The Existence of Rg Family and Tg Family, and Their Geometric Invariants

mathematics

Article The Existence of rG Family and tG Family, and Their Geometric Invariants

Norio Ejiri 1,† and Toshihiro Shoda 2,*,‡ 1 Department of Mathematics, Meijo University, Tempaku, Nagoya 468-8502, Japan; [email protected] 2 Faculty of Education, Saga University, 1 Honjo-machi, Saga 840-8502, Japan * Correspondence: [email protected] † Partly supported by JSPS Grant-in-Aid for Scientiﬁc Research (C) 15K04859. ‡ Partly supported by JSPS Grant-in-Aid for Scientiﬁc Research (C) 20K03616.

Received: 2 September 2020; Accepted: 21 September 2020; Published: 2 October 2020

Abstract: In the 1990s, physicists constructed two one-parameter families of compact oriented embedded minimal surfaces in flat three-tori by using symmetries of space groups, called the rG family and tG family. The present work studies the existence of the two families via the period lattices. Moreover, we will consider two kinds of geometric invariants for the two families, namely, the Morse index and the signature of a minimal surface. We show that Schwarz P surface, D surface, Schoen’s gyroid, and the Lidinoid belong to a family of minimal surfaces with Morse index 1.

Keywords: minimal surfaces; ﬂat tori; Morse index; signature

1. Introduction

A triply periodic minimal surface in R3 is concerned with natural phenomena, and it has been studied in physics, chemistry, crystallography, and so on. It can be replaced by a compact oriented minimal surface in a ﬂat three-torus f : M R3/Λ and the conformal structure induced by the → immersion f makes M a Riemann surface. We usually call such f a conformal minimal immersion. Applying the complex function theory gives us a useful description of a triply periodic minimal surface. In fact, the following is one of basic tools.

Theorem 1 (Weierstrass representation formula). Let f : M R3/Λ be a conformal minimal immersion. → Then, up to translations, f can be represented by the following path-integrals:

p t f (p) = (ω1, ω2, ω3) mod Λ, < Zp0 where p0 is a ﬁxed point on M and the ωi’s are holomorphic differentials on M satisfying the following three conditions.

2 2 2 ω1 + ω2 + ω3 = 0,

ω1, ω2, ω3 have no common zeros,

t (ω1, ω2, ω3) C H1(M, Z) is a sublattice of Λ. < ∈ ZC

Conversely, the real part of path-integrals of holomorphic differentials satisfying the above three conditions deﬁnes a conformal minimal immersion.

Mathematics 2020, 8, 1693; doi:10.3390/math8101693 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 1693 2 of 39

In 1970, Schoen [1] discovered a triply periodic minimal surface in R3 which is called the gyroid, and it has been studied in natural sciences. In 1993, Fogden, Haeberlein, and Hyde [2] constructed the rG family and the tG family which are two one-parameter families including the gyroid. They gave their representation formulae as follows.

Example 1 (rG family). For a = seiφ (0 < s < 1, φ ( π , π ]), let M be the hyperelliptic Riemann surface ∈ − 6 3 p dz 2 = ( 3 3)( 3 + 1 ) ( ) = iθ t(1 2, (1 + 2), 2 ) of genus three deﬁned by w z z a z a3 . Set f p e z i z z . − < p0 − w π Z For (φ, s, θ) = ( 3 , 0.494722 ... , 0), f belongs to the rPD family. f must be the Lidinoid for (φ, s, θ) = ( π , 0.5361, 0.335283 . . .). Moreover, f must be the gyroid for (φ, s, θ) = (0, 1 , 0.663483 . . .). 6 √2

Example 2 (tG family). For b = seiφ (0 < s < 1, φ < π ), let M be the hyperelliptic | | 4 2 = 8 + 4 + 1 = ( 4 + 4)( 4 + 1 ) ( ) = Riemann surface of genus three defined by w z az z b z b4 . Set f p p iθ t 2 2 dz √3 1 ie (1 z , i(1 + z ), 2z) . For (φ, s, θ) = (0, − , 0.663483 ...), f must be the gyroid. < − w √2 Zp0 f tends to be an element of the tD family as (φ, s, θ) ( π , 0.431882, π ). → − 4 2 Note that we have to determine φ, s, θ such that f is well-defined in Examples1 and2. Fogden, Haeberlein, and Hyde used the computer simulation to show that s and θ are functions of φ, that is, s = s(φ) and θ = θ(φ). Hence, the rG family and the tG family can be considered as one-parameter families by φ. After that, Fogden and Hyde [3] reconstructed the two families via symmetries of space groups. In this paper, we will consider the existence of them by period calculations and obtain sufficient conditions for it (see Theorems2 and3). As their applications, we have numerical evidence that the rG family and the tG family are defined as one-parameter families of compact oriented embedded minimal surfaces of genus three in flat three-tori (Theorem5). Next, we will consider the Morse index and the signature of the rG family and the tG family. The Morse index (resp. the nullity) of a compact oriented minimal surface in a flat three-torus is defined as the number of negative eigenvalues (resp. zero eigenvalues) of the area, counted with multiplicities. In this case, it is easy to verify that the Morse index is greater than or equal to 1. Ros [4] proved that every compact oriented minimal surface in a flat three-torus with Morse index 1 has genus three. This result suggests that compact oriented minimal surfaces of genus three in flat three-tori are important objects. Recently, the first author has established an algorithm to compute the Morse index and the nullity of a minimal surface (see [5,6], see also [7] for an explanation of the algorithm). By applying the algorithm to the genus three case, the Morse index can be translated into the number of negative eigenvalues of an 18 18 real symmetric matrix W W defined in Section3 and the × 2 − 1 nullity can be translated into the number of zero eigenvalues of a 9 9 Hermitian matrix W defined × in Section3, counted with multiplicities. The two key matrices consist of periods of the Abelian differentials of the second kind on a minimal surface. We now call a pair of the number of positive eigenvalues and that of negative eigenvalues of W, counted with multiplicities the signature of a minimal surface. In the previous paper [8], we carried out the algorithm for five one-parameter families (the H family, the rPD family, the tP family, the tD family, the tCLP family) and computed their Morse indices and signatures. We will continue this work and give explicit descriptions of W and W W 2 − 1 for the rG family and the tG family in Section3. We then numerically compute the Morse indices and the signatures of the rG family and the tG family (Theorems7 and9). Moreover, we can show that there exists a path with the following two properties (Theorem 10): (i) it connects the P surface, the D surface, the gyroid, the Lidinoid (G L P D), (ii) it consists of minimal surfaces with Morse index → → → 1. Recall that this family is called the rGPD union in [3], and our result implies that the rGPD union consists of minimal surfaces with Morse index 1. The outline of the paper is as follows. In Section2, we discuss of well-definedness and embeddedness for the rG family and the tG family. We shall give sufficient conditions that the rG family and the tG family are defined as two families of compact oriented minimal surfaces of genus Mathematics 2020, 8, 1693 3 of 39 three in flat three-tori. After that, we use numerical arguments and show that the two families are defined as one-parameter families of embedded minimal surfaces as in Fogden—Haeberlein—Hyde’s paper. Section3 gives explicit descriptions of the two key matrices to compute the Morse indices and the signatures for the rG family and the tG family. Moreover, we shall numerically compute the Morse indices and the signatures of the rG family and the tG family. The above two sections need period calculations of the two families and finally in AppendixA there is a collection of the details of period calculations. All figures except for Figures 6, 9 and 10 and all numerical arguments in Sections2 and3 are generated by Mathematica [9].

2. The Existence and Embeddedness of the rG Family and tG Family In this section, we shall give two sufﬁcient conditions that (i) the rG family can be obtained as a deformation from an element of the rPD family, (ii) the tG family can be obtained as a deformation from the gyroid. As their applications, we can show the existence of the rG family and the tG family by numerical arguments. Moreover, we will discuss the embeddedness of the rG family and the tG family. We ﬁrst review fundamental arguments. Recall that a 3 6 real matrix L generates a lattice in × R3 if and only if there exist a 3 3 real regular matrix X, a 3 6 matrix g with integer entries, and a × × 6 3 matrix h with integer entries such that Lh = X and Xg = L. Thus, we have the following key × fact (see § 6 in [5] and § 6.2.2 in [6]).

Proposition 1. Let L be a 3 6 real matrix which generates a lattice in R3. Suppose that L(t) is a smooth × 3 deformation of L with L(0) = L and rankR(L(t)) = 3. Then, L(t) generates a lattice in R if and only if there exist a 3 6 matrix g with integer entries and a 6 3 matrix h with integer entries such that gh = E and × × 3 L(t)hg = L(t).

Next we shall consider the existence of the rG family and the tG family by Proposition1.

2.1. rG Family iφ π π For a = se (0 < s < 1, φ ( 6 , 3 ]), let M be the hyperelliptic Riemann surface of genus three ∈ − 2 ( + 2) deﬁned by w2 = z(z3 a3)(z3 + 1 ). Deﬁne G = t( 1 z dz, i 1 z dz, 2z dz) as a column vector which − a3 1 −w w w consists of a basis of holomorphic differentials on M. We now use the notation as in Appendix A.1 of AppendixA. Let A , B 3 be the canonical homology basis on M as in Appendix A.1.1 of { j j}j=1 AppendixA, and set

iθ L = e G1 G1 G1 G1 G1 G1 . (1) < A A A B B B Z 1 Z 2 Z 3 Z 1 Z 2 Z 3 π We assume θ = 0 for φ = 3 . In this case,

p iθ e G1 < Zp0 deﬁnes a conformal minimal immersion of M into a ﬂat three-torus R3/Λ. It belongs to the rPD family and Λ is given by

3α 3α 2α Λ = √3α √3α 0 ,  −  0 0 β  −    1 1 1 s2t2 1 t α = − dt, β = 4 dt .  1 1  s Z0 t(1 t3) s3t3 + Z0 t(1 t3) s3t3 + − s3 − s3  p q p q  Mathematics 2020, 8, 1693 4 of 39

π i π i In fact, by setting (Z, W) = (e 3 z, e 6 w), M and G1 can be rewritten as

√ 1 3 0 1 Z2 1 2 2 dZ 2 3 3 2 −√3 − 1 − 2 W = Z(Z s ) Z + 3 , G1 = i  0  i(1 + Z ) . − s 2 − 2 W 0 0 1 2Z      −    So it belongs to the rPD family (see § 1 in [8]). By choosing the branch 3 3 + 1 = a t a3 3 3 1 q i s t + 3 iR>0 and (A12) from AppendixA, we ﬁnd − s ∈ − q α 0 0 α 3α 2α − L = √3α 0 0 √3α √3α 0 −  β 3β 3β 2β 0 β  − − − −    π for φ = 3 . Thus, setting

0 0 0 0 0 0 1 1 1 0 1 0   − − 0 0 0 g =  0 1 1 1 0 0 , h =   , (2) − − 0 1 0 1 3 3 2 0 1  −   −  1 0 0     0 2 1     we have Lh = Λ, Λg = L, and gh = E . We now apply Proposition1 to L given by (1) for φ ( π , π ]. 3 ∈ − 6 3 Set L = (L , L , L , L , L , L ), where L is a 3 1 real matrix. Then, by (A12), we obtain 1 2 3 4 5 6 j × Lhg = L L = L + L , L = L = L + L L ⇐⇒ 1 − 5 6 2 − 3 − 4 5 − 6 eiθ(A + √3 iB) = eiθ(i C + D) = 0 ⇐⇒ <{ } <{ } (A + √3 iB) (i C + D) tan θ = < = < . ⇐⇒ (A + √3 iB) (i C + D) = = Therefore, we ﬁnd the following result.

Theorem 2. Let φ ( π , π ]. Suppose that there exist s and φ such that ∈ − 6 3 (A + √3 iB) (i C + D) (A + √3 iB) (i C + D) = 0. (3) {< }{= } − {= }{< }

(A+√3 iB) 2 Set θ = arctan < , λ = ( A tan θ A), µ = (iC) + tan θ (iC) for s and φ as the (A+√3 iB) √3 = < − = −< = above. Then p iθ e G1 < Zp0 defines a conformal minimal immersion of M into a flat three-torus R3/Λ, where 3λ 3λ 2λ Λ = cos θ √3λ √3λ 0 . In particular, for φ = π , it belongs to the rPD family.  −  3 0 0 µ  −    Mathematics 2020, 8, 1693 5 of 39

Proof. It is sufficient to show that L defined by (1) can be transformed to Λ by g and h as in (2). From (A12), we have

2 iB 2 A 2 iB A iB 2 A √3A 3 iB 2 A 2 i B L − √3 − − √3 − √3 − √3 − = 2A 0 A + √3 iB 2√3 iB A √3 iB 0 cos θ < − − −  i C 2 i C + D 2 i C D i C D 0 D  − − − −    2 iB 2 A 2 iB A iB 2 A √3A 3 iB 2 A 2 i B − √3 − − √3 − √3 − √3 − tan θ 2A 0 A + √3 iB 2√3 iB A √3 iB 0 . − = − − −  i C 2 i C + D 2 i C D i C D 0 D  − − − −    (A+√3 iB) (i C+D) By tan θ = < = < , L can be rewritten as (A+√3 iB) (i C+D) = = λ 0 0 λ 3λ 2λ − L = cos θ √3λ 0 0 √3λ √3λ 0 . −  µ 3µ 3µ 2µ 0 µ  − − − −    So the theorem follows.

By using Mathematica, we find the locus of (φ, s) which satisfies (3) as in Figure1. π π Set = (φ, s, x3) φ ( , ], 0.4 < s < 1, x3 R . The green domain is the plane x3 = 0 and D { | ∈ − 6 3 ∈ } the blue domain is the graph of x = (A + √3 iB) (i C + D) (A + √3 iB) (i C + D) 3 {< }{= } − {= }{< } in . Hence, their intersection is precisely the locus of (φ, s) which satisfies (3) in the (φ, s)-plane. D Thus we obtain a numerical evidence for the existence of (φ, s) which satisfies (3), and rG family can be defined as one-parameter family of compact oriented minimal surfaces in flat three-tori, that is, s = s(φ) and θ = θ(φ). For φ = 0 and s = 1 , we find θ = 0.663483 ... and it leads to the surface √2 π known as the gyroid. For φ = 6 and s = 0.5361, we have θ = 0.335283 ... and it leads to the surface known as the Lidinoid. Remark that the intersection curve in Figure1 is given in [ 2] (see p. 2381 in [2]). π For φ = 3 , we find s = 0.494722. Numerical Result 2 in [8] implies that it corresponds to a minimal surface with nullity 4 which belongs to the rPD family. π We finally note that the green domain intersects the blue domain at the line φ = 3 in the (φ, s)-plane, and it corresponds to the rPD family.

Figure 1. The locus of (φ, s) which satisﬁes (3) (the intersection curve of the green domain and the blue domain). Mathematics 2020, 8, 1693 6 of 39

2.2. tG Family iφ π For b = se (0 < s < 1, φ < 4 ), let M be the hyperelliptic Riemann surface of genus three | | 2 2 deﬁned by 2 = 8 + 4 + 1 = ( 4 + 4)( 4 + 1 ). Deﬁne = t( 1 z , i(1+z ) , 2z ) as a column w z az z b z b4 G1 −w dz w dz w dz vector which consists of a basis of holomorphic differentials on M. We now use the notation as in Appendix A.2 of AppendixA. Let A , B 3 be the canonical homology basis on M given in { j j}j=1 Appendix A.2, and set

iθ L = i e G1 G1 G1 G1 G1 G1 . (4) < A A A B B B Z 1 Z 2 Z 3 Z 1 Z 2 Z 3

√3 1 B We assume s = − and θ = arctan 0.663483 . . . for φ = 0. In this case, √2 A ≈ p iθ i e G1 < Zp0 defines a conformal minimal immersion of M into a flat three-torus R3/Λ known as the gyroid, and Λ 1 1 0 is given by Λ = AB 1 1 0 . In fact, we first observe that √A2+B2  −  1 1 2     ∞ dx 1 1 A = 4 x = t + , 0 √16x4 16x2 + 2 + a 2 t Z − ∞ dx 1 1 B = 4 x = t + , √ 4 2 2 − t Z0 16x + 16x + 2 + a ∞ dx 1 t2 C = 4 x = − . 0 (a + 2)x4 + ( 2a + 12)x2 + 2 + a 1 + t2 Z − p √3 1 Hence, if we substitute s = − and φ = 0, that is, a = 14, then we find A = C and B = D. √2 B By (A32) from AppendixA and tan θ = A , we have

1 1 1 1 2 1 − L = B cos θ 1 1 1 1 0 1 − − −  1 1 1 1 0 1  − − −    for φ = 0. Thus, setting

0 1 1 1 0 1 0 0 0 0 1 1 − −  0 0 0 g = 1 1 1 1 1 0  , h =   , (5) −  0 0 0  0 1 1 1 1 1    − − − −   1 0 0       0 0 0      we have Lh = Λ, Λg = L, and gh = E . We now apply Proposition1 to L given by (4) for φ ( π , π ). 3 ∈ − 4 4 Set L = (L , L , L , L , L , L ), where L is a 3 1 real matrix. Then, by (A32), we obtain 1 2 3 4 5 6 j × Lhg = L L = L L , L = L = L ⇐⇒ 1 5 − 6 2 − 3 4 eiθ(iA + B) = eiθ(i C + D) = 0 ⇐⇒ <{ } <{ } A + B C + D tan θ = −= < = −= < . ⇐⇒ A + B C + D < = < = Mathematics 2020, 8, 1693 7 of 39

Therefore, we ﬁnd the following result.

Theorem 3. Let φ < π . Suppose that there exist s and φ such that | | 4 ( A + B)( C + D) ( A + B)( C + D) = 0. (6) −= < < = − < = −= < A+ B Set θ = arctan −=A+

λ λ 0 deﬁnes a conformal minimal immersion of M into a ﬂat three-torus R3/Λ, where Λ = cos θ λ λ 0 .  −  µ µ 2µ   In particular, for φ = 0, it must be the gyroid.  

Proof. It is sufﬁcient to show that L deﬁned by (4) can be transformed to Λ by g and h as in (5). From (A32), we have

B iA BB 2BB B iA BB 2BB L − − − − = iA B iA B 0 B tan θ iA B iA B 0 B . cos θ <  − − −  − =  − − −  D D D i C 0 i C D D D i C 0 i C  −   −      A+ B C+ D By tan θ = −=A+

By using Mathematica, we find the locus of (φ, s) which satisfies (6) as in Figure2. π For = (φ, s, x3) φ < , 0.4 < s < 1, x3 R , the green domain is the plane x3 = 0 and the blue D { | | | 4 ∈ } domain is the graph of x = ( A + B)( C + D) ( A + B)( C + D) in . Hence the 3 −= < < = − < = −= < D intersection of them is precisely the locus of (φ, s) which satisfies (6) in the (φ, s)-plane. Thus we obtain a numerical evidence for the existence of (φ, s) which satisfies (6), and tG family can be defined as one-parameter family of compact oriented minimal surfaces in flat three-tori, that is, s = s(φ) and √3 1 π θ = θ(φ). For φ = 0 and s = − , we find θ = 0.663483 ... and it must be the gyroid. For φ √2 → − 4 and s 0.431882 (s4 + 1 28.7783), we have θ π and it belongs to the tD family. Numerical Result → s4 ≈ → 2 3 in [8] implies that it corresponds to a minimal surface with nullity 4 which belongs to the tD family. Remark that Figure2 is given in [2] (see p. 2381 in [2]).

Figure 2. The locus of (φ, s) which satisﬁes (6) (the intersection curve of the green domain and the blue domain). Mathematics 2020, 8, 1693 8 of 39

2.3. Embeddedness The embeddedness of the rG family and the tG family is an immediate consequence of the following theorem, which has essentially been proven by Meeks (see the proof of Theorem 7.1 in [10]).

Theorem 4. If a one-parameter family of compact oriented minimal surfaces of genus three in flat three-tori contains an embedded minimal surface in a flat three-torus, then every element of the one-parameter family must be an embedded minimal surface in a flat three-torus.

Recall that the rPD family consists of only embedded minimal surfaces in ﬂat three-tori because it satisﬁes the assumption of Theorem 7.1 in [10]. The rG family contains an element of the rPD family π (the case φ = 3 ), and also, the rG family meets the tG family at the gyroid. Therefore, we conclude

Theorem 5. The rG family and the tG family are one-parameter families of compact oriented embedded minimal surfaces of genus three in ﬂat three-tori.

This gives an alternative proof of the following result which is obtained in [11]: the gyroid and the Lidinoid are embedded minimal surfaces in ﬂat three-tori. In fact, the rG family contains the gyroid π (the case φ = 0) and the Lidinoid (the case φ = 6 ).

3. The Morse Indices of rG Family and tG Family In this section, we shall consider the Morse indices of the rG family and the tG family. The Morse index can be translated into the number of negative eigenvalues of a 18 18 real symmetric matrix × counted with multiplicities and the nullity can be translated into the number of zero eigenvalues of a 9 9 Hermitian matrix counted with multiplicities (see Theorems6 and8). We will describe explicitly × the two key matrices for the rG family and the tG family by periods of the Abelian differentials of the second kind. After that, we can compute the Morse index of the rG family and the tG family by numerical arguments. Let indexa denote the Morse index of a minimal surface.

3.1. rG Family We can apply the same arguments as the rPD family to the rG family since a type of a Riemann surface of the rG family coincides with that of the rPD family (see § 3.2 in [8]). We shall use the notation as in Appendix A.1 of AppendixA. Set

1 i 0 0 0 0 2 − 2  1  0 0 0 0 0 2   1 0 1  1 i  −  0 0 0 0 P1 = i 0 i  , P2 = − 2 − 2  ,  1 i  0 2 0  0 0 0 0    2 − 2       0 0 0 0 0 1    1 i   0 0 0 0  − 2 − 2    5 1 1 1 1 1 1 1 1 a2 + a + 1 + 6 2 3 2 i 4 2 i 5 2 6 − ai − 2ai − 6ai ai ! ai ! ai !  1 1 1 1 1 1 1 1 1 =  3 + 2 + +  Pai  2 ai 3 ai 4 ai 5  ,  6 − 2ai − 6a 2 a ! 2 a ! 2 a !  i i i i     ai 1 1 1 4 1 1 3 1 1 2 1   ai + 2 ai + 3 ai + 4   6 2 − 6ai 2 a ! 2 a ! 2 a !  i i i    Mathematics 2020, 8, 1693 9 of 39

G1 G1 G1 G1 G1 G1 ΩrG = ZA1 G2! ZA2 G2! ZA3 G2! ZB1 G2! ZB2 G2! ZB3 G2!!

2 πi 4 πi 2 πi and choose a1 = a, a2 = e 3 a, a3 = e 3 a, a4 = 1/a, a5 = e 3 /a. We deﬁne 5 − − 5 1 Tj = := P1Paj P2ΩrG . { }j 1 2 j=1 SettingnC and C areo 3 3 complex matrices given by 1 2 × 2 iB 2 A 2 iB A iB 2 A √3A 3 iB 2 A 2 i B − √3 − − √3 − √3 − √3 − (C1, C2) = 2A 0 A + √3 iB 2√3 iB A √3 iB 0 , − − −  i C 2 i C + D 2 i C D i C D 0 D  − − − −    1 we have the Riemann matrix τ = C1− C2 and deﬁne

0 1 0 T := (C , C ), T := 1 0 0 (C , C ), 6 1 2 7 −  1 2 0 0 0   0 0 1   0 0 0 T8 :=  0 0 0 (C1, C2), T9 := 0 0 1 (C1, C2). 1 0 0 0 1 0 −   −      Decompose T = (Z , Z ) into two 3 3 complex matrices. We introduce j j1 j2 × η((Z , Z ), (Z , Z )) = i tr(tZ Z tZ Z ) j1 j2 j1 j2 − j2 j1 − j1 j2 and set W = (η(T , T )). W is a 9 9 Hermitian matrix and one of the two key matrices. Let (p, q) i j × denote the signature of W. For a decomposition Tj = (Zj1, Zj2), we set

1 1 K0 = Z + i Z τ Z ( τ)− , K00 = (iZ ) + i (iZ ) τ (iZ ) ( τ)− . j < j1 {< j1< − < j2} = j < j1 {< j1 < − < j2 } = Deﬁne

T (1 j 9) (K0, K0τ)(1 j 9) U = j ≤ ≤ , V = j j ≤ ≤ j j  (iTj 9 (10 j 18) (K00j 9, K00j 9τ)(10 j 18) − ≤ ≤  − − ≤ ≤ and W = ( (η(U , U ))), W = ( (η(V , V ))). W W is a 18 18 real symmetric matrix and the 1 < i j 2 < i j 2 − 1 × another key matrix. The next theorem follows from Theorem 7.10 and § 14 in [5], Theorems 5.4, 5.5, and 6.3 in [6] (see also § 2.6 in [7]).

Theorem 6. The nullity of the rG family is equal to the number of zero eigenvalues of W, counted with multiplicities plus 3. Moreover, if the number of zero eigenvalues of W W is equal to 8, then the Morse index 2 − 1 of the rG family is equal to the number of negative eigenvalues of W W , counted with multiplicities plus 1. 2 − 1 π π For = (φ, s, x3) φ ( 6 , 3 ], 0.4 < s < 1, x3 R , by using Mathematica, we obtain the D { | ∈ − 3 ∈ } graph of x3 = det(W) and the plane x3 = 0 in R (see Figures3 and4). The intersection of them − consists of the points whose nullities are at least 4. Mathematics 2020, 8, 1693 10 of 39

The intersection of the three graphs of x = det(W), the plane x = 0, and x = (A + 3 − 3 3 {< √3 iB) (i C + D) (A + √3 iB) (i C + D) in is given in Figure5. }{= } − {= }{< } D From Figure5, we can show that the locus of (φ, s) which satisﬁes (3) contains two connected components (see Figure6). In fact, we will numerically compute the eigenvalues of W and W W . 2 − 1 We ﬁrst consider the eigenvalues of W. Substituting (φ, s) = ( π , 0.45), ( π , 0.6), − 6.01 − 6.01 ( π , 0.9) to W yields the following sets of the eigenvalues: − 6.01 152.006, 83.3364, 71.4652, 61.5538, 4.94045, 3.35369, 0.49079, 0.144368, 0.0369606 , { − − − − } 188.14, 111.028, 110.934, 94.906, 4.81546, 4.40164, 1.81147, 0.204606, 0.0535919 , { − − − − − } 369.902, 254.671, 215.325, 179.236, 51.9581, 38.125, 24.2347, 1.33455, 0.354338 . { − − − }

Figure 3. The graph of x = det W (the red domain) and the plane x = 0 (the green domain) in . 3 − 3 D They intersect at a curve which looks like a straight line.

Figure 4. The graph of x = det W (the red domain) and the plane x = 0 (the green domain) in . 3 − 3 D The former seems to be tangent to the latter at some curves which look like parabolic curves. Mathematics 2020, 8, 1693 11 of 39

Figure 5. The graph of x = det W (the red domain), the plane x = 0 (the green domain), and x = 3 − 3 3 (A + √3 iB) (i C + D) (A + √3 iB) (i C + D) (the blue domain) in . {< }{= } − {= }{< } D

Next, we substitute (φ, s) = (0, 0.45), (0, 0.9) to W and obtain the following sets of the eigenvalues:

152.402, 82.803, 70.8832, 61.6518, 4.7351, 3.21415, 0.459665, 0.163676, 0.0419736 , { − − − − } 265.411, 151.782, 145.946, 134.336, 7.37701, 3.5439, 2.57836, 0.805086, 0.181996 . { − − − − − } π π π Moreover, substituting (φ, s) = ( 6 , 0.45), ( 6 , 0.6), ( 6 , 0.9) to W, we have the following sets of the eigenvalues:

145.716, 73.5343, 59.8217, 51.8676, 5.45245, 3.07576, 0.836401, 0.106013, 0.0253394 , { − − − − } 188.14, 111.028, 110.934, 94.906, 4.81547, 4.40165, 1.81147, 0.204605, 0.0535916 , { − − − − − } 369.906, 254.676, 215.327, 179.237, 51.961, 38.1271, 24.236, 1.33459, 0.354349 . { − − − } π π Finally, we substitute (φ, s) = ( 3 , 0.45), ( 3 , 0.9) to W and ﬁnd the following sets of the eigenvalues:

145.971, 73.3025, 59.586, 51.9706, 5.34193, 3.0093, 0.815662, 0.113851, 0.0272483 , { − − − − } 265.411, 151.782, 145.946, 134.336, 7.37701, 3.5439, 2.57836, 0.805086, 0.181996 . { − − − − − } We now consider the eigenvalues of W W . Substituting (φ, s) = ( π , 0.45), ( π , 0.6), 2 − 1 − 6.01 − 6.01 ( π , 0.9) to W W yields the following sets of the eigenvalues: − 6.01 2 − 1 14.5758, 14.549, 10.0379, 9.97935, 8.89303, 0.709172, 0.543324, 0.245049, 0.128432, { − 0.0581318, 0, 0, 0, 0, 0, 0, 0, 0 , } 16.6566, 16.5459, 14.8343, 10.5963, 10.293, 2.65305, 1.62215, 0.520714, 0.2287, { 0.0775775, 0, 0, 0, 0, 0, 0, 0, 0 , } 146.521, 129.728, 114.312, 106.675, 70.2955, 70.1246, 4.00972, 1.3252, 1.06449, { − 0.357892, 0, 0, 0, 0, 0, 0, 0, 0 . − } Mathematics 2020, 8, 1693 12 of 39

π π π Moreover,Next, substituting we substitute (φ, s ()ϕ, = s ()0, 0.45 =) (, (60,, 0.90.)45)to W, ( 6 W, 0.and6), obtain( 6 , 0 the.9) following to W2 sets ofW1, we 2 − 1 − havethe the eigenvalues: following sets of the eigenvalues:

14.1866, 14.1478, 9.71217, 9.70707, 8.64341, 0.670382, 0.496307, 0.269501, 0.131996, 14.5758, 14{ .549, 10.0379, 9.97935, 8.89303− , 0.709172, 0.543324, 0.245049, 0.128432, { 0.0715017, 0, 0, 0, 0, 0, 0, 0, 0 , − } 0.058131718.0216,, 0 14.5156,, 0, 0, 0 14.1481,, 0, 0, 13.3706,0, 0 , 10.1915, 8.73066, 1.82703, 1.44482, 0.580926, { } 16.6566, 16.54590.434436,, 14 0,. 0,8343 0, 0, 0,, 0,10 0,. 05963. , 10.293, 2.65306, 1.62215, 0.520714, 0.228698, { } 0.077577Moreover,, 0 substituting, 0, 0, 0, 0(φ,,0s,)0 =, (0π ,, 0.45), ( π , 0.6), ( π , 0.9) to W W , we have the following 6} 6 6 2 − 1 146sets.529 of eigenvalues:, 129.736, 114.319, 106.682, 70.2996, 70.1288, 4.00972, 1.3252, 1.06451, { − 14.5758, 14.549, 10.0379, 9.97935, 8.89303, 0.709172, 0.543324, 0.245049, 0.128432, 0.3579{ , 0, 0, 0, 0, 0, 0, 0, 0 . − − 0.0581317, 0, 0, 0, 0, 0, 0, 0, 0}, } 16.6566, 16.5459, 14.8343, 10.5963, 10.293,π 2.65306, 1.62215,π 0.520714, 0.228698, Finally, we{ substitute (ϕ, s) = ( 3 , 0.45), ( 3 , 0.9) to W2 W1 and ﬁnd the 0.077577, 0, 0, 0, 0, 0, 0, 0, 0 , − following sets of the eigenvalues:} 146.529, 129.736, 114.319, 106.682, 70.2996, 70.1288, 4.00972, 1.3252, 1.06451, { − 14.1866, 14.14780.3579,, 9.71217 0, 0, 0, 0,, 0,9 0,.70707 0, 0 . , 8.64341, 0.670382, 0.496307, 0.269501, 0.131996, { − } − Finally, we substitute (φ, s) = ( π , 0.45), ( π , 0.9) to W W and ﬁnd the following sets of 0.0715017, 0, 0, 0, 0, 0, 0, 0,30 , 3 2 − 1 the eigenvalues: } 18.0216, 14.5156, 14.1481, 13.3706, 10.1915, 8.73066, 1.82703, 1.44482, 0.580926, { 14.1866, 14.1478, 9.71217, 9.70707, 8.64341, 0.670382, 0.496307, 0.269501, 0.131996, 0.434436{ , 0, 0, 0, 0, 0, 0, 0, 0 . − 0.0715017, 0, 0, 0, 0, 0, 0, 0, 0 , }} 18.0216, 14.5156, 14.1481, 13.3706, 10.1915, 8.73066, 1.82703, 1.44482, 0.580926, Therefore,{ we conclude that (i) (p, q) = (6, 3) and indexa = 3 on the domain 0.434436, 0, 0, 0, 0, 0, 0, 0, 0 . A, (ii) (p, q) = (4, 5) and index} a = 1 on the domain B, (iii) (p, q) = (5, 4) and indexa Therefore,= 2 on we the conclude domain that (i) (pC, q)in = (6, Figure 3) and index 3.4.a = 3 Hence,on the domain weA obtain, (ii) (p, q) the = (4, following 5) result.and indexa = 1 on the domain B, (iii) (p, q) = (5, 4) and indexa = 2 on the domain C in Figure6. Hence, we obtain the following result. Theorem 3.2. The rG family contains minimal surfaces with (p, q) = (6, 3), Theorem 7. The rG family contains minimal surfaces with (p, q) = (6, 3), indexa = 3 and minimal surfaces indexwitha(=p, q 3) =and (4, 5) minimal, indexa = 1. surfaces with (p, q) = (4, 5), indexa = 1.

A A

Figure 6. The locus det(W) = 0 and that of (φ, s) which satisﬁes (3) in (φ, s) φ ( π , π ], 0.4 < s < 1 . { | ∈ − 6 3 } Figure 3.4: The locus det(W ) = 0 and that of (ϕ, s) which satisﬁes (2.3) in (ϕ, s) ϕ ( π , π ], 0.4 < s < 1 . { | ∈ − 6 3 }

Remark 3.3. It is clear to see that the graph of x3 = det(W ) intersects the plane x3 = 0 at a curve looks like a straight line as in Figure 3.1 by Mathematica. The intersection curve is the locus at which the jump of Morse index is equal to

14 Mathematics 2020, 8, 1693 13 of 39

Remark 1. It is clear to see that the graph of x3 = det(W) intersects the plane x3 = 0 at a curve which looks like a straight line as in Figure3 by Mathematica. The intersection curve is the locus at which the jump of the Morse index is equal to 1. On the other hand, the situation of Figure4 is quite different. We obtain numerical evidence of the existence of two kinds of minimal surfaces, namely, the minimal surface with Morse index 1 and the minimal surface with Morse index 3. So there must exist a boundary at which the jump of the Morse index is equal to 2, and it remains an important problem to study such boundaries, which are more complicated than the above. Hence there might be error terms of the numerical approximation, and it is not clear to check the graph of x3 = det(W) is tangent to the plane x3 = 0 at some curves which look like parabolic curves as in Figure4 by Mathematica.

3.2. tG Family We can apply the same arguments as the tP family to the tG family since a type of Riemann surface of the tG family coincides with that of the tP family (see § 3.3 in [8]). We shall use the notation as in Appendix A.2 of AppendixA. Set

1 i 0 0 0 0 2 − 2  1  0 0 0 0 0 2   1 0 1  1 i  −  0 0 0 0 P1 = i 0 i  , P2 = − 2 − 2  ,  1 i  0 2 0  0 0 0 0    2 − 2       0 0 0 0 0 1    1 i   0 0 0 0  − 2 − 2  3 1 1  a a a  + a3 + a2 + a 4 2 3 2 i 2 i 3 i − ai − 2ai − 4ai ai 2ai 2ai  1 1 1 a 1 a a  P = + a3 + a2 , ai  4 − 2a − 4a2 − 2 − a4 2a i 2a2 i   i i i i i   a 1 1 a 1 a 1 a   i a + a3   4 2 − 4a − 2 i − 3 − 2 − 4 2a i   i ai ai i    G1 G1 G1 G1 G1 G1 ΩtG = ZA1 G2! ZA2 G2! ZA3 G2! ZB1 G2! ZB2 G2! ZB3 G2!!

π i 3 πi π i 3 πi π i and choose a1 = e 4 b, a2 = e 4 b, a3 = e− 4 b, a4 = e− 4 b, and a5 = e 4 /b. We deﬁne 5 5 1 Tj = := P1Paj P2ΩtG . { }j 1 2 j=1 SettingnC and C areo 3 3 complex matrices given by 1 2 × iB A iB iB 2iB i B − − − − − (C , C ) = A iB A iB 0 i B , 1 2  − −  iD iD iD C 0 C − −    1 we have the Riemann matrix τ = C1− C2 and deﬁne

0 1 0 T := (C , C ), T := 1 0 0 (C , C ), 6 1 2 7 −  1 2 0 0 0   0 0 1   0 0 0 T8 :=  0 0 0 (C1, C2), T9 := 0 0 1 (C1, C2). 1 0 0 0 1 0 −   −      Mathematics 2020, 8, 1693 14 of 39

Decompose T = (Z , Z ) into two 3 3 complex matrices. We introduce j j1 j2 × η((Z , Z ), (Z , Z )) = i tr(tZ Z tZ Z ) j1 j2 j1 j2 − j2 j1 − j1 j2 and set W = (η(Ti, Tj)). Let (p, q) denote the signature of W. For a decomposition Tj = (Zj1, Zj2), we set

1 1 K0 = Z + i Z τ Z ( τ)− , K00 = (iZ ) + i (iZ ) τ (iZ ) ( τ)− . j < j1 {< j1< − < j2} = j < j1 {< j1 < − < j2 } = Deﬁne

T (1 j 9) (K0, K0τ)(1 j 9) U = j ≤ ≤ , V = j j ≤ ≤ j j  (iTj 9 (10 j 18) (K00j 9, K00j 9τ)(10 j 18) − ≤ ≤  − − ≤ ≤ and W = ( (η(U , U ))), W = ( (η(V , V ))).  1 < i j 2 < i j Theorem 8. The nullity of the tG family is equal to the number of zero eigenvalues of W, counted with multiplicities plus 3. Moreover, if the number of zero eigenvalues of W W is equal to 8, then the Morse index 2 − 1 of the tG family is equal to the number of negative eigenvalues of W W , counted with multiplicities plus 1. 2 − 1 π For = (φ, s, x3) φ < , 0.4 < s < 1, x3 R , by using Mathematica, we obtain the graph D { | | | 4 ∈ } of x = det(W) and the plane x = 0 in (see Figure7). Their intersection consists of the points 3 3 D whose nullities are at least 4.

Figure 7. The graph of x = det W (the red domain) and the plane x = 0 (the green domain) in . 3 3 D They intersect at a curve which looks like a straight line and curves which look like parabolic curves.

The intersection of the three graphs of x3 = det(W), the plane x3 = 0, and x = ( A + B)( C + D) ( A + B)( C + D) in is given in Figure8. 3 −= < < = − < = −= < D From Figure8, we can show that the locus of (φ, s) which satisﬁes (6) contains two connected components. In fact, we will numerically compute the eigenvalues of W and W W . 2 − 1 We ﬁrst consider the eigenvalues of W. Substituting (φ, s) = ( π , 0.4), ( π , 0.5), ( π , 0.8) ± 4.1 ± 4.1 ± 4.1 to W yields the following sets of the eigenvalues:

35.6189, 22.208, 20.1174, 15.8479, 2.53191, 1.47707, 0.0950535, 0.042007, 0.00973469 , { − − − − } 51.9452, 38.3399, 31.0118, 27.6411, 2.29972, 2.08639, 0.350738, 0.143488, 0.0187563 , { − − − − − } 140.144, 119.05, 87.6919, 73.3903, 9.2786, 7.68567, 3.04764, 1.18144, 0.372079 . { − − − − } Mathematics 2020, 8, 1693 15 of 39

Next, we substitute (φ, s) = ( π , 0.4), ( π , 0.5), ( π , 0.8) to W and obtain the following sets ± 8 ± 8 ± 8 of the eigenvalues:

35.6372, 22.1863, 20.103, 15.8582, 2.52123, 1.47505, 0.0946025, 0.0420784, 0.00980311 , { − − − − } 52.0196, 38.1231, 30.8936, 27.6668, 2.27561, 2.0269, 0.347043, 0.143621, 0.0201313 , { − − − − − } 131.882, 96.7425, 81.38, 65.5126, 5.37746, 4.23261, 1.94126, 1.42518, 0.0835739 . { − − − }

Figure 8. The graph of x3 = det W (the red domain), the plane x3 = 0 (the green domain), and x = ( A + B)( C + D) ( A + B)( C + D) (the blue domain) in . 3 −= < < = − < = −= < D Moreover, substituting (φ, s) = (0, 0.4), (0, 0.5), (0, 0.8) to W, we have the following sets of eigenvalues:

35.6188, 22.2081, 20.1175, 15.8479, 2.53197, 1.47708, 0.0950562, 0.0420066, 0.00973429 , { − − − − } 51.9447, 38.3412, 31.0125, 27.641, 2.29986, 2.08675, 0.350761, 0.143487, 0.0187482 , { − − − − − } 140.187, 119.276, 87.7299, 73.461, 9.32263, 7.72632, 3.0591, 1.17902, 0.379217 . { − − − − } We now consider the eigenvalues of W W . Substituting (φ, s) = ( π , 0.4), ( π , 0.5), 2 − 1 ± 4.1 ± 4.1 ( π , 0.8) to W W yields the following sets of eigenvalues: ± 4.1 2 − 1 7.77508, 4.38731, 4.37729, 3.7917, 2.34541, 0.145468, 0.112206, 0.102314, 0.0342688, { − 0.0177381, 0, 0, 0, 0, 0, 0, 0, 0 , } 8.96534, 6.10147, 6.05441, 4.04023, 2.79856, 0.544306, 0.394305, 0.367723, 0.0998041, { 0.0287088, 0, 0, 0, 0, 0, 0, 0, 0 , } 24.074, 21.1387, 19.5594, 19.2666, 14.565, 8.19564, 7.48133, 0.960304, 0.71742, { − 0.568975, 0, 0, 0, 0, 0, 0, 0, 0 . } Mathematics 2020, 8, 1693 16 of 39

Next, we substitute (φ, s) = ( π , 0.4), ( π , 0.5), ( π , 0.8) to W W and obtain the following ± 8 ± 8 ± 8 2 − 1 sets of the eigenvalues:

7.75667, 4.37796, 4.368, 3.78497, 2.34032, 0.144906, 0.111943, 0.102423, 0.0339594, { − 0.0180341, 0, 0, 0, 0, 0, 0, 0, 0 , } 8.83532, 6.02283, 5.97696, 3.98793, 2.77818, 0.539496, 0.389623, 0.366682, 0.0950817, { 0.032076, 0, 0, 0, 0, 0, 0, 0, 0 , } 13.5834, 10.9767, 10.9737, 10.1408, 8.8041, 5.17527, 4.61406, 2.22136, 0.489929, { − 0.115717, 0, 0, 0, 0, 0, 0, 0, 0 . − } Moreover, substituting (φ, s) = (0, 0.4), (0, 0.5), (0, 0.8) to W W , we have the following sets 2 − 1 of the eigenvalues:

7.77519, 4.38737, 4.37734, 3.79174, 2.34544, 0.145471, 0.112207, 0.102313, 0.0342706, { − 0.0177364, 0, 0, 0, 0, 0, 0, 0, 0 , } 8.96612, 6.10194, 6.05487, 4.04054, 2.79869, 0.544335, 0.394332, 0.367729, 0.0998308, { 0.0286902, 0, 0, 0, 0, 0, 0, 0, 0 , } 24.1893, 21.2568, 19.6545, 19.3614, 14.6312, 8.22985, 7.5127, 0.924861, 0.71973, { − 0.598225, 0, 0, 0, 0, 0, 0, 0, 0 . }

Therefore, we conclude that (i) (p, q) = (6, 3) and indexa = 3 on the domain A, (ii) (p, q) = (5, 4) and indexa = 2 on the domain B, (iii) (p, q) = (4, 5) and indexa = 1 on the domain C in Figure9. Hence, we obtain the following result.

Theorem 9. The tG family contains minimal surfaces with (p, q) = (5, 4), indexa = 2 and minimal surfaces with (p, q) = (4, 5), indexa = 1.

AA BBB

Figure 9. The locus det(W) = 0 and that of (φ, s) which satisﬁes (6) in (φ, s) φ < π , 0.4 < s < 1 . { | | | 4 } 3.3. A One-Parameter Family Which Contains P Surface, D Surface, Gyroid, Lidinoid Schröder-Turk, Fogden, and Hyde [12] gave Figure 10 as a correlation diagram for the seven one-parameter families, namely, the H family, the rPD family, the tP family, the tD family, the tCLP family, the rG family, the tG family. Every line implies a one-parameter family, for example, the line labeled H indicates the H family, and so on. From Numerical Result 1, Numerical Result 2, Numerical Result 3 in [8], we can see the minimal surfaces with indexa = 1 for the H family, the rPD family, the tP family, the tD family. Combining these results, Theorems7 and9 yield the red lines in Figure 10. Therefore we have Mathematics 2020, 8, 1693 17 of 39

Theorem 10. The P surface, the D surface, the gyroid, the Lidinoid are contained in a one-parameter family which consists of minimal surfaces with indexa = 1. Mathematics 2020, xx, 5 17 of 37

tCLP

P: the P surface H tP D: the D surface G G: the gyroid tD L: the Lidinoid rG L tG

D rPD P

Figure 10. The diagram of the H family, the rPD family, the tP family, the tD family, the tCLP family, Figure 3.8. The diagram of the H family, the rPD family, the tP family, the tD family, the tCLP family, the rG family, the tG family (the red lines indicate minimal surfaces with indexa = 1). the rG family, the tG family (the red lines indicate minimal surfaces with indexa = 1) Author Contributions: Conceptualization, N.E. and T.S.; methodology, N.E. and T.S.; software, N.E. and T.S.; 2 validation, N.E. and T.S.;t formal1 z2 analysis,i(1+z N.E.) and2z T.S.; investigation, N.E. and T.S.; data curation, N.E. and T.S.; writing—originalDefine G1 draft:= preparation,( −w dz, N.E.w anddz T.S.;, w writing—reviewdz) as a column and editing, vector N.E. and which T.S.; visualization, consists N.E.of a basis of holomorphicand T.S.; project differentials administration, on N.E.M. and Up T.S. to All exact authors one-forms, have read and the agreed Abelian to the differentials published version of the of the second kind manuscript. are given by Funding: This research received no external funding. 1 z2 i(1 + z2) 2z z4 z6 i(z4 + z6) z5 Conflicts of Interest: The− authorsdz, declare no conflictdz, of interest.dz, − dz, dz, dz w w w w3 w3 w3 Appendix A. List of Period Calculations4 6 t z4 z6 i(z +z ) z5 3 and we set G2 = ( −3 dz, 3 dz, 3 dz). Let Aj, Bj = be a canonical homology basis on M. We shall determinew canonicalw homologyw bases and{ present}j period1 calculations of the Abelian We shall determine a complex 6 6 matrix given by differentials of the second kind along× the canonical homology bases for the rG family and the tG family as an appendix. G G G G G G 1 1 1 1 1 1 . Appendix A.1. rG Family ZA1 G2! ZA2 G2! ZA3 G2! ZB1 G2! ZB2 G2! ZB3 G2!! For a = s eiφ (0 < s < 1, φ ( π , π ]), let M be a hyperelliptic Riemann surface of genus three ∈ − 6 3 Notedefined that theby upper2 = ( complex3 3)( 3 +3 1 )6. matrix w z z a z a3 − × 2 t 1 z2 i(1+z ) 2z Define G1 := ( −w dz, w dz, w dz) as a column vector which consists of a basis of holomorphic differentials on MG.1 Up toG exact1 one-forms,G1 theG1 AbelianG1 differentialsG1 of the second A A A B B B kind are given by Z 1 Z 2 Z 3 Z 1 Z 2 Z 3

238 is the complex period1 matrixz2 oni(1 M+ z. 2) 2z z4 z6 i(z4 + z6) z5 − dz, dz, dz, − dz, dz, dz Consider the followingw isometriesw onwM: w3 w3 w3 

4 6 i(z4+z6) 5 and we set = t( z z , 2 π, z ). Let , 3 be1 a canonical homology basis on . G2 w−3 dz wπ3i dz iw3 dz Aj Bj j=1 w M φ (z, w) = (e 3 z, e 3 w), φ2(z,{w) =} , , j(z, w) = (z, w). We shall determine1 a complex 6 6 matrix given by − z z4 − × G G G G G G It is straightforward to1 show that1 1 1 1 1 . ZA1 G2! ZA2 G2! ZA3 G2! ZB1 G2! ZB2 G2! ZB3 G2!! 1 √3 0 0 0 0  2 2  √3 1 0 0 0 0   − 2 2  G1  0 0 1 0 0 0  G1 φ∗ =  −  , 1 G  1 √3  G 2!   2!  0 0 0 0   2 2   √3 1   0 0 0 0     − 2 2   0 0 0 0 0 1  −    Mathematics 2020, 8, 1693 18 of 39

Note that the upper complex 3 6 matrix ×

G1 G1 G1 G1 G1 G1 A A A B B B Z 1 Z 2 Z 3 Z 1 Z 2 Z 3 is the complex period matrix on M. Consider the following isometries on M:

2 πi π i 1 w ϕ (z, w) = (e 3 z, e 3 w), ϕ2(z, w) = , , j(z, w) = (z, w). 1 − z z4 − It is straightforward to show that

1 √3 0 0 0 0  2 2  √3 1 0 0 0 0   − 2 2  G1  0 0 1 0 0 0  G1 ϕ∗ =  −  , 1 G  1 √3  G 2!   2!  0 0 0 0   2 2   √3 1   0 0 0 0     − 2 2   0 0 0 0 0 1  −   

1 0 0 0 0 0 − 0 1 0 0 0 0   G1 0 0 1 0 0 0 G1 G1 G1 ϕ2∗ =  −  , j∗ = . G2!  0 0 0 1 0 0  G2! G2! − G2!  −   0 0 0 0 1 0     0 0 0 0 0 1    −  Remark A1. For φ = 0, there exists an isometry deﬁned by ϕ3(z, w) = (z, w) and we ﬁnd

1 0 0 0 0 0 0 1 0 0 0 0  −  G1 0 0 1 0 0 0 G1 ϕ3∗ =   . G2! 0 0 0 1 0 0 G2!   0 0 0 0 1 0  −  0 0 0 0 0 1     We now construct M as a two-sheeted branched cover of C. Let π : M C be the two-sheeted → covering deﬁned by (z, w) z which is branched at the following eight ﬁxed points of j: 7→

π i π i 2 πi 2 πi 1 e− 3 e 3 (0, 0), (∞, ∞), (a, 0), (ae 3 , 0), (ae− 3 , 0), , 0 , , 0 , , 0 . − a a a ! !

We prepare two copies of C and take two closed curves passing through the eight points, respectively. So we can divide C into two domains and label “ + ” and “ ” (see Figure A1). Slit them − along the thick lines. Identifying each of the upper (resp. lower) edges of the thick lines in (i) with each of the lower (resp. upper) edges of the thick lines in (ii), we obtain the hyperelliptic Riemann surface M of genus three. Mathematics 2020, 8, 1693 19 of 39

π ϕ ( 6 , 0] (i) π i ∈ − π i (ii) e 3 /a e 3 /a

2 πi 2 πi ae 3 + ae 3 + O O a a

2 πi 2 πi 1/a ae 3 − 1/a ae 3 − − − − − π i π i e− 3 /a e− 3 /a

ϕ (0, π ] (i) π i 6 π i (ii) e 3 /a ∈ e 3 /a + + 1/a 2 πi 1/a 2 πi − ae 3 − ae 3 a a O O

2 πi 2 πi ae− 3 − ae− 3 −

π i π i e− 3 /a e− 3 /a

ϕ ( π , π ] (i) ∈ 6 3 (ii) 1/a 1/a − a − a 2 2 3 πi π i 3 πi π i ae e 3 /a ae e 3 /a O O − − + 2 πi + 2 πi ae− 3 ae− 3 π i π i e− 3 /a e− 3 /a

(ii) −

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a

(i) +

Figure A1. M represented as a two-sheeted branched cover of C.

Remark that j can be represented as the 180◦ rotation around the middle axis between (i) and (ii) (see Figure A2).

(ii) −

⟳

(i) +

Figure A2. The action of j on M. Mathematics 2020, 8, 1693 20 of 39

Appendix A.1.1. A Canonical Homology Basis We ﬁrst take the following three key paths.

π π 2 C = (z, w) = (e 3 it/a, e 6 i t(t3 1) a3 + t3/a3/a ) 1 t ∞ , 1 { − − | ≤ ≤ } q p C = (z, w) = (at, ia2 t(1 t3) a3t3 + 1/a3 ) 0 t 1 , 2 { − | ≤ ≤ } q p C = (z, w) = (it, t (1/a3 a3)t3 i(1 + t6) ) 0 t ∞ , 3 { { − − } | ≤ ≤ } q 3 where we choose the branches as follows: t(t3 1) > 0 and arg a3 + t [ π , π ] for C . − a3 ∈ − 2 4 1 t(1 t3) > 0 and arg a3t3 + 1 [ π ,pπ ] for C . For C , arg qt ( 1 a3)t3 i(1 + t6) − a3 ∈ − 2 4 2 3 { a3 − − } ∈ [ π , π ]. q q p− 2 4 Remark A2. Straightforward calculations yield

t3 t3 t3 a3 + = s3 + cos(3φ) + i s3 sin(3φ), a3 s3 − s3 1 1 1 a3t3 + = s3t3 + cos(3φ) + i s3t3 sin(3φ), a3 s3 − s3 1 1 1 a3 t3 i(1 + t6) = s3 t3 cos(3φ) i 1 + t6 + + s3 t3 sin(3φ) . a3 − − s3 − − s3

3 For C , we have arg(a3 + t ) [ π, π ]. For C , arg(a3t3 + 1 ) [ π, π ] holds. For C , we ﬁnd 1 a3 ∈ − 2 2 a3 ∈ − 2 3 3 arg ( 1 a3)t3 i(1 + t6) [ π, π ]. Hence a3 + t , a3t3 + 1 , ( 1 a3)t3 i(1 + t6) have { a3 − − } ∈ − 2 a3 a3 a3 − − single-valued branches for φ ( π , π ]. q q q ∈ − 6 3 We now consider two cases, namely, the case φ ( π , π ) and the case φ [ π , π ]. We assume ∈ − 6 6 ∈ 6 3 φ ( π , π ), and introduce the following two paths. ∈ − 6 6

π π 2 C = (z, w) = (e 3 it/a, ie 6 i t(1 t3) a3 + t3/a3/a ) 0 t 1 , 4 { − | ≤ ≤ } q p C = (z, w) = (at, a2 t(t3 1) a3t3 + 1/a3 ) 1 t ∞ , 5 { − | ≤ ≤ } q p 3 where we choose the branches as follows: t(1 t3) > 0 and arg a3 + t [ π , π ] for C . − a3 ∈ − 4 4 4 t(t3 1) > 0 and arg a3t3 + 1 [ π , π ]pfor C . q − a3 ∈ − 4 4 5 Using C and C , weq shall describe a canonical homology basis on M by C 3 . To start with, p 4 5 { j}j=1 let us choose C4 as in Figure A3. Mathematics 2020, 8, 1693 21 of 39

(i) π i π i (ii) e 3 /a e 3 /a

C 2 πi 4 2 πi ae 3 + ae 3 + O O a a

2 πi 2 πi 1/a ae 3 − 1/a ae 3 − − − − − π i π i e− 3 /a e− 3 /a

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a C (i) 4 +

Figure A3. C4 in (i).

There exist l, m, n 0, 1 such that jl(C ) + C is homologous to jm(C ) + jn(C ) (write jl(C ) + ∈ { } 1 4 2 5 1 C jm(C ) + jn(C )) for an arbitrary φ ( π , π ). Setting φ = 0, we have l = m = n = 0 (see § 5.2.1 4 ∼ 2 5 ∈ − 6 6 in [8]). Thus, l = m = n = 0 hold for φ ( π , π ) (see Figure A4). In the process we have also shown ∈ − 6 6 that the equation

1 a2 t2 1 a2 + t2 √3 − dt i dt 3 3 Z0 t(1 t3) a3 + t − Z0 t(1 t3) a3 + t − a3 − a3 p 1 q 1 a2t2 p 1 q 1 + a2t2 + √3 i − dt dt = 0 (A1) Z0 t(1 t3) a3t3 + 1 − Z0 t(1 t3) a3t3 + 1 − a3 − a3 q q π π p p is satisﬁed for φ ( 6 , 6 ). ∈ − l 2z 2z By choosing a suitable l 0, 1 , we ﬁnd C1 + j(C4) j (C3). From dz = l dz, ∈ { } ∼ C1+j(C4) w j (C3) w we have R R 1 t 1 t ∞ t dt + i dt = ( 1)l dt. 3 Z0 t(1 t3) a3 + t Z0 t(1 t3) a3t3 + 1 − Z0 t ( 1 a3)t3 i(1 + t6) − a3 − a3 { a3 − − } p q p q q (A2)

Setting φ = 0, we obtain l = 0. It follows that C + j(C ) C holds for φ ( π , π ) 1 4 ∼ 3 ∈ − 6 6 (see Figure A5). Mathematics 2020, 8, 1693 22 of 39

(i) (ii) C1 π i π i e 3 /a e 3 /a

2 πi 2 πi ae 3 + ae 3 + O O a a C2 2 πi C5 2 πi 1/a ae 3 − 1/a ae 3 − − − − − π i π i e− 3 /a e− 3 /a

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a

C C2 (i) 5 + C1

Figure A4. C1, C2, C5 in (i).

(i) π i π i (ii) e 3 /a e 3 /a C3

2 πi 2 πi ae 3 + ae 3 + O O a a

2 πi 2 πi 1/a ae 3 − 1/a ae 3 − − − − − π i π i e− 3 /a e− 3 /a

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a

(i) + C3

Figure A5. C3 in (i).

1 z2 1 z2 i(1+z2) i(1+z2) By − dz = − dz, dz = dz, and (A2), we ﬁnd C1+j(C4) w C3 w C1+j(C4) w C3 w R R R R 1 1 + t2 2 i dt Z0 t ( 1 a3)t3 i(1 + t6) { a3 − − } q1 1 1 a2 t2 √3 1 a2 + t2 = − dt + i dt a − 2 t3 2 t3 Z0 t(1 t3) a3 + Z0 t(1 t3) a3 + − a3 − a3 i 1 p 1 a2qt2 √3 1 p 1 + a2qt2 + − dt dt , (A3) 2 0 3 3 3 1 − 2 0 3 3 3 1 Z t(1 t ) a t + 3 Z t(1 t ) a t + 3 − a − a 1 a2 +pt2 q 1 a2 t2p q i dt √3 − dt 3 3 − Z0 t(1 t3) a3 + t − Z0 t(1 t3) a3 + t − a3 − a3 p1 1q+ a2t2 p 1 q 1 a2t2 + dt + √3 i − dt = 0, (A4) Z0 t(1 t3) a3t3 + 1 Z0 t(1 t3) a3t3 + 1 − a3 − a3 p q p q Mathematics 2020, 8, 1693 23 of 39

∞ t dt Z0 t ( 1 a3)t3 i(1 + t6) { a3 − − } q 1 t 1 t = dt + i dt, (A5) 3 Z0 t(1 t3) a3 + t Z0 t(1 t3) a3t3 + 1 − a3 − a3 p q p q for φ ( π , π ). Combining (A1) and (A4), we have ∈ − 6 6 1 a2 t2 1 1 + a2t2 √3 − dt = dt, (A6) 3 Z0 t(1 t3) a3 + t Z0 t(1 t3) a3t3 + 1 − a3 − a3 1 p 1 a2qt2 1 p a2 +qt2 √3 − dt = dt, (A7) 3 Z0 t(1 t3) a3t3 + 1 Z0 t(1 t3) a3 + t − a3 − a3 q q π π p p for φ ( 6 , 6 ). ∈ − l 2 Similarly, there exists l 0, 1 such that C2 + j(C5) j (ϕ1(C3)). 2z 2z ∈ { } ∼ From dz = l 2 dz and (A5), we ﬁnd l = 0 (see Figure A6). C2+j(C5) w j (ϕ1(C3)) w R R

2 Figure A6. C2, ϕ1(C3), j(C5) in (i).

By C + j(C ) C , we obtain ϕ2(C ) + j(ϕ2(C )) ϕ2(C ) (see Figure A7). 1 4 ∼ 3 1 1 1 4 ∼ 1 3

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a 2 j(φ (C4)) + 1 (i) 2 φ1(C1)

2 2 Figure A7. ϕ1(C1), j(ϕ1(C4)) in (i).

l m Choosing suitable l, m 0, 1 , we have j (ϕ1(C2)) + j (ϕ1(C5)) C3. 2z ∈2z { } ∼ From l m dz = dz and (A5), we ﬁnd l = 1, m = 0 (see Figure A8). j (ϕ1(C2))+j (ϕ1(C5)) w C3 w R R Mathematics 2020, 8, 1693 24 of 39

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a j(φ1(C2)) (i) + φ1(C5)

Figure A8. j(ϕ1(C2)), ϕ1(C5) in (i).

To determine ϕ1(C3) and j(ϕ1(C3)), we introduce the following new path.

C := (z, w) = (eit, e2it 2i sin(3t) + 1/a3 a3 ) π/2 t 7π/6 , 6 { − − | ≤ ≤ } q where we choose the branch such that arg 2i sin(3t) + 1/a3 a3 ( π/4, π/4). − ∈ − p Remark A3. For C6, we obtain

w2 1 1 1 1 = z3 + a3 = s3 cos(3φ) + i 2 sin(3t) + s3 sin(3φ) . z4 − z3 a3 − s3 − − s3 

2 Thus, arg w ( π , π ) holds for φ ( π , π ) and we may choose the branch such that arg w ( 3 π, 5 π). z4 ∈ − 2 2 ∈ − 6 6 z2 ∈ 4 4 Substituting t = π/2, we have w = √1/a3 a3 2i, where arg √1/a3 a3 2i ( π/4, π/4). − − − − ∈ − By Remark A3, C C = ∅ holds. Substituting t = 7 π to (z, w) in C , 3 ∩ 6 6 6 6 2 π π 1 1 we ﬁnd (z, w) = ie 3 i, e 3 i a3 2i = j ϕ i, a3 2i j(ϕ (C )). − a3 − − 1 a3 − − ∈ 1 3 Hence j(ϕ (C )) C = (see Figures A9q and A10). q 1 3 ∩ 6 6

(i) π i π i (ii) e 3 /a e 3 /a C3

2 πi 2 πi ae 3 3 + ae + C6 O O a 1/a a − C6 2 πi 2 πi ae− 3 1/a ae− 3 − − − j(φ1(C3)) π i π i e− 3 /a e− 3 /a

(ii) j(φ1(C3)) − C6

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a

(i) C6 + C3

Figure A9. C6 in M. Mathematics 2020, 8, 1693 25 of 39

(ii) j(φ1(C3)) −

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a

(i) + φ1(C3)

Figure A10. ϕ1(C3), j(ϕ1(C3)) in M.

By C + j(C ) C , we obtain ϕ (C ) + j(ϕ (C )) ϕ (C ), and so Figure A11 follows. 1 4 ∼ 3 1 1 1 4 ∼ 1 3

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a j(φ (C )) (i) 1 4 + φ1(C1)

Figure A11. ϕ1(C1), j(ϕ1(C4)) in (i).

From j(ϕ (C )) + ϕ (C ) C , we have j(ϕ2(C )) + ϕ2(C ) ϕ (C ) (see Figure A12). 1 2 1 5 ∼ 3 1 2 1 5 ∼ 1 3

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a j(φ2(C )) (i) 1 2 + 2 φ1(C5)

2 2 Figure A12. j(ϕ1(C2)), ϕ1(C5) in (i).

Therefore, we can describe a canonical homology basis as follows (see Figure A13).

A = C + j(C ), A = C j(C ) ϕ (C ) + j(ϕ (C )) C + j(C ), 1 − 2 2 2 1 − 1 − 1 2 1 2 − 3 3 A = ϕ2(C ) + j(ϕ2(C )) A A , B = C + j(C ) + ϕ2(C ) j(C ), 3 − 1 1 1 1 − 1 − 2 1 − 1 1 1 3 − 3 B = C + ϕ2(C ), B = ϕ (C ) + ϕ2(C ). 2 − 3 1 3 3 − 1 3 1 3

(ii)

A3 A1 A2 (i) B1 B3 B2

Figure A13. A canonical homology basis on M.

Next we assume φ [ π , π ], and introduce the following two paths. ∈ 6 3

π π 2 C = (z, w) = (e 3 it/a, ie 6 i t(1 t3) a3 + t3/a3/a ) 0 t 1 , 7 { − | ≤ ≤ } q p C = (z, w) = (at, a2 t(t3 1) a3t3 + 1/a3 ) 1 t ∞ , 8 { − | ≤ ≤ } q p Mathematics 2020, 8, 1693 26 of 39

3 where we choose the branches as follows: t(1 t3) > 0 and arg a3 + t [ 3 π, π ] for C . − a3 ∈ − 4 − 4 7 t(t3 1) > 0 and arg a3t3 + 1 [ 3 π, pπ ] for C . q − a3 ∈ − 4 − 4 8 Using C and C , weq shall describe a canonical homology basis on M by C 3 . To start with, p 7 8 { j}j=1 let us choose C3 as in Figure A14.

(i) (ii) 1/a C3 1/a − a − a 2 2 3 πi π i 3 πi π i ae e 3 /a ae e 3 /a O O − − + 2 πi + 2 πi ae− 3 ae− 3 π i π i e− 3 /a e− 3 /a

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a

(i) + C3

Figure A14. C3 in (i).

There exist l, m 0, 1 such that jl(C ) + jm(C ) C . So we ﬁnd ∈ { } 2 8 ∼ 3 2z 2z dz = dz, l m Zj (C2)+j (C8) w ZC3 w that is,

∞ t dt Z0 t ( 1 a3)t3 i(1 + t6) { a3 − − } q 1 t 1 t = ( 1)li dt + ( 1)m+1 dt. 3 − Z0 t(1 t3) a3t3 + 1 − Z0 t(1 t3) a3 + t − a3 − a3 q q π p p Setting φ = 3 , we have l = m = 1 (see Figure A15).

Figure A15. Cont. Mathematics 2020, 8, 1693 27 of 39

Figure A15. j(C2), j(C8) in M.

Thus, we obtain

∞ t 1 t 1 t dt = dt i dt. (A8) 3 Z0 t ( 1 a3)t3 i(1 + t6) Z0 t(1 t3) a3 + t − Z0 t(1 t3) a3t3 + 1 { a3 − − } − a3 − a3 q p q p q 1 z2 1 z2 By − dz = − dz, we ﬁnd j(C2)+j(C8) w C3 w R R 1 1 + t2 1 1 1 a2t2 1 a2 t2 2 dt = − dt i − dt . (A9) 0 1 3 3 6 a 0 3 3 3 1 − 0 3 3 t3 Z t ( 3 a )t i(1 + t ) Z t(1 t ) a t + 3 Z t(1 t ) a + { a − − } − a − a3 q p q p q ( + 2) ( + 2) From i 1 z dz = i 1 z dz, we have j(C2)+j(C8) w C3 w R R 1 1 + a2t2 1 a2 + t2 dt = i dt. (A10) 3 Z0 t(1 t3) a3t3 + 1 − Z0 t(1 t3) a3 + t − a3 − a3 p q p q To determine j(ϕ1(C3)), we introduce the following path.

C = (z, w) = (eit, e2it 2i sin(3t) + 1/a3 a3 ) π/2 t 7π/6 , 9 { − − | ≤ ≤ } q where we choose the branch such that arg 2i sin(3t) + 1/a3 a3 [ 3π/4, π/4]. − ∈ − − p Remark A4. For C9, we obtain

w2 1 1 = s3 cos(3φ) + i 2 sin(3t) + s3 sin(3φ) . z4 s3 − − s3 

2 Thus, arg w [ 3 π, π ] holds for φ [ π , π ] and we may choose the branch such that z4 ∈ − 2 − 2 ∈ 6 3 arg w [ π , 3 π]. Substituting t = π/2, we have w = √1/a3 a3 2i, where arg √1/a3 a3 2i z2 ∈ 4 4 − − − − ∈ [ 3π/4, π/4]. − − By Remark A4, C C = ∅ holds. Substituting t = 7 π to (z, w) in C , 3 ∩ 9 6 6 9 2 π π 1 1 we ﬁnd (z, w) = ie 3 i, e 3 i a3 2i = j ϕ i, a3 2i j(ϕ (C )). − a3 − − 1 a3 − − ∈ 1 3 Hence j(ϕ (C )) C = ∅ (see Figure A16q). q 1 3 ∩ 9 6 Mathematics 2020, 8, 1693 28 of 39

(i) (ii) C 1/a 9 C3 1/a − a − a 2 2 3 πi π i C9 3 πi π i ae e 3 /a ae e 3 /a O O j(φ1(C3)) − − + 2 πi + 2 πi ae− 3 ae− 3 π i π i e− 3 /a e− 3 /a

(ii) j(φ1(C3)) − C9 π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a

+ (i) C9

Figure A16. j(ϕ1(C3)), C9 in M.

From j(C ) + j(C ) C , we have ϕ (C ) + ϕ (C ) j(ϕ (C )) (see Figure A17). 2 8 ∼ 3 1 2 1 8 ∼ 1 3 (i) (ii) 1/a 1/a − a − a 2 πi 2 ae 3 3 πi π i π i ae e 3 /a e 3 /a φ1(C8) O φ1(C2) O − − + 2 πi + 2 πi ae− 3 ae− 3 π i π i e− 3 /a e− 3 /a

(ii) φ (C ) − 1 2 φ1(C8) π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a

Figure A17. ϕ1(C2), ϕ1(C8) in (ii).

l 2 m 2 Choosing suitable l, m 0, 1 , we obtain j (ϕ1(C1)) + j (ϕ1(C7)) j(ϕ1(C3)). 2z ∈ { 2z } ∼ By l 2 m 2 dz = dz and (A8), we ﬁnd l = m = 1 (see Figure A20). j (ϕ1(C1))+j (ϕ1(C7)) w j(ϕ1(C3)) w R R Mathematics 2020, 8, 1693 29 of 39

(i) (ii) 1/a 1/a − a − a 2 2 3 πi π i 3 πi π i ae e 3 /a ae e 3 /a O 2 O j(φ1(C7)) − − + 2 πi + 2 πi ae− 3 ae− 3 π i π i e− 3 /a 2 e− 3 /a 2 j(φ1(C1)) j(φ1(C1)) (ii) 2 j(φ (C7)) 1 −

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a 2 2 Figure A18. j(ϕ1(C1)), j(ϕ1(C7)) in (ii).

By similar arguments as j(ϕ (C )) C = ∅, we obtain j(ϕ (C )) j(ϕ (C )) = ∅ and 1 3 ∩ 9 6 1 3 ∩ 1 9 6 ϕ2(C ) j(ϕ (C )) = ∅ (see Figure A19). 1 3 ∩ 1 9 6

(i) (ii) 1/a 1/a − a − a 2 2 3 πi π i 3 πi π i ae e 3 /a ae e 3 /a O j(φ (C )) O 2 1 3 φ (C3) 1 − + − + 2 πi 2 πi ae− 3 j(φ (C )) ae− 3 1 9 j(φ1(C9)) π i π i e− 3 /a e− 3 /a

(ii)

j(φ1(C9)) −

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a

(i) 2 + φ1(C3)

2 Figure A19. ϕ1(C3), j(ϕ1(C9)) in M.

From j(ϕ2(C )) + j(ϕ2(C )) j(ϕ (C )), we have C + C j(ϕ2(C )) (see Figure A20). 1 1 1 7 ∼ 1 3 1 7 ∼ 1 3 Mathematics 2020, 8, 1693 30 of 39

(ii) C7 −

π i π i e 3 2 πi 1 2 πi e− 3 a O ae 3 ae− 3 ∞ a − a a C7 (i) + C1

Figure A20. C1, C7 in M.

Therefore, we can obtain the same canonical homology basis as for φ ( π , π ) (see Figure A13). ∈ − 6 6 Finally, by Figures A15 and A20, we ﬁnd C1 C2 + j(C7) j(C8) 0. Combining (A10) and 1 z2 − − ∼ C C +j(C ) j(C ) −w dz = 0 implies 1− 2 7 − 8 R 1 1 + a2t2 √3 1 1 a2t2 1 a2 t2 dt = i − dt − dt . (A11) 0 3 3 3 1 2 0 3 3 3 1 − 0 3 3 t3 Z t(1 t ) a t + 3 Z t(1 t ) a t + 3 Z t(1 t ) a + − a − a − a3 p q p q p q Appendix A.1.2. Periods of the Abelian Differentials of the Second Kind Let A , B 3 be the canonical homology basis on M which is given in the previous subsection. { j j}j=1 We ﬁrst calculate the complex period matrix

G1 G1 G1 G1 G1 G1 . A A A B B B Z 1 Z 2 Z 3 Z 1 Z 2 Z 3 Straightforward calculations yield

1 z2 √3 1 1 + a2t2 i 1 1 a2t2 − dz = dt + − dt, ZC1 w − 2a Z0 t(1 t3) a3t3 + 1 2a Z0 t(1 t3) a3t3 + 1 − a3 − a3 i(1 + z2) 1 1 p 1 + qa2t2 √3 i p1 1 qa2t2 dz = dt + − dt, ZC1 w 2a Z0 t(1 t3) a3t3 + 1 2a Z0 t(1 t3) a3t3 + 1 − a3 − a3 2z 1 p t q p q dz = 2i dt, ZC1 w − Z0 t(1 t3) a3t3 + 1 − a3 1 z2 i p1 1q a2t2 − dz = − dt, ZC2 w − a Z0 t(1 t3) a3t3 + 1 − a3 i(1 + z2) 1 1p 1 +qa2t2 dz = dt, ZC2 w a Z0 t(1 t3) a3t3 + 1 − a3 2z 1 p t q dz = 2i dt. ZC2 w − Z0 t(1 t3) a3t3 + 1 − a3 p q Mathematics 2020, 8, 1693 31 of 39

From (A3), (A6), (A7), (A9) and (A11), we ﬁnd

1 z2 1 1 + t2 − dz = 2i dt ZC3 w Z0 t ( 1 a3)t3 i(1 + t6) { a3 − − } 2 √3 1 q 1 + a2t2 1 1 a2t2 = dt + i − dt , a − 3 0 3 3 3 1 0 3 3 3 1 Z t(1 t ) a t + 3 Z t(1 t ) a t + 3 − a − a i(1 + z2) p 2z q ∞ pt q dz = 0, dz = 2 dt. ZC3 w ZC3 w − Z0 t ( 1 a3)t3 i(1 + t6) { a3 − − } q Therefore, setting

1 1 1 + a2t2 1 1 1 a2t2 A := dt, B := − dt, a Z0 t(1 t3) a3t3 + 1 a Z0 t(1 t3) a3t3 + 1 − a3 − a3 1 p tq ∞ p qt C := 4 dt, D := 4 dt, Z0 t(1 t3) a3t3 + 1 Z0 t ( 1 a3)t3 i(1 + t6) − a3 { a3 − − } p q q we have

G1 G1 G1 G1 G1 G1 A A A B B B Z 1 Z 2 Z 3 Z 1 Z 2 Z 3 2 iB 2 A 2 iB A iB 2 A √3A 3 iB 2 A 2 iB − √3 − − √3 − √3 − √3 − = 2A 0 A + √3 iB 2√3 iB A √3 iB 0 (A12) − − −  i C 2 i C + D 2 i C D i C D 0 D  − − − −    for φ ( π , π ]. ∈ − 6 3 Next we shall calculate

G2 G2 G2 G2 G2 G2 . A A A B B B Z 1 Z 2 Z 3 Z 1 Z 2 Z 3 The following lemma appears in Section 3.2 of [8].

Lemma A1. For an arbitrary one-cycle γ, we have

z4 a6 z3 1 a6 dz dz = 2 dz + − , (A13) γ w3 3(z6 + 1)2 γ w a3 γ w Z Z Z z5 a6 z4 1 a6 z dz = 2 dz + − dz , (A14) γ w3 (z6 + 1)2 γ w a3 γ w Z Z Z z6 5a6 z5 1 a6 z2 dz = 2 dz + − dz . (A15) γ w3 3(z6 + 1)2 γ w a3 γ w Z Z Z Mathematics 2020, 8, 1693 32 of 39

Note that M can be constructed via a glueing the following two curves by the relation x = 1/z, − y = w/z4: 1 1 w2 = z(z3 a3) z3 + y2 = x(x3 a3) x3 + . − a3 ←→ − a3 1 w x= , y= −|z{z} z4 We also have Lemma A1 in the coordinates (x, y). We now assume φ ( π , π ). C and C can be rewritten as ∈ − 6 6 1 5

π π i 2 i 3 3 3 3 C = (x, y) = (ae− 3 t, a e− 6 t(1 + t )( a t + 1/a )) 1 t 0 , 1 { − − | − ≤ ≤ } q C = (x, y) = (t/a, t(1 + t3)(a3 t3/a3)/a2) 1 t 0 . 5 { − − | − ≤ ≤ } q Recall that C + C C C 0 and C j(C ) + C j(C ) C + j(C ) C + j(C ) is a 1 4 − 2 − 5 ∼ 1 − 1 4 − 4 − 2 2 − 5 5 closed curve on M. Hence we obtain

z4 3 dz C j(C )+C j(C ) C +j(C ) C +j(C ) w Z 1− 1 4− 4 − 2 2 − 5 5 z4 z4 z4 z4 = 3 dz + 3 dz 3 dz 3 dz C j(C ) w C j(C ) w − C j(C ) w − C j(C ) w Z 1− 1 Z 4− 4 Z 2− 2 Z 5− 5 x6 z4 z4 x6 = 3 dx + 3 dz 3 dz 3 dx = 0. C j(C ) y C j(C ) w − C j(C ) w − C j(C ) y Z 1− 1 Z 4− 4 Z 2− 2 Z 5− 5 Combining this equation and Lemma A1 yields

1 6 3 1 6 2 6 5 2 π i 1 a 2t 2 π i (1 a )t 2a t a e− 3 − − dt + 5a e 6 − − − dt 3 Z0 t(1 t3) a3 + t Z0 t(1 t3) a3t3 + 1 − a3 − a3 1p 1 a6q+ 2a6t3 1 p(1 a6)t2 q2t5 + i − dt 5 − − dt = 0. (A16) 3 Z0 t(1 t3) a3t3 + 1 − Z0 t(1 t3) a3 + t − a3 − a3 p q p q Similarly, C C + j(C ) 0 and C j(C ) C + j(C ) C + j(C ) is a closed curve on M. 1 − 3 4 ∼ 1 − 1 − 3 3 − 4 4 z4 By Lemma A1 and C j(C ) C +j(C ) C +j(C ) 3 dz = 0, we have 1− 1 − 3 3 − 4 4 w R ∞ t4 i dt 0 3 Z t ( 1 a3)t3 i(1 + t6) { a3 − − } q π e 6 ia4 1 1 a6 2t3 1 (1 a6)t2 + 2a6t5 = 6 2 i − − dt 5 − dt . (A17) 3(a + 1) 0 3 3 t3 − 0 3 3 3 1 Z t(1 t ) a + Z t(1 t ) a t + 3 − a3 − a p q p q z6 From Lemma A1 and C j(C ) C +j(C ) C +j(C ) 3 dz = 0, we ﬁnd 1− 1 − 3 3 − 4 4 w R ∞ t6 i dt 0 3 Z t ( 1 a3)t3 i(1 + t6) { a3 − − } q π i 2 1 6 2 5 1 6 6 3 e− 6 a (1 a )t 2t 1 a + 2a t = 6 2 5i − − dt − dt . (A18) 3(a + 1) 0 3 3 t3 − 0 3 3 3 1 Z t(1 t ) a + Z t(1 t ) a t + 3 − a3 − a p q p q Mathematics 2020, 8, 1693 33 of 39

z5 By Lemma A1 and C j(C ) C +j(C ) C +j(C ) 3 dz = 0, we obtain 1− 1 − 3 3 − 4 4 w R ∞ t5 dt 0 3 Z t ( 1 a3)t3 i(1 + t6) { a3 − − } q a3 1 (1 a6)t 2t4 1 (1 a6)t + 2a6t4 = 6 2 − − dt + i − dt . (A19) (a + 1) 0 3 3 t3 0 3 3 3 1 Z t(1 t ) a + Z t(1 t ) a t + 3 − a3 − a p q p q From (A17) (A18) = 0, we have − 1 6 3 1 6 2 6 5 2 π 1 a 2t (1 a )t + 2a t a e 6 i i − − dt 5 − dt 0 3 3 t3 − 0 3 3 3 1 Z t(1 t ) a + Z t(1 t ) a t + 3 − a3 − a 1 q 6 6 3 1 q 6 2 5 π ip 1 a + 2a t p (1 a )t 2t e− 6 − dt + 5i − − dt = 0. (A20) − − 0 3 3 3 1 0 3 3 t3 Z t(1 t ) a t + 3 Z t(1 t ) a + − a − a3 p q p q Combining (A16) and (A20) yields

a2 1 1 a6 2t3 − − dt 2 3 Z0 t(1 t3) a3 + t − a3 p 5 q1 (1 a6)t2 + 2a6t5 1 1 1 a6 + 2a6t3 = a2 − dt + − dt, (A21) − 2√3 Z0 t(1 t3) a3t3 + 1 √3 Z0 t(1 t3) a3t3 + 1 − a3 − a3 1 (1 a6)t2 2pt5 q p q 5 − − dt 3 Z0 t(1 t3) a3 + t − a3 p 10 q1 (1 a6)t2 + 2a6t5 1 1 1 a6 + 2a6t3 = a2 − dt + − dt. (A22) − √3 Z0 t(1 t3) a3t3 + 1 √3 Z0 t(1 t3) a3t3 + 1 − a3 − a3 p q p q Next we assume φ [ π , π ]. C and C can be rewritten as ∈ 6 3 1 8

π π i 2 i 3 3 3 3 C = (x, y) = (ae− 3 t, a e− 6 t(1 + t )( a t + 1/a )) 1 t 0 , 1 { − − | − ≤ ≤ } q C = (x, y) = (t/a, t(1 + t3)(a3 t3/a3)/a2) 1 t 0 . 8 { − − | − ≤ ≤ } q C C + j(C ) j(C ) 0 and C j(C ) C + j(C ) C + j(C ) + C j(C ) is a closed curve 1 − 2 7 − 8 ∼ 1 − 1 − 2 2 − 7 7 8 − 8 z4 on M. From Lemma A1 and C j(C ) C +j(C ) C +j(C )+C j(C ) 3 dz = 0, we ﬁnd 1− 1 − 2 2 − 7 7 8− 8 w R 1 6 3 1 6 2 6 5 2 π i 1 a 2t 2 π i (1 a )t 2a t a e− 3 − − dt + 5a e 6 − − − dt 3 − Z0 t(1 t3) a3 + t Z0 t(1 t3) a3t3 + 1 − a3 − a3 1 p 1 a6 +q2a6t3 1 (1 pa6)t2 2qt5 + i − dt + 5 − − dt = 0. (A23) 3 Z0 t(1 t3) a3t3 + 1 Z0 t(1 t3) a3 + t − a3 − a3 p q p q Mathematics 2020, 8, 1693 34 of 39

j(C ) C + j(C ) 0 and C + j(C ) C + j(C ) C + j(C ) is a closed curve on M. 2 − 3 8 ∼ − 2 2 − 3 3 − 8 8 z4 By Lemma A1 and C +j(C ) C +j(C ) C +j(C ) 3 dz = 0, we obtain − 2 2 − 3 3 − 8 8 w R ∞ t4 i dt 0 3 Z t ( 1 a3)t3 i(1 + t6) { a3 − − } q a2 1 (1 a6)t2 2t5 1 1 a6 + 2a6t3 = 6 2 5 − − dt + i − dt . (A24) 3(a + 1) − 0 3 3 t3 0 3 3 3 1 Z t(1 t ) a + Z t(1 t ) a t + 3 − a3 − a p q p q z6 From Lemma A1 and C +j(C ) C +j(C ) C +j(C ) 3 dz = 0, we have − 2 2 − 3 3 − 8 8 w R ∞ t6 i dt 0 3 Z t ( 1 a3)t3 i(1 + t6) { a3 − − } q a4 1 1 a6 2t3 1 (1 a6)t2 + 2a6t5 = 6 2 − − dt 5i − dt . (A25) 3(a + 1) 0 3 3 t3 − 0 3 3 3 1 Z t(1 t ) a + Z t(1 t ) a t + 3 − a3 − a p q p q z5 By Lemma A1 and C +j(C ) C +j(C ) C +j(C ) 3 dz = 0, we ﬁnd − 2 2 − 3 3 − 8 8 w R ∞ t5 dt 0 3 Z t ( 1 a3)t3 i(1 + t6) { a3 − − } q a3 1 (1 a6)t 2t4 1 (1 a6)t + 2a6t4 = 6 2 − − dt i − dt . (A26) (a + 1) 0 3 3 t3 − 0 3 3 3 1 Z t(1 t ) a + Z t(1 t ) a t + 3 − a3 − a p q p q From (A24) (A24) = 0, we have − 1 1 a6 2t3 1 (1 a6)t2 + 2a6t5 a2 − − dt + 5ia2 − dt 3 − Z0 t(1 t3) a3 + t Z0 t(1 t3) a3t3 + 1 − a3 − a3 p1 1 qa6 + 2a6t3 p1 (1 a6q)t2 2t5 + i − dt 5 − − dt = 0. (A27) 3 Z0 t(1 t3) a3t3 + 1 − Z0 t(1 t3) a3 + t − a3 − a3 p q p q Combining (A23) and (A27) yields

a2 1 1 a6 2t3 − − dt 2 3 Z0 t(1 t3) a3 + t − a3 q1 6 2 6 5 1 6 6 3 p 5 2 (1 a )t + 2a t 1 2 π 1 a + 2a t = a − dt + e 3 i − dt, (A28) − 2√3 Z0 t(1 t3) a3t3 + 1 √3 Z0 t(1 t3) a3t3 + 1 − a3 − a3 1 (1 a6)t2 2pt5 q p q 5 − − dt 3 Z0 t(1 t3) a3 + t − a3 q 1 6 2 6 5 1 6 6 3 p 10 π 2 (1 a )t + 2a t 1 1 a + 2a t = e 3 ia − dt + − dt. (A29) √3 Z0 t(1 t3) a3t3 + 1 √3 Z0 t(1 t3) a3t3 + 1 − a3 − a3 p q p q Mathematics 2020, 8, 1693 35 of 39

For φ ( π , π ], straightforward calculations yield ∈ − 6 3 π z4 10 e 6 ia4 1 (1 a6)t2 + 2a6t5 dz = − dt, 3 ( 6 + )2 1 ZC1 j(C1) w − 3 a 1 Z0 t(1 t3) a3t3 + − − a3 z5 2 i a3 1 p(1 a6)t +q2a6t4 dz = − dt, 3 ( 6 + )2 1 ZC1 j(C1) w − a 1 Z0 t(1 t3) a3t3 + − − a3 6 π i 2 1 6 q 6 3 z 2 e− 6 a p 1 a + 2a t dz = − dt, 3 ( 6 + )2 1 ZC1 j(C1) w 3 a 1 Z0 t(1 t3) a3t3 + − − a3 z4 2 i a2 p1 1 aq6 + 2a6t3 dz = − dt, 3 ( 6 + )2 1 ZC2 j(C2) w − 3 a 1 Z0 t(1 t3) a3t3 + − − a3 z5 2 i a3 1 p(1 a6)t +q2a6t4 dz = − dt, 3 ( 6 + )2 1 ZC2 j(C2) w − a 1 Z0 t(1 t3) a3t3 + − − a3 z6 10 i a4 1p (1 a6)qt2 + 2a6t5 dz = − dt. 3 ( 6 + )2 1 ZC2 j(C2) w − 3 a 1 Z0 t(1 t3) a3t3 + − − a3 p q Moreover, by (A17)–(A19), (A21), (A22), (A24)–(A26), (A28) and (A29), we ﬁnd

z4 z6 3 dz = 3 dz ZC3 w − ZC3 w π i 2 1 6 2 6 5 1 6 6 3 2 e 3 a 2 (1 a )t + 2a t π 1 a + 2a t = 5 a − dt + e 3 i − dt , 3√3(a6 + 1)2 − 0 3 3 3 1 0 3 3 3 1 Z t(1 t ) a t + 3 Z t(1 t ) a t + 3 − a − a z5 ∞ t5p q p q 3 dz = 3 dt C w − 0 Z 3 Z t ( 1 a3)t3 i(1 + t6) { a3 − − } q for φ ( π , π ]. ∈ − 6 3 Therefore, setting

10 a4 1 (1 a6)t2 + 2a6t5 2 a2 1 1 a6 + 2a6t3 E := − dt, F := − dt, ( 6 + )2 1 ( 6 + )2 1 3 a 1 Z0 t(1 t3) a3t3 + 3 a 1 Z0 t(1 t3) a3t3 + − a3 − a3 2 a3 1 p(1 a6)tq+ 2a6t4 ∞ p t5 q H := 6 2 − dt, I := 2 3 dt, (a + 1) 0 3 3 3 1 0 1 3 3 6 Z t(1 t ) a t + 3 Z t ( a )t i(1 + t ) − a { a3 − − } p q q we have

2 π 2 2 π π 3 πi 3 i i( E + F) (e 3 iE e 3 i F) e E e F √3 √−3 − − π π G2 G2 G2 =  3 i 3 i  , (A30) A A A E F 0 e− E + e F Z 1 Z 2 Z 3 − −  i H 2 i H + I 2 i H I   − −   + π π 2 π π E F √3(e 3 iE + e 3 i F) (e 3 iE + e 3 i F) √3 − √3 − π i π i G2 G2 G2 = √3 i(E F) e 3 E + e− 3 F 0  (A31) B1 B2 B3 − Z Z Z i H I 0 I  − −    for φ ( π , π ]. ∈ − 6 3 Mathematics 2020, 8, 1693 36 of 39

Appendix A.2. tG Family Set a = b4 + 1/b4, where b = s eiφ (0 < s < 1, φ < π ). Let M be a hyperelliptic Riemann | | 4 surface of genus three deﬁned by 2 = 8 + 4 + 1 = ( 4 + 4)( 4 + 1 ). w z az z b z b4 2 t 1 z2 i(1+z ) 2z Deﬁne G1 := ( −w dz, w dz, w dz) as a column vector which consists of a basis of holomorphic differentials on M. Up to exact one-forms, the Abelian differentials of the second kind are given by

1 z2 i(1 + z2) 2z z4 z6 i(z4 + z6) z5 − dz, dz, dz, − dz, dz, dz w w w w3 w3 w3 4 6 ( 4+ 6) 5 and we set G = t( z z dz, i z z dz, z dz). Let A , B 3 be a canonical homology basis on M. 2 w−3 w3 w3 { j j}j=1 We shall determine a complex 6 6 matrix given by × G G G G G G 1 1 1 1 1 1 . ZA1 G2! ZA2 G2! ZA3 G2! ZB1 G2! ZB2 G2! ZB3 G2!!

Consider the following isometries on M:

j(z, w) = (z, w), ϕ(z, w) = (iz, w). − It is straightforward to show that

0 1 0 0 0 0 1 0 0 0 0 0 −  G1 G1 G1 0 0 1 0 0 0 G1 j∗ = , ϕ∗ =  −  . G2! − G2! G2!  0 0 0 0 1 0  G2!    0 0 0 1 0 0   −   0 0 0 0 0 1    −  Remark A5. For φ = 0, there exists an isometry deﬁned by ψ(z, w) = (z, w) and we ﬁnd

1 0 0 0 0 0 0 1 0 0 0 0  −  G1 0 0 1 0 0 0 G1 ψ∗ =   . G2! 0 0 0 1 0 0 G2!   0 0 0 0 1 0  −  0 0 0 0 0 1     We now construct M as a two-sheeted branched cover of C. Let π : M C be the two-sheeted → covering deﬁned by (z, w) z which is branched at the following eight ﬁxed points of j: 7→ π i 3 πi π i 3 πi e± 4 e± 4 (be± 4 , 0), (be± 4 , 0), , 0 , , 0 . ( b ! b !)

We prepare two copies of C and take two closed curves passing through the eight points, respectively. So we can divide C into two domains and label “ + ” and “ ” (see Figure A21). − Slit them along the thick lines. Identifying each of the upper (resp. lower) edges of the thick lines in (i) with each of the lower (resp. upper) edges of the thick lines in (ii), we obtain the hyperelliptic Riemann surface M of genus three. Mathematics 2020, 8, 1693 37 of 39 Mathematics 2020, xx, 5 35 of 37

(i) (ii)

3 π 3 π e 4 i/b e 4 i/b

− π − π e 4 i/b e 4 i/b + π + π 4 i 4 i 3 π b e 3 π b e b e 4 i b e 4 i π π 3 πi 4 i 3 πi 4 i b e− 4 b e− b e− 4 b e−

3 πi π i 3 πi π i e− 4 /b e− 4 /b e− 4 /b e− 4 /b

(ii) −

π i π i π i π i 3 πi 3 πi 3 πi 3 πi e 4 /b b e 4 b e− 4 e− 4 /b e− 4 /b b e− 4 b e 4 e 4 /b

(i) +

FigureFigure A21. A21.MMrepresentedrepresented as a as two-sheeted a two-sheeted branched branched cover cover of C. of C

RemarkWe can A7. applyStraightforward the same arguments calculations as in yield[8] (see § 5.3 in [8]). We ﬁrst introduce the following key one-cycles.

8 4 8 4 1 4 t + at + 1 = t + 8 b +4 t + 1 C1 = (z, w) = (t, t + at b+4 1 ) 0 t ∞ { | ≤ ≤ } (z, wp) = ( it,1 t8 + at4 + 1 ) ∞ t 1 0 , ∪ {= t8 + s4−+ t4 cos(4ϕ) +|1 −+ i ≤ ≤ +}s4 t4 sin(4ϕ), s4 − s4 C = (z, w) = ( it, t8 + atp4+ 1 ) 1 t 1 2 { − | − ≤ ≤ } 4 1 1 4 a + 2 cos(4t) = s +p it cos2(it4ϕ) + 2 cos(4t) + i + s sin(4ϕ). (z, w) =s (4e , e a + 2 cos(4t)) π−/2s4 t π/2 , ∪ { − | − ≤ ≤ } q 311 π 8 4 √ 8 4 π π whereϕ < 4 weimplies choosearg(t + theat + branches1), arg(a as+ 2 cos follows:(4t)) ( argπ, πt).+ Soat we+ may1 choose( the2 branches, 2 ) for such | | π π π π ∈ − π π ∈ − 312 Cthat, argarg√√t8t+8 +at4at+4 1+ 1,( arg, a),+ and2 cos arg(4t)a +(2 cos,(4t)). ( , ) for C . 1 ∈ − 2 2 ∈ − 2 2 ∈ − 2 2 2 p p RemarkA canonical A6. Straightforward homology calculations basis on M yieldis given as follows.

A = C , A = φ12(C ) φ3(C ) + φ3(C ), A = φ2(C ), t8 + at41 + 1 =2 t8 +2 b4 −+ 1t4 +−1 1 2 3 2 b4 3 2 B1 = C1, B2= B1 + φ (C1), B3 = B2 + φ (C1). 1 1 = t8 + s4 + t4 cos(4φ) + 1 + i + s4 t4 sin(4φ), s4 − s4 Set 1 1 + ( ) = 4 + 2 ( ) + ( ) + + 4 ( ) a 2 cos 4t 1s 14 cost 4φ 2 cos 14t i dt4 s sin 4φ . A = 2 s − dt + 4 − s , 8 4 4 2 Z0 √t + at + 1 Z0 √16t 16t + 2 + a π 8 4 2 − φ < 4 implies arg(t + at + 1)1, arg(1a + 2t cos(4t)) ( π, π)1. So we mayt choose the branches such that | | √ 8 4 B = 2 π π dt, ∈C−= 8 dt, arg t + at + 1, arg a + 2 cos(4t)8 ( 42 , 2 ). 8 4 Z0 √t ∈+ at− + 1 Z0 √t + at + 1 p 1 dt A canonical homologyD = 8 basis on M is given as follows. , 0 (2 + a)t4 + (2a 12)t2 + 2 + a Z − A = C , A1 = 4ϕ2(C6 ) ϕ3(C ) +1ϕ3(C ), A = ϕ2(C ), 1 2 2p −t t 1 − 1 2 dt3 2 E = 2 − 3 3 dt + 4 2 3 , B1 = C1, Z0B2 √=tB8 1++atϕ4 +(C1), B3 =Z0B2√+16ϕt4(C1)16. t2 + 2 + a − 1 t4 + t6 1 t5 F = 2 3 dt, H = 4 3 dt, Z0 √t8 + at4 + 1 Z0 √t8 + at4 + 1 1 (1 + t2)2 I = 4 3 dt, Z0 (2 + a)t4 + (2a 12)t2 + 2 + a − p Mathematics 2020, 8, 1693 38 of 39

Set

1 1 t2 1 dt A = 2 − dt + 4 , 0 √t8 + at4 + 1 0 √16t4 16t2 + 2 + a Z Z − 1 1 + t2 1 t B = 2 dt, C = 8 dt, 8 4 8 4 Z0 √t + at + 1 Z0 √t + at + 1 1 dt D = 8 , 0 (2 + a)t4 + (2a 12)t2 + 2 + a Z − 4 6 1 p t t 1 dt E = 2 − 3 dt + 4 3 , Z0 √t8 + at4 + 1 Z0 √16t4 16t2 + 2 + a − 1 t4 + t6 1 t5 F = 2 3 dt, H = 4 3 dt, Z0 √t8 + at4 + 1 Z0 √t8 + at4 + 1 1 (1 + t2)2 I = 4 3 dt, Z0 (2 + a)t4 + (2a 12)t2 + 2 + a − where we choose the branches asp follows: π π π π arg 16t4 16t2 + 2 + a ( , ), arg (2 + a)t4 + (2a 12)t2 + 2 + a ( , ). − ∈ − 2 2 − ∈ − 2 2 p q Remark A7. Straightforward calculations yield

1 16t4 16t2 + 2 + a = 16t4 16t2 + 2 + b4 + − − b4 1 1 = 16t4 16t2 + 2 + s4 + cos(4φ) + i + s4 sin(4φ), − s4 − s4 (2 + a)t4 + (2a 12)t2 + 2 + a = 2t4 12t2 + 2 + a(t4 + 2t2 + 1) − − 1 1 = 2t4 12t2 + 2 + s4 + cos(4φ)(t2 + 1)2 + i + s4 (t2 + 1)2 sin(4φ). − s4 − s4 φ < π implies arg(16t4 16t2 + 2 + a), arg((2 + a)t4 + (2a 12)t2 + 2 + a) ( π, π). So we may | | 4 − − ∈ − choose the branches such that arg √16t4 16t2 + 2 + a, arg (2 + a)t4 + (2a 12)t2 + 2 + a ( π , π ). − − ∈ − 2 2 p Then

G G G G G G 1 1 1 1 1 1 . ZA1 G2! ZA2 G2! ZA3 G2! ZB1 G2! ZB2 G2! ZB3 G2!! is obtained by iB A iB iB 2iB iB − − − − − A iB A iB 0 iB  − −  iD iD iD C 0 C − −  . (A32)  iF E iF iF 2iF iF  − − − − −   E iF E iF 0 iF  − −   0   iI iI iI H H   − −  Mathematics 2020, 8, 1693 39 of 39

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