multiple warped product metrics into quadrics - mirandola, vitorio

7/27/2019 Multiple Warped Product Metrics Into Quadrics - Mirandola, Vitorio

1/18

arXiv:1210

.1812v1

[math.DG

]5Oct2012

MULTIPLE WARPED PRODUCT METRICS INTO QUADRICS

H. MIRANDOLA AND F. VITORIO

Abstract. In this paper, we will construct smooth isometric immersions ofmultiple warped products metrics in quadrics of semi-Euclidean spaces. Ourmain theorem generalizes previous results as given by Blanusa, Rozendorn,

Henke and Azov.

1. Introduction

By fundamental works of Nash [18] and Gromov and Rokhlin [11] we knowthat every m-dimensional smooth Riemannian manifold admits a smooth isometricimmersion in an n-dimensional Euclidean space Rn, for some n max{m(m +5)/2, m(m + 3)/2 + 5}, see Gunther [12]. Since then the problem of finding thelowest codimension possible is one of the major open problem in the theory ofisometric immersions. For books and surveys about this subject see, for instance,Gromov and Rokhlin [11], Jacobowitz [17], Poznyak and Sokolov [21], Aminov [1],Dajczer [8], Borisenko [6] and Han and Hong [13]. On the other hand, since that theresults of Nash and Gromov and Rokhlin arrive from existence results for certainPDEs, it is also a very interesting problem the construction of isometric immersionsfor each Riemannian metric given in a Euclidean space, mainly if the codimensionattained is lower than the one given by the theorems of Nash [ 18] and Gromov andRokhlin [11]. This is the point of view of the present paper.

Blanusa [4, 5] gave a beautiful method to construct smooth isometric embeddingsof the hyperbolic plane H2 in the Euclidean space R6 and in the standard sphericalspace S8. Pozniak [20] wrote about the Blanusas surface: There is no doubt thatthis result is one of the most elegant in the theory of immersions of two-dimensionalmanifolds in Euclidean space. Also in [4, 5], Blanusa constructed smooth isometricimmersions of an infinite Mobius band with hyperbolic metric in R8 and in S10,and an isometric immersion of the hyperbolic space Hn in R6n5. His methodwas used and modified in further works: (i) Rozendorn [22] constructed smoothisometric immersions of the plane R2 with a warped product metric of the formd2 = dt2 + f(t)2dx2 in the Euclidean space R5; such class of metrics includes thehyperbolic one which is isometric to the metric d2 = dt2 + e2tdx2. A celebratedtheorem of Hilbert [16] states that H2 cannot be isometrically immersed in R3

and the existence of an isometric immersion ofH2 in R4 it is already an unsolvedproblem (for a partial answer see Sabitov [23]). (ii) Henke [14] exhibited isometricimmersions ofHn in R4n3 and in S4n3. (iii) Henke and Nettekoven [15] showedthat Hn can be isometrically embedded in R6n6 whose image is the graph of asmooth map g : Rn R5n6. (iv) Azov [2, 3] considered the Euclidean spaceRn = R Rn1 endowed with one of the two class of warped metrics: d2 =

2010 Mathematics Subject Classification. Primary 53C20; Secondary 31C05.The second author was partially supported by CNPq-Brazil.
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2 H. MIRANDOLA AND F. VITORIO

dt2 + f(t)2n1

j=1 dx2j and d

2 = g(t)2n

j=1 dx2j and solve the problem of their

immersibility in Euclidean and spherical spaces.We will consider the product manifold Mn = I Rn1, where I is an open

interval, endowed with a multiple warped product metric of the form:

d2 = (t)2 dt2 + 1(t)2dx21 + . . . + n1(t)

2dx2n1(1)

where , j : I (0, ), with j = 1, . . . , n 1, are positive smooth functions.We will use the Blanusa method to exhibit isometric immersion of these metrics inquadrics of semi-Euclidean spaces. There exists a wide literature about aspects ofrigidity and non immersibility of these spaces (see for instance Nolker [19], Chen[7], Florit [10], Dajczer and Tojeiro [9] and references therein).

We recall that the semi-Euclidean space Rna , with 0 a n integers, is simplythe Euclidean space Rn endowed with the inner product of signature (a, na) givenby

(2) , = dx21 . . . dx2a + dx

2a+1 + . . . + dx

2n,

where dxj , with j = 1, . . . , n, denote the canonical coframes ofRn. Given c > 0,let Sna (c) and H

na (c) be the following quadrics hypersurfaces:

Sna(c) =

x Rn+1a

x, x = 1c

;

Hna(c) =

x Rn+1a+1

x, x = 1c

.

Both hypersurfaces are semi-Riemannian manifolds with signature (a, n a) andhave constant curvatures c and c, respectively. If a = 0 then Sn0 (c) = S

n(c) isthe standard sphere and Hn0 (c) = H

n(c) is the hyperbolic space. If a = 1 thenthe semi-Riemannian universal covering spaces ofSn1 (1) and H

n1 (1) are called de

Sitter dSn and anti-de Sitter AdSn spaces, respectively.Our main result is the following theorem.

Theorem 1.1. Let Mn = I Rn1, where I is an open interval, be the productmanifold endowed with a multiple warped product metric as given in (1). Then, forall c > 0 and all non-negative integer 0 a n 1, the manifold M admits:

(i) smooth isometric immersions inR4n2a3a ;(ii) smooth isometric immersions inS4n2a3a (c);

(iii) smooth isometric immersion inH4na3a (c);(iv) smooth isometric embeddings inR8n6a7a ;(v) smooth isometric embeddings inS8n6a5a (c);

(vi) smooth isometric embeddings inH8n5a7a (c);(vii) smooth isometric embeddings inR5n3a4a , if inf

tIa+j(t) > 0, for all j 1.

All the immersions and embedding above are given explicitly. Ifa = n 1 then the

immersions of items (i) and (iii) are actually embeddings and all the immersionsand embedding above are analytic provided that , are analytic functions.

The two-dimensional version of Theorem 1.1 is the following result.

Corollary 1.1. LetM2 = IR be the product surface, whereI is an open interval,endowed with the metric d2 = (t)2dt2 + (t)2dx2, where , : I (0, ) arepositive smooth (analytics) functions. Then, for allc > 0, the surface M admits:

(i) smooth isometric immersions inR5, S5(c) andH5(c);


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MULTIPLE WARPED PRO DUCT METRICS IN QUADRI CS 3

(ii) smooth isometric embeddings inR9, S11(c) andH9(c);(iii) smooth isometric embeddings inR6, if inftI(t) > 0.(iv) smooth (analytic) isometric embeddings inR3

1;

(v) smooth (analytic) isometric immersion inS31(c);(vi) smooth (analytic) isometric embeddings inH51(c);

Remark 1. About Theorem 1.1, if the metric d2 is of the form:

(3) d2 = dt2 + (t)2 or d2 = (t)2(dt2 + dx2)

then the smooth isometric immersion ofM2 in R5 was given by Rozendorn [22] andthe smooth isometric immersion ofM2 in S5(1) was given by Azov [2]. Henke andNettekoven [15] showed that if the metric of M2 is of the form

(4) d2 = dt2 + ((t)2 + 1)dx2

then it admits an isometric embedding in R6. Note that the metrics of the form (3)and (4) include the hyperbolic ones. The isometric embedding of the hyperbolic

metric d2

= dt2

+ cosh(t)2

dx2

= dt2

+ (sinh(t)2

+ 1)dx2

was given by Blanusa [4].Let fl : (Ml, gl) Rnl , with l = 1, . . . , k, be isometric immersions of the Rie-

mannian manifold (Ml, gl) in the Euclidean spaces Rnl . Let , l : I (0, ) be

positive smooth functions, where I is an open interval. It is simple to show thatthe product manifold M = I M1 . . . Mk with the warped product metric

(5) g = (t)2dt2 + 1(t)2g1 + . . . + k(t)

2gk

is isometrically immersed in the product space I Rn1 . . . Rnk endowed withthe warped product metric

d2 = (t)2dt2 + 1(t)2(dx211 + . . . + dx

21n1) + . . . + k(t)

2(dx2k1 + . . . + dx2knk

),

where xl = (xl1, . . . , xlnl) denotes the coordinates ofRnl . Thus it is a consequence

of Theorem 1.1 the following result.

Corollary 1.2. With the notation above, the manifold (M, g) admits:

(i) smooth isometric immersions inR4n2a+1a , S4n2a+1a (c) andH

4na+1a (c);

(ii) smooth isometric embeddings inR8n6a+1a , S8n6a+3a (c) andH

8n5a+1a (c),

provided that fl is an embedding, for all l = 1, . . . , k;(iii) smooth isometric embeddings inR5n3a+1a , provided thatfl is an embedding

and inftIl(t) > 0, for all l = 1, . . . , k.

2. Two-dimensional warped metrics in tridimensional quadrics andthe Blanusas functions.

We will to consider some situations where is relatively simple to construct iso-metric immersions of warped product surfaces in tridimensional quadrics. Cylinderover plane curves and surfaces of revolutions are the simplest class of warped prod-

uct metrics isometrically immersed in R3. Furthermore, the hyperbolic plane H2

can be seen as a connected component ofH2(1) R31 or as a totally geodesicembedded surface in H3(1). Further cases are given in Proposition 2.1 below.

Proposition 2.1. Let M2 = I J the product surface endowed with the warpedproduct metric:

d2 = (t)2dt2 + (t)2dx2.

Then, for all c > 0, the surface M admits:


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(a) isometric immersions inR3, if (t)2 < (t)2, for all t I;(b) isometric immersions inS3(c), if (t)2 < 1

c 1

c(t)2, for all t I.

(c) isometric immersions inH

3

(c), if

(t)2

< (t)2

(1

c + (t)2

), for all t I;(d) isometric embeddings inR31(e) isometric immersions inS31(c)(f) isometric immersions inH31(c), if (t)

2 < 1c

; or (t)2 > (t)2( 1c

+ (t)2),for all t I;

Moreover, if the functions , are analytic then the immersions above are analytic.

Remark 2. About Items (a), (c) and (f) in Proposition 2.1, the warped productmetrics of the form:

(t)2dt2 + (t)2dx2 and(t)2

1c

+ (t)2dt2 + (t)2dx2

have constant Gaussian curvatures 0 and c, respectively. In particular, the warpedmetrics dt2 + t2dx2 and dt2 + sinh(t)2dx2, with t > 0, have constant curvatures 0

and 1, respectively.

Proof of Proposition 2.1. First we will prove (a). To do this, we assume that(t)2 < (t)2, for all t I. Let f : I J R3 be the smooth function

f(t, x) = t

t0

()2 ()2d , (t)g(x)

,

where t0 I and g : R R2 is the map given by

(6) g(u) = (cos(u), sin(u)).

Note that f is obtained by the revolution of the curve (t) = (

(t)2 (t)2, (t)),with t I, in the yz-plane around the z-axis. A simple computation shows thatthe metric induced by f satisfies f( , ) = (t)2dt2 + (t)2dx2, which implies that

f : M R

3

is an isometric immersion. This concludes (a).To show (c), we assume that (t)2 < (t)2( 1c

+ (t)2), for all t I. Let fh :I J R21 R

2 be the smooth map given by

fh(t, x) =1

c+ (t)2 h(h(t)) , (t)g(x)

,

where h : R R21 is the map given by

(7) h(u) = (cosh(u), sinh(u))

and h : I R is the function given by

h(t) =

tt0

1

1c

+ ()2

()2

()2

1c

+ ()2

d.

The function h is well defined since (t)2 >

(t)

2

1c

+(t)2 , for all t I. Furthermore,since fh(t, x), fh(t, x) =

1c

, the image of fh is contained in H3(c). Using that

h(u), h(u) = 1, h(u), h(u) = 0 and h(u), h(u) = 1, by a direct computation,it follows that

fh( , ) =

(t)2(t)2

1c

+ (t)2+1

c+ (t)2

h(t)

2 + (t)2

dt2 + (t)2dx2

= (t)2dt2 + (t)2dx2.


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Thus, fh : M H3(c) is an isometric immersion, which concludes (c).To show (b), we assume that 1

(t)2 + c(t)2 < 1, for all t I. Let fs : I J R

4

be the map given by

fs(t, x) =

1

c (t)2 g(s(t)) , (t)g(x)

,

where s : I R is the function defined by

s(t) =

tt0

1

1c

(t)2

(t)2

1c

1c

(t)2

.

The function fs is well defined since1

(t)2+ c(t)2 < 1, which implies:

1

c

(t)2 >1

c(t)2

> 0 and1c

1

c (t)2

< (t)2.

By a direct computation, fs ( , ) = d2. Since fs(t, x), fs(t, x) =

1c

we conclude

that fs : M2 S3(c) is an isometric immersion. Item (b) is proved.

To show (d) we consider the map f : I J RR21 given by

f(t, x) =

tt0

()2 + ()2d , (t)h(x)

,

where h : R R21 is the map as given in (7). By a direct computation we obtain thatf( , ) = d2. This implies that f : M2 R31 = RR

21 is an isometric immersion.

Furthermore, it is easy to show that f is an embedding. This follows easily from

the facts that the functions t I

t

t0 ()2 + (t)2d and x J sinh(x)

are diffeomorphism onto their corresponding images. This concludes (d).We will prove (e). Let fs : I J R2 R21 be the smooth function given by

fs(t, x) =

1

c+ (t)2 g(s(t)) , (t)h(x)

,

where s : I R is the function given by

s(t) =

tt0

1

1c

+ ()2

()2 + ()2

()2()2

1c

+ ()2

d.

The function s is well defined since (t)2 (t)2(t)2

1+(t)2 =(t)2

1+(t)2 0, for all t I.

Since fs(t, x), fs(t, x) = 1c , for all (t, x) I J, the image of fs is contained inthe quadric S31(c). Furthermore, the induced metric of fs satisfies:

fs ( , ) =

(t)2(t)2

1c

+ (t)2+1

c+ (t)2

s(t)

2 (t)2

dt2 + (t)2dx2

= (t)2dt2 + (t)2dx2.

Thus fs : M2 S31 is an isometric immersion. This concludes (e).


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Finally, we will show (f). First, we assume that (t)2 < 1c

, for all t I. Let

fh : I J R21 R


fh(t, x) =1

c (t)2 h(h(t)) , (t)h(x)

,

where h : I R is the function defined by

h(t) =

tt0

1

1c

()2

()2 +

()2

1c

()2

d.

A standard computation shows that

fh(t, x), fh(t, x)

= 1c

and fh( , ) = d2.

It is not difficult to show that fh is an embedding. Thus, fh : M2 H3(c) is an

isometric embedding.Now, we assume that (t)2 > (t)2( 1

c+ (t)2), for all t I. Let fh : I J

R22 R


fh(t, x) =

1

c+ (t)2 cos(h(t)),

1

c+ (t)2 sin(h(t)), (t)g(x)

,

where h : I R is the function given by

h(t) =

tt0

1

1c

+ ()2

()2

1c

+ ()2 ()2

By a direct computation we obtain that

fh, fh

= 1

cand fh( , ) = d

2. Thus,

we have that fh : M H31(c) is an isometric immersion. Item (f) and Proposition

2.1 are proved.

Lemma 2.1 below was proved by Blanusa [4]. We will recall its proof for the sake

of completeness.Lemma 2.1. There exist smooth functions 1, 2 : R R with the followingproperties:

(i) 1(x)2 + 2(x)2 = 1, for all x R and j = 1, 2;(ii) j(x) = j(x + 2), for all x R and j = 1, 2;

(iii) The derivatives (k)1 (2l + 1) = 0, for all k 0 and l integers ;

(iv) The derivatives (k)2 (2l) = 0, for all k 0 and l integers.

Proof. Consider the function : R R given by

(x) =

sin(x)e1

(sin(x))2 , for all x non-integer;

0, otherwise.

Set A =1

0(t)dt and let 1, 2 : R [0, ) the non-negative functions defined

by

1(x) =

1

A

x+10

(t)dt and 2(x) =

1

A

x0

(t)dt.

It is simple to see that 1 and 2 are well defined, sincex

0(t)dt 0, for all x. Since

(x) = (x + 2), for all x and2

0 (t)dt = 0 we have that j(x) = j(x + 2), for all


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x and j {1, 2}. Since (t + 1) = (t), for all t, using change of variable, we have

A 1(x)2 =

x+1

0(t)dt = A +

x+1

1(t)dt = A +

x

0( + 1)d = A

x

0()d =

A A 2(x)2

. Thus it holds that 1(x)2

+ 2(x)2

= 1, for all x.We will show that 1 and 2 are smooth and that all the derivatives of 1 and

2 satisfy (k)1 (2l + 1) =

(k)2 (2l) = 0, for all k 0 and l integers. In fact, it is

simple to see that is smooth and the all derivatives (k)(l) = 0, for all k 0and l integers. Thus 21,

22 are smooth functions. If x = 2l, for all l integer, then

2(x) = 0, hence 2 is smooth at x. Since 2(x) = 2(x + 2), to show that 2 issmooth at the points of the form x = 2l, with l integer, it suffices to show that 2is smooth at the point x = 0. Using that sin(t)2 (x)2, for all |t| |x| < 1/2,

we obtain that |2(x)|

|x|e1|x| , for all |x| < 1. This implies that

limx0

xk2(x) limx0

|x|k+12 e

1|x| = 0,

for all k 0 integer. We conclude that (k)

2(0) = 0, for all k 0 integer. A similar

argument shows that 1 is smooth and that (k)(2l + 1) = 0, for all k 0 and lintegers. Lemma 2.1 is proved.

Let S1, S2 : R (0, ) be positive step functions satisfying:

(I) S1(x) is constant on each interval [2l + 1, 2l + 3), for all l integer;(II) S2(x) is constant on each interval [2l, 2l + 2), for all l integer.

Let I be an open interval and I : I R an analytic increasing diffeomorphism.For each j = 1, 2, we consider the following functions:

Ij = j I and SIj = Sj

I.

We also consider the points tl = (I)1(l) I, with l integer. Using Lemma 2.1,

(I) and (II) we obtain

(8)

I1 (t)2 + I2 (t)

2 = 1, for all t I;

Ij

(I)1(x)

= Ij

(I)1(x + 2)

, for all x R and j = 1, 2;

The derivatives (I1 )(k)(t2l+1) = 0, for all k 0 and l integers

The derivatives (I2 )(k)(t2l) = 0, for all k 0 and l integers.

SI1 is constant on each interval [t2l+1, t2l+3), for all l integer.SI2 is constant on each interval [t2l, t2l+2), for all l integer.

The following lemma follows easily from (8).

Lemma 2.2. For any smooth function C(I), the functions(t)Ij (t)

SIj

(t)with

t I and j = 1, 2, are smooth and all their derivatives satisfy

d k

dtk

(t)Ij (t)

SIj (t)

=

dk

dtk

(t)Ij (t)

SIj (t)

,

for all k 0 integer.


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Consider the map = (1, 2) : I R R4 where j : I R R2, withj = 1, 2, is the map given by

(9) j(t, x) = Ij (t)

SIj (t)

cos(SIj (t)x), sin(S

Ij (t)x)

.

The following result is essentially proved in [15].

Lemma 2.3. Let : I R be a smooth function and : I (0, ) be a positivecontinuous function. Then for each integer l we can choose the step functionsSI1 |[t2l+1,t2l+3) and S

I2 |[t2l,t2l+2) sufficiently large so that the pull-back metric induced

by the map : I J R4 satisfies:

(10) ()( , ) = (t)2dt2 + (t)2dx2,

where = : I R is a smooth function satisfying 0 (t) < (t), for all t I.

Proof. Using Lemma 2.2, the partial derivatives of j , with j = 1, 2, satisfy

(t)j(t, x)t

=(t)Ij (t)

SIj (t)

cos(SIj (t)x), sin(S

Ij (t)x)

(t)j(t, x)

x= (t)Ij (t)

sin(SIj (t)x), cos(S

Ij (t)x)

.

Thus, denoting by , the Euclidean metric ofR2, we obtain that

(t)j(t, x)

t,

(t)j(t, x)

t

=

(t)Ij (t)

SIj (t)

2,

(t)j(t, x)

t,

(t)j(t, x)

x

= 0,(11)

(t)j(t, x)

x,

(t)j(t, x)

x

= (t)2Ij (t)

2.

Using (11) and the fact that I1 (t)2 + I2 (t)

2 = 1, for all t I we have that thepull-back metric induced by satisfies

() , = (t)2dt2 + (t)2

I1 (t)2 + I2 (t)

2

dx2

= (t)2dt2 + (t)2dx2,

where = : I [0, ) is the smooth function given by

(12) (t)2 =

(t)I1 (t)

2SI1 (t)

2+

(t)I2 (t)

2SI2 (t)

2

Since t I

(t)Ij (t)

, with j = 1, 2, are smooth functions and : I R isa positive continuous function, we can choose for each integer l the step functionsSI1 |[t2l+1,t2l+3) and S

I2 |[t2l,t2l+2) sufficiently large so that

(t)I1 (t) 2

0, for all t I. Claim 3.2 is proved.

Note that fs(t, x), fs(t, x) =1c

. Using that g(u), g(u) = 0 and |g(u)|2 = 1,by a standard computation,

fs ( , ) =

b(t)2

4

1c

b(t) + 1

c b(t)

s(t)

2

dt2

+ ( h)( , ) + ( )( , )

= (t)2dt2 + 1(t)2dx21 + . . . + a+b(t)

2dx2a+b.

Thus, fs : Mn S4n2a3a (c) is an isometric immersion. Item (ii) is proved.

Let fh : IRn1 R4na2a+1 = R

21 R

a (R21)a R4b be the map

fh(t, x) =

1

c+ a(t) h(h(t)) , (t) , h ,

,

where h : I R is the function given by

h(t) =

tt0

11c

+ a(t)

(t)2 (t)2 +

a(t)2

4

1c

+ a(t).

Note that h is well defined since (t)2 (t) > 0. It is easy to see that

fh(t, x), fh(t, x) = 1c + a(t) + |(t)|2 + b(t) = 1c .Using that h(u), h(u) = 1 and h(u), h(u) = 0 and h(u), h(u) = 1, for allu R, by a standard computation,

fh( , ) =

a(t)2

1c

+ a(t)

+

1

c+ a(t)

h(t)

2 + |(t)|2

dt2

+( h)( , ) + ( )( , )

= (t)2(dt)2 + 1(t)2(dx1)

2 + . . . + a+b(t)2(dxa+b)

2.

Thus fh : Mn H4na3a is an isometric immersion. Item (iii) is proved.

We will prove item (iv). Let T1 : R (0,2 ) and T2 : R R be the analytic

functions

(21) T1(u) = 4

(1 + tanh(u)) and T2(u) = u0

1 T1()2d.Note that T2 is analytic since T1(u) =

4 sech

2(u) 4 < 1.Consider the map = (11, 21, 12, 22) : IR R8, where jk : IR R2,

with j, k = 1, 2, are the maps given by

(22) jk(t, u) =Ij (t)

SIj (t)

cos

Tk(S

Ij (t)u)

, sin

Tk(S

Ij (t)u)

.


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It follows from (8) that the maps r(t)jk(t, u), with j, k = 1, 2, r = a + 1, . . . , a + band (t, u) I R, are smooth and their partial derivatives satisfy:

(r(t)jk (t, u))t

=r(t)Ij (t)

SIj (t)

cos

Tk(S

Ij (t)u)

, sin

Tk(S

Ij (t)u)

;

(r(t)jk (t, u))

u= r(t)

Ij (t)T

k(S

Ij (t)u)

sin

Tk(S

Ij (t)u)

, cos

Tk(S

Ij (t)u)

.

Thus, it holds that(r(t)jk(t, u))

t,

(r(t)jk (t, u))

t

=

(r(t)

Ij (t))

(SIj (t))

2;

(r(t)jk(t, u))

t,

(r(t)jk (t, u))

x

= 0;

(r(t)jk(t, u))u

, (r(t)jk (t, u))u

= ((t))2(Ij (t))2(Tk(Sj(t)u))2.Since T1(t)

2 + T2(t)2 = I1 (t)

2 + I2 (t)2 = 1, the pull-back metric induced by the

map ( ) : IRn1 R8b satisfies

(23) ( )( , ) = 2(t)2 dt2 + a+1(t)2dx2a+1 + . . . + a+b(t)

2dx2a+b,

where = : I [0, ) is the function given as in (15) and satisfying that

0 (t) < (t)2

2 , for all t I.

Let f : IRn1 R8n6a7a = R (R21)

a R8b be the map

f(t, x) =

tt0

()2 + |()|2 2()2 d , h (t, x) , (t, x)

.

Using (23), it is easy to see that f : M R8n6a7

is an isometric immersion.

Claim 3.3. The immersion f is injective.

In fact, assume that f(t1, x1) = f(t2, x

2), for some t1, t2 I and x1, x2 Rn1.

We write xj = (xj1, . . . , xja+b), with j = 1, 2. Using that the first coordinate of

f(t, x) is the strictly increasing function

s(t) =

tt0

()2 + |()|2 2()2 d,

we obtain that t1 = t2. Since I1 (t1)2 + I2 (t1)

2 = I1 (t2)2 + I2 (t2)

2 = 1 we canassume, without loss of generality, that I1 (t1) = 0. Using that i(t) > 0, for

all i = 1, . . . , a + b and f(t1, x1) = f(t1, x2), we have that h(x1k) = h(x2k) and

11(t1, x

1

r) = 11(t1, x

2

r), for all k = 1, . . . , a and r = a + 1, . . . , a + b. These implythat

sinh(x1k) = sinh(x2k) and sin(T1(S1(t1)x

1r)) = sin(T1(S1(t1)x

2r)),

for all k = 1, . . . , a and r = a + 1, . . . , a + b. Since S1(t1) > 0 and the functionssinh(u) and sin(T1(u)), with u R, are injective we obtain that x1 = x2. Claim3.3 is proved.

Claim 3.4. f is an embedding.


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In fact, we need to show that the inverse map f1 : f(I Rn1) I Rn1

is continuous. In fact, let yn = f(tn, xn) be a sequence that converges to a point

y0 =

f(t0, x0

). Thus we have:(a) lim s(tn) = s(t0);(b) lim jk(tn, x

n) = jk(t0, x0).

Since s(t) is a diffeomorphism of I onto its image we obtain that from (a) thatlim tn = t0. Using that

I1 (t0)

2 + I2 (t0)2 = 1, we can assume, without loss of

generality, that I1 (t0) = 0. Since SI1 is a positive step function and

I1 (t0) > 0 we

obtain that SI1 is a positive constant function in a neighborhood oft0. This impliesthat S1(tn) = S1(t0) > 0, for n sufficiently large. Since limn Ij (tn) =

Ij (t0) > 0,

for all j = 1, 2 we obtain from (b) and (22) that

limn

cos(T1(SIj (t0)xn) = lim

n

SIj (tn)

Ij (tn)P j1(tn, x

n) =SIj (t0)

Ij (t0)P j1(t0, x

0)

= cos(T1(SIj (t0)x0)),

where P : R2 R is the projection P(u, v) = u for all u, v R. Again using thatS1(t0) > 0 and since cos(T1(u)) is a diffeomorphism ofR onto (0, 1), it follows thatlimn xn = x0. Claim 3.4 and item (iv) are proved.

We will prove item (v). Let fs : I Rn1 R8n6a4a = R

4 (R21)a R8b be

the map

(24) fs(t, x) =

1

2c+

1

2|(t)|2 a(t) C(s(t)) , h (t, x) , (t, x)

,

where C(u) =

cos(T1(u), sin(T1(u)), cos(T2(u)), sin(T2(u))

, with u R, and the

function s : I R is given by

s(t) =

tt0

G()

12c +

12 |()|

2 a()d,

where G : I R is given by G(t) =

(t)2 + |(t)|2 2(t)2

((a(t)+ 12 |(t)|2)

)2

2( 12c+12 |(t)|

2a(t))

.

Claim 3.5. For each integer l, we can choose the step functions SI1 |[t2l+1,t2l+3) and

SI2 |[t2l,t2l+2) sufficiently large so that the function s : I R is well defined.

In fact, it follows from (18) that 12c +12 |(t)|

2 a(t) 12c a(t) >1

4c . Note that

a(t) + 12 (t), (t)

2

= (a

(t) +

(t), (t))

2

(t) + |(t)|

2

|

(t)|

2

,

where (t) = a(t)2 + 2|a(t)||(t)||(t)|. Thus, it holds thata(t) + 12 |(t)|

22

2

12c +

12 |(t)|

2 a(t) 2c (t) + |(t)|2 |(t)|2

|(t)|2 +

1c

2a(t)

2c (t) + |(t)|2


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This implies that

(t)2 + |(t)|2 2(t)2 a(t) + 12 |(t)|22

2

12c +

12 |(t)|

2 a(t) (t)2 2(t)2 2c (t)

Since (t) 2(t) > 0, using (19) we also can choose for each integer l the stepfunctions SI1 |[t2l+1,t2l+3) and S

I2 |[t2l,t2l+2) sufficiently large satisfying that (t)

2

2(t)2 2c(t) > 0. This implies that G(t)12c+

12 |(t)|

2a(t)> 0. Claim 3.5 is proved.

Note that

fs(t, x), fs(t, x)

= 1c

since |C(s(t))|2 = 2, | (t, x)|2 = 2a(t) and

h (t, x) , h (t, x) = |(t)|2. Thus the image fs(I Rn1) S8n6a5a (c).Furthermore, a direct computation shows that

fs ( , ) =

a(t) + 12 |(t)|2

2

2 12c + 12 |(t)|2 a(t) +

1

2c+

1

2|(t)|2 a(t)

s(t)

2

dt

2

+( h)( , ) + ( )( , )

= (t)2dt2 + 1(t)2dx21 + . . . + a+b(t)

2dx2a+b.

This implies that fs : I Rn1 S8n6a5a (c) is an isometric immersion.

Claim 3.6. The immersion fs is injective.

In fact, assume that fs(t1, x1) = fs(t2, x

2), for some t1, t2 I and x1, x2 Rn1.In particular, using (24), we have

1

2c+

1

2|(t1)|2 a(t1) C(s(t1)) =

1

2c+

1

2|(t2)|2 a(t2) C(s(t2)).

Since |C(s(t))| = 2, for all t I, we obtain that |(t1)|2 2a(t1) = |(t2)|2 2a(t2),

thus C(s(t1)) = C(s(t2)). This implies that cos(T1(s(t1))) = cos(T1(s(t2))).

Since the functions s, cos(T1(u)) are diffeomorphism ofI, R onto its correspondingimages we obtain that t1 = t2. The argument to show that x1 = x2 is similar toClaim 3.3.

Claim 3.7. fs is an embedding.

In fact, we need to prove that the inverse map (fs)1 : fs(IR

n1) IRn1

is continuous. To do this, let yn = fs(tn, xn) be a sequence that converges to apoint y0 = f(t0, x0). In particular, using (24), we obtain

limn 12c + 12 |(tn)|2 a(tn) C(s(tn)) = 12c + 12 |(t0)|2 a(t0) C(s(t0)).Since |C(s(t))| = 2, for all t I, it follows that lim |(tn)|

2 2a(tn) = |(t0)|2

2a(t0). Since1

2c +12

|(t0)|2 a(t0) > 0 we obtain that lim C(s(tn)) = C(s(t0)).

Thus, lim cos(T1(s(tn))) = cos(T1(s(t0))). Using that cos(T1(s(t))) is a diffeo-morphism of I onto its image it follows that lim tn = t0. The argument to showthat lim xn = x0 is similar to Claim 3.4. This concludes Claim 3.7. Item (v) isproved.


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Finally we will prove (vi). Let fh : IRn1 R8n5a6a+1 = R

21 R

a(R21)aR8b

be the map

(25) fh(t, x) =1

c+ 2a(t) h(h(t)) , (t) , h (t, x) , (t, x)

,

where a : I [0, ) is defined in (18) and h : I R is the function given by

(26) h(t) =

tt0

1

1c

+ 2a()

()2 2()2 +

a()2

1c

+ 2a()

d

The function h : I R is well defined since (t)2 2(t)2 > 0. Note that

fh(t, x)fh(t, x)

=

1c

+ 2a(t)

+ |(t)|2 |(t)|2 + 2a(t) = 1c

. Thus the image

fh(IRn1) H8n5a7a (c). Furthermore, by a standard computation, we obtain

(fh)( , ) =

a(t)2

1

c + 2a(t)2

+1

c

+ 2a(t) h(t)2 + |(t)|2 dt2+ ( h)( , ) + ( )( , )

= (t)2dt2 + 1(t)2dx21 + . . . + a+b(t)

2dx2a+b

This implies that fh : M R8n5a7a is an isometric immersion.

Claim 3.8. The immersion fh is injective.

In fact, assume that fh(t1, x1) = fh(t2, x

2), for some t1, t2 I and x1, x2 Rn1.

Since

h(h(t)), h(h(t))

= 1 it follows from (25) that a(t1) = a(t2), hence

h(h(t1)) = h(h(t2)). Thus sinh(h(t1)) = sinh(h(t2)). Using that sinh(h(t)),with t I, is injective we obtain that t1 = t2. The argument to show that x1 = x2

is similar to Claim 3.3. This proves Claim 3.8.

Claim 3.9. fh is an embedding.

In fact, it suffices to prove that the inverse map (fh)1 : fh(IRn1) IRn1

is continuous. To do this, let yn = fh(tn, xn) be a sequence convergent to a point

y0 = fh(t0, x0). In particular, we obtain that

limn

1

c+ 2a(tn) h(h(tn)) =

1

c+ 2a(t0) h(h(t0)).

Since

h(h(t)), h(h(t))

1 we obtain that lim 1c

+ a(tn) =1c

+ a(t0) > 0, hence

lim h(h(tn)) = h(h(t0)), which implies that lim sinh(h(tn)) = sinh(h(t0)). Since

sinh(h(t)) is a diffeomorphism ofI onto its image, we obtain that lim tn = t0. Theargument to prove that lim xn = x0 is similar to Claim 3.4. This concludes Claim

3.9. Theorem 1.1 is proved.

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[2] Azov, D. G.: Imbedding by the Blanusa method of certain classes of complete n-dimensionalRiemannian metrics in Euclidean spaces, Vestnik Moskov. Univ. Ser. I Mat. Mekh. 1985, no.

5, 70 74; English transl., Moscow Univ. Math. Bull. 40:5 (1985), 64 66.


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[3] Azov, D. G.: Isometric embedding of n-dimensional metrics into Euclidean and sphericalspaces. (Russian) Vestnik Chelyabinsk. Univ. Ser. 3 Mat. Mekh. 1994, no. 1(2), 12 16.

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3(339), 3 78; translation in Russian Math. Surveys 56 (2001), no. 3, 425 497.

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[8] Dajczer, M. Submanifolds and isometric immersions. Based on the notes prepared by Mauri-cio Antonucci, Gilvan Oliveira, Paulo Lima-Filho and Rui Tojeiro. Mathematics Lecture

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Manuscripta Math. 34 (1981), 265 278.[15] Henke, W. and Nettekoven, W. : The hyperbolic n-space as a graph in Euclidean (6n 6)-

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[16] Hilbert, D.: Uber Flachen von konstanter Krummung. Trans. Amer. Math. Soc. 2 (1901),87-99.

[17] Jacobowitz, H.: Implicit function theorems and isometric embeddings. Ann. of Math. (2) 95(1972), 191 225.

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space. Russian Math. Surv. 28 (4), 47 77 (1973) (Russian)

[21] Poznyak, G. and Sokolov, D. D.: Isometric immersions of Riemannian spaces in Euclidean

spaces, Algebra, Geometry, and Topology, vol. 15 (Itogi Nauki i Tekhniki), VINITI, Moscow1977, pp. 173 211; English transl., J. Soviet Math. 14 (1980), 1407 1428.

[22] Rozendorn, E. R.: A realization of the metric ds2 = du2 + f2(u)dv2 in five-dimensionalEuclidean space, Dokl. Akad. Nauk Armenian SSR 30:4 (1960), 197 199. (Russian)

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30:5 (1989), 179 186; English transl., Siberian Math. J. 30 (1989), 805 811.

Heudson Mirandola

Universidade Federal do Rio de Janeiro

Instituto de Matematica

21945-970 Rio de Janeiro-RJ

Brazil

[email protected]


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Feliciano Vitorio

Universidade Federal de Alagoas

Instituto de Matematica

57072-900 Maceio-ALBrazil

[email protected]

multiple warped product metrics into quadrics - mirandola, vitorio

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