5. Construction of the closed differential ideal associated with M

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

$$d\\$\_1 + (\alpha\_1 + \mathcal{Q} \* \omega\_{12}) \land \S\_1 = 0.\tag{29}$$

The structure equation for the ϑ<sup>i</sup> will be needed,

$$d\theta\_1 = \vartheta\_{12} \land \theta\_2 = -\*\vartheta\_{12} \land \vartheta\_1. \tag{30}$$

From the second equation in Eq. (26), we have �ω12−d logA þ α<sup>1</sup> ¼ �dψ, and putting this in the first equation of Eq. (26), we find

$$-\*\,\mathfrak{d}\_{12} + \mathfrak{a}\_1 + \mathfrak{d}\_{12}\*\omega\_{12} = \mathbf{2}\,d\,\log A.\tag{31}$$

Using Eq. (31) in Eq. (30),

ϑ<sup>1</sup> ¼ Að cos ðψÞ ω<sup>1</sup> þ sin ðψÞω2Þ; ϑ<sup>2</sup> ¼ Að− sin ðψÞ ω<sup>1</sup> þ cos ðψÞ ω2Þ;

The forms ωi, ϑi, α<sup>i</sup> define the same structure on M and we let ω12, ϑ12, α<sup>12</sup> be the connection forms associated to the coframes ω1;ω2; ϑ1;ϑ2; α1;α2. The next theorem is crucial for what

Proof: Each of the transformations which yield the ϑ<sup>i</sup> and α<sup>i</sup> in the form (25) can be thought of as a composition of the two transformations which occur in the Theorems 3.1

get the ϑ<sup>i</sup> equations in Eq. (25). Invoking Theorems 3.1 and 4.1 in turn, the first result is

ϑ<sup>12</sup> ¼ dψ þ ω<sup>12</sup> þ � dlog A:

α<sup>12</sup> ¼ −dψ þ ω<sup>12</sup> þ � dlog A:

This implies �dlog A ¼ α<sup>12</sup> þ dψ−ω12. When replaced in the first equation of (26), the second equation appears. Note that from Theorem 3.2, α<sup>12</sup> ¼ α2, so the second equation can be given

Differentiating the second equation in Eq. (14) and using dα<sup>1</sup> ¼ 0, it follows that

Lemma 4.1 The angle ψ is a harmonic function d � dψ ¼ 0 and moreover, d � ϑ<sup>12</sup> ¼ 0.

d � dψ ¼ 0:

This states that ψ is a harmonic function. Equation (28) also implies that d � ϑ<sup>12</sup> ¼ 0.

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

Exterior differentiation of this equation using d � ω<sup>12</sup> ¼ 0 immediately gives

and 4.1. First apply the transformation ω<sup>i</sup> ! Aω<sup>i</sup> and τ ! −ψ with ω�

The transformation to the α<sup>i</sup> is exactly similar except that τ ! ψ, hence

follows.

obtained

as ϑ<sup>12</sup> ¼ 2dψ þ α2.

Theorem 4.2

28 Manifolds - Current Research Areas

<sup>α</sup><sup>1</sup> <sup>¼</sup> <sup>A</sup><sup>ð</sup> cos <sup>ð</sup>ψ<sup>Þ</sup> <sup>ω</sup>1<sup>−</sup> sin <sup>ð</sup>ψ<sup>Þ</sup> <sup>ω</sup>2Þ; <sup>α</sup><sup>2</sup> <sup>¼</sup> <sup>A</sup><sup>ð</sup> sin <sup>ð</sup>ψ<sup>Þ</sup> <sup>ω</sup><sup>1</sup> <sup>þ</sup> cos <sup>ð</sup>ψ<sup>Þ</sup> <sup>ω</sup>2Þ: (25)

ϑ<sup>12</sup> ¼ dψ þ ω<sup>12</sup> þ �d log A ¼ 2dψ þ α12: (26)

d � ω<sup>12</sup> ¼ 0: (27)

�ϑ<sup>12</sup> ¼ �ω<sup>12</sup> þ �dψ−dlogA ¼ 2 � dψ−α1: (28)

<sup>i</sup> ! ϑ<sup>i</sup> in Eq. (15), we

$$d\theta\_1 + (a\_1 + \mathcal{Z} \* a\_{12}) \wedge \theta\_1 = \mathcal{Z} \, d\log A \wedge \theta\_1. \tag{32}$$

Replacing dϑ<sup>1</sup> by means of Eq. (29) implies the following important result

$$d\log A \wedge \theta\_1 = 0.\tag{33}$$

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

$$d\log A = B\vartheta\_1.\tag{34}$$

This is the first in a series of results which relates many of the variables in question such as J, B, and ϑ<sup>12</sup> directly to the one-form ϑ1. To show this requires considerable work. The way to proceed is to use the forms α<sup>i</sup> in Theorem 3.2 because their exterior derivatives are known. For an arbitrary function on M, define

$$df = f\_1 \alpha\_1 + f\_2 \alpha\_2. \tag{35}$$

Differentiating Eq. (35) and extracting the coefficient of α1∧α2, we obtain

$$f\_{21} \! \! - f\_{12} \! + f\_2 = 0.\tag{36}$$

In terms of the αi, �dψ ¼ ψ1α2−ψ2α1, Lemma 4.1 yields

$$
\psi\_{11} + \psi\_{22} + \psi\_1 = 0.\tag{37}
$$

Finally, since �ϑ<sup>12</sup> ¼ 2 � dψ−α1, substituting for �dψ, we obtain that

$$
\omega \ast \theta\_{12} = -(2\psi\_2 + 1)\alpha\_1 + 2\psi\_1 \ a\_2. \tag{38}
$$

Differentiating structure equation (30) and using Lemma 4.1,

$$\*\vartheta\_{12} \wedge d\vartheta\_1 = 0,$$

so,

$$\ast \vartheta\_{12} \wedge \vartheta\_{12} \wedge \vartheta\_2 = 0$$

This equation implies that either ϑ<sup>12</sup> or �ϑ<sup>12</sup> is a multiple by a function of the form ϑ2. Hence, for some function p,

$$\begin{aligned} \mathfrak{P}\_{12} &= \neg p \mathfrak{P}\_2, & \ast \mathfrak{P}\_{12} &= p \mathfrak{P}\_1, \\\\ \mathfrak{P}\_{12} &= p \mathfrak{P}\_1, & \ast \mathfrak{P}\_{12} &= p \mathfrak{P}\_2, \end{aligned} \tag{39}$$

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

$$d\theta\_1 = 0.\tag{40}$$

The second line yields simply dϑ<sup>1</sup> ¼ pϑ1∧ϑ2. Only the first case is examined now. Substituting Eq. (40) into Eq. (29), the following important constraint is obtained

$$(\alpha\_1 + 2 \* \omega\_{12}) \wedge \theta\_1 = 0.\tag{41}$$

Theorem 5.1 The function ψ satisfies the equation

$$2\psi\_1 \cos\left(2\psi\right) + \left(2\psi\_2 + 1\right)\sin\left(2\psi\right) = 0.\tag{42}$$

Proof: By substituting �dψ into Eq. (28) we have

$$
\omega \* \mathfrak{G}\_{12} = \mathcal{Q} \* (\psi\_1 \alpha\_1 + \psi\_2 \alpha\_2) \text{--} \alpha\_1 = -(2\psi\_2 + 1)\alpha\_1 + 2\psi\_1 \alpha\_2. \tag{43}
$$

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

$$\*\omega\_{12} = \*\ $\_{12} - \*\,d\psi + d\log A = \*\$ \_{12} - \*d\psi + B\-1 = \*d\psi - \alpha\_1 + B\beta\_1.$$

This can be put in the equivalent form

$$
\Delta \ast \omega\_{12} + \alpha\_1 = 2 \ast d\psi \text{--} \alpha\_1 + 2B\vartheta\_1. \tag{44}
$$

Taking the exterior product with ϑ<sup>1</sup> and using dψ1, we get

$$\begin{aligned} (\alpha\_1 + 2 \ast \omega\_{12}) \wedge \mathfrak{F}\_1 &= (2 \ast d \psi \neg \alpha\_1) \wedge \mathfrak{F}\_1 = (2 \psi\_1 \ast \alpha\_1 + 2 \psi\_2 \ast \alpha\_2 - \alpha\_1) \wedge \mathfrak{F}\_1 \\ &= (2 \psi\_1 \cos(2 \psi) + (2 \psi\_2 + 1) \sin(2 \psi)) \mathfrak{F}\_2 \wedge \mathfrak{F}\_1. \end{aligned}$$

Imposing the constraint (41), the coefficient of ϑ1∧ϑ<sup>2</sup> can be equated to zero. This produces the result (42).

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

$$2\psi\_1 = \mathbb{C}\sin\left(2\psi\right), \qquad 2\psi\_2 + 1 = -\mathbb{C}\cos\left(2\psi\right). \tag{45}$$

Differentiation of each of these with respect to the α<sup>i</sup> basis, we get for i ¼ 1; 2 that

$$2\psi\_{1i} = \mathbb{C}\_i \sin\left(2\psi\right) + 2\psi\_i \gets \cos\left(2\psi\right), \qquad 2\psi\_{2i} = -\mathbb{C}\_i \cos\left(2\psi\right) + 2\psi\_i \gets \sin\left(2\psi\right).$$

Substituting f ¼ ψ into Eq. (36) and using the fact that ψ satisfies Eq. (37) gives the pair of equations

$$\begin{array}{c} -\mathbb{C}\_{1}\cos\left(2\psi\right) - \mathbb{C}\_{2}\sin\left(2\psi\right) + 2\psi\_{1}\mathbb{C}\sin\left(2\psi\right) - \left(2\psi\_{2} + 1\right)\mathbb{C}\cos\left(2\psi\right) - 1 = 0, \\\mathbb{C}\_{1}\sin\left(2\psi\right) - \mathbb{C}\_{2}\cos\left(2\psi\right) + 2\psi\_{1}\mathbb{C}\cos\left(2\psi\right) + \left(2\psi\_{2} + 1\right)\mathbb{C}\sin\left(2\psi\right) = 0. \end{array}$$

This linear system can be solved for C<sup>1</sup> and C<sup>2</sup> to get

$$\mathbb{C}\mathbb{C}\_1 + \mathbb{C}(2\psi\_2 + 1) + \cos(2\psi) = 0, \quad \mathbb{C}\_2 - 2\mathbb{C}\psi\_1 + \sin(2\psi) = 0. \tag{46}$$

By differentiating each of the equations in (46), it is easy to verify that C satisfies Eq. (36), namely, C12−C21−C<sup>2</sup> ¼ 0. Hence, there exist harmonic functions which satisfy Eq. (42). The solution depends on two arbitrary constants, the values of ψ and C at an initial point.

#### Lemma 5.1

�ϑ<sup>12</sup> ¼ −ð2ψ<sup>2</sup> þ 1Þα<sup>1</sup> þ 2ψ<sup>1</sup> α2: (38)

<sup>ϑ</sup><sup>12</sup> <sup>¼</sup> <sup>p</sup>ϑ1; �ϑ<sup>12</sup> <sup>¼</sup> <sup>p</sup>ϑ2, (39)

dϑ<sup>1</sup> ¼ 0: (40)

ðα<sup>1</sup> þ 2 � ω12Þ∧ϑ<sup>1</sup> ¼ 0: (41)

2ψ<sup>1</sup> cos ð2ψÞþð2ψ<sup>2</sup> þ 1Þ sin ð2ψÞ ¼ 0: (42)

2 � ω<sup>12</sup> þ α<sup>1</sup> ¼ 2 � dψ−α<sup>1</sup> þ 2Bϑ1: (44)

�ϑ<sup>12</sup> ¼ 2 � ðψ1α<sup>1</sup> þ ψ2α2Þ−α<sup>1</sup> ¼ −ð2ψ<sup>2</sup> þ 1Þα<sup>1</sup> þ 2ψ1α2: (43)

Differentiating structure equation (30) and using Lemma 4.1,

so,

for some function p,

30 Manifolds - Current Research Areas

�ϑ12∧ dϑ<sup>1</sup> ¼ 0;

�ϑ12∧ϑ12∧ϑ<sup>2</sup> ¼ 0

This equation implies that either ϑ<sup>12</sup> or �ϑ<sup>12</sup> is a multiple by a function of the form ϑ2. Hence,

ϑ<sup>12</sup> ¼ −pϑ2; �ϑ<sup>12</sup> ¼ pϑ1;

The second line yields simply dϑ<sup>1</sup> ¼ pϑ1∧ϑ2. Only the first case is examined now. Substituting

�ω<sup>12</sup> ¼ �ϑ12− � dψ þ dlog A ¼ �ϑ12− � dψ þ Bϑ<sup>1</sup> ¼ �dψ−α<sup>1</sup> þ Bϑ1:

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

Eq. (40) into Eq. (29), the following important constraint is obtained

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

Taking the exterior product with ϑ<sup>1</sup> and using dψ1, we get

Theorem 5.1 The function ψ satisfies the equation

Proof: By substituting �dψ into Eq. (28) we have

This can be put in the equivalent form

$$d\mathbb{C} = (\mathbb{C}^2 - 1)\theta\_1, \qquad \*\,\theta\_{12} = \mathbb{C}\theta\_1. \tag{47}$$

Proof: It is easy to express the ϑ<sup>i</sup> in terms of the αi,

$$
\Psi\_1 = \cos(2\psi)a\_1 + \sin(2\psi)a\_2, \qquad \Psi\_2 = -\sin(2\psi)a\_1 + \cos(2\psi)a\_2. \tag{48}
$$

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

$$d\mathbb{C} = \mathbb{C}\_1 \alpha\_1 + \mathbb{C}\_2 \alpha\_2 = (\mathbb{C}^2 - 1)(\cos(2\psi)\alpha\_1 + \sin(2\psi)\alpha\_2) = (\mathbb{C}^2 - 1)\mathbb{S}\_1.$$

Using Eq. (45), it follows that

$$\begin{aligned} \ast \ast\_{12} &= -(2\psi\_2 + 1)\alpha\_1 + 2\psi\_1\alpha\_2 = \mathcal{C}\cos\left(2\psi\right)\alpha\_1 + \mathcal{C}\sin\left(2\psi\right)\alpha\_2 \\ &= \mathcal{C}(\cos\left(2\psi\right)\alpha\_1 + \sin\left(2\psi\right)\alpha\_2) = \mathcal{C}\mathfrak{d}\_1. \end{aligned}$$

This implies that ϑ<sup>12</sup> ¼ −Cϑ2.

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

$$(\log A)\_1 = B \cos \left(2\psi\right), \qquad (\log A)\_2 = B \sin \left(2\psi\right). \tag{49}$$

Differentiating each of these in turn, we obtain for i ¼ 1; 2,

$$(\log A)\_{1i} = B\_i \cos \left(2\psi \right) - 2B\psi\_i \sin \left(2\psi \right), \qquad (\log A)\_{2i} = B\_i \sin \left(2\psi \right) + 2B\psi\_i \cos \left(2\psi \right). \tag{50}$$

Taking f ¼ logA in Eq. (36) produces a first equation for the Bi,

$$B\_1 \sin\left(2\psi\right) + 2B\psi\_1 \cos\left(2\psi\right) - B\_2 \cos\left(2\psi\right) + 2B\psi\_2 \sin\left(2\psi\right) + B\sin\left(2\psi\right) = 0.\tag{51}$$

If another equation in terms of B<sup>1</sup> and B<sup>2</sup> can 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

$$d\alpha\_{12} = -ac \,\, a\_1 \wedge \alpha\_2 = -ac \,\, A^{-2} \,\, \alpha\_1 \wedge \alpha\_2 \,.$$

Solving Eq. (26) for ω12, we have

$$
\omega\_{12} = d\psi + \alpha\_2 + (\log A)\_2 \alpha\_1 - (\log A)\_1 \alpha\_2 \dots
$$

The exterior derivative of this takes the form,

$$d\omega\_{12} = [1 - (\log A)\_{11} - (\log A)\_{22} - (\log A)\_1] \alpha\_1 \wedge \alpha\_2 \dots$$

Putting this in the Gauss equation,

$$-(\log A)\_{11} - (\log A)\_{22} + \{-(\log A)\_1 + 1\} + acA^{-2} = 0.1$$

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

$$-B\_1 \cos\left(2\psi\right) - B\_2 \sin\left(2\psi\right) + B\langle 2\psi\_1 \sin\left(2\psi\right) - \left(2\psi\_2 + 1\right)\cos\left(2\psi\right) \rangle + 1 + acA^{-2} = 0.\tag{52}$$

Solving Eqs. (51) and (52) together, the following expressions for B<sup>1</sup> and B<sup>2</sup> are obtained

$$B\_1 + B(2\psi\_2 + 1) - (1 + acA^{-2})\cos\left(2\psi\right) = 0, \qquad B\_2 - 2B\psi\_1 - (1 + acA^{-2})\sin\left(2\psi\right) = 0. \tag{53}$$

Given these results for B<sup>1</sup> and B2, it is easy to produce the following two Lemmas.

#### Lemma 5.2

$$dB = (BC + 1 + acA^{-2})\mathfrak{d}\_1, \qquad d\log I = (C + 2B)\mathfrak{d}\_1. \tag{54}$$

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

$$dB = B\_1\alpha\_1 + B\_2\alpha\_2 = (BC + 1 + acA^{-2})(\cos(2\psi)\alpha\_1 + \sin(2\psi)\alpha\_2) = (BC + 1 + acA^{-2})\ \ \theta\_1.$$

Moreover,

It is possible to obtain formulas for B1; B2. Using Eq. (48) in Eq. (34), the derivatives of logA can

ðlogAÞ1<sup>i</sup> ¼ Bi cos ð2ψÞ−2Bψ<sup>i</sup> sin ð2ψÞ; ðlogAÞ2<sup>i</sup> ¼ Bi sin ð2ψÞ þ 2Bψ<sup>i</sup> cos ð2ψÞ: (50)

B<sup>1</sup> sin ð2ψÞ þ 2Bψ<sup>1</sup> cos ð2ψÞ−B<sup>2</sup> cos ð2ψÞ þ 2Bψ<sup>2</sup> sin ð2ψÞ þ B sin ð2ψÞ ¼ 0: (51)

If another equation in terms of B<sup>1</sup> and B<sup>2</sup> can 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)

<sup>d</sup>ω<sup>12</sup> <sup>¼</sup> <sup>−</sup>ac <sup>ω</sup>1∧ω<sup>2</sup> <sup>¼</sup> <sup>−</sup>ac A<sup>−</sup><sup>2</sup> <sup>α</sup>1∧α2:

ω<sup>12</sup> ¼ dψ þ α<sup>2</sup> þ ðlogAÞ2α1−ðlogAÞ1α2:

dω<sup>12</sup> ¼ ½1−ðlogAÞ11−ðlogAÞ22−ðlogAÞ1�α1∧α2:

<sup>−</sup>ðlogAÞ11−ðlogAÞ<sup>22</sup> <sup>þ</sup> {−ðlogAÞ<sup>1</sup> <sup>þ</sup> 1} <sup>þ</sup> acA<sup>−</sup><sup>2</sup> <sup>¼</sup> <sup>0</sup>:

<sup>−</sup>B<sup>1</sup> cos <sup>ð</sup>2ψÞ−B<sup>2</sup> sin <sup>ð</sup>2ψÞ þ <sup>B</sup>{2ψ<sup>1</sup> sin <sup>ð</sup>2ψÞ−ð2ψ<sup>2</sup> <sup>þ</sup> <sup>1</sup><sup>Þ</sup> cos <sup>ð</sup>2ψÞ} <sup>þ</sup> <sup>1</sup> <sup>þ</sup> acA<sup>−</sup><sup>2</sup> <sup>¼</sup> <sup>0</sup>: (52)

<sup>Þ</sup> cos <sup>ð</sup>2ψÞ ¼ <sup>0</sup>; <sup>B</sup>2−2Bψ1−ð<sup>1</sup> <sup>þ</sup> acA<sup>−</sup><sup>2</sup>

Þ sin ð2ψÞ ¼ 0: (53)

Þϑ1; dlogJ ¼ ðC þ 2BÞϑ1: (54)

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

Solving Eqs. (51) and (52) together, the following expressions for B<sup>1</sup> and B<sup>2</sup> are obtained

Given these results for B<sup>1</sup> and B2, it is easy to produce the following two Lemmas.

dB ¼ ðBC <sup>þ</sup> <sup>1</sup> <sup>þ</sup> acA<sup>−</sup><sup>2</sup>

Differentiating each of these in turn, we obtain for i ¼ 1; 2,

Taking f ¼ logA in Eq. (36) produces a first equation for the Bi,

ðlogAÞ<sup>1</sup> ¼ B cos ð2ψÞ; ðlogAÞ<sup>2</sup> ¼ B sin ð2ψÞ: (49)

be written down

32 Manifolds - Current Research Areas

which we put in the form

Solving Eq. (26) for ω12, we have

Putting this in the Gauss equation,

<sup>B</sup><sup>1</sup> <sup>þ</sup> <sup>B</sup>ð2ψ<sup>2</sup> <sup>þ</sup> <sup>1</sup>Þ−ð<sup>1</sup> <sup>þ</sup> acA<sup>−</sup><sup>2</sup>

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

Lemma 5.2

The exterior derivative of this takes the form,

$$\begin{split} d\text{logJ} = \alpha\_1 + \text{2} \ast \omega\_{12} &= \alpha\_1 + \text{2}(\ast \aleph\_{12} - \ast d\psi + d\text{log}A) = \alpha\_1 + \text{2} \ast \aleph\_{12} - \text{2} \ast d\psi + \text{2}d\text{log}A \\ &= \ast \aleph\_{12} + \text{2}d\text{log}A = \mathbb{C}\mathbb{S}\_1 + \mathbb{Z}\mathbb{S}\_1. \end{split}$$

Lemma 5.3

$$d\psi = -\frac{1}{2}\sin\left(2\psi\right)\mathfrak{d}\_1 - \frac{1}{2}\left(\mathbb{C} + \cos\left(2\psi\right)\right)\mathfrak{d}\_2.\tag{55}$$

Proof:

$$\begin{array}{l} \mathsf{2}d\psi = 2\psi\_{1}\alpha\_{1} + 2\psi\_{2}\alpha\_{2} = \mathsf{C}\sin\left(2\psi\right)\alpha\_{1} - \left(\mathsf{C}\cos\left(2\psi\right) + 1\right)\alpha\_{2} \\ = \mathsf{C}\sin\left(2\psi\right)\left(\cos\left(2\psi\right)\theta\_{1} - \sin\left(2\psi\right)\theta\_{2}\right) - \left(\mathsf{C}\cos\left(2\psi\right) + 1\right)\left(\sin\left(2\psi\right)\theta\_{1} + \cos\left(2\psi\right)\theta\_{2}\right) \\ = -\sin\left(2\psi\right)\theta\_{1} - \left(\mathsf{C} + \cos\left(2\psi\right)\right)\theta\_{2}. \end{array}$$

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

Theorem 5.2 The function B satisfies Eq. (36) provided ψ satisfies both Eqs. (37) and (41).

Proof: Differentiating B<sup>1</sup> and B<sup>2</sup> given by Eq. (53), the left side of Eq. (36) is found to be

$$B\_{21} - B\_{12} + B\_2 = 2B\_1\psi\_1 + B\_2(2\psi\_2 + 1) + 2B(\psi\_{11} + \psi\_{22} + \psi\_1) + A^{-2}((ac)\_1\sin(2\psi) - (ac)\_2\sin(2\psi))$$

$$= 2acBA^{-2}(\cos(2\psi)\sin(2\psi) - \sin(2\psi)\cos(2\psi)) + (1 + acA^{-2})(2\psi\_1\cos(2\psi) + (2\psi\_2 + 1)\sin(2\psi))$$

$$= 2(1 + acA^{-2})(2\psi\_1\cos(2\psi) + (2\psi\_2 + 1)\sin(2\psi)) + A^{-2}((ac)\_1\sin(2\psi) - (ac)\_2\cos(2\psi)).$$

10 To simplify this, Eq. (37) has been substituted. Using Eq. (48) and �dðacÞ¼ðacÞ1α2−ðacÞ2α1, it follows that

$$\wedge \ast d(ac) \wedge \theta\_2 = ((ac)\_1 \sin \left( 2\psi \right) - (ac)\_2 \cos \left( 2\psi \right)) \alpha\_1 \wedge \alpha\_2 \dots$$

12 Note that the coefficient of α1∧α<sup>2</sup> in this appears in the compatibility condition. To express it in 13 another way, begin by finding the exterior derivative of 4ac ¼ ð<sup>a</sup> <sup>þ</sup> <sup>c</sup><sup>Þ</sup> 2 −ða−cÞ 2 ,

$$4d(ac) = 2(a+c)(a-c)\,\partial\_1 - 2(a-c)^2(\alpha\_1 + 2\*a\nu\_{12})\dots$$

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

$$2\*\frac{d(ac)}{a-c} = (a+c)\,\mathfrak{G}\_2 - (a-c)\,(\alpha\_2 - 2\omega\_{12})\,.$$

15 Consequently, we can write

$$-\frac{2}{\left(a-c\right)^{2}}\,\,\ast\,d(ac)\wedge\vartheta\_{2} = (\alpha\_{2}-2\omega\_{12})\wedge\vartheta\_{2} = -(2\psi\_{1}\cos\left(2\psi\right)+(2\psi\_{2}+1)\sin\left(2\psi\right))\alpha\_{1}\wedge\alpha\_{2}\,.$$

16 Therefore, it must be that

$$-(ac)\_1 \sin\left(2\psi\right) + (ac)\_2 \cos\left(2\psi\right) = -\frac{1}{2}(a-c)^2(2\psi\_1 \cos\left(2\psi\right) + (2\psi\_2 + 1)\sin\left(2\psi\right)).$$

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

$$[(1+H^2A^{-2})[2\psi\_1\cos(2\psi)+(2\psi\_2+1)\sin(2\psi)]=0.4$$

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