6. Intrinsic characterization of M

During the prolongation of the exterior differential system, the additional variables ψ, A, B, and C have 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 6.1 The differential system generated in terms of the differentials of the variables ψ, A, B, and C is closed. The variables H; J; A; B;C remain constant along the ϑ2-curves so ϑ<sup>1</sup> ¼ 0. Hence, an isometry that preserves H must map the ϑ1, ϑ<sup>2</sup> curves onto the corresponding ϑ� <sup>1</sup>, ϑ� <sup>2</sup> curves of the associated surface M� which is isometric to M.

Along the ϑ1, ϑ<sup>2</sup> curves, consider the normalized frame,

$$\zeta\_1 = \cos\left(\psi\right)e\_1 + \sin\left(\psi\right)e\_2, \qquad \zeta\_2 = -\sin\left(\psi\right)e\_1 + \cos\left(\psi\right)e\_2. \tag{56}$$

The corresponding coframe and connection form are

$$\xi\_1 = \cos\left(\psi\right)\omega\_1 + \sin\left(\psi\right)\omega\_2, \quad \xi\_2 = -\sin\left(\psi\right)\omega\_1 + \cos\left(\psi\right)\omega\_2, \quad \xi\_{12} = d\psi + \omega\_{12}.\tag{S7}$$

Then ϑ<sup>1</sup> can be expressed as a multiple of ξ<sup>1</sup> and ϑ2;ϑ<sup>12</sup> in terms of ξ2, and the differential system can be summarized here:

$$\begin{array}{ll} \mathfrak{G}\_{1} = A\xi\_{1}, & \mathfrak{G}\_{2} = A\xi\_{2}, & \mathfrak{G}\_{12} = \xi\_{12} + \*d\log A = -\complement A\xi\_{2}, \\\ d\log A = AB\xi\_{1}, & dB = A(BC + 1 + acA^{-2})\xi\_{1}, & d\mathbb{C} = A(\mathbb{C}^{2} - 1)\xi\_{1}, \\\ d H = A\{\xi\_{1}, \quad d\} = A\{(2B + \mathbb{C})\xi\_{1}. \end{array} \tag{58}$$

The condition dϑ<sup>1</sup> ¼ 0 is equivalent to

$$dA \wedge \xi\_1 + Ad\xi\_1 = 0.$$

This implies that dξ<sup>1</sup> ¼ 0 since dA is proportional to ξ1. Also, d � ϑ<sup>12</sup> ¼ 0 is equivalent to d � ξ<sup>12</sup> ¼ 0.

Moreover, d � ξ<sup>12</sup> ¼ 0 is equivalent to the fact that the ξ1;ξ<sup>2</sup> curves can be regarded as coordinate curves parameterized by isothermal parameters. Therefore, along the ξ1;ξ<sup>2</sup> curves, orthogonal isothermal coordinates denoted ðs;tÞ can be introduced. The first fundamental form of M then takes the form,

$$I = \xi\_1^2 + \xi\_2^2 = E(s)(ds^2 + dt^2). \tag{59}$$

Now suppose we set <sup>e</sup>ðsÞ ¼ ffiffiffiffiffiffiffiffiffi <sup>E</sup>ðs<sup>Þ</sup> <sup>p</sup> , then

$$\xi\_1 = e(\mathbf{s}) \, d\mathbf{s}, \qquad \xi\_2 = e(\mathbf{s}) \, dt, \qquad \xi\_{12} = \frac{e'(\mathbf{s})}{e^2(\mathbf{s})} \xi\_2 = \frac{e'(\mathbf{s})}{e(\mathbf{s})} \, dt. \tag{60}$$

This means such a surface is isometric to a surface of revolution. Since ψ, d� ξ<sup>12</sup> ¼ 0, Eq. (57) implies that d� ω<sup>12</sup> ¼ 0. This can be stated otherwise as the principal coordinates are isothermal and so M is an isothermic surface.

Since A; B;C; H, and J are functions of only the variable s, this implies that H and J, or H and K, are constant along the t curves where s is constant. This leads to the following proposition.

### Proposition 6.2

−ðacÞ<sup>1</sup> sin ð2ψÞþðacÞ<sup>2</sup> cos ð2ψÞ ¼ −

<sup>ð</sup><sup>1</sup> <sup>þ</sup> <sup>H</sup><sup>2</sup>

6. Intrinsic characterization of M

<sup>1</sup>, ϑ�

system can be summarized here:

The condition dϑ<sup>1</sup> ¼ 0 is equivalent to

Along the ϑ1, ϑ<sup>2</sup> curves, consider the normalized frame,

The corresponding coframe and connection form are

<sup>d</sup>log<sup>A</sup> <sup>¼</sup> ABξ1; dB <sup>¼</sup> <sup>A</sup>ðBC <sup>þ</sup> <sup>1</sup> <sup>þ</sup> acA<sup>−</sup><sup>2</sup>

to the constraint (41).

34 Manifolds - Current Research Areas

justify the following.

corresponding ϑ�

d � ξ<sup>12</sup> ¼ 0.

It follows that when f ¼ B, Eq. (36) finally reduces to the form

A<sup>−</sup><sup>2</sup>

1 2 ða−cÞ 2

Þ½2ψ<sup>1</sup> cos ð2ψÞþð2ψ<sup>2</sup> þ 1Þ sin ð2ψÞ� ¼ 0:

The first factor is clearly nonzero, so the second factor must vanish. This of course is equivalent

During the prolongation of the exterior differential system, the additional variables ψ, A, B, and C have 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

Proposition 6.1 The differential system generated in terms of the differentials of the variables ψ, A, B, and C is closed. The variables H; J; A; B;C remain constant along the ϑ2-curves so ϑ<sup>1</sup> ¼ 0. Hence, an isometry that preserves H must map the ϑ1, ϑ<sup>2</sup> curves onto the

<sup>2</sup> curves of the associated surface M� which is isometric to M.

ξ<sup>1</sup> ¼ cos ðψÞω<sup>1</sup> þ sin ðψÞω2; ξ<sup>2</sup> ¼ − sin ðψÞω<sup>1</sup> þ cos ðψÞω2; ξ<sup>12</sup> ¼ dψ þ ω12: (57)

Then ϑ<sup>1</sup> can be expressed as a multiple of ξ<sup>1</sup> and ϑ2;ϑ<sup>12</sup> in terms of ξ2, and the differential

ϑ<sup>1</sup> ¼ Aξ1; ϑ<sup>2</sup> ¼ Aξ2; ϑ<sup>12</sup> ¼ ξ<sup>12</sup> þ �d logA ¼ −CAξ2;

dH ¼ AJξ1; dJ ¼ AJð2B þ CÞξ1:

dA∧ξ<sup>1</sup> þ Adξ<sup>1</sup> ¼ 0:

This implies that dξ<sup>1</sup> ¼ 0 since dA is proportional to ξ1. Also, d � ϑ<sup>12</sup> ¼ 0 is equivalent to

ζ<sup>1</sup> ¼ cos ðψÞe<sup>1</sup> þ sin ðψÞe2; ζ<sup>2</sup> ¼ − sin ðψÞe<sup>1</sup> þ cos ðψÞe2: (56)

<sup>Þ</sup>ξ1; dC <sup>¼</sup> <sup>A</sup>ðC<sup>2</sup>

−1Þξ1;

(58)

ð2ψ<sup>1</sup> cos ð2ψÞþð2ψ<sup>2</sup> þ 1Þ sin ð2ψÞÞ:

$$dH \wedge dK = 0, \qquad \xi\_{12} = -(C+B)A\xi\_2. \tag{61}$$

This is equivalent to the statement M is a Weingarten surface.

Proof: The first result follows from the statement about the coordinate system above. Since <sup>ϑ</sup><sup>12</sup> <sup>¼</sup> <sup>ξ</sup><sup>12</sup> þ �dlog<sup>A</sup> <sup>¼</sup> <sup>−</sup>CAξ<sup>2</sup> and dA <sup>¼</sup> <sup>A</sup><sup>2</sup> Bξ1,

$$d\xi\_{12} = -\mathbb{C}A\xi\_2 - \*d\log A = -\mathbb{C}A\xi\_2 - \*A^{-1}dA = -\mathbb{C}A\xi\_2 - AB\*\xi\_1 = -(\mathbb{C}+B)A\xi\_2$$

Consequently, the geodesic curvature of each ξ<sup>2</sup> curve, s constant, is

$$\frac{e'(s)}{e^2(s)} = -A(B+C),$$

which is constant.

To express the ω<sup>i</sup> in terms of ds and dt, start by writing ω<sup>i</sup> in terms of the ξ<sup>i</sup> and then substituting Eq. (60),

$$
\omega\_1 = \cos\left(\psi\right)e\,ds - \sin\left(\psi\right)e\,dt,\qquad \omega\_2 = \sin\left(\psi\right)e\,ds + \cos\left(\psi\right)e\,dt.\tag{62}
$$

Subscripts <sup>ð</sup>s;t<sup>Þ</sup> denote differentiation and Hs <sup>¼</sup> <sup>H</sup>′ is used interchangeably. Beginning with dH <sup>¼</sup> <sup>H</sup>′ ds and using Eq. (62), we have

$$dH = H\_1 \omega\_1 + H\_2 \omega\_2 = \left( H\_1 \cos\left(\psi\right) + H\_2 \sin\left(\psi\right) \right) \varepsilon \, ds + \left( -H\_1 \sin\left(\psi\right) + H\_2 \cos\left(\psi\right) \right) \varepsilon \, dt = H' \, ds.$$

Equating coefficients of differentials, this implies that

$$H\_1 e \cos\left(\psi\right) + H\_2 e \sin\left(\psi\right) = H', \qquad -H\_1 \sin\left(\psi\right) + H\_2 \cos\left(\psi\right) = 0.$$

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

$$H\_1 = \frac{H'}{\mathcal{e}} \cos \left( \psi \right), \qquad H\_2 = \frac{H'}{\mathcal{e}} \sin \left( \psi \right). \tag{63}$$

Noting that u ¼ H1=J and v ¼ H2=J, using Eq. (57) the forms α<sup>i</sup> can be expressed in terms of ds;dt

$$a\_1 = \frac{H'}{J}(\cos(2\psi)\,ds - \sin(2\psi)\,dt), \qquad a\_2 = \frac{H'}{J}(\sin(2\psi)\,ds + \cos(2\psi)\,dt). \tag{64}$$

Substituting ξ<sup>1</sup> from Eq. (60) into dH ¼ AJξ1,

$$dH = H^{'}ds = A f \xi\_1 = A f \ e(s) \ ds.$$

Therefore, <sup>H</sup>′ <sup>¼</sup> AJe <sup>&</sup>gt; 0 and so <sup>H</sup>ðs<sup>Þ</sup> is an increasing function of <sup>s</sup>. Now define the function QðsÞ to be

$$Q = \frac{H}{J} = A \cdot e > 0.\tag{65}$$

Substituting Eq. (65) into Eq. (64), α<sup>i</sup> is expressed in terms of Q as well. Equations (20) in Theorem 3.2 can easily be expressed in terms of ψ and Q.

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

$$\begin{aligned} \sin\left(2\psi\right)(\log(Q))\_s + 2\cos\left(2\psi\right)\psi\_s - 2\sin\left(2\psi\right)\psi\_t &= 0, \\ \cos\left(2\psi\right)(\log(Q))\_s - 2\sin\left(2\psi\right)\psi\_s - 2\cos\left(2\psi\right)\psi\_t &= Q. \end{aligned} \tag{66}$$

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

$$
\psi\_s = -\frac{1}{2}Q\sin\left(2\psi\right), \qquad \psi\_t = \frac{1}{2}(\log(Q))\_s - \frac{1}{2}Q\cos\left(2\psi\right). \tag{67}
$$

System (67) can be thought of as a type of Lax pair. Moreover, Eq. (67) implies that ψ is harmonic as well. Differentiating ψ<sup>s</sup> with respect to s and ψ<sup>t</sup> with respect to t, it is clear that ψ satisfies Laplace's equation in the ðs;tÞ variables ψss þ ψtt ¼ 0. This is another proof that ψ is harmonic.

Theorem 6.2 The function QðsÞ satisfies the following second-order nonlinear differential equation

An Intrinsic Characterization of Bonnet Surfaces Based on a Closed Differential Ideal http://dx.doi.org/10.5772/67008 37

$$Q''(s)Q(s) - (Q'(s))^2 = Q^4(s). \tag{68}$$

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

$$\left(Q^{'}(\mathbf{s})^{2} = Q(\mathbf{s})^{4} + \kappa Q(\mathbf{s})^{2}\right), \quad \kappa \in \mathbb{R}. \tag{69}$$

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

$$\psi\_{st} = -\frac{Q}{2}\cos\left(2\psi\right)((\log Q)\_s - Q\cos\left(2\psi\right)), \quad \psi\_{ts} = \frac{1}{2}(\log Q)\_{ss} - \frac{1}{2}Q\_s\cos\left(2\psi\right) + Q\sin\left(2\psi\right)\psi\_s.$$

Equating derivatives ψst ¼ ψts, the required (68) follows.

Differentiating both sides of Eq. (69) we get

$$Q''(\mathbf{s}) = 2Q(\mathbf{s})^3 + \kappa Q(\mathbf{s}).\tag{70}$$

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

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

#### Corollary 6.1

<sup>H</sup>1<sup>e</sup> cos <sup>ð</sup>ψÞ þ <sup>H</sup>2<sup>e</sup> sin <sup>ð</sup>ψÞ ¼ <sup>H</sup>′

<sup>H</sup><sup>1</sup> <sup>¼</sup> <sup>H</sup>′ e

<sup>J</sup> <sup>ð</sup> cos <sup>ð</sup>2ψ<sup>Þ</sup> ds<sup>−</sup> sin <sup>ð</sup>2ψ<sup>Þ</sup> dtÞ; <sup>α</sup><sup>2</sup> <sup>¼</sup> <sup>H</sup>′

dH <sup>¼</sup> <sup>H</sup>′

Theorem 3.2 can easily be expressed in terms of ψ and Q.

ψ<sup>s</sup> ¼ − 1 2

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

<sup>Q</sup> sin <sup>ð</sup>2ψÞ; <sup>ψ</sup><sup>t</sup> <sup>¼</sup> <sup>1</sup>

cos <sup>ð</sup>ψÞ; <sup>H</sup><sup>2</sup> <sup>¼</sup> <sup>H</sup>′

Noting that u ¼ H1=J and v ¼ H2=J, using Eq. (57) the forms α<sup>i</sup> can be expressed in terms of

ds ¼ AJξ<sup>1</sup> ¼ AJ eðsÞ ds:

Therefore, <sup>H</sup>′ <sup>¼</sup> AJe <sup>&</sup>gt; 0 and so <sup>H</sup>ðs<sup>Þ</sup> is an increasing function of <sup>s</sup>. Now define the function

Substituting Eq. (65) into Eq. (64), α<sup>i</sup> is expressed in terms of Q as well. Equations (20) in

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

sin ð2ψÞðlogðQÞÞ<sup>s</sup> þ 2 cos ð2ψÞψs−2 sin ð2ψÞψ<sup>t</sup> ¼ 0;

2

System (67) can be thought of as a type of Lax pair. Moreover, Eq. (67) implies that ψ is harmonic as well. Differentiating ψ<sup>s</sup> with respect to s and ψ<sup>t</sup> with respect to t, it is clear that ψ satisfies Laplace's equation in the ðs;tÞ variables ψss þ ψtt ¼ 0. This is another proof that ψ is

Theorem 6.2 The function QðsÞ satisfies the following second-order nonlinear differential

cos <sup>ð</sup>2ψÞðlogðQÞÞs−2 sin <sup>ð</sup>2ψÞψs−2 cos <sup>ð</sup>2ψÞψ<sup>t</sup> <sup>¼</sup> <sup>Q</sup>: (66)

1 2

Q cos ð2ψÞ: (67)

ðlogðQÞÞs−

<sup>Q</sup> <sup>¼</sup> <sup>H</sup>′

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

Substituting ξ<sup>1</sup> from Eq. (60) into dH ¼ AJξ1,

ds;dt

QðsÞ to be

harmonic.

equation

<sup>α</sup><sup>1</sup> <sup>¼</sup> <sup>H</sup>′

36 Manifolds - Current Research Areas

; −H<sup>1</sup> sin ðψÞ þ H<sup>2</sup> cos ðψÞ ¼ 0:

sin ðψÞ: (63)

<sup>J</sup> <sup>ð</sup> sin <sup>ð</sup>2ψ<sup>Þ</sup> ds <sup>þ</sup> cos <sup>ð</sup>2ψ<sup>Þ</sup> dtÞ: (64)

<sup>J</sup> <sup>¼</sup> <sup>A</sup> � <sup>e</sup> <sup>&</sup>gt; <sup>0</sup>: (65)

e

$$C = \left(\frac{1}{Q}\right)^{\cdot}.\tag{71}$$

Proof: Using ϑ<sup>i</sup> from Eq. (58) in Lemma 5.3, in the s;t coordinates

$$2d\psi = -\sin\left(2\psi\right)Ae\,d\mathfrak{s} - \left(\mathbb{C} + \cos\left(2\psi\right)\right)Ae\,dt = \psi\_s ds + \psi\_t \,dt$$

Hence using Eq. (67), this implies that 2ψ<sup>s</sup> ¼ − sin ð2ψÞ Ae ¼ −Q sin ð2ψÞ, hence Q ¼ Ae. The second equation in Eq. (67) for ψ<sup>t</sup> implies that ðC þ cos ð2ψÞÞ Ae ¼ Q cos ð2ψÞ−ðlogQÞ ′ . Replacing Ae ¼ Q, this simplifies to the form (71).
