О континууме топологически различных полных связных строго выпуклых гиперповерхностей в гиперболическом пространстве, полученных склеиванием двух копий фиксированного многообразия с краем Текст научной статьи по специальности «Математика»

UDC

515.163+514.13+514.17

ON CONTINUUM MANY TOPOLOGICALLY DIFFERENT COMPLETE CONNECTED STRICTLY CONVEX HYPERSURFACES IN THE HYPERBOLIC SPACE THAT ARE OBTAINED BY GLUING TWO COPIES OF A FIXED MANIFOLD WITH BOUNDARY

A.V. Kuzminykh

We prove the existence of a family M{n) (n > 2) of complete connected strictly convex ??,-dimensional surfaces in the n + 1-dimensional hyperbolic space Hra+1 (boundaries of strictly convex bodies in Hra+1) such that 1) card M{n) = c (where c is the cardinality of the continuum); 2) if Mi € M{n) (i = 1,2), Mi / M2, and, for each i, Si c Mi is a compact set such that card Si < N0, then Mi \ Si and M2 \ 52 are not homeomorphic; 3) there is a connected ??,-dimensional topological manifold G(n) with boundary (in case n > 3 the boundary of G(n) is connected) such that each M e M{n) is a topological manifold obtained by gluing two copies of G(n) along their boundaries.

To the memory of my Teacher Alexander Danilovich Alexandrov

Introduction

In this paper we consider the topological diversity of complete connected strictly convex ??,-dimensional (n >

2) surfaces in the n + 1-dimensional hyperbolic space

H«+i

As usual, a convex body B c Hra+1 (respectively,

B c Era+1, where Era+1 is the n + 1-dimensional Euclidean space) is a closed convex n+ 1-dimensional set.

A convex body B is called strictly convex if the boundary dB of the body B does not contain any straight-line segment and dB ^ 0.

In [3,4] we have shown how topologically different unbounded convex (and strictly convex) bodies in Hra+1 (whose boundaries are not connected) might be.

A.V. Kuzminykh. 1970

Copyright © 2012 A.V. Kuzminykh Purdue University (USA)

E-mail: [email protected]

We call a set M c Hra+1 (respectively, M c Era+1) a strictly convex n-dimensional surface if there is a strictly convex body B{M) c Hra+1 (respectively, B(M) c Era+1) such that M c dB(M), the set M is open (in the topology of dB(M)), and М/

It follows from [1, p. 73] that every complete strictly convex n-dimensional surface in the Euclidean space Era+1 is homeomorphic either to the sphere S™ or to the space E™.

Note that all strictly convex гг-dimensional surfaces of the family M{n) (see the statement of the Theorem) are (connected) boundaries of strictly convex bodies in Hra+1; hence they are complete.

As usual, convS is a convex hull of a set S; cardS is the cardinality of S; c := card R, where R is the set of real numbers; := card N, where N is the set of positive integers.

Let us fix a hyperplane P0 c Hra+1. Let Hi and H2 (%i ф %) be closed half-spaces of Hra+1 such that P0 is their common boundary.

Theorem 1. There is a family M(n), n > 2, of complete connected strictly convex n-dimensional surfaces in the space Hra+1 such that

1) card Ai(n) = c;

2) for each M є M(n), convM is a strictly convex body and M is the boundary of convM;

3) if Мі є M(n) (i = 1,2), Mi ф M2, and, for each i, Si c Mi is a compact set such that card Si < then Mi \ Si and M2 \ S2 are not homeomorphic;

4) there is a connected n-dimensional topological manifold G(n) with boundary such that

a) each M є M(n) is a topological manifold obtained by gluing two copies of G(n) along their boundaries; moreover, for each M є M(n), the intersections M П 1-Li and M П %2 are homeomorphic to G(n);

b) in case n>3 the boundary of G(n) is connected.

Remark 1. Thus, on the one hand, the manifolds from M(n) are, in particular, pairwise not homeomorphic but, on the other hand, they are topologically ’’similar” (they are obtained by gluing two copies of G(n) along their boundaries).

Proof. As usual, a region in the space R™ is a nonempty connected open set. We need the following lemma.

Lemma 1. There is a family U(n), n > 2, of regions in R™ such that

1) card U(n) = c;

2) if Ui є U(n) (i = 1,2), Піф U2, and, for each i, Si c Ui is a compact set such that card Si < then Ui\Si and U2 \ S2 are not homeomorphic;

3) there is a connected n-dimensional topological manifold G(n) c R™ with boundary such that

a) each U eU(n) is a topological manifold obtained by gluing two copies of G(n) along their boundaries; moreover, for each U eU(n), the intersections

U П {(жі,..., xn) : Xi < 0} and U П {(жі,..., xn) : Xi > 0}

are homeomorphic to G(n);

b) in case n>3 the boundary of G(n) is connected.

Proof. In [5, p. 112-113] we have defined points of rank k (k e N U {0}) of a topological space and, for each k G N, have defined a set A(k) c [0,1].

Let C C [0,1] be the Cantor perfect set. Put C~ := {—x : x G C} and B(k) := C-U#).

Let Ik C [0,1] (fc G N) be open intervals such that C = [0,1] \ Ufcli h', denote by X the set of all intervals Ik, k G N. Let C* c C be the set of all points of C which are not the ends of the intervals Ik, k G N, and are not the ends of [0,1].

Let us fix a point xq g C*.

Like in [2, p. 101], for each point x G C*, there is a homeomorphism

4>x : [0,xo] ->• [0,x]

such that (f)x(0) = 0, <px(xo) = x, and <px(C n [0,xo]) = C n [0,x] (note that this statement is a consequence of the existence of a bijective mapping <p*x of the set of all intervals from I contained in [0,xo] onto the set of all intervals from I contained in [0,x] such that <p*x preserves the order of the intervals).

Hence, for each point x G C*, there is a homeomorphism : R —> R such that rtpx(C) = C, rtpx(xo) = x, for each I el, the restriction of rtpx to the interval I is a superposition of a linear mapping and a translation, and outside the open interval ]0,1[ the mapping rtpx is the identity.

Let n : R —> R™ be the isometry such that /i(0) = (0,...,0) and n{ 1) = (0,1,0,..., 0) (so n{R) is the x2-axis). Let p : R™ —> /i(R) be the orthogonal projection.

For each k G N, let sk C {(xi,...,xn) : xx > 0,x3 = 0,...,xn = 0} be a closed straight line interval such that sk is parallel to //(/&), p(sk) C /i(Ik), and the distance between sk and n(h) is less than the length of Ik-

For each k G N, let rk : [—1,1] —> sk be a bijective superposition of a linear mapping and a translation.

Hence, for each k G N, the set rk(B(k)) is compact and nowhere dense (in sk).

Put Q := \J^=lTk(B(k)). Put G(n) := {(xi, ...,xn) : xx > 0} \ (fi(C) U Q).

Obviously, G(n) (with the topology induced from Rra) is an n-dimensional connected topological manifold with boundary.

Note that the boundary of G(n) is {(xi, ...,xn) : x\ = 0} \ 11(C).

Put C*(x0) := {x : x G C*,x > Xo}.

For each x G C*(x0), let us define a homeomorphism : R™ —> R™ as follows:

^X((xi,x2,x3, ...,xn)) = (xi/ipx(x2),x3, ...,xn).

Let rq : R™ —> R™ be the orthogonal symmetry with respect to the hyperplane x\ = 0.

For each x G C*(x0), put U(x) := R™ \ (11(C) U r](Q) U ^x(fl)).

Let i c f/(x) be a compact set such that card A < K0. Obviously, the set U(x)\A is open and connected (in particular, U(x) is a region).

It is easy to see (considering, for instance, the surface of a corresponding parallelepiped) that, for any points X* g /i(C) U r/(Q) U ^(Q) U A, i = 1,2, Xi ^ X2, there is an (n — l)-dimensional topological sphere S c U(x)\A (i.e. a set homeomorphic to Sra_1) such that S separates Xx from X2 (i.e. Xx and X2 belong to different connected components of (U(x) \ A) \ S).

Lemma 2. Let Xi G C*(x0) (i = 1,2), x\ < x2, and, for each i, let Ai C U(xi) be a compact set such that cardAi < K0. Then the sets U(xi) \ A\ and U(x2) \ A2 are not homeomorphic.

Proof. Put Dn+l := {(xi, ...,xn+i) : x\ + ... + x2n + (xn+i — l)2 < 1} (hence Dn+l is an open ball in Rra+1). Let Sn be the boundary of Dn+l. Let O0 G Sn be the point (0, ...,0,2).

Let £ : Sn \ {O0} —> R-™ (we identify R™ with {(xi,..., xn+i) : xn+i = 0}) be the stereographic projection (for each X e Sn\ {O0}, the points O0,X, and £(X) are collinear).

For each i (i = 1,2), put Ti := {O0} U £_1(/i(C') U rj(Q) U ^(H) U Ai).

Obviously, Ti is a compact set. Note that O0 is an isolated point of Ti.

Suppose that U(Xi) \ Ai and U(X2) \^42 are homeomorphic. Hence there is a homeomorphism / : Sn \ T\ —> Sn \ T2.

We call a sequence {Yk}^=1 of points of the sphere Sn i-regular (i = 1,2) if, for each k G N, Yk G Sn \ Ti, and there is a point Y* G T such that Y* = lim^oo (in the topology of Sn).

Let us show that if a sequence {Yk}^=l is 1-regular, then the sequence {f(Yk)}T= i is 2-regular.

Suppose that {fiYk^kLi is not 2-regular. Since the set T2 is compact and / is a homeomorphism, there are two points Y% G T2 [i = 1, 2) each of which is a limit point of {f{Yk)}f=1.

Like it was stated earlier for U(x), there is an (n— l)-dimensional topological sphere S C Sn \ T2 such that S separates Y1 from Y2. The set /_1(5') c Sn\Ti is compact. It follows from the structure of T\ that there is an open set V C Sn\f~1(S) such that Y* g V and the set Fyj7! is connected; thus f(V\T\) is connected.

There are ki G N (i = 1,2) such that, for each i, Yk. G V \ T\ and the point f(Yki) is so close to Y% that S separates f(Ykl) from f(Yk2).

Hence f(V \ Ti) nS/0, a contradiction.

Therefore the sequence {/Q'fc)}fcli is 2-regular.

Analogously, if a sequence {Yl}^=1 is 2-regular, then is 1-regular.

We define a mapping g := T\ —> T2 as follows: if Y G Jj, a sequence {Yk}^=1 is 1-regular, and Y = lim*,.*» Yk, then g(Y) := lirrifc^oo/(yfc).

Hence the mapping g is defined correctly and g is a bijection.

Suppose that the mapping g is not continuous. Hence there are points Z% G T\ (i = 1,2), Zl ^ Z2, and there is a sequence {Zk}^= 1 of points of T\ such that Z1 := linifc^oo Zk but linifc^oo g(Zk) = g(Z2). For each k G N, let {Zm(k)}^=l be a 1-regular sequence such that Zk = linim^oo Zm(k). Hence, for each k G N, there is an element Z*(k) of the sequence {Zm(k)}^=l such that the sequence {Z*(k)}%°=1

is 1-regular, Z1 := limk^00Z*(k) but lim^oo f(Z*(k)) = g(Z2) ^ g(Zl), a contradiction.

Thus the mapping g is continuous; since T\ is a compact set, g is a homeomor-phism.

Let T = (T, T) be a topological space (where T is the topology on the set T). We define a type (do not confuse with a rank) of a point I 6 T in the following way.

We say that a point X eT is

1) a point of type 1 of T if there is U eT such that X e U and card U < H0;

2) a point of type 2 of T if, for each U eT such that X e U, there is V eT such that X e V, V C U, card V = c, and V does not have isolated points;

3) a point of type 3 of T if, for each U eT such that X e U, there is V eT such that X e V, V c U, and we have: each point Y e V \ {X} is either a point of type 1 of T or a point of type 2 of T, and, for each i (i = 1,2), there is a point of type i of T in V \ {X};

4) a point of type 4 of T if, for each j (j = 1,2,3), X is not a point of type j of T.

The mapping g := T\ —>• T2 is a homeomorphism. Hence, for each number j (j = 1,2, 3,4), if X G Ti is a point of type j of T\ (we consider the set T\ with the topology induced from Sn and, as usual, use the same notation for the set T\ and for the corresponding topological space), then g(X) is a point of type j of T2, and conversely.

For each i (i = 1,2), put

CO

A* := Ai U |J(ri(Tk(A(k) \ {0})) U '&Xi(rk(A(k) \ {0}))) fc=i

and

CO

b: ■= UfoMC- \ {o})) u \ {o})».

fc=i

For each i (i = 1,2) and for each k e N, define a set Bi(k) c Ti as follows:

a point X e Ti belongs to Bi(k) if, and only if, either X = rj(rk(0)) or X =

^(^(0)). Obviously, for each i (i = 1,2) and for each he N, the set Bi(k) consists of two points.

For each k e N, the number 0 is a limit point of the set of isolated points of

A(k). Hence, for each i (i = 1,2), each point of n(C) is a limit point of the set of

isolated points of Ti.

Taking this into account, it is easy to see that, for each i (i = 1,2), a point X g Ti is

a) a point of type 1 of Ti if, and only if, X e A*;

b) a point of type 2 of T if, and only if, X e B*;

c) a point of type 3 of T if, and only if, X g Ufcli Bi(k);

d) a point of type 4 of Ti if, and only if, X g h(C).

Note, by the way, that it is not true that, for any topological space T, any point X eT is a point of at most one type of T (but for T\ and T2 it is true).

Since g is a homeomorphism, g(A\) = A2, g(B\) = B2, g(n(C)) = n(C), and

Since g is a homeomorphism, if X G T\ is a point of rank k (k e N U {0}) of T\ (see [5, p. 112]), then g(X) is a point of rank k of T2, and conversely.

Hence, for each k e N, g(B\(k)) = B2{k).

By construction, there are increasing sequences {ak}^=l and {bk}^=1 of positive integers such that /i(x0) = lim^oo rak (0) and /i(xi) = lim^oo rbk(0).

For each % (i = 1,2) and for each k e N, Bi(ak) = {r](rak (0)), ^xiijau(0))} and Biih) = {v{nk{0)),^Xi{nk{0))}.

Hence giJCi) = K2.

Obviously, the set /Ci has three limit points (n(x0), n(xi), and ^X1(/i(xi))) but the set 1C2 has four limit points (n(xo), n(xi), /i(x2), and ^X2(/i(xi))), a contradiction.

Put U(n) := {U(x) : x G C'*(x0)}. It is easy to check that the family U(n) enjoys properties 1) - 3) (see the statement of Lemma 1).

We use the model of the hyperbolic space Hra+1 in the open ball Dn+1 called Cayley-Klein (or Beltrami-Klein) model (in particular, in this model straight lines of the space Hra+1 are open chords of the ball Dn+l).

Let x G C*{xQ). In the proof of Lemma 1 a region U(x) was defined.

Put T(x) := Sn\^~l{U(x)). Obviously, T(x) is a compact set.

Let D0 c Rra+1 be a closed (n + l)-dimensional ball such that {O0} £ D0 and D0 c Dn+l U {O0}.

Let us define a strictly convex body B0(x) c Rra+1 as follows: B0(x) is the intersection of all closed balls in Rra+1 each of which has radius 2 and contains the compact set T(x) U D0.

Put B(x) := B0(x) Pi Dn+l; hence B(x) is a strictly convex body in Hra+1.

Let M(x) be the boundary (in the topology of Hra+1) of the body B(x) (obviously, the topology of Hra+1 coincides with the topology of the open ball Dn+l).

Let P0* be the hyperplane xx = 0 in R™.

Put T-L\ := {(xi, ...,xn) : Xi < 0}, W2 := {(xi, ...,xn) : Xi > 0}.

Without loss of generality, assume that the hyperplane P0 c Hra+1 coincides with Dn+l n conv^~l(Po), and, for each i (i = 1, 2), Hi = Dn+l n conv^~l{l-L*).

Put M(n) := {M(x) : x G C*(x0)}.

For each x G C*(x0), let us define a mapping hx : M(x) —> C~1(U(x)) as follows:

For each i (i = 1,2), put

1U := \J(Bt(ak)UBt(bk)).

k= 1

Lemma 2 is proved.

Lemma 1 is proved.

for each point Y є M(x), hx{Y) є C~1(U(x)) is a point such that the points Oo,Y,hx(Y) are collinear.

Obviously, the mapping hx is defined correctly and is a homeomorphism; the intersection М(х)Г\'Ні is homeomorphic to U(x) Г\ {(xi, ...,xn) : xx < 0}, and the intersection M(x) n%2 is homeomorphic to U(x) П {(xi,...,xn) : x\ > 0}.

Since the set T[x) is nowhere dense in Sn, for each point X є intB(x) (where intB(x) is the interior of B(x)), there are points Xj Є C~1(U(x)) (i = 1, 2) such that X є conv{Xi,X2}. Hence there are points X[ є M(x) П conv{Xi,X2} (i = 1,2) such that X є conv{X[, X’2}. Therefore convM(x) = B(x).

Now it is easy to check that the family M(n) enjoys properties 1) - 4) (see the statement of the Theorem).

The Theorem is proved. ■

Remark 2. Let us show that, for each x є C*(x0), the strictly convex surface M(x) is smooth, i.e., for each X є M(x), there is the unique hyperplane Px С jjra+i SUpp0rt of M(x) at X.

Let x Є C*(xq), let X є M(x).

Let Xk є Dn+1 \ convM(x) (k Є N) be such that X = Ит^ооХд,.

Denote by V the family of all closed balls in Rra+1 each of which has radius 2 and contains the set T(x) U D0. For each к Є N, there is D(k) Є V such that Xk ф D(k). Let O* be a limit point of the set of the centers of D(k), к Є N. Let D* be the closed ball of radius 2 with center O*. It is easy to see that D* є V and X є dD*.

Suppose that there are hyperplanes Pi С Hra+1 (г = 1,2) of support of M(x) at X, Pi ф P2. Hence there is a straight line segment I* с D* such that X is an end of I* and (/* \ {X}) П convM(x) = 0.

Let Yk є I* \ {X} (к Є N) be such that X = lim^oo yfc. For each к Є N, there is a ball D\k) Є V such that (I* \ [X,Yk]) П D\k) = 0 (where [X, Yk] is the closed interval with ends X and Yk). Let O** be a limit point of the set of the centers of D'{k), к Є N. Let D** be the closed ball of radius 2 with center O**.

Obviously, D** є V, X є dD**, and D** ф D*. Hence there is a ball Dx є V

such that X є dDi and D* П D** \ {X} с intDi. Hence T(x) U D0 С intDi,

therefore there is a ball D2 Є V such that X ^ D2, a contradiction.

Remark 3. It immediately follows from property 3 (see the statement of the Theorem) that if Mj Є M(n) (і = 1,2), Mi Ф M2, and, for each i, S- С convMi is a compact set such that card S[ < K0, then (convMi) \ S[ and (convM2) \ S'2 are not homeomorphic.

References

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2. Gelbaum B.R., Olmsted М.Н. Counterexamples in Analysis. San Francisco, London,

Amsterdam : Holden-Day, Inc., 1964. 194 p.

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