of automorphisms (holomorphic isomorphisms) without fixed point of W. Hence we have a complex manifold

Transcription

1 Tohoku Math. Journ. 26 (1974), AUTOMORPHISM GROUPS OF HOPF SURFACES MAKOTO NAMBA (Received April 27, 1973) Introduction. Let GL(2, C) be the group of non-singular (2 ~2)- matrices. An element of GL(2, C) operates on C2 as follows: Let M be a subset of GL(2, C) defined by Then M is a complex manifold. Let 0 be the origin of C2. We put W = C2-0. Let u M. Then u defines a properly discontinuous group of automorphisms (holomorphic isomorphisms) without fixed point of W. Hence we have a complex manifold Vu is easily seen to be compact. It is called a Hopf surface. It can be shown that the collection forms a complex analytic family (X, Ĕ; M). We denote by Aut (Vu) the group of automorphisms of Vu. The purpose of this note is prove the following theorem. THEOREM. The disjoint union admits a (reduced) analytic space structure such that 1) Ď: A M is a surjective holomorphic map, where Ď is the canonical projection, 2) the map defined by

2 134 M. NAMBA is holomorphic, where the fiber product of A and X over M, 3) the map MA defined by is holomorphic, where 1, is the identity map of Vu, 4) the map defined by is holomorphic, where the fiber product of A and A over M. 1. The complex analytic family of Hopf surfaces. By a complex analytic family of compact complex manifolds, we mean a triple (X, ƒî, M) of complex manifolds X and M and a proper holomorphic map of X onto M which is of maximal rank at every point of X, i.e., for all P X, where J(f)P is the Jacobian matrix of f at P. In this case, each fiber ƒî-1(u), u M, is a compact complex manifold. M is called the parameter space of the family (X,ƒÎ; M). Now, let and we define a holomorphic map by

3 AUTOMORPHISM GROUPS OF HOPF SURFACES 135 Then ƒå is an automorphism, for ƒå-1 is given by We put G = {ƒån n Z}. LEMMA 1. G is a properly discontinuous group of automorphisms without fixed point of M ~ W. PROOF. We assume that (u, unx) = (u, x) for an integer n 0. Then unx = x. We write and x = (z, w). Then Since 0 < ƒ < 1 and 0 < ƒà < 1, unx = x implies that w = 0, so that z = 0, a contradiction. Hence G has no fixed point. In order to show that G is a properly discontinuous group, it is enough to show that, for a compact subset K1 of M and a compact subset K2 of W, is a finite set. There are positive constants c and d such that for all. We define a norm in C2 by Then there are positive constants a and b such that for all x K2. Now where, x = (z, w) K2 and Hence, for a positive integer n,

4 136 M. NAMBA as n +. Thus there is a positive integer N such that for all n N. Next, we show that there is a positive integer N' such that for all n N' and for all (u, x) KI ~ K2. We assume the converse. Then there are a sequence of points {(uđ,xđ)}đ=1,2, c of KI ~ K2 and a sequence of integers such that We put yđ = u-nđđxđ, đ =1, 2, E E E. Then xđ = unđđyđ, đ =1, 2, E E E. We put Then where Hence as đ +, This contradicts to Now is contained in By Lemma 1, the quotient space q.e.d.

5 AUTOMORPHISM GROUPS OF HOPF SURFACES 137 is a complex manifold. Let ƒî: M ~ W M be the canonical projection. Then ƒîƒå = ƒî. Hence there is a holomorphic map such that the diagram is commutative, where p is the canonical projection. Since p is a covering map, ƒî is a surjective holomorphic map of maximal rank at every point of X. LEMMA 2. ƒî is a proper map. PROOF. Let K be a compact subset of M. We show that ƒî-1(k) is compact. Let {PƒË}ƒË=1,2, c be a sequence of points in ƒî-1(k). We want to choose a subsequence of {PƒË}ƒË=1,2, c converging to a point of ƒî-1(k). We may assume that {ƒî(pƒë)}ƒë=1,2, c converges to a point u K. We put uƒë = ƒî (PƒË), ƒë =1, 2, E E E. We put and Then ƒ ƒë ƒ, ƒàƒë --> ƒà and tƒë t as ƒë +.We may assume that there are positive constants c1, c2 and d such that for all ƒë. Let xƒë, ƒë =1, 2,..., be points of W such that We put We define a norm in C2 by First, we assume that there is a subsequence

6 138 M.NAMBA such that Then zƒëk 0, k =1, 2, E E E. Thus there are integers nk, k =1, 2, E E E, such that We put z'ƒëk = ƒ nkƒëkzƒëk, k = 1, 2, E E E. We put x'ƒëk = (z'ƒëk,0), k =1,2, E E E. Then X'ƒËk = unkƒëkxƒëk, k =1, 2, E E E. Hence Since c1 x'ƒëk 1, k =1, 2, E E E, we may assume that {x'ƒëk}k=1,2,... converges to a point x W. Then {PƒËk}k=1,2,... converges to p(u, x). Now, we may assume that wƒë 0, ƒë =1, 2, E E E. Since there are integers nƒë, ƒë =1, 2, E E E, such that we may assume that (We use unƒëƒëxƒë instead of xƒë.) Hence we may assume that {wƒë}ƒë=1,2,... converges to a point w C, c1 w 1. Since the Riemann sphere C is compact, we may assume that {zƒë}ƒë=1,2,... converges to a point z of C. If z, then x = (z, w) is a point of W and {PƒË}ƒË=1,2,... converges to p(u,x). If z =, we may assume that Then there is a sequence of positive integers {nƒë}ƒë=1,2,... such that Let N be a positive integer such that for all n N. We may assume that z1 is so large that Then Hence

7 AUTOMORPHISM GROUPS OF HOPF SURFACES 139 This shows that Hence We put where Then Hence we may assume that {z'đ}đ=1,2,... converges to a point z' C, c1/2 z' 1 + c1/2. Since wđ 1, đ =1, 2, E E E, We may assume that {Ĉnđđwđ}đ=1,2,... converges to w' C. We put x = (z', w') W. Then converges to p(u, x), q.e.d. Lemma 2 shows that (X, Ĕ, M) is a complex analytic family of compact complex manifolds. Each fiber Ĕ-1(u), u M, is called a Hopf surface. Each fiber can be written as where and A similar but simpler argument to the proof of Lemma 1 shows that G is a properly discontinuous group of automorphisms without fixed point

8 140 M. NAMBA of W. Henceforth, we identify ƒî-1(u) with Vu. 2. Automorphism groups of Hopf surfaces. Let u M. Let Vu be the corresponding Hopf surface. Let Aut (Vu) be the group of automorphisms of Vu. Let Then Cu is a complex Lie subgroup of GL(2, C). We define a homomorphism by where v is an automorphism of Vu defined by for all x W, where p: WVu is the canonical projection. Since uv = vu, v is well defined. LEMMA 3. ker (hu) = Gu. PROOF. Let un Gu. Then Hence Gu ¼ ker (hu). Conversely, let v ker (hu). Then for all x W. Hence, for each x W, there is an integer k(x) such that We show that if c C and c 0. In fact so that Since Gu operates on W without fixed point, Thus we may consider k to be a Z -valued function on P1(C), the 1-di-

9 AUTOMORPHISM GROUPS OF HOPF SURFACES 141 mensional projective space. Since the cardinal number of the set P1(C) is greater than that of Z, there are distinct points L1 and 112 in P1(C) such that k(l1) = k(l2). We put k = k(l1) = k(l2). Let x1 and x2 be points in W such that x1 L1 and x2 L2. Then, for any point x W, there are complex numbers a and b such that Here we regard x, x1 and x2 as vectors 0x, 0x1, and 0x2 respectively. Then Hence v = uk, q.e.d. Now, we determine Aut (Vu) following the argument in [1]. Let be an automorphism. Since W is the universal covering space of Vu, there is an automorphism such that the diagram is commutative where p is the canonical projection. Moreover f satisfies for all x W, where ug is a generator of Gu, (ug = u or u-1). We show ug = u. By Hartogs's theorem, f is extended to an automorphism which maps 0 to 0. If for all x W, then for n = 1, 2, E E E and for all x W. We fix x = (z, w) W. We put.then where

10 142 M. NAMBA Hence as n +. Hence as n +, for the extended map f maps 0 to 0. On the other hand, there is an integer N such that for all n N. In fact, it is enough to take N such that Hence as n +. This contradicts to Hence for all x W. We write the extended automorphism f : C2 C2 as Then the above condition is written as We expand g and h in the power series of z and w at the origin: Case 1. ƒà=ƒ and t=0; In this case, above equations reduce to

11 AUTOMORPHISM GROUPS OF HOPF SURFACES 143 Since 0 < ƒ <1, we get if p + q > 1. Hence Since f is an automorphism, the matrix is non-singular. Thus for all. Case 2. ƒà = ƒ and t 0; In this case, above equations reduce to From the last equation, we have Hence Hence, from the first equation, we have Thus Since, we have for all, t 0.

12 144 M. NAMBA Case 3. ƒà ƒ and t=0; In this case, above equations reduce to Hence, if p > 0 and q > 0, then If p = 0, then Hence c01 = 0 and doq = 0, if q > 1. If q = 0, then Hence d10 = 0 and cp0 = 0, if p > 1. Case 3 is thus divided as follows. Case 3-A. ƒàq = ƒ for some q 2; In this case, f is generally written as where ad ~ 0 and b is arbitrary. We note that We introduce a group operation in the set Cu ~ C as follows: where and. By this group operation, Cu ~ C becomes a complex Lie group. Cu is then isomorphic to the complex Lie subgroup Cu ~ 0 of Cu ~ C. The group Cu ~ C is isomorphic to the group of automorphisms f of W such that fu = uf. The isomorphism is given by where

13 AUTOMORPHISM GROUPS OF HOPF SURFACES 145 Hence there is a surjective homomorphism We show that ker (gu) is equal to Gu ~ 0. First, for an integer n, un ~ 0 corresponds to the automorphism of W which corresponds to the identity map of Vu. Next, let be an automorphism of W which corresponds to the identity map of Vu. Then, for each x = (z, w) W, there is an integer k(x) such that In particular, let x W' where Then, by the second equation, d = ƒàk(x). Hence k(x) = k is constant for x W'. By the first equation, a = ƒ k and b = 0. Hence ker (gu) is equal to Gu ~ 0. Thus for all, q 2. We note that the center of the group Cu ~ C is Hence Gu ~ 0 is contained in the center. Case 3-B. ƒ p = ƒà for some p 2; In this case, f is generally written as where ad 0 and b is arbitrary. We note that We introduce a group operation in the set Cu ~ C as follows:

14 146 M. NAMBA where and. By this group operation, Cu ~ C becomes a complex Lie group. Cu is then isomorphic to the complex Lie subgroup Cu ~ 0 of Cu ~ C. By a similar argument to Case 3-A, we have for all, p 2. We note that the center of the group Cu ~ C is Hence Gu ~ 0 is contained in the center. Case 3-C. ƒàq ƒ for any positive integer q and ƒ p ƒà for any positive integer p. In this case, f is generally written as where ad 0. We note that Thus for all such that ƒàq ƒ for any positive integer q and ƒ p ƒà for any positive integer p. Case 4. ƒ ƒà and t 0, Let and. Then and u = yuy-1. Thus y induces a holomorphic isomorphism defined by where pu: W Vu and pu : W Vu are canonical projections. Hence

15 AUTOMORPHISM GROUPS OF HOPF SURFACES 147 by the correspondence Thus Case 4 reduces to Case 3. We note that, in Case 4, Case 4-A. ƒàq = ƒ for some q 2 and t 0. where the group operation in Cu ~ C is defined as in Case 3-A: where, e = ((a - d)/(ƒ - ƒà))t and, e' = ((a' - d')/(ƒ - ƒà))t. Case 4-B. ƒ p = ƒà for some p 2 and t 0. where the group operation in Cu x C is defined as in Case 3-B: where, e = ((a - d) /(ƒ - ƒà))t and, e' = ((a' - d')/(ƒ - ƒà))t. Case 4-C. t 0, ƒàq ƒ for any positive integer q and ƒ p ƒà for any positive integer p. 3. Proof of Theorem. In 2, we have shown that Aut (Vu) is isomorphic to Cu ~ C/Gu ~ 0 if u is in one of Case 3-A, Case 3-B, Case 4-A and Case 4-B, and is isomorphic to Cu/Gu if a is in one of other cases. We introduce an analytic space structure in the disjoint union of these quotient groups. If this is done, an analytic space structure in u M Aut(Vu) is induced by it. We consider closed subvarieties U of M ~GL(2, C) ~ C defined by

16 148 M. NAMBA for k = 2, 3, E E E, where, and for k = 2, 3, E E E, where. It is clear that X2, X3, E E E, Y2, Y3, are mutually disjoint, while each of them intersects Z0. Let Z E E E be the union of these subvarieties: LEMMA 4. Z is a closed subvariety of M ~ GL(2, C) ~ C. PROOF. First, we show that Z is closed in M ~ GL(2, C) ~ C. Let {(uƒë, vƒë, bƒë)}ƒë=1,2,... be a sequence of points in Z converging to a point (u, v, b) K ~ GL(2f C) ~ C. Since uƒëvƒë = vƒëuƒë, ƒë =1,2,..., we have uv = vu. We put. We assume that i, e., ƒ k ƒà,ƒàk ƒ for any k 2. Since ƒ k and ƒàk converge to 0 as k +, there is a positive number such that (1) for all k 2. We may assume that (2) We put, ƒë = 1, 2,... Then ƒ ƒë ƒ, ƒàƒëƒà and tƒë t as +. Hence there is an integer No such that ƒë (3) for all ƒë N0. Now we show that there is an integer N, N N0, such that (4) for all k 2 and for all ƒë N. We show the first half of (4). The second half is shown in a similar way. We assume the converse. Then there are a sequence N0 ƒë1 < ƒë2 <... of integers and a sequence k1, k2, E E E of integers each of which is greater than 1 such that

17 AUTOMORPHISM GROUPS OF HOPF SURFACES 149 for n =1, 2, E E E. If {k1, k2, E E E} is bounded, then there is a subsequence kn1, kn2, E E E such that Then for m =1, 2, E E E. On the other hand, ƒ kƒënm ƒ k as m +, a contradiction. Hence we may assume that Then (by (3)). The right hand side converges to 0 as n + by (2), a contradiction. This shows (4). By (1), (3) and (4), for all k 2 and for all ƒë N. This proves that for any ƒë N. Hence (uƒë, vƒë, bƒë) Z0 for all ƒë N. Hence bƒë = 0 for all N so that b = 0, i. e., (u, v, b) = (u, v, 0) Z0. Hence Z is closed. ƒë Next, let (u, v, b) Xk. We put. Then ƒàk = ƒ. We show that there is a positive number such that (5) where ƒê: M ~ GL(2, C) ~ C M is the canonical projection and It is enough to claim that there is a positive number ƒã such that (6) for any k' k, k' 1, for any k" 1 and for any ƒà' and ƒ ' with ƒà' < ƒã and ƒ ' < ƒã. (It is enough to prove (6) for any k' k, k' 2 and for

18 150 M. NAMBA any k" 2 for our present purpose. But we use the case k' = k" = 1 afterwards.) We show the first half of (6). The second half is shown in a similar way. We assume the converse. Then there are sequences {ƒ 'ƒë}ƒë=1,2,..., {ƒà'ƒë}ƒë=1,2,... such that for ƒë =1, 2, E E E, and a sequence k1, k2,... of positive integers each of which is different from k such that (7) for v =1, 2, E E E. If {k1, k2, } is bounded, then there is a subsequence kn1, kn2, E E E. such that Then for m = 1, 2, E E E. The left hand side converges to ƒàk' as m +, while the right hand side converges to ƒ. Hence ƒàk' = ƒ, a contradiction. Hence we may assume that Then 0 as ƒë +. Hence the left hand side of (7) converges to 0 as ƒë +, while the right hand side of (7) converges to a, a contradiction. Hence (5) is proved. Let (u, v, b) Z0 Xk. Then (5) shows that Z coincides with Z0 ¾ Xk in a neighbourhood of (u, v, b). Let (u, v, b) Xk - Z0. Then b 0. The open subset of ƒê-1(n(u, ƒã)) does not intersect Z0, and (8) Thus Z coincides with Xk in a neighbourhood of (u, v, b). In a similar way to (5), we can show that, for every point (u, v, b) Yk, there is a positive number such that (9)

19 AUTOMORPHISM GROUPS OF HOPF SURFACES 151 Let (u, v, b) Z0 Yk. Then (9) shows that Z coincides with Z0 ¾ Yk in a neighbourhood of (u, v, b). Let (u, v, b) Yk - Z0. Then b 0 and (10) where N is the open subset of ƒê-1(n(u, ƒã)) defined above. Hence Z coincides with Yk in a neighbourhood of (u, v, b). Finally, let (u, v, b) Z0 - (Uk 2Xk) ¾ (Uk 2Y(k). Then b = 0 and uv = vu. A similar proof to the proof of (5) shows that there is a positive number such that (11) This means that Z coincides with Z0 in a neighbourhood of (u, v, 0). This completes the proof of Lemma 4. q.e.d. Let be an automorphism defined by where uv is the product of matrices u and v and u = t. We note that ƒä: Z0 Z0 and ƒä: Xk Xk (resp. ƒä: Z0 Z0 and ƒä: Yk Yk) coincide on Z0 Xk (resp. Z0 Yk). The inverse is given by We put LEMMA 5. H is a properly discontinuous group of automorphisms without fixed point of Z. PROOF. Let (u, v, b) Z. We assume that ƒän(u, v, b) = (u, v, b) for an integer n. Then unv = v. Hence un =1 so that n = 0. Next, we

20 152 M. NAMBA show that, for any compact set K in Z, is a finite set. Let p and R be positive numbers such that for all (u, v, b) K, where det u is the determinant of u. Then there is a positive integer no such that Then, for any positive integer n n0, and Hence is contained in q.e.d. By Lemma 5, the quotient space is an analytic space such that the canonical projection is a covering map. Let be the restriction to Z of the projection map Then ĎĊ = Ď. Hence there is a holomorphic map such that the diagram

21 AUTOMORPHISM GROUPS OF HOPF SURFACES 153 is commutative. Since (u, 1, 0) Z0 ¼ Z, where 1 is the identity matrix of GL(2, C),ƒÉ is surjective, so that is surjective. By the construction above, each fiber ƒé-1(u) is naturally isomorphic to if u is in one of Case 3-A, Case 3-B, Case 4-A and Case 4-B, and is isomorphic to if u is in one of other cases. Now, we prove 1)-4) of the theorem. 1) is already done. Next, we show 2). We define a holomorphic map by where p: M ~ W X is the canonical projection. Then r is a covering map. Let ((u, v, b), (u, x)) Z ~ (M ~ W). Let f be the automorphism of W corresponding to (u, v, b), see 2. Let f be the automorphism of Vu corresponding to q(u, v, b). Since the diagram where P = p(u, x), is commutative, and since r and p are covering maps, it is enough to show that J(x) depends holomorphically on (u, v, b, x). Since the problem is local, it is enough to show that f(x) depends holomorphically on (u, v, b, x) in a neighbourhood of any point (u0, v0, b0, x0). Case A. In this case, by (11) in the proof of Lemma 4, there is a positive number such that Let (u, v, 0) Z ƒê-1(n(u0,ƒã)) = Z0 ƒê-1(n(u0,ƒã)}. Let f be the automor-

22 154 M. NAMBA phism of W corresponding to (u, v, 0). Then for all x W, as the argument in 2 shows. v(x) depends holomorphically on (v, x). Case B. (u0, v0, b0) Xk - Z0. In this case, by (8) in the proof of Lemma 4, where N = {(u, v, b) ƒê-1(n(u0,ƒã)) b 0}. Let (u, v, b) Z N = Xk N. Then b 0 and ƒàk = a where. Let f be the automorphism of W corresponding to (u, v, b). Let x = (z, w) W. Then f(x) is written as where and, e = ((a - d)/(ƒ - ƒà))t. In fact, f = y~1gy where and g(z, w) = (az + bwk, dw), (see Case 4-A in 2). Hence f(x) depends holomorphically on (u, v, b, x) (Z N) ~ W. Case C. (u0, v0, b0) Xk Z0. In this case, by (5) in the proof of Lemma 4, Let Let f be the automorphism of W corresponding to (u, v, b). Let x = (z, w) W. Then it is easy to see that f(x) is written as for all (u, v, b, x) (Z ƒê-1(n(u0, ƒã))) ~ W, where and, e = ((a - d)/(ƒ - ƒà))t. (We note that ƒ ƒà in Z ƒê-1(n(u0, ƒã)) = (Xk ¾ Z0) ƒê-1(n(u0, ƒã)) by (5) of the proof of Lemma 4.) This shows that f(x) depends holomorphically on (u, v, b, x). Case D. (u0, v0, b0) Yk - Z0. In this case, by (10) in the proof of Lemma 4,

23 AUTOMORPHISM GROUPS OF HOPF SURFACES 155 where N = {(u, v, b) ƒê-1(n(u0, ƒã)) b 0}. Let (u, v, b) Z N = Yk N. Then b 0 and ƒ k = ƒà where. Let f be the automorphism of W corresponding to (u, v, b). Let x = (z, w) W. Then f(x) is written as where and,e= ((a - d)/(ƒ - ƒà))t. In fact, f =y-1gy where and g(z, w) = (ax, dw + bzk), (see Case 4-B in 2). Hence f(x) depends holomorphically on (u, v, b, x) (Z N) ~ W. Case E. (u0, vo, bo) Yk Z0. In this case, by (9) in the proof of Lemma 4, Let f be the automorphism of W corresponding to (u, v, b). Let x = (z, w) W. Then for all (u, v, b, x) (Z ƒê-1(n(uo,ƒã))) ~ W, where and, e = ((a - d)/(ƒ - ƒà))t. Hence f(x) depends holomorphically on (u, v, b, x). This completes the proof of 2) of the theorem. Next, we prove 3) of the theorem. Let 1 be the (2 ~ 2)-identity matrix. Then the map is holomorphic. Hence the map is holomorphic. It is clear that q(u, 1, 0) corresponds to the identity map of Tic. Finally we show 4) of the theorem. We define a holomorphic map

24 156 M. NAMBA by Then s is a covering map. Let ((u a product, v, b), (u, v', b')) Z ~ Z. We define by (1) if u is in Case 3-A or 4-A of 2, where, e = ((a - d)(ƒ - ƒà))t and, e' = ((a' - d')/(ƒ - ƒà))t, (2) if u is in Case 3-B or 4-B of 2, where, e = ((a - d)/(ƒ - ƒà))t and, e' = ((a' - d')/(ƒ - ƒà))t, (3) if u is in one of other cases. Then, as in the proof of 2) of the theorem, by diving in various cases, we can easily see that the map is holomorphic. We define a product by This is well defined, as is easily shown by dividing in various cases. Since the map s defined above and the map q are covering maps, the map is holomorphic. It is clear that q(u, v', b')q(u, v, b) corresponds to the composition gf of automorphisms g and f of Vu corresponding to q(u, v', b') and q(u, v, b) respectively. Now, we define a holomorphic map by

25 AUTOMORPHISM GROUPS OF HOPF SURFACES 157 where and, e = ((a - d)/(ƒ - ƒà))t. We note that ƒæ: Z0 Z0 and ƒæ: Xk Xk (resp. ƒæ: Z0 Z0 and ƒæ: Yk Yk) coincide on Z0 Xk (resp. Z0 Yk). It is easy to see that ƒæƒä = ƒäƒæ. Hence we can define a map by Since q is a covering map, B is holomorphic. It is clear that ƒæ(q(u, v, b)) corresponds to the inverse f-1 of the automorphism f of Vu corresponding to q(u, v, b). This completes the proof of 4) of the theorem. REFERENCE [1] K. KODAIRA AND D. C. SPENCER, On deformations of complex analytic structures I and U, Ann. of Math., 67(1958), MATHEMATICAL INSTITUTE TOHOKU UNIVERSITY SENDAI, JAPAN