Open access peer-reviewed chapter

# An Intrinsic Characterization of Bonnet Surfaces Based on a Closed Differential Ideal

By Paul Bracken

Submitted: November 21st 2016Reviewed: November 22nd 2016Published: January 18th 2017

DOI: 10.5772/67008

## Abstract

The structure equations for a two‐dimensional manifold are introduced and two results based on the Codazzi equations pertinent to the study of isometric surfaces are obtained from them. Important theorems pertaining to isometric surfaces are stated and a theorem due to Bonnet is obtained. A transformation for the connection forms is developed. It is proved that the angle of deformation must be harmonic, and that the differentials of many of the important variables generate a closed differential ideal. This implies that a coordinate system exists in which many of the variables satisfy particular ordinary differential equations, and these results can be used to characterize Bonnet surfaces.

### Keywords

• manifold
• differential form
• closed
• isometric
• differential equation
• Bonnet surface

## 1. Introduction

Bonnet surfaces in three‐dimensional Euclidean space have been of great interest for a number of reasons as a type of surface [1, 2] for a long time. Bonnet surfaces are of nonconstant mean curvature that admits infinitely many nontrivial and geometrically distinct isometries, which preserve the mean curvature function. Nontrivial isometries are ones that do not extend to isometries of the whole space E3. Considerable interest has resulted from the fact that the differential equations that describe the Gauss equations are classified by the type of related Painlevé equations they correspond to and they are integrated in terms of certain hypergeometric transcendents [35]. Here the approach first given by Chern [6] to Bonnet surfaces is considered. The development is accessible with many new proofs given. The main intention is to end by deriving an intrinsic characterization of these surfaces which indicates they are analytic. Moreover, it is shown that a type of Lax pair can be given for these surfaces and integrated. Several of the more important functions such as the mean curvature are seen to satisfy nontrivial ordinary differential equations.

Quite a lot is known about these surfaces. With many results the analysis is local and takes place under the assumptions that the surfaces contain no umbilic points and no critical points of the mean curvature function. The approach here allows the elimination of many assumptions and it is found the results are not too different from the known local ones. The statements and proofs have been given in great detail in order to help illustrate and display the interconnectedness of the ideas and results involved.

To establish some information about what is known, consider an oriented, connected, smooth open surface Min E3with nonconstant mean curvature function H. Moreover, suppose Madmits infinitely many nontrivial and geometrically distinct isometries preserving H. Suppose Uis the set of umbilic points of Mand Vthe set of critical points of H. Many global facts are known with regard to U,Vand H, and a few will now be mentioned. The set Uconsists of isolated points, even if there exists only one nontrivial isometry preserving the mean curvature, moreover, UV[7, 8]. Interestingly, there is no point in VUat which all order derivatives of Hare zero, and Vcannot contain any curve segment. If the function by which a nontrivial isometry preserving the mean curvature rotates the principal frame is considered, as when there are infinitely many isometries, this function is a global function on Mcontinuously defined [911]. As first noted by Chern [6], this function is harmonic. The analysis will begin by formulating the structure equations for two‐dimensional manifolds.

## 2. Structure equations

Over M, there exists a well‐defined field of orthonormal frames, which is written as x, e1,e2,e3such that xM, e3is the unit normal at x, and e1,e2are along principal directions [12]. The fundamental equations for Mhave the form

dx=ω1e1+ω2e2,de1=ω12e2+ω13e3,de2=ω12e1+ω23e3,de3=ω13e1ω23e2.E1

Differentiating each of these equations in turn, results in a large system of equations for the exterior derivatives of the ωiand ωij, as well as a final equation which relates some of the forms [13]. This choice of frame and Cartan's lemma allows for the introduction of the two principal curvatures which are denoted by aand cat xby writing

ω12=hω1+kω2,ω13=aω1,ω23=cω2.E2

Suppose that a>cin the following. The mean curvature of Mis denoted by Hand the Gaussian curvature by K. They are related to aand cas follows

H=12(a+c),K=ac.E3

The forms which appear in Eq. (1) satisfy the fundamental structure equations which are summarized here [14],

dω1=ω12ω2,dω2=ω1ω12dω13=ω12ω23dω23=ω13ω12,dω12=acω2ω1=Kω1ω2.E4

The second pair of equations of (4) is referred to as the Codazzi equation and the last equation is the Gauss equation.

Exterior differentiation of the two Codazzi equations yields

(da(ac)hω2)ω1=0,(dc(ac)kω1)ω2=0.E5

Cartan's lemma can be applied to the equations in (5). Thus, there exist two functions uand vsuch that

1acdahω2=(uk)ω1,1acdckω1=(vh)ω2.E6

Subtracting the pair of equations in (6) gives an expression for dlog(ac)

dlog(ac)=(u2k)ω1(v2h)ω2.E7

Define the variable Jto be

J=12(ac)>0.E8

It will appear frequently in what follows. Equation (7) then takes the form

dlogJ=(u2k)ω1(v2h)ω2.E9

The ωiconstitute a linearly independent set. Two related coframes called ϑiand αican be defined in terms of the ωiand the functions uand vas follows,

ϑ1=uω1+vω2,ϑ2=vω1+uω2,α1=uω1vω2,α2=vω1+uω2.E10

These relations imply that ϑ1=0is tangent to the level curves specified by Hequals constant and α1=0is its symmetry with respect to the principal directions.

Squaring both sides of the relation 2H=a+cand subtracting the relation 4K=4acyields 4(H2K)=(ac)2. The Hodge operator, denoted by *, will play an important role throughout. It produces the following result on the basis forms ωi,

*ω1=ω2,*ω2=ω1,*2=1.E11

Moreover, adding the expressions for daand dcgiven in Eq. (6), there results

1ac(da+dc)=(uk)ω1+hω2+(vk)ω2+kω1=uω1+vω2=ϑ1.E12

Finally, note that

α1+2*ω12=uω1vω2+2*(hω1+kω2)=(u2k)ω1(v2h)ω2=dlogJ.E13

Therefore, the Codazzi equations (12) and (13) can be summarized using the definitions of Hand Jas

dH=Jϑ1,dlogJ=α1+2*ω12.E14

## 3. A theorem of Bonnet

Suppose that M*is a surface which is isometric to Msuch that the principal curvatures are preserved [1012]. Denote all quantities which pertain to M*with the same symbols but with asterisks, as for example

a*=a,c*=c.

The same notation will be applied to the variables and forms which pertain to Mand M*. When Mand M*are isometric, the forms ωiare related to the ωi*by the following transformation

ω1*=cosτω1sinτω2,ω2*=sinτω1+cosτω2.E15

Theorem 3.1Under the transformation of coframe given by Eq. (15), the associated connection forms are related by

ω12*=ω12dτ.E16

Proof:Exterior differentiation of ω1*produces

dω1*=sinτdτω1+cosτdω1cosτdτω2sinτdω2=dτ(sinτω1cosτω2)+cosτω12ω2sinτω1ω12=(dτ+ω12)ω2*.

Similarly, differentiating ω2*gives

dω2*=cosτω1+sinτdω1sinτdτω2+cosτdω2=dτ(cosτω1sinτω2)+sinτω12ω2+cosτω1ω12=ω1*(dτ+ω12).

There is a very important result that can be developed at this point. In the case that a=a*and c=c*, the Codazzi equations imply that

α1+2*ω12=dlog(ac)=dlog(a*c*)=α1*+2*ω12*.

Apply the operator *to both sides of this equation, we obtain

α22ω12=α2*2ω12*.

Substituting for ω12*from Theorem 3.1, this is

2dτ=α2α2*.E17

Lemma 3.1

ϑ1=ϑ1*.

Proof:This can be shown in two ways. First from Eq. (15), express the ωiin terms of the ωi*

ω1=cosτω1*+sinτω2*,ω2=sinτω1*+cosτω2*.E18

Therefore,

ϑ1=uω1+vω2=u(cosτω1*+sinτω2*)+v(sinτω1*+cosτω2*)=u*ω1*+v*ω2*=ϑ1*,

where u*=ucosτvsinτand v*=usinτ+vcosτ.

Lemma 3.1 also follows from the fact that dH=dH*and Eq. (8).

Lemma 3.2

α2*=sin(2τ)α1+cos(2τ)α2.

Proof:

α2*=(usinτ+vcosτ)(cosτω1sinτω2)+(ucosτvsinτ)(sinτω1+cosτω2)=(usin(2τ)+vcos(2τ))ω1+(vsin(2τ)+ucos(2τ))ω2=sin(2τ)α1+cos(2τ)α2.

Substituting α2*from Lemma 3.2 into Eq. (13), dτcan be written as

dτ=12(α2sin(2τ)α1cos(2τ)α2)=12((1cos(2τ))α2sin(2τ)α1).E19

Introduce the new variable t=cot(τ)so dt=csc2(τ)dτand sinτ=11+t2, cosτ=11+t2, hence the following lemma.

Lemma 3.3

dt=tα1α2.

This is the total differential equation which must be satisfied by the angle τof rotation of the principal directions during the deformation. If the deformation is to be nontrivial, it must be that this equation is completely integrable.

Theorem 3.2A surface Madmits a nontrivial isometric deformation that keeps the principal curvatures fixed if and only if

dα1=0,dα2=α1α2.E20

Proof:Differentiating both sides of Lemma 3.3 gives

dtα1+tdα1dα2=(tα1α2)α1+tdα1dα2=0.E96

Equating the coefficients of tto zero gives the result (20).

This theorem seems to originate with Chern [6] and is very useful because it gives the exterior derivatives of the αi. When the mean curvature is constant, dH=0, hence it follows from Eq. (14) that ϑ1=0. This implies that u=v=0, and so α1and α2must vanish. Hence, dt=0which implies that, since the αiis linearly independent, tequals a constant. Thus, we arrive at a theorem originally due to Bonnet.

Theorem 3.3A surface of constant mean curvature can be isometrically deformed preserving the principal curvatures. During the deformation, the principal directions rotate by a fixed angle.

## 4. Connection form associated to a coframe and transformation properties

Given the linearly independent one forms ω1,ω2, the first two of the structure equations uniquely determine the form ω12. The ω1,ω2is called the orthonormal coframe of the metric

ds2=ω12+ω22,

and ω12is the connection form associated with it.

Theorem 4.1Suppose that A>0is a function on M. Under the change of coframe

ω1*=Aω1,ω2*=Aω2,E21

the associated connection forms are related by

ω12*=ω12+*dlogA.E22

Proof:The structure equations for the transformed system are given as

dω1*=ω12*ω2*,dω2*=ω1*ω12*.

Using Eq. (21) to replace the ωi*in these, we obtain

dlogAω1+dω1=ω12*ω2,dlogAω2+dω2=ω1ω12*.

The ωisatisfy a similar system of structure equations, so replacing dωihere yields

(ω12*ω12)ω2=dlogAω1,(ω12*ω12)ω1=dlogAω2.

Since the form ωisatisfies the equations *ω1=ω2and *ω2=ω1, substituting these relations into the above equations and using Ωk(*Θk)=Θk(*Ωk), we obtain that in the form

ω1*(ω12*ω12)=ω1dlogA,ω2*(ω12*ω12)=ω2dlogA.

Cartan's lemma can be used to conclude from these that there exist functions fand gsuch that

*(ω12*ω12)=dlogAfω1,*(ω12*ω12)=dlogA+gω2.

Finally, apply *to both sides and use *2=1to obtain

ω12*ω12=*dlogA+fω2,ω12*ω12=*dlogA+gω1.

The forms ωiare linearly independent, so for these two equations to be compatible, it suffices to put f=g=0, and the result follows.

For the necessity in the Chern criterion, Theorem 3.2, no mention of the set Vof critical points of His needed. In fact, when His constant, this criterion is met and the sufficiency also holds with τconstant. However, when His not identically constant, we need to take the set Vof critical points into account for the sufficiency. In this case, MVis also an open, dense, and connected subset of M. On this subset J>0and the function Acan be defined in terms of the functions uand vas

A=+u2+v2>0.E23

To define more general transformations of the ωi, define the angle ψas

u=Acos(ψ),v=Asin(ψ).E24

This angle, which is defined modulo 2π, is continuous only locally and could be discontinuous in a nonsimply connected region of MV. With Aand ψrelated to uand vby Eq. (24), the forms ϑiand αican be written in terms of Aand ψas

ϑ1=A(cos(ψ)ω1+sin(ψ)ω2),ϑ2=A(sin(ψ)ω1+cos(ψ)ω2),α1=A(cos(ψ)ω1sin(ψ)ω2),α2=A(sin(ψ)ω1+cos(ψ)ω2).E25

The forms ωi, ϑi, αidefine the same structure on Mand we let ω12, ϑ12, α12be the connection forms associated to the coframes ω1,ω2; ϑ1,ϑ2; α1,α2. The next theorem is crucial for what follows.

Theorem 4.2

ϑ12=dψ+ω12+*dlogA=2dψ+α12.E26

Proof:Each of the transformations which yield the ϑiand αiin the form (25) can be thought of as a composition of the two transformations which occur in the Theorems 3.1 and 4.1. First apply the transformation ωiAωiand τψwith ωi*ϑiin Eq. (15), we get the ϑiequations in Eq. (25). Invoking Theorems 3.1 and 4.1 in turn, the first result is obtained

ϑ12=dψ+ω12+*dlogA.

The transformation to the αiis exactly similar except that τψ, hence

α12=dψ+ω12+*dlogA.

This implies *dlogA=α12+dψω12. When replaced in the first equation of (26), the second equation appears. Note that from Theorem 3.2, α12=α2, so the second equation can be given as ϑ12=2dψ+α2.

Differentiating the second equation in Eq. (14) and using dα1=0, it follows that

d*ω12=0.E27

Lemma 4.1The angle ψis a harmonic function d*dψ=0and moreover, d*ϑ12=0.

Proof:From Theorem 4.2, it follows by applying *through Eq. (26) that

*ϑ12=*ω12+*dψdlogA=2*dψα1.E28

Exterior differentiation of this equation using d*ω12=0immediately gives

d*dψ=0.

This states that ψis a harmonic function. Equation (28) also implies that d*ϑ12=0.

## 5. Construction of the closed differential ideal associated with M

Exterior differentiation of the first equation in (14) and using the second equation produces

dϑ1+(α1+2*ω12)ϑ1=0.E29

The structure equation for the ϑiwill be needed,

dϑ1=ϑ12ϑ2=*ϑ12ϑ1.E30

From the second equation in Eq. (26), we have *ω12dlogA+α1=*dψ, and putting this in the first equation of Eq. (26), we find

*ϑ12+α1+2*ω12=2dlogA.E31

Using Eq. (31) in Eq. (30),

dϑ1+(α1+2*ω12)ϑ1=2dlogAϑ1.E32

Replacing dϑ1by means of Eq. (29) implies the following important result

dlogAϑ1=0.E33

Equation (33) and Cartan's lemma imply that there exists a function Bsuch that

dlogA=Bϑ1.E34

This is the first in a series of results which relates many of the variables in question such as J, B, and ϑ12directly to the one‐form ϑ1. To show this requires considerable work. The way to proceed is to use the forms αiin Theorem 3.2 because their exterior derivatives are known. For an arbitrary function on M, define

df=f1α1+f2α2.E35

Differentiating Eq. (35) and extracting the coefficient of α1α2, we obtain

f21f12+f2=0.E36

In terms of the αi, *dψ=ψ1α2ψ2α1, Lemma 4.1 yields

ψ11+ψ22+ψ1=0.E37

Finally, since *ϑ12=2*dψα1, substituting for *dψ, we obtain that

*ϑ12=(2ψ2+1)α1+2ψ1α2.E38

Differentiating structure equation (30) and using Lemma 4.1,

*ϑ12dϑ1=0,

so,

*ϑ12ϑ12ϑ2=0

This equation implies that either ϑ12or *ϑ12is a multiple by a function of the form ϑ2. Hence, for some function p,

ϑ12=pϑ2,*ϑ12=pϑ1,ϑ12=pϑ1,*ϑ12=pϑ2,E39

Substituting the first line of Eq. (39) back into the structure equation, we have

dϑ1=0.E40

The second line yields simply dϑ1=pϑ1ϑ2. Only the first case is examined now. Substituting Eq. (40) into Eq. (29), the following important constraint is obtained

(α1+2*ω12)ϑ1=0.E41

Theorem 5.1The function ψsatisfies the equation

2ψ1cos(2ψ)+(2ψ2+1)sin(2ψ)=0.E42

Proof:By substituting *dψinto Eq. (28) we have

*ϑ12=2*(ψ1α1+ψ2α2)α1=(2ψ2+1)α1+2ψ1α2.E43

Substituting Eq. (43) into Eq. (26) and solving for *ω12, we obtain that

*ω12=*ϑ12*dψ+dlogA=*ϑ12*dψ+Bϑ1=*dψα1+Bϑ1.

This can be put in the equivalent form

2*ω12+α1=2*dψα1+2Bϑ1.E44

Taking the exterior product with ϑ1and using dψ1, we get

(α1+2*ω12)ϑ1=(2*dψα1)ϑ1=(2ψ1*α1+2ψ2*α2α1)ϑ1=(2ψ1cos(2ψ)+(2ψ2+1)sin(2ψ))ϑ2ϑ1.

Imposing the constraint (41), the coefficient of ϑ1ϑ2can be equated to zero. This produces the result (42).

As a consequence of Theorem 5.1, a new function Ccan be introduced such that

2ψ1=Csin(2ψ),2ψ2+1=Ccos(2ψ).E45

Differentiation of each of these with respect to the αibasis, we get for i=1,2that

2ψ1i=Cisin(2ψ)+2ψiCcos(2ψ),2ψ2i=Cicos(2ψ)+2ψiCsin(2ψ).

Substituting f=ψinto Eq. (36) and using the fact that ψsatisfies Eq. (37) gives the pair of equations

C1cos(2ψ)C2sin(2ψ)+2ψ1Csin(2ψ)(2ψ2+1)Ccos(2ψ)1=0,C1sin(2ψ)C2cos(2ψ)+2ψ1Ccos(2ψ)+(2ψ2+1)Csin(2ψ)=0.

This linear system can be solved for C1and C2to get

C1+C(2ψ2+1)+cos(2ψ)=0,C22Cψ1+sin(2ψ)=0.E46

By differentiating each of the equations in (46), it is easy to verify that Csatisfies Eq. (36), namely, C12C21C2=0. Hence, there exist harmonic functions which satisfy Eq. (42). The solution depends on two arbitrary constants, the values of ψand Cat an initial point.

Lemma 5.1

dC=(C21)ϑ1,*ϑ12=Cϑ1.E47

Proof:It is easy to express the ϑiin terms of the αi,

ϑ1=cos(2ψ)α1+sin(2ψ)α2,ϑ2=sin(2ψ)α1+cos(2ψ)α2.E48

Therefore, using Eqs. (45) and (46), it is easy to see that

dC=C1α1+C2α2=(C21)(cos(2ψ)α1+sin(2ψ)α2)=(C21)ϑ1.

Using Eq. (45), it follows that

*ϑ12=(2ψ2+1)α1+2ψ1α2=Ccos(2ψ)α1+Csin(2ψ)α2=C(cos(2ψ)α1+sin(2ψ)α2)=Cϑ1.

This implies that ϑ12=Cϑ2.

It is possible to obtain formulas for B1,B2. Using Eq. (48) in Eq. (34), the derivatives of logAcan be written down

(logA)1=Bcos(2ψ),(logA)2=Bsin(2ψ).E49

Differentiating each of these in turn, we obtain for i=1,2,

(logA)1i=Bicos(2ψ)2Bψisin(2ψ),(logA)2i=Bisin(2ψ)+2Bψicos(2ψ).E50

Taking f=logAin Eq. (36) produces a first equation for the Bi,

B1sin(2ψ)+2Bψ1cos(2ψ)B2cos(2ψ)+2Bψ2sin(2ψ)+Bsin(2ψ)=0.E51

If another equation in terms of B1and B2can be found, it can be solved simultaneously with Eq. (51). There exists such an equation and it can be obtained from the Gauss equation in (4) which we put in the form

dω12=acω1ω2=acA2α1α2.

Solving Eq. (26) for ω12, we have

ω12=dψ+α2+(logA)2α1(logA)1α2.

The exterior derivative of this takes the form,

dω12=[1(logA)11(logA)22(logA)1]α1α2.

Putting this in the Gauss equation,

(logA)11(logA)22+{(logA)1+1}+acA2=0.

Replacing the second derivatives from Eq. (50), we have the required second equation

B1cos(2ψ)B2sin(2ψ)+B{2ψ1sin(2ψ)(2ψ2+1)cos(2ψ)}+1+acA2=0.E52

Solving Eqs. (51) and (52) together, the following expressions for B1and B2are obtained

B1+B(2ψ2+1)(1+acA2)cos(2ψ)=0,B22Bψ1(1+acA2)sin(2ψ)=0.E53

Given these results for B1and B2, it is easy to produce the following two Lemmas.

Lemma 5.2

dB=(BC+1+acA2)ϑ1,dlogJ=(C+2B)ϑ1.E54

Proof:Substituting Eq. (53) into dB, we get

dB=B1α1+B2α2=(BC+1+acA2)(cos(2ψ)α1+sin(2ψ)α2)=(BC+1+acA2)ϑ1.

Moreover,

dlogJ=α1+2*ω12=α1+2(*ϑ12*dψ+dlogA)=α1+2*ϑ122*dψ+2dlogA=*ϑ12+2dlogA=Cϑ1+2Bϑ1.

Lemma 5.3

dψ=12sin(2ψ)ϑ112(C+cos(2ψ))ϑ2.E55

Proof:

2dψ=2ψ1α1+2ψ2α2=Csin(2ψ)α1(Ccos(2ψ)+1)α2=Csin(2ψ)(cos(2ψ)ϑ1sin(2ψ)ϑ2)(Ccos(2ψ)+1)(sin(2ψ)ϑ1+cos(2ψ)ϑ2)=sin(2ψ)ϑ1(C+cos(2ψ))ϑ2.

In the interests of completeness, it is important to verify the following theorem.

Theorem 5.2The function Bsatisfies Eq. (36) provided ψsatisfies both Eqs. (37) and (41).

Proof:Differentiating B1and B2given by Eq. (53), the left side of Eq. (36) is found to be

B21B12+B2=2B1ψ1+B2(2ψ2+1)+2B(ψ11+ψ22+ψ1)+A2((ac)1sin(2ψ)(ac)2sin(2ψ))2acBA2(cos(2ψ)sin(2ψ)sin(2ψ)cos(2ψ))+(1+acA2)(2ψ1cos(2ψ)+(2ψ2+1)sin(2ψ))=2(1+acA2)(2ψ1cos(2ψ)+(2ψ2+1)sin(2ψ))+A2((ac)1sin(2ψ)(ac)2cos(2ψ)).

To simplify this, Eq. (37) has been substituted. Using Eq. (48) and *d(ac)=(ac)1α2(ac)2α1, it follows that

*d(ac)ϑ2=((ac)1sin(2ψ)(ac)2cos(2ψ))α1α2.

Note that the coefficient of α1α2in this appears in the compatibility condition. To express it in another way, begin by finding the exterior derivative of 4ac=(a+c)2(ac)2,

4d(ac)=2(a+c)(ac)ϑ12(ac)2(α1+2*ω12).

Applying the Hodge operator to both sides of this, gives upon rearranging terms

2*d(ac)ac=(a+c)ϑ2(ac)(α22ω12).

Consequently, we can write

2(ac)2*d(ac)ϑ2=(α22ω12)ϑ2=(2ψ1cos(2ψ)+(2ψ2+1)sin(2ψ))α1α2.

Therefore, it must be that

(ac)1sin(2ψ)+(ac)2cos(2ψ)=12(ac)2(2ψ1cos(2ψ)+(2ψ2+1)sin(2ψ)).

It follows that when f=B, Eq. (36) finally reduces to the form

(1+H2A2)[2ψ1cos(2ψ)+(2ψ2+1)sin(2ψ)]=0.

The first factor is clearly nonzero, so the second factor must vanish. This of course is equivalent to the constraint (41).

## 6. Intrinsic characterization of M

During the prolongation of the exterior differential system, the additional variables ψ, A, B, and Chave been introduced. The significance of the appearance of the function C, is that the process terminates and the differentials of all these functions can be computed without the need to introduce more functions. This means that the exterior differential system has finally closed.

The results of the previous section, in particular, the lemmas, can be collected such that they justify the following.

Proposition 6.1The differential system generated in terms of the differentials of the variables ψ, A, B, and Cis closed. The variables H,J,A,B,Cremain constant along the ϑ2‐curves so ϑ1=0. Hence, an isometry that preserves Hmust map the ϑ1, ϑ2curves onto the corresponding ϑ1*, ϑ2*curves of the associated surface M*which is isometric to M.

Along the ϑ1, ϑ2curves, consider the normalized frame,

ζ1=cos(ψ)e1+sin(ψ)e2,ζ2=sin(ψ)e1+cos(ψ)e2.E56

The corresponding coframe and connection form are

ξ1=cos(ψ)ω1+sin(ψ)ω2,ξ2=sin(ψ)ω1+cos(ψ)ω2,ξ12=dψ+ω12.E57

Then ϑ1can be expressed as a multiple of ξ1and ϑ2,ϑ12in terms of ξ2, and the differential system can be summarized here:

ϑ1=Aξ1,ϑ2=Aξ2,ϑ12=ξ12+*dlogA=CAξ2,dlogA=ABξ1,dB=A(BC+1+acA2)ξ1,dC=A(C21)ξ1,dH=AJξ1,dJ=AJ(2B+C)ξ1.E58

The condition dϑ1=0is equivalent to

This implies that dξ1=0since dAis proportional to ξ1. Also, d*ϑ12=0is equivalent to d*ξ12=0.

Moreover, d*ξ12=0is equivalent to the fact that the ξ1,ξ2curves can be regarded as coordinate curves parameterized by isothermal parameters. Therefore, along the ξ1,ξ2curves, orthogonal isothermal coordinates denoted (s,t)can be introduced. The first fundamental form of Mthen takes the form,

I=ξ12+ξ22=E(s)(ds2+dt2).E59

Now suppose we set e(s)=E(s), then

ξ1=e(s)ds,ξ2=e(s)dt,ξ12=e(s)e2(s)ξ2=e(s)e(s)dt.E60

This means such a surface is isometric to a surface of revolution. Since ψ, d*ξ12=0, Eq. (57) implies that d*ω12=0. This can be stated otherwise as the principal coordinates are isothermal and so Mis an isothermic surface.

Since A,B,C,H, and Jare functions of only the variable s, this implies that Hand J, or Hand K, are constant along the tcurves where sis constant. This leads to the following proposition.

Proposition 6.2

dHdK=0,ξ12=(C+B)Aξ2.E61

This is equivalent to the statement Mis a Weingarten surface.

Proof:The first result follows from the statement about the coordinate system above. Since ϑ12=ξ12+*dlogA=CAξ2and dA=A2Bξ1,

ξ12=CAξ2*dlogA=CAξ2*A1dA=CAξ2AB*ξ1=(C+B)Aξ2

Consequently, the geodesic curvature of each ξ2curve, sconstant, is

e(s)e2(s)=A(B+C),

which is constant.

To express the ωiin terms of dsand dt, start by writing ωiin terms of the ξiand then substituting Eq. (60),

ω1=cos(ψ)edssin(ψ)edt,ω2=sin(ψ)eds+cos(ψ)edt.E62

Subscripts (s,t)denote differentiation and Hs=His used interchangeably. Beginning with dH=Hdsand using Eq. (62), we have

dH=H1ω1+H2ω2=(H1cos(ψ)+H2sin(ψ))eds+(H1sin(ψ)+H2cos(ψ))edt=Hds.

Equating coefficients of differentials, this implies that

H1ecos(ψ)+H2esin(ψ)=H,H1sin(ψ)+H2cos(ψ)=0.

Solving this as a linear system we obtain H1, H2,

H1=Hecos(ψ),H2=Hesin(ψ).E63

Noting that u=H1/Jand v=H2/J, using Eq. (57) the forms αican be expressed in terms of ds,dt

α1=HJ(cos(2ψ)dssin(2ψ)dt),α2=HJ(sin(2ψ)ds+cos(2ψ)dt).E64

Substituting ξ1from Eq. (60) into dH=AJξ1,

dH=Hds=AJξ1=AJe(s)ds.

Therefore, H=AJe>0and so H(s)is an increasing function of s. Now define the function Q(s)to be

Q=HJ=Ae>0.E65

Substituting Eq. (65) into Eq. (64), αiis expressed in terms of Qas well. Equations (20) in Theorem 3.2 can easily be expressed in terms of ψand Q.

Theorem 6.1Equation (20) is equivalent to the following system of coupled equations in ψand Q:

sin(2ψ)(log(Q))s+2cos(2ψ)ψs2sin(2ψ)ψt=0,cos(2ψ)(log(Q))s2sin(2ψ)ψs2cos(2ψ)ψt=Q.E66

Moreover, Eq. (66) is equivalent to the following first‐order system

ψs=12Qsin(2ψ),ψt=12(log(Q))s12Qcos(2ψ).E67

System (67) can be thought of as a type of Lax pair. Moreover, Eq. (67) implies that ψis harmonic as well. Differentiating ψswith respect to sand ψtwith respect to t, it is clear that ψsatisfies Laplace's equation in the (s,t)variables ψss+ψtt=0. This is another proof that ψis harmonic.

Theorem 6.2The function Q(s)satisfies the following second‐order nonlinear differential equation

Q(s)Q(s)(Q(s))2=Q4(s).E68

There exists a first integral for this equation of the following form

Q(s)2=Q(s)4+κQ(s)2,κR.E69

Proof:Equation (68) is just the compatibility condition for the first‐order system (67). The required derivatives are

ψst=Q2cos(2ψ)((logQ)sQcos(2ψ)),ψts=12(logQ)ss12Qscos(2ψ)+Qsin(2ψ)ψs.E135

Equating derivatives ψst=ψts, the required (68) follows.

Differentiating both sides of Eq. (69) we get

Q(s)=2Q(s)3+κQ(s).E70

Isolating κQ(s)from Eq. (69) and substituting it into Eq. (70), Eq. (68) appears.

It is important to note that the function Cwhich appears when the differential ideal closes can be related to the function Q.

Corollary 6.1

C=(1Q).E71

Proof:Using ϑifrom Eq. (58) in Lemma 5.3, in the s,tcoordinates

2dψ=sin(2ψ)Aeds(C+cos(2ψ))Aedt=ψsds+ψtdt

Hence using Eq. (67), this implies that 2ψs=sin(2ψ)Ae=Qsin(2ψ), hence Q=Ae. The second equation in Eq. (67) for ψtimplies that (C+cos(2ψ))Ae=Qcos(2ψ)(logQ). Replacing Ae=Q, this simplifies to the form (71).

## 7. Integrating the Lax pair system

It is clear that the first‐order equation in (67) for Q(s)is separable and can be integrated. The integral depends on whether Kis zero or nonzero:

Q(s)=1εs+γ,K=0;log(2(K+KQ2+K)Q)=εKs+γ,K0.E72

Here ε=±1and γis the last constant of integration. Taking specific choices for the constants, for example, eγ=2Kwhen K0and a=K, the set of solutions (72) for Q(s)can be summarized below.

Dom(s)Q(s)Dom(s)Q(s)s>01ss<01s0<s<πaasin(as)πa<s<0asin(as)s>0asinh(as)s<0asinh(as)E73

It is presumed that other choices of the constants can be geometrically eliminated in favor of Eq. (73). The solutions (73) are then substituted back into linear system (67). The first equation in (67) implies that either

ψ0,modπ2;2ψssin(2ψ)=Q.E74

Substitute ψ0into the second equation in (67). It implies that (logQ)s=Qand ψ=π/2gives (logQ)s=Q. In both cases Q(s)is a solution which already appears in Eq. (73).

For the second case in Eq. (74), the equation can be put in the form

(log|tan(ψ)|)s=Q.

Integrating we have for some function y(t)to be determined,

tan(ψ)=eQ(s)dsy(t).E75

Therefore, tan(ψ)can be obtained by substituting for Q(s)for each of the three cases in Eq. (73). The upper sign holds for s>0and the lower sign holds if s<0.

1. Q(s)=±s1, Q(s)ds=log|s|and

tan(ψ)=sy(t).E76

• Q(s)=±asin(as), Q(s)ds=log|csc(as)cot(as)|and

tan(ψ)=(tan(as2))y(t).E77

• Q(s)=±asinh(as), Q(s)ds=arctanh(eas), and

tan(ψ)=(tanh(as2))y(t).E78

• In case (ii), if s>0and y(t)=±1then ψ=±12(as+π), modπ, and if s<0and y(t)=±1, then ψ=±12as, modπ.

It remains to integrate the second equation of the Lax pair (67) using solutions for both Q(s)and tan(ψ). The first case (i)is not hard and will be shown explicitly here. The others can be done, and more complicated cases are considered in the Appendix.

(i)Consider Q(s)=s1and tan(ψ)=s1y(t). The second equation in (67) simplifies considerably to yt=1, therefore,

y(t)=(t+σ),tan(ψ)=(t+σ)s.E79

For Q(s)=s1and tan(ψ)=sy(t), the second equation of (67) becomes yt=y2, therefore,

y(t)=1t+σ,tan(ψ)=st+σ.E80

## 8. A third‐order equation for Hand fundamental forms

Since ξ12=(loge(s))dt, using Eq. (60) ω12can be written as

ω12=ξ12dψ=(loge(s))dtdψ.E81

Using Eqs. (14) and (64) for α1, it follows that

dlog(J)=Q(cos(2ψ)dssin(2ψ)dt)2*(ψtdt+ψsds)+2*(log(e(s)))dt.

when ωiare put in the s,tcoordinates, using *ω1=ω2, it can be stated that *ds=dtand *dt=ds. Consequently, dlog(J)simplifies to

dlog(J)=(Qcos(2ψ)+2ψt2(log(e(s))))ds+(Qsin(2ψ)2ψs)dt.E82

First‐order system (67) permits this to be written using e(s)=E(s)as

(log(J))+(log(E))=(log(Q)).E83

Hence, there exists a constant τindependent of ssuch that EJ=τQor

E=τQJ=τQ2H.E84

This result (84) for Eis substituted into the Gauss equation −((log(E))ss+(log(E))tt)=2E(H2J2) giving

(log(E))=2(log(Q))(log(Hs))=2Q2(HH).E85

Therefore, the Gauss equation transforms into a third‐order differential equation in the svariable,

(HH)+2τH=2Q2(1+τH2H).E86

Thus, a characterization of Bonnet surfaces is reached by means of the solutions to these equations. This equation determines the function H(s)and after that the functions J(s)and E(s). Therefore, Bonnet surfaces have as first fundamental form the expression

I=E(s)(ds2+dt2),E(s)=τQ2(s)H(s).E87

Since ψis the angle from the principal axis e1to the scurve with tequals constant, the second fundamental form is given by

II=Lds2+2Mdsdt+Ndt2.E88

where the coefficients L,M,Nare given by

L=E(H+Jcos(2ψ))=EH+τQcos(2ψ),M=EJsin(2ψ)=τQsin(2ψ),N=E(HJcos(2ψ)).E89

Appendix

It is worth seeing how the second equation in (67) can be integrated for cases (ii) and (iii). Only the case s>0will be done with Q(s)taken from Eq. (73).

(a)Differentiating tan(ψ)given in Eq. (77), we obtain that

ψt=tan(as2)tan2(as2)+y2yt(t).

The following identities are required to simplify the result,

tan(as)=2tan(as2)1tan2(as2),cos(2ψ)=tan2(as2)y2tan2(as2)+y2.

Substituting ψtinto Eq. (67), we obtain

2tan(as2)tan2(as2)+y2yt=acot(as)asin(as)tan2(as2)y2tan2(as2)+y2.

Simplifying this, we get

4ayt=12(1tan2(as2))12(cot2(as2)1)y2sec2(as2)+csc2(as2)y2.

This simplifies to the elementary equation,

yt=a2(y21),y(t)=tanh(at2+η).

Here ηis an integration constant. To summarize then,

tan(ψ)=tanh(at2+η)tan(as+π2).

(b)Consider now s>0and take Q(s)from the last line of Eq. (73). Differentiating tan(ψ)from (78), we get

ψt=coth(as2)1+coth2(as2)y2yt(t).

In this case, the following identities are needed,

tanh(as)=2tanh(as2)1+tanh2(as2),cos(2ψ)=1coth2(as2)y21+coth2(as2)y2.

Therefore, Eq. (67) becomes

2coth(as2)1+coth2(as2)y2yt=acoth(as)asinh(as)tanh2(as2)y2tanh2(as2)+y2.

This reduces to

4ayt=(1+tanh2(as2)+sech2(as2))+(coth2(as2)+1csch2(as2))y2.

Simplifying and integrating, it has been found that

yt=a2(1+y2),y(t)=tan(at2+η).

To summarize then, it has been shown that,

tan(ψ)=cot(at2+η)coth(as2).

These results apply to the case s>0and similar results can be found for the case s<0as well.

MSCs: 53A05, 58A10, 53B05

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Paul Bracken (January 18th 2017). An Intrinsic Characterization of Bonnet Surfaces Based on a Closed Differential Ideal, Manifolds - Current Research Areas, Paul Bracken, IntechOpen, DOI: 10.5772/67008. Available from:

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