# A PDE for Smooth Isometric Immersion of an Open Set of the Hyperbolic Planeℍ2 into ℝ4

Authors: Mor Dione; Bira Gaye Diop; Athoumane Niang
Journal: Journal of Mathematics
 (2021)

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## Abstract

A simple interpretation of the PDE for isometric immersion of the hyperbolic planeℍ2 into ℝ4 is given. Thus, necessary and sufficient conditions are obtained. And, under natural additional conditions, we show that there are no complete solutions, but we can have special local solutions.


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## Body

## 1. Introduction

LetM,g and M′,g′ be two smooth Riemannian manifolds. The solutions to differential equations f∗g′=g of maximal rank are the isometric immersions. This equation is not formally integrable. One must impose some conditions of compatibility like the Gauss and Weingarten equation (see E-Cartan [1]). IfM is an open set U of the hyperbolic plane ℍ2 and M′,g′ is the Euclidean space ℝ4, the conditions of compatibility can be reduced to the equality of the mixed derivatives (as fuv=fvu, for u,v∈U), where ℍ2 is set u,v∈ℝ×ℝ+∗ with the Riemann metric g=d2u+d2v/v2.Hilbert and Uebu Flachenhave shown in [2] that the hyperbolic plane ℍ2 cannot be immersed in ℝ3 [3]. Hilbert’s statement was generalized by Efimov [4] (and reviewed by Milnor [5]). From [6], Roserdorn has immersed ℍ2 into ℝ5. More generally, the hyperbolic plane ℍn can be isometrically immersed into ℝ4n−3. Gromov in [7] has given analytic imbedding of ℍ2 in ℝ8. In [8], smooth isometric immersions of open set of form ℝ×c,+∞,c∈ℝ+∗, of ℍ2 into ℝ4 are given, but these immersions cannot be extended to the whole ℍ2. It is not yet known that if the hyperbolic plane can be immersed isometrically in the Euclidean space ℝ4.The differential equationf∗g′=g for the isometric immersion is generally very difficult. To avoid this difficulty one can first try to find particular solutions. For example, in [9], surfaces of rotations of constant curvature K=−1 have been studied. However, only particular domains in ℍ2 can be immersed by rotations of ℝ4. That is, there is no complete surface of rotation in ℝ4 of constant curvature K=−1.Inℝ3, like in ℝ4, there are surfaces with curvature K=−1 (surfaces of rotation for example), but if you try to extend them to be complete, they have singularities such as a cusp of revolution. Hilbert’s proof consists in showing that traveling far enough along a geodesic, one is forced to develop such a singularity. In the problem of the existence (or not) of isometric immersion of the hyperbolic plane ℍ2 in ℝ4, one can first try to see the problem for particular open sets of ℍ2 like u,v/a<u<b,v>0 or like ℝ×0,c.Before giving our main results, we will put some notations and relations.LetU be an open set of ℍ2 and let f be a smooth immersion; f:U⊂ℍ2⟶ℝ4,u,v↦fu,v.For a maph from U into ℝ, the notation h=c, where c∈ℝ, means the set of point u,v∈U such as hu,v=c. When h is a real or vector valued function on U. We will denote ∂h/∂uu,v and ∂h/∂vu,v by hu and hv, respectively. The map u,v↦1/v2 will be denoted l.So,f:U⊂ℍ2↦ℝ4 is an isometric immersion if only if(1)fu⋅fu=fv⋅fv=l2,(2)fu⋅fv=0,on U, with ⋅ denoting the Euclidean dot product in ℝ4.If we denote byX,Y,A, and B the components of the map f, the relation in ( 1)and (2) becomes(3)Xu2+Yu2+Au2+Bu2=l2iXv2+Yv2+Av2+Bv2=l2iiXuXv+YuYv+AuAv+BuBv=0iii.Equation (i) (resp (ii)) above has a natural interpretation. For example, let us consider equation (i). It means that there is a smooth mapθ:U⊂ℍ2⟶0,π/2⊂ℝ such that(4)cosθ=1lXu2+Yu2sinθ=1lAu2+Bu2.We define the following closed setsF1 and F2 of U by(5)F1=θ=0,F2=θ=π2,and we put F=F1∪F2.OnU∖F, the system in (4) is equivalent to the existence of two smooth maps φ1andφ2 on U∖F⟶0,2π such that(6)fu=lcosθcosφ1cosθsinφ1sinθcosφ2sinθsinφ2.By this interpretation of (i) in (3) with the notation and relations (4)–(6) above, we have the following.Theorem 1.
Iff:U⊂ℍ2⟶ℝ4is a smooth isometric immersion, then there exist three smooth real-valued functionsλ,μ1, andμ2onU∖Fsatisfying(7)λ2+μ1cosθ2+μ2sinθ2=l2onU∖F,such that the maps φ1,φ2:U∖F⟶0,2π are solutions of systems S1 and S2 below:(8)S1lcosθφ1v+λsinθφ1u=μ1cosθu−μ1cosθφ1u=lcosθv+λsinθu,(9)S2lsinθφ2v−λcosθφ2u=μ2sinθu−μ2sinθφ2u=lsinθv−λcosθu,on U∖F.A similar interpretation of (ii) in (3) leads to the existence of a map which we denote also by θ from U into 0,π/2 such as(10)cosθ=1lXv2+Yv2,sinθ=1lAv2+Bv2.Likewise, we define the closed setsF1 and F2 by(11)F1=θ=0,F2=θ=π2,and F=F1∪F2. So, we get, from (10), the existence of two smooth maps denoted also by φ1,φ2 on U∖F with values in 0,2π such as(12)fv=lcosθcosφ1cosθsinφ1sinθcosφ2sinθsinφ2.From (10)–(12), we have the following.Theorem 2.
Iff:U⊂ℍ2⟶ℝ4is a smooth isometric immersion, then there exist three smooth real-valued functionsλ,μ1, andμ2onU∖Fsatisfying(13)λ2+μ1cosθ2+μ2sinθ2=l2onU∖F,such that the maps φ1,φ2:U∖F⟶0,2π are solutions of systems S1′ and S2′ below:(14)S1′lcosθφ1u+λsinθφ1v=μ1cosθv−μ1cosθφ1v=lcosθu+λsinθv,(15)S2′lsinθφ2u−λcosθφ2v=μ2sinθv−μ2sinθφ2v=lsinθu−λcosθv,on U∖F.As we will see below, the existence of the four functionsθ,λ,μ1, and μ2:ℍ2⟶ℝ satisfying λ2+μ1cosθ2+μ2sinθ2=l2 on ℍ2 and two functions φ1,φ2:ℍ2⟶ℝ satisfying (8) and (9) (resp (14) and (15)) are sufficient to find an isometric immersion of ℍ2 into ℝ4.Proposition 1.
(i)
In the case of (4) and (5), system (3), with the relation defining the closed set F1 or the closed set F2, has solutions that are not extendible to the whole ℍ2(ii)
In the case of (10) and (11), system (3), with the condition defining the closed set F1 or the closed set F2, has no solution defined in an open setNow, we give sufficient conditions for the existence of a smooth isometric immersion ofℍ2 into ℝ4. We have the following proposition.Proposition 2.
Let us assume there are four smooth functionsλ,μ1,μ2, andθfromℍ2intoℝwithθtaking values in0,π/2and satisfying the relation in(7)(or in (13)) onℍ2.
If the systemsS1 and S2 (resp the systems S1′ and S2′) have two smooth solutions φ1,φ2 on ℍ2, then one can immerse smoothly ℍ2 into ℝ4.In the view of the EDPS1 and S2 of Theorem 1 and (7), it seems natural to ask the following questions: does there exists complete solution under the assumption that the functions λ,μ1,μ2, and θ depend only on the variable v ?We give the answer to this question in the following proposition.Proposition 3.
If the functionsλ,μ1,μ2, andθdepend only onvthen, we have(i)
The PDES1 and S2 of Theorem 1 under condition (7) have no solution defined in an open set of the form ℝ×0,c (for c>0)(ii)
One can find solution defined an open sets of the fromℝ×c,+∞ for c>0This paper is a first step for coming works in the existence or nonexistence of isometric immersions of particular open setU of ℍ2 into ℝ4. We will expect that this work can help to say more on this problem.Some examples of isometric immersion of open set ofℍ2 of the form ℝ×c,+∞ are given at the end of the paper.


## 2. Proof of the Main Results

### 2.1. Proof of Theorem1 and Theorem 2

We keep all the notations and the hypothesis given above. Let us show some facts. Letφ be the map defined by the relation fu=lφ. Then, with (6), we obtain(16)φ=cosθcosφ1,cosθsinφ1,sinθcosφ2,sinθsinφ2,onU∖F.Since the three vectors’ valued functions∂φ/∂θ,∂φ/∂φ1, and ∂φ/∂φ2 are mutually orthogonal on U∖F, the equation in (2) is equivalent to the existence of three smooth functions λ,μ1,andμ2 from U∖F into ℝ such that(17)fv=λ∂φ∂θ+μ1∂φ∂φ1+μ2∂φ∂φ2,onU∖F.That is,(18)fv=λ−sinθcosφ1−sinθsinφ1cosθcosφ2cosθsinφ2+μ1−cosθsinφ1cosθcosφ100+μ200−sinθsinφ2sinθcosφ2on U∖F.By (18), the relation fv⋅fv=l2 in (1) becomes (7) of Theorem 1.Now, let us show the relations in (8) and (9).The necessary conditions for the existence of the isometric immersionf:U⊂ℍ2⟶ℝ4 are fuv=fvu.By (6) and (18), this condition is equivalent to the four following relations:α:lcosθcosφ1v=−λsinθcosφ1−μ1cosθsinφ1uβ:lcosθsinφ1v=−λsinθsinφ1+μ1cosθcosφ1uγ:lsinθcosφ2v=λcosθcosφ2−μ2sinθsinφ2uδ:lsinθsinφ2v=λcosθsinφ2+μ2sinθcosφ2uIf we replace inα,φ1 by φ1−π/2, one gets the relation in β. And, if we replace in γ,φ2 by φ2−π/2, we get δ. By these remarks, an easy computation shows that α,β,γ, and δ become, respectively, α′,β′,γ′, and δ′:α′:lcosθv+λsinθu+μ1cosθφ1ucosφ1−lcosθφ1v+λsinθφ1u−μ1cosθusinφ1=0β′:lcosθv+λsinθu+μ1cosθφ1usinφ1+lcosθφ1v+λsinθφ1u−μ1cosθucosφ1=0γ′:lsinθv−λcosθu+μ2sinθφ2ucosφ2−lsinθφ2v−λcosθφ2u−μ2sinθusinφ2=0δ′:lsinθv−λcosθu+μ2sinθφ2usinφ2+lsinθφ2v−λcosθφ2u−μ2sinθucosφ2=0The two relationsα′,β′ together with γ′ and δ′ lead to the system in (8) and (9).The proof of Theorem2 can be obtained by commuting u to v.


### 2.2. Proof of Proposition 1

For (a), we can consider only system (3) under the condition that defines the closed set F2. So, we consider system (3), with the relation Xu=0andYu=0.Thus, locally, in some open setV, the maps X and Y depend only on v. Then, we denote Xv and Yv by X′ and Y′, respectively. In V, system (3) becomes(19)X′2+Y′2+Av2+Bv2=l2Au2+Bu2=l2AuAv+BuBv=0.The second equation in (19) means that there exist on V a smooth map Ω:V⟶ℝ such that Au=lcosΩ,Bu=lsinΩ.Then, the last relation in (19) shows that there exist a map μ:V⟶ℝ such that Av=−μlsinΩandBv=μlcosΩ. One gets easily with (19) that(20)X′2+Y′2+μ2l2=l2.By (20), we see that the function μ depends only on v.With the expression ofAv and Bv given above, we use the conditions Auv=Avu and Buv=Bvu, and we obtain(21)lv+μlΩucosΩ−lΩvsinΩ=0lv+μlΩusinΩ+lΩvcosΩ=0.The above system is justlv+μlΩu=0 and Ωv=0. From (20), we see that the function μl depends only on v. Thus, we get Ωu=cc∈ℝ∗. Finally, we get μ=1/cv and Ω=cu+c0,c0∈ℝ.Then, we obtainAu=lcoscu+c0,Bu=lsincu+c0, Av=−1/cl2sincu+c0,andBv=1/cl2coscu+c0. And, the condition in (20) becomes(22)X′2+Y′2+1c2l4=l2.From this, one gets the solutionsf (by an easy integration) depending on c∈ℝ, defined on ℝ×1/c,+∞. By (22), such solutions cannot be extended to the whole ℍ2. We get the same result if we consider system (3) with the relation that defined the closed set F1. Thus, we get (a) of the proposition.For (b), we can just consider system (3) under the condition defining F2, that is, Xv=0andYv=0.Then, locally, the functionsX and Y depend only on u in some open set V1. We put Xu=X˙ and Yu=Y˙ and system (3) becomes(23)X˙2+Y˙2+Au2+Bu2=l2Av2+Bv2=l2AuAv+BuBv=0.SinceAv2+Bv2=l2, then there exist a smooth map ω:V1⟶ℝ such that Av=lcosωandBv=lsinω. With these relations, the last equation in (23) shows that there is a function μ:V1⟶ℝ such as Au=−μlsinωandBu=μlcosω. Then, we obtain(24)X˙2+Y˙2+μ2l2=l2.By settingAuv=AvuandBuv=Bvu, we get the relations:(25)lu+μlωvcosω−lωu−lμvsinω=0,lu+μlωvsinω+lωu−lμvcosω=0.These relations becomelu+μlωv=0 and lωu−lμv=0.Since the functionl depends only on v, we get ωv=0. We get lμ=εl2−X˙2+Y˙2 with ε=∓1, this leads to lμv=llv/l2−X˙2+Y˙2. Then, lw˙=llv/l2−X˙2+Y˙2. This gives w˙l2−X˙2+Y˙2=lv2. By differentiating this last equation in v, we have w˙2llv=lvlvv, which can be satisfied only on some curves. This gives the proof of Proposition 1.


## 2.1. Proof of Theorem1 and Theorem 2

We keep all the notations and the hypothesis given above. Let us show some facts. Letφ be the map defined by the relation fu=lφ. Then, with (6), we obtain(16)φ=cosθcosφ1,cosθsinφ1,sinθcosφ2,sinθsinφ2,onU∖F.Since the three vectors’ valued functions∂φ/∂θ,∂φ/∂φ1, and ∂φ/∂φ2 are mutually orthogonal on U∖F, the equation in (2) is equivalent to the existence of three smooth functions λ,μ1,andμ2 from U∖F into ℝ such that(17)fv=λ∂φ∂θ+μ1∂φ∂φ1+μ2∂φ∂φ2,onU∖F.That is,(18)fv=λ−sinθcosφ1−sinθsinφ1cosθcosφ2cosθsinφ2+μ1−cosθsinφ1cosθcosφ100+μ200−sinθsinφ2sinθcosφ2on U∖F.By (18), the relation fv⋅fv=l2 in (1) becomes (7) of Theorem 1.Now, let us show the relations in (8) and (9).The necessary conditions for the existence of the isometric immersionf:U⊂ℍ2⟶ℝ4 are fuv=fvu.By (6) and (18), this condition is equivalent to the four following relations:α:lcosθcosφ1v=−λsinθcosφ1−μ1cosθsinφ1uβ:lcosθsinφ1v=−λsinθsinφ1+μ1cosθcosφ1uγ:lsinθcosφ2v=λcosθcosφ2−μ2sinθsinφ2uδ:lsinθsinφ2v=λcosθsinφ2+μ2sinθcosφ2uIf we replace inα,φ1 by φ1−π/2, one gets the relation in β. And, if we replace in γ,φ2 by φ2−π/2, we get δ. By these remarks, an easy computation shows that α,β,γ, and δ become, respectively, α′,β′,γ′, and δ′:α′:lcosθv+λsinθu+μ1cosθφ1ucosφ1−lcosθφ1v+λsinθφ1u−μ1cosθusinφ1=0β′:lcosθv+λsinθu+μ1cosθφ1usinφ1+lcosθφ1v+λsinθφ1u−μ1cosθucosφ1=0γ′:lsinθv−λcosθu+μ2sinθφ2ucosφ2−lsinθφ2v−λcosθφ2u−μ2sinθusinφ2=0δ′:lsinθv−λcosθu+μ2sinθφ2usinφ2+lsinθφ2v−λcosθφ2u−μ2sinθucosφ2=0The two relationsα′,β′ together with γ′ and δ′ lead to the system in (8) and (9).The proof of Theorem2 can be obtained by commuting u to v.


## 2.2. Proof of Proposition 1

For (a), we can consider only system (3) under the condition that defines the closed set F2. So, we consider system (3), with the relation Xu=0andYu=0.Thus, locally, in some open setV, the maps X and Y depend only on v. Then, we denote Xv and Yv by X′ and Y′, respectively. In V, system (3) becomes(19)X′2+Y′2+Av2+Bv2=l2Au2+Bu2=l2AuAv+BuBv=0.The second equation in (19) means that there exist on V a smooth map Ω:V⟶ℝ such that Au=lcosΩ,Bu=lsinΩ.Then, the last relation in (19) shows that there exist a map μ:V⟶ℝ such that Av=−μlsinΩandBv=μlcosΩ. One gets easily with (19) that(20)X′2+Y′2+μ2l2=l2.By (20), we see that the function μ depends only on v.With the expression ofAv and Bv given above, we use the conditions Auv=Avu and Buv=Bvu, and we obtain(21)lv+μlΩucosΩ−lΩvsinΩ=0lv+μlΩusinΩ+lΩvcosΩ=0.The above system is justlv+μlΩu=0 and Ωv=0. From (20), we see that the function μl depends only on v. Thus, we get Ωu=cc∈ℝ∗. Finally, we get μ=1/cv and Ω=cu+c0,c0∈ℝ.Then, we obtainAu=lcoscu+c0,Bu=lsincu+c0, Av=−1/cl2sincu+c0,andBv=1/cl2coscu+c0. And, the condition in (20) becomes(22)X′2+Y′2+1c2l4=l2.From this, one gets the solutionsf (by an easy integration) depending on c∈ℝ, defined on ℝ×1/c,+∞. By (22), such solutions cannot be extended to the whole ℍ2. We get the same result if we consider system (3) with the relation that defined the closed set F1. Thus, we get (a) of the proposition.For (b), we can just consider system (3) under the condition defining F2, that is, Xv=0andYv=0.Then, locally, the functionsX and Y depend only on u in some open set V1. We put Xu=X˙ and Yu=Y˙ and system (3) becomes(23)X˙2+Y˙2+Au2+Bu2=l2Av2+Bv2=l2AuAv+BuBv=0.SinceAv2+Bv2=l2, then there exist a smooth map ω:V1⟶ℝ such that Av=lcosωandBv=lsinω. With these relations, the last equation in (23) shows that there is a function μ:V1⟶ℝ such as Au=−μlsinωandBu=μlcosω. Then, we obtain(24)X˙2+Y˙2+μ2l2=l2.By settingAuv=AvuandBuv=Bvu, we get the relations:(25)lu+μlωvcosω−lωu−lμvsinω=0,lu+μlωvsinω+lωu−lμvcosω=0.These relations becomelu+μlωv=0 and lωu−lμv=0.Since the functionl depends only on v, we get ωv=0. We get lμ=εl2−X˙2+Y˙2 with ε=∓1, this leads to lμv=llv/l2−X˙2+Y˙2. Then, lw˙=llv/l2−X˙2+Y˙2. This gives w˙l2−X˙2+Y˙2=lv2. By differentiating this last equation in v, we have w˙2llv=lvlvv, which can be satisfied only on some curves. This gives the proof of Proposition 1.


## 3. Proof of Proposition2

Let us simply assume that there exist four smooth functionsθ,λ,μ1, and μ2:ℍ2⟶ℝ4 satisfying the equation in (7), such that φ1 and φ2 are solutions of the systems S1 and S2 in (8) and (9), respectively.Let us findf:ℍ2⟶ℝ4 satisfying (6) and (8). Since ℍ2 is star shaped from any given point p0∈ℍ2, then the existence of f is guaranteed by the relation fuv=fvu which is the equation of S1 and S2 from (6) and (18).By the same way, when the relation in (7) and the systems S1′ and S2′ are satisfied in ℍ2, then one will get isometric immersion of ℍ2 into ℝ4 in a smooth way.


## 4. Proof of Proposition 3

Proof .
of (a): we will argue by contradiction.
We assume that the functionsλ,μ1,μ2, and θ depend only on the variable v and that the EDP S1 and S2 have solutions φ1 and φ2 defined on ℝ×0,c, and condition (7) is satisfied for all v∈0,c.
Thus, the EDPS1 and S2 become(26)S1″lcosθφ1v+λsinθφ1u=0−μ1cosθφ1u=lcosθv,(27)S2″lsinθφ2v−λcosθφ2u=0−μ2sinθφ2u=lsinθv.
And, we have condition (7):(28)λ2+μ1cosθ2+μ2sinθ2=1v2.
By differentiating the second equation ofS1″ (resp S2″) with respect to the variable u we obtain(29)μ1cosθφ1uu=0,μ2sinθφ2uu=0.
So, we consider the following cases:Case 1.
We assumeμ1=0.
Then, the second equation ofS1″ gives(30)lcosθ=k0∈ℝ+∗.
Then, we havecosθ=vk0, which 0<v≤1/k0≤c.
Thus, the second equation ofS2″ becomes(31)−μ2sinθφ2u=1−v2k02v2v.
From (31), we get φ2u=k2∈ℝ+∗, and μ2sinθ=1/k2v21−v2k02.
By using (7), we have(32)μ2sinθ2=1k22v41−v2k02≤1v2,which gives the inequality 1≤k22v41−v2k02, for v∈0,1/k0.
This leads to a contradiction whenv goes to 0.
Similarly, if we assume thatμ2=0, we also get a contradiction. It remains the following case.Case 2.
We assumeφ1uu=0 and φ2uu=0 on ℝ×0,c.
Then,φ1u=k1∈ℝ∗ and φ2u=k2∈ℝ∗; then, we get from (26)(33)μ1cosθ=−1k1lcosθv,μ2sinθ=−1k2lsinθv.
Then, (7) becomes(34)λ2+1k12lcosθv2+1k22lsinθv2=1v2.
This equation gives(35)−1v≤1k1lcosθv≤1v,−1v≤1k2lsinθv≤1v.
We putz=lcosθ. And, we have lsinθ=l2−z2.
Thus, (35) becomes(36)−1v≤1k1zv≤1v,−1v≤1k2l2−z2v≤1v.
The relation−1/v≤1/k1zv≤1/v means that there is a function t:0;c⟶0,1 such that 1/k1zv=1/v2t−1; then, we have(37)−1v≤1v2t−1≤1v.
By integration, we obtain(38)−logcv≤∫vc1τ2tτ−1dτ≤logcv.
The last relation shows there is a functionu:0,c⟶0,1 such that(39)1k1zc−zv=∫vc1τ2tτ−1dτ=logcv2u−1.
Thus, we have(40)z=zc−k1logcv2u−1.
If we putα=zc and X=logc/v, i.e., c/v=expX, we obtain(41)z=α−k1X2u−1.
Likewise, we prove that there is a functionw:0,c⟶0,1 such that(42)l2−z2=β−k2X2w−1,with β=1/c2−zc.
From (41) and (42), we obtain(43)α−k1X2u−12+β−k2X2w−12=l2.
From this equivalent, we obtain(44)α2+β2−l2−2Xk1α2u−1+k2β2w−1+X2k122u−12+k222w−12=0,which is equivalent to(45)α2+β2expX−1c2−2XexpXk1α2u−1+k2β2w−1+X2expXk122u−12+k222w−12=0.
Whenv goes to zero, X goes to +∞; we get a contradiction from this last equation.Proof .
of (b): By assumption,μ1, μ2, and θ depend only on the variable v. We put(46)cosθ=ℒ1v,sinθ=ℒ2v.
Equations (8) and (9) take the form(47)alℒ1φ1v=−λℒ2φ1u−μ1ℒ1φ1u=lℒ1v,blℒ2φ2v=λℒ1φ2u−μ2ℒ2φ2u=lℒ2v.
We write them as follows:(48)a′φ1u=−lℒ1vμ1ℒ1,φ1v=λℒ2lℒ1lℒ1vμ1ℒ1,b′φ2u=−lℒ2vμ2ℒ2,φ2v=−λℒ1lℒ2lℒ2vμ2ℒ2.
Under the assumption above, (48) gives(49)a″φ1=−lℒ1vμ1ℒ1u+M1v−lℒ1vμ1ℒ1vu+M1′v=λℒ2lℒ1lℒ1vμ1ℒ1,b″φ2=−lℒ2vμ1ℒ2u+M2v−lℒ2vμ2ℒ2vu+M2′v=−λℒ1lℒ2lℒ2vμ2ℒ2,where M1 and M2 are functions which depend only on v.
Equation (49) is satisfied when there are two real numbers k1 and k2 such that(50)lℒ1vμ1ℒ1=−k1andM1′=−λℒ2lℒ1k1,lℒ2vμ2ℒ2=−k2andM2′=λℒ1lℒ2k2.
In order to satisfy (50) (and therefore (47)–(49)), we put(51)ℒ1=vR1G1,ℒ2=vR2G2,μ1=−lG1′k1G1,μ2=−lG2′k2G2.where G1 and G2 are two functions that depend only on v and R1andR2 are positive real numbers.
The relation in (51) solves (48) (i.e., (8) and (9)). Now, it remains to show that G1 and G2 can be chosen such that (7) is satisfied.
Sinceℒ1=cosθ and ℒ2=sinθ, by (51), we have(52)R12G12+R22G22=1v2.
Let us assume that1/v2−R12G12 is not identically zero.
From (52), we obtain(53)f1:R22G22=1v2−R12G12,R22G2′2=1/v3+R12G1′G121/v2−R12G12.
Thus, the relation in (7) becomes(54)λ2=1v2−R12k12G1′2+R22k22G2′2.
Let us introduce the variablex=1/v. And, we put G1v=Hx=xr2+1, when H is the expression of G1 in term of the variable x. We denote dH/dx by H˙.
Thus, we will assumeR1>0 and k1>0. Let us introduce the variable x=1/v. And, we put G1v=Hx=xr2+1, where H is the expression of G1 in term of the variable x. We denote H/dx by H˙. Thus, we have(55)G1v=Hx;G1′=−x2H˙,G1′G1=−x2HH˙.
The relation in (52) becomes(56)R22G22=x2−R12H2.
The relation (56) defines the function G2 when(57)0<Hx<xR1bull;withr2+1<1R1.
Therefore, the relations in (56) and (57) shows that (52) is satisfied.
In order to find a functionλ in (53) not identically zero, we must solve. Now, it remains to show that (53) is satisfied, i.e.,(58)1v2−R12k12G1′2+R22k22G2′2>0,for v∈0,k1/R1r2+1.
By using (53) and (55), the inequality in (58) becomes (after an easy computation)(59)k22x2−R12H2x2−R12k12x4H˙2−x3−x2R12HH˙2>0.
One can show that (59) is satisfied when x<x1; x1=k2/1−R12r2+1+k22R12r2+1/k12. That is, v>1/x1.
Consequently, an open set ofℍ2 of the form ℝ×c,+∞c>0 can be immersed isometrically in ℝ4.



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*Source: 2012354-2021-10-28.xml*