and 1T1(K2 - J2) ;22 ;fl1(K' - J) we can lift f to a homeomorphism : : K-F + K-F2 such that qli = Eq. Since :(F) = F2 and E is continuous this is true on all of K. Now, q' is a covering projection and the diagram below commutes _ h __ K'___E..__)K \ K2(__t_ K' I 2 2 / qI q. q' 1 ql K/ t .49 K/

    <_ t K/ hence for some i, E-llfé E = h-i. Also, since qh = h'q and qlh2 = h'2 ql, it follows that: qhi=h'iq. Hence qhi = h-iq = E- h.ZEq = E-lh-qut: = E-lql h2: = q=t:- h2 I. Thus, if xe F, then h i+p(x) = h inc), note that here qIF is a homeomorphism. If x4 F, then hi+p(x) = :"1h2?(x) or hi(x) = :-1h2:(x). Therefore h ~ 11 . w 2 18 Finally in order for the proof of theorem 1 to be complete we prove the following. Lemma 5. Let h : K + K be a periodic homeomorphism of . . . p 1 1 period 2p, p odd prime, with F(h ):~s tJS . Then h ~wll3. Proof Let K3 = K/ . By lemma 3, K3 ~ A. As in lemma 4 we obtain that h ~ 11 . U w 3 This finishes case 2 and completes the proof of theorem 1. CHAPTER TWO 22p - ACTIONS ON THE SOLID KLEIN BOTTLE K Section 2.1. FREE ACTIONS ON K. Z 2k+l Lemma 1. If h : K + K is a (PL) homeomorphism of period 2k-+l, k 21, on K, then Fi.xtl is either w or 1 z 8 . Proof . t t Let n = 2k + l . n can be written as n = pll.p22 .pgml where pl, ..., pm are distinct odd primes and t1, .., tm are positive integers. If m = 1, then h is of period pffi. on K which is a homology 1- sphere. Hence by [4] Fi.xll is either a or a homology l-sphere. F(h) cannot be 2-dimensional because psi. # 2. So F(h) is either ¢ or 3813 If m = 2, then t n = pltl p2t2 and hpl 1 is of period p2t2 . As above t t F(hpll) is either 95 or ~81 . If F(hpl 1) = fl , t F(h) = fl. Since F(hpl 1) is invariant under h, then t if F(hpl 1)::51 and F(h) a! o, h is of period pltl t on F(hplil) le a homology 1- sphere and again by [4] F(h)ssSl. Hence if m = 2, F(h) = H or z 81. 19 20 Now assume that the result is proven for m = i. t t. . Let c = p1 l...pi i. Let the period of h be ti+l ti+l pi+l 1+1 . Then h is of period c on K so by ti+l . . pi+l l theinduction hypothesis F(h ) is either a or ::S . ti+1 ti+1 pi+l pi+l 1 If F(h ) = fl , F(h) = g. If F(h ) z S , then by [4] F(h) z S1 or fl. 1 Hence F(h) = g or 2:8 . U Remark. The proof above shows that if F(h) z 51, then F(hl) = F(h) z s1 for all 1 acts freely on K, then K/::K. Proof Let B = K/ and let p : K + B be the orbit map. Since acts freely on K, K is a regular 2k + 1 covering of B [11, theorem 8.2, Chapter 5]. Hence p#(fl1(K))ssZ5 is a normal subgroup of index 2k + l of nl(B). So we have a short exact sequence 21 O. ..). O + Z5 H (B) 1 252k + 1 Since B is covered by a contractible space and no nontrivial finite group can act freely on a finite dimensional, contractible space [6; page 287], “1(8) has no torsion subgroup. Let a = a(generator of Z” and b be such that B(b) is a generator of 252k + 1' Since p#(fll(K)) is normal in N1(B), béib-lé . 1 -l _1 -1. If béib- = a , then So baib = a or a hl(B)/[fl1(B), Wl(B)]; H (B) is finite (for the coset l E = b + [fl1(B), fll(B)] is of order 2k + 1). Hence 3 i X(B) = Z (-l) p. = l + 0 because 0 = O = p , o = O . i 2 l 3 3 i=0 because B is nonorientable. But this implies that X(B)£Zl contradicting the fact that x(B) = 0. Hence -1 -l -1 bab 7! a . So we must have bab = a and so fll(B) is abelian. From the principal theorem for abelian groups “1(8) = 2Z+ Tor(TT1(B)) = Z3 + O = Z . Hence “1(8) = Z . Note that B is compact, nonorientable, irreducible with K (2-dim Klein bottle) as its boundary. Moreover, . . . . 2 B contains no 2-Sided prOJective planes P , because . . 2 -l 2 . if B contains such P , then p (P ) will be a 2- sphere S, and since K is irreducible S bounds a 22 a 3-cell C. Then p(C) is a 3-manifold bounded by P2. Now by [5] theorem 11.7 BRsK. U Proposition 3. Up to weak equivalence there is exactly one free > . ZZK + l (k _ 1) action on K. Proof. Let hl : K-+K be defined by h1([z, t]) = 2k . . . [z, t + EEII ]. hl is a homeomorphism of period 2k-tl and F(hl) = g for all l. z K. Now let h : K + K be any homeomorphism of period 2x + 1 such that F(hl) o, lssK. Let pl : K + K/

    and p : K + K/ be the orbit maps. p1 and p are (2k + l) - coverings of K/ and K/ respectively. Let t : K/ + K/ be a homeomorphism. Since tp1 and p are (2k + l) - covering projections of K and since Nl(K/) ; Z! has a unique normal subgroup of .\ index 2k + l, 3 a homeomorphism t : K + K making the K K/ l_; K/ diagram r*l 23 commutative, i.e such that tpl = pt. Now'.as in proposition 1, section 1.1, h~h1l for some 1.Si 52k. Hence any two free Z§2k+—l actions on K are weakly equivalent. U Section 2.2. 212p - ACTIONS ON K In this section we classify all Z: - actions on 29 the solid Klein bottle K, up to weak equivalence, where p is a prime number. The case p = 2 is studied in [13], so we study here the case where p is an odd prime. Our main result is the following Theorem 1. If h : K-*K is a homeomorphism of period 2p on K, p an odd prime, then F(h) = fl and F(hz) = fl and h is weakly equivalent to one of the following period 2p homeomorphisms on K depending on whether F(hp)£aM, A or S1 respectively: hl([z, t]) [-z, t + —E— J, F(hf’) z Mobius band M p-l ], F(h;)) : Annulus A II r—I NI n + h2([zI t]) ll r—a I N (1' + 9-1 9 ~ h3([z. t]) J. F(h3 ) ~ 3 The rest of this section is devoted to the proof of theorem 1. 24 Let h : K + K be of period 2p. Since (h2)p = 1d, h2 is of period p on K. Since K is a homology l-sphere, then [4] gives that F(hz) is a homology r-sphere, where r s l and l - r is even. Hence r is either -1 or 1. p # 2, so F(hz) is either 1 2 2 a or 2:8 . Note that F(h) g F(h ). If F(h ) = fl, then F(h) = fl. If F(hz) z 51, then since h is 2 an involution on F(h ) z 81, it follows from Smith [15] that F(h) is a homology r-sphere, r S 1. Hence r = -l, O, l i.e F(h) is one of the sets: 0 . . fl, 8 , 51. So we conSider the following cases: Case 1: F(hz) H, hence F(h) = fi- 1 O 1 Case 2: F(hz) S and F(h) is a, S or 22 22 U) Case 1: F(hz) fl. In this case h is determined up to weak equivalence by the F(hp). We prove the following Proposition 2. Let h : K + K be of period 2p on K, p odd 2 prime, with F(h ) = fl. Then h is weakly equivalent to one of the homeomorphisms hl' 2, h3 depending on whether F(hp) z M, A or z Sl respectively. (The maps hl' 2, h3 are the ones in theorem 1). 25 Proof of Proposition 2. hp is an involution on K, hence by [13] F(hp) . . . 2. is homeomorphic to one of the sets: I Upt, D UI , M, A, 51. Since F(h) = fl and F(hp) is invariant under h, F(hp) cannot be IOpt or D20 I, otherwise in the first case h(pt) = pt and h(I) = I so h would have at least 2 fixed points. In the other case h(Dz) = D2 and h(I) = I and Brouwer fixed point theorem implies that h has at least 2 fixed points. Hence . . l F(hp) is homeomorphic to one of the sets M, A or S . The map h induces a homeomorphism h- of period p (n1 K/ defined by h-q = qh where q : K-+K/ is the orbit map. Note that acts freely on K/. From [13] if F(hp)fivsl or M, then K/:sK and if F(hp) z A, then K/ a D2 x 81, the solid torus. Now, we show that any homeomorphism h of period 2p with F(h2) = fl is weakly equivalent to one of the homeomorphism h 1'2’3 (i) Assume F(hp)::M. We show that h is weakly equivalent to hl. Note that F(hl) = F(hf) = fl and K/zK, also since F(hp) zM, K/zK. The induced maps h]. and h are free on K/ and K/ respectively. Since K/ 3 K 3 K/ and hl and h are free of odd period p it follows from 26 lemma 2, Section 2.1 that (K/ c: K and that 1 (K/)/ 3 K. Let Kl = K/ and K2 = K/. Let ql : K + Kl, q2 : Kl + Kl/, q : K + K2 and q' : K2 + K2/ ~ K/ be the orbit maps. Since hlp and hp are involutions, ql(F(h1p)) = M1 is a Mobius band in 3(K1) and q(F(hp)) = M2 is a Mobius band in 3(K2). F(hlp) is invariant under h l and F(hp) is invariant under h, hence M1 and M2 are invariant under h3_ and h- respectively. Since 3(K/) and 3(K/) are 2-dim Klein bottles, I - . q2(M1) and q:2(M2) are Mobius bands in 8(K/) and 3(K/) respectively. P hl 3 a meridional disk D in K which is invariant under is an involution on K so by [9] and [10] h p. Cut K along D we obtain K' z D2 x I and by l [9] hlp/K' ~ f , f(z, t) = (Z, 1 — t). Let '5 be an hlp invariant disk in K', then q1(5) = E1 is a meridional disk in Kl. h- acts freely on K1 so - p-l . . . El, h].(El)' ..., h1 (E1) are mutually diSjOint. Hence q2(El) is a disk in K/. Similarly for hp we obtain E and q'(E) a meridional disk in K/. Let cl = 3(q2(El)).- c = 3(q'(E)) and el = 3(q2(Ml)). e = 3(q'(M2)). el and e separate 3(K/

    ) and 3(K/) respectively each into two Mobius bands. Also 01 and el meet in 2 points so do (2 and e. 27 Note that 3(K/) -c Mel. and 3(K/) -c -e each 1 consists of two open rectangles. Now let t1 :clue1 g c ue and extend tl to t2 : 3(K/

    ) + 8(K/) such that t2q2(M1)_=q (M2). Then extend to t3 on q2(El), finally accross the open 3-cell K/ - 3(K/

    ) - q (E ). The final map is a homeomorphism 2 l : < > < > ' = I . t K/ h1 + K/ h with t(q2(Ml)) q (M2) K t ) K ql l'q F . K1 7 K2 q2 Jrq t a K/ .7 K/ are free so g2 and q' are p- covering projections and since hl(K);;Z: has a unique normal subgroup of index p, t can be lifted to a homeomorphism t : K + K such that tq2 = q'? and 1 2 {(M1) = M2. Now q1 is a double cover: K - F(hlp)-> Kl - M1 and q a double covering projection: P ~ _ K F(h ) + K2 M2. But 1T1(Kl M1) = 25, t can be lifted to I : K - F(hlp) + K - F(hp). 23 By continuity t can be extended to all of K. So qt = tql. Now as in the proof of lemma 4, section 1.2 we see that h Nw h1° (ii) Assume F(hp) z A, an annulus. We show that N . _ 2 _ p 2 l h w h2. Note that F(hz) — F(h2 ) — fl, K/

    D ){S , a solid torus. Similarly K/ z DZJcSl. Since h2 and h- act freely on K/ and K/ respectively and p is odd, then by [7] Kl/ z D2 x S1 and - 2 . . . K2/ z D 1:51, notations as in (i) above. h hp are involutions imply that ql(F(h2P)) = A1 and q(F(hp)) = A2 are annuli in 3K1 and 3K2 respectively. As in (i) q2(A1) and q‘(Az) are annuli in 3(K/

    ) and 3(K/) the latters are homeomorphic to 51)(51. 112 and h free imply that g2 and q' are .p-covering projections. Note that Kl/==K/

    and K2/ z K/. Hence K/

    z D2 x S1 x K/. As in (i) there are meridional disks -El and E in K1 and K2 respectively such that q2(E1) and q'(E) are meridional disks in K/ and K/ respectively. Let cl = 3(q2El)r1q2Al, c}. the complement of c1 in 3(q2E1), c2 = 3(q'E)rlq'A2, c3 its complement in 3(q'E). Let t : c1 + c2 be a homeomorphism. Extend t to a homeomorphism : q2(Al) + q'(Az) then on c3_ on to CE , then on qu1 on to q'E, next on the open rectangle 3(K/

    ) - qu1 on to 3(K/) - 29 q'(Az). finally across the remaining 3-cell in K/. Still call the new homeomorphism t. Note that t(q2Al) 2. As in (i) lift t to E : Kl + K2 such that '-= .' TTK-A~1T -A ~ ‘ q t tq2 Since 1( 1 1):: 1(K2 2) = Z5 lift 3 to E : K - F(th) + K - F(hp), by continuity extend q'A "H on all of K. So qt = tql on K. Hence as in (i) (iii) Assume F(hp) z 51. We show that h ~ h We use the same notations of (1). By [13] K =us:K P D P _ < > _ < -> _ where Kl K/ hi , K2 K/ h . q1(F(h3 )) c and q(F(hp)) = c are simple closed curves in the interiors 2 of K1 and K2 respectively. By lemma 2 (section 2.1) <->z z <-> l . . . Kl/ h3 K K2/ h . qzcl and q c2 are s c c in the interiors ofK/

    and K/ respectively. Let t : K/

    + K/ be a homeomorphism mapping q2(cl) on to q'(cz). As in (1) lift t to E : K + K 1 2 mapping cl on to c2 and q'? = tq2. By [13] F(h3p) and F(hp) are the "cores" of K, so c1 and c2 are the cores of K1 and K2 respectively. Hence "1(K1 — cl) ; ”1(K2), K2 the 2-dim Klein bottle. "1(K ) has a unique normal subgroup of index 2 . 2 . _ corresponding to K [14]. So we can lift t to n" K - F(h3p) + K - F(hp). As in (i) we conclude that h is weakly equivalent to h3. This completes the proof of proposition 2. D This takes care of case 1. 3O 2 Case 2. F(h ) z 51 and F(h) a or S0 or 1:51. We consider the following 3 subcases: Subcase 2.1. F(hz) z S1 and F(h) 5:80. We show that this case cannot happen. Proposition 3. There is no homeomorphism h of period 2p, p odd prime, on K with F(hz) z 81 and F(h) : SO. Proof hp is an involution on K. Hence by [13] F(hp) is homeomorphic to one of the sents: I Opt, D2 OI, M, A, 31. Note that F(h) c F(hp). If F(hp) 2 M, A or 81, then F(hp) is a homology l-sphere. Since F(hp) is invariant under h, h is a period p homeomorphism on F(hp). Hence by [4] F(h) is a homoloyg r-sphere with. r'Sl and l - r is even. So r = -1 or 1, i.e F(h) is either ¢ or 2 S1 contradicting our assumption that F(h) z SO. So F(hp) cannot be M, A or 51. If F(hp) z D2 0 I, then since h(F(hp)) = F(hp), h(DZ) = 02 and h(I) = 1. So h has one fixed point in D2 and one fixed point x in I. So h must interchange the sides of I - x and since p is odd hp interchanges the sides of I - contradicting the fact that If: F(hp). Hence F(hp) i DZIJI. Finally if F(hp) z I Opt, then since F(hp) is invariant under h, we have h(I) = I so h must have X 31 a fixed point in I. We get a contradiction as above. U Subcase 2.2. F(hz) 3 S1 and F(h) z 81. 2 Note that F(h) = F(h ) n F(hp) and hp is an involution on K. Since F(hp) is invariant under h and F(h) z 81, F(hp) cannot be 2-dimensiona1, other- wise h must interchange the two components of U - F(h), where UCF(hp) is a small neighborhood of x in F(h) and since p is odd this implies that hp interchanges the two components of U- F(h) contradicting the fact that UCF(hp). So F(hp) cannot be 02 01, M or A. Since F(h)c=F(hp) and F(h) z 81, F(hp) cannot be zIIth. Hence the only possibility is the F(hp) z 81. So now we have F(h) = F(hl) z S1 for all 1 siK. Proof h induces an orientation preserving homeomorphism h~ on the orientable double cover D2 x S1 of K with l . period 2p and qh~ = hq where q : D x S + K is the covering projection. By definition of h~, F(hN) # B and F(h”) = q-1(F(h)). By [7] F(h”) z s and F(h~) = N1 . ~ . . . F(h ), 1 <1 <2p. Hence h p is an involution on 32 2 1 . . . . D x S With fixed pOint set a Simple closed curve. Tollefson [17] shows that if an orientation preserving . 2 . homeomorphism f on D x S1 has a Simple closed curve as its fixed point set, then f is equivalent to a . 1 2 rotation about the core 0 x S of D x 81. In particular F(f) = O x S1 i.e it is unknotted. So Fi.x(hp) = O x S1 and since F(h~p) = F(h”), F(hN):3 ~ 2 O x 51. Hence by [7] D2 x Sl/ z D x 81. Now consider the diagram .0 :02 WV ' ZJL 1 2 1 p D xS /~D xS B< where B = K/, q : K + B the orbit map, a 2 2 . 2 l D x S1 -* D x S1 the orbit map of D x S and 2 1 . . . . p : D x S + K the coverning prOJection of the orientable 2 2 . double cover D x S1 of K. p' : D x S1 + B is defined by p'q = qp. Since QI(K - F(h)) is a 2p:- 1 sheeted cover of B - q(F(h)) and END2 x S F(h~)) is a 2p — cover of D2 x S1 - q(F(h~)) and since p is a covering map, then given be" B-q(F(h)) we can find an open neighborhood Vb such that p'-1(Vb) = 010 U2 with 33 U. open and p'(Ui) 2 V i = i, 2. Note also that i b' F(h”) is a double cover of F(h) and q(F(h)) z F(h) ~ ~ ~ 2 and q(F(h )) z F(h ). Hence D x S1 double covers B. Since B is non orientable and BB = K2 and is double 2 1 covered by D x S , B 3 K. D The proof of lemma 4 can be used to prove the following: Corollary 5. If h : K + K is a homeomorphism of period n, where n is an odd integer, with F(h) a simple closed curve z core of K, then K/ 3 K. Proof Since n is odd and F(h) z 81, then by the remark following lemma 1, section 2.1, F(h) = F(hi) for all 1 z K. Now h2 is of period p and F(hz) 2 S1 and by the above remark F(hz) = core of K. Hence to be done with this subcase we prove the following 34 Proposition 6. There is no homeomorphism h : K + K of period p, p an odd prime, with F(h) = core of K z 81. Proof Let q : K + Kl = K/ be the orbit map. By corollary 4, K1 3 K, q(F(h)) = Core of K1. Let D be a meridional disk in K/ such that Dr1q(F) is a point (Note that q(F) is the core of K1). -1— q (D) is either one disk or n disks meeting at a . . . 2 common interior pOint. Note that 3K = 3K1 = K the 2 2-dim Klein bottle. qIK is a p- sheeted cover of 2 — . 2 3Kl = K . 3D is a Simple loop (1: I + K . Let eO = a(0) = a(l) and let 306:q-1(e0). Let a = [a], and let 8 be an orientation reversing loop which meets a transversaly once at eO and let b = [B]. So 1 -1 >. fll(3Kl, e0) = . Let n : I + K be a loop which wraps once around the component of q-1(BD) = q-l(a) containing 30, with n(0) = n(1) = 30' Then P (qIBK)#( [0]) = a and so q|n(I) covers a(I) = SD . . -l - . . . in a p to 1 fashion. Hence q (D) is a meridional disk D in K. Since hq (o) = q'1(5), h(D) = o i.e D is invariant under h. Now cut K along D and K1 along D to obtain D x I and D1 x I respectively, where D z D z D , 35 the standard disk. 3(D x I) contains two copies D' and D" of D and 3(Dl x I) contains two copies 0' and D” of D. Let e be an arc 21 in D joining the center of D(= Dr1q(F(h))) and SD. So 0' contains a copy e' of e and D" contains another copy e" of e. Let FCZD x I be F(h) after cutting K and FCDl x I be q(h(F)) after cutting Kl. Let E be a disk in D1 x I bounded by e', e", F and an arc in (SDI) x I joining the point e'rlD' and the point e” n D" . See the figure below .‘I ,’//‘./_,', ,I / t I . 1 I . ,l I' I ,l ' ‘ . " ..... / ’ ', _I . I i . f 1'; / ,l '1; i I I I f I . 'I // ‘ / 1/ // /,' "I / ‘ , ‘l' / ,' ‘r .- ,l . l '/'/ /,’ //J [ll/1’" ..- q-1(E) consists of p disks E‘l), i = l, ..., p having F as a common edge. Let 2’(i) = E(l)rlD' and 2” = E(1)0 D" in D x I. 2'. and l" . are (i) (1) (1) pairwise disjoint arcs for all i joining F to BD' and 3D" respectively. Let BN1] be the part of D' and BD'. that lies between 1'. and 2'. (1) (1+1) Define B"(i) similarly. See the figure below (p =3). 36 é— — 9—4 III—- r—"" —- fi-I- '— " m -— H . / / l ’ / (3’/ r / / l’ I , y / / / / / I I ‘DXI The case p = 3 Let N be the identification map of D x I and w' the identification map of D x I. W' identifies e l with e" and D' with D" so that Dl x I/W' = Kl. So identifies 2' . with 2". i = l ... . q) (l) (l), I I p So it must identify 3' . with B". for i = l, (1) (1) ., p. Hence w identifies D' with D" in an . . . 2 l orientation preserVing manner. Hence D x I/w = D x S . 80 K cannot be restored from D x I. But D x I/w must yield K. D Remark 7. In the above proposition p may be replaced by any odd positive integer. 37 This follows from corollary 5, section 2.2 and lemma 1 section 2.1 and the proof of proposition 6. This completes subcase 2.2. Now we handle the last subcase Subcase 2.3. F(h2) z S1 and F(h) = fl. Since hp is an involution on K by [13] F(hp) . . 2 . l , is RIIupt, D u I, M, A, S . As before, Since p . 2. F(h) = fl, F(h ) cannot be ILth or D LJI. So we are left with the 3 cases F(hp) z M, A, 81. We shall Show that none of these cases is possible. First we prove the following: Lemma 8. Let F = F(hz) and let X06 F. Let i : F + K be (K! 1(0)): . . . 1* the incluSion map. Then image(fll(F, x0)____,wl n '. h , . l(K, x0) 1 e F generates 1(K x0) Proof We follow here a method of proof used by Conner and 2 Raymond [2]. Let f = h . Let (K*, q) be the universal cover of K and let fioe'q-l(x0). Note that h2 = f is of period p on K. Lift the Zip action of f on K to a Zip action on K*. From covering space theory f induces f on K* with period p and qf = fq. Let E = F(f). From Smith theory E is connected and acyclic and since E is connected, q(E) = F. Let defil(K, x0) 38 and let A(t) be a loop in K based at xO representing a. Lift A(t) into a covering path 1(t) with 1(0) = I and 1(1) = o(§o)e E (for O qcx = q implies q(a(§0)) = q(§0) = x0, Ed = a? implies fa(x0) = af(x0) = a(x0), hence a(xo)e E). 3 a path A'(t) in E with 17(0) = :0 and 1'(1) = 1(1) = G(;O). Then 1(2t) OStSk W(t) = 1’(2-2t) L:Stsl is a loop based at K0 representing [W] e fil(K*, ;O) = 0. But q(AYt)) represents Be'fll(F, x0) (for q(1'(0)) = q(ZO) = x q(1'(1)) = q(1(1)) = q(a(§0)) 0' A'CIE implies q(1'(t))CF) and [W] = 0 implies B = a in fll(K, x0), for A ~ 1' implies q(A) = q(A') i.e (1= 8. Now we are ready to handle the remaining cases for F(hp). (i) F(hp) 2 51. From Chapter 1, section 1.2, we wee that F(hp) and F(h2) are in the interior of K which has IR3 as a universal cover. hp induces an . . 3 . p involution a on ER with qa = h q, where q : 1R3 ‘+ Int(K) is the covering projection. Since 39 1T1(F(hp), x ~ 1r (K, x) (lemma 8), F(OL) z IR. Let ) = l E = qfl(F(h2)). By lemma 8 we obtain the fact that Fa IR. Since F(h2)nF(hp) F(h) =9), F(OL)nF=¢, otherwise iezF(d)r)E would imply p(§)e F(hp)r1F(h2). It is easy to see that a(E) = E. Hence a is an involution on IR3 - E 5:: 1R3 - 1R . By Alexander's duality ~ 3 3 l theorem [3] we have Hi(fli - HI) = Hi(S -S ). But Hr(53 -Sl) = 0 for r > 1 and 2Z for r = 0, 1. In other words 1R3 - 1R 8: 1R3 - E is a homology l-sphere. Now F(a) g HKB- E is a homology r—sphere with r = -l, O, l by [15]. That is F(a) is a homology 1-sphere, 80, fl. But F(a) z HI has the homology of a point. This contradiction shows that F(hp) cannot be 2:81. (ii) F(hp) z A, an annulus. By [9] and [10] 3 an hp-invariant meridional disk D embedded in K and D in general position with respect to F(hp). So F(hp)r1D is a properly embedded arc I in D. Cut K along D to obtain a component U z D x I and U contains two copies of D, say D' and D" each contains a copy of I, say IKID' and I'W:D". Let F be F(hp) after cutting K. Fer' = I' and Fr1D" = I". By Brown [1] F is an unknotted disk in U. Let F be the image of F(hz) in U. Since F(h2)r)F(hp) o, ErlF = fl. K is obtained from U by identifying (x,O) in D' with (¢(x), l) in D" 40 where a is an orientation reversing homeomorphism ._ . ' 2 D' + D". But FrlF = a so in order to obtain F(h ) we have to let F travel twice around K, otherwise 2 p — . . . . F(h )r)F(h ) ¢ ¢. But F under the identification 2 . . . . 2 . must be F(h ), but it is not, it is 2F(h ). Since 2 . F(h ) generates fl1(K, x), lemma 8, this cannot happen. Hence F(hp) cannot be A. (iii) F(hp) : M, Mobius band. As in (ii) this cannot occur. This completes case 2 and finishes the proof of Theorem 1. Section 2.3. SEMI-FREE ACTIONS ON K In this section we study the semi-free actions on the solid Klein bottle K. We state our results in the following. Theorem 1. Let h : K + K be a periodic homeomorphism of period n acting semi-freely on K. Then (1) There is no such h if n is even and n >2. (2) There are 5 such h up to equivalence (3) There is no such h if n is odd and F(h) = 41 (4) There is a unique such h, up to weak equivalence if n is odd and F(h) = a. Proof Case 1. n is odd, of course n > 1. By lemma 1, section 2.1, F(h) is either g or a simple closed curve ‘381. If F(h) = fl, then by definition of h, F(hi) = U for all l;$i‘ acts freely on K. By proposition 3, section 2.1, where h : K + K is 1' l n-l n h is weakly equivalent to h defined by hl([z, t]) = [z, t + ] . This proves (4). If F(h) zsl, then F(hl) ~51, 15i 2, then n has one of the two a O. . forms : n = 2 , a > 1 or n = 2 m, m is an odd positive integer. If n = 2 , a positive integer > 1, then by [13] such an h doesn't exist. 42 01 Finally, n = 2 m, m positive integer > 1. If F(h) = 2, then this cannot happen, otherwise since 2 B . 2 2 . h acts semi-freely, F(h ) = fl. But h is an involution on K and by [13] there is no involution on K with empty fixed point set. 0. Now, assume F(h) # 2. Since h2 has an odd (1 period m, lemma 1, section 2.1 gives F(h2 ) z 81 and since h acts semi-freely on K, F(h) z 81. Hence F(hl) = F(h) z S1 for all 1