An Explicit Formula for the Discrete Power Function Associated with Circle Patterns of Schramm Type

Transcription

1 Funkcialaj Ekvacioj, 57 (2014) 1 41 An Explicit Formula for the Discrete Power Function Associated with Circle Patterns of Schramm Type By Hisashi Ando1, Mike Hay2, Kenji Kajiwara1 and Tetsu Masuda3 (Kyushu University1, Japan, Roma Tre University2, Italy, and Aoyama Gakuin University3, Japan) Abstract. We present an explicit formula for the discrete power function introduced by Bobenko, which is expressed in terms of the hypergeometric t functions for the sixth Painlevé equation. The original definition of the discrete power function imposes strict conditions on the domain and the value of the exponent. However, we show that one can extend the value of the exponent to arbitrary complex numbers except even integers and the domain to a discrete analogue of the Riemann surface. Moreover, we show that the discrete power function is an immersion when the real part of the exponent is equal to one. Key Words and Phrases. Discrete conformal mapping, Hypergeometric function, Painlevé VI equation, Circle patterns, Discrete di erential geometry Mathematics Subject Classification Numbers. 33C05, 33E17, 34M55, 52C26, 53A Introduction The theory of discrete analytic functions has been developed in recent years based on the theory of circle packings or circle patterns, which was initiated by Thurston s idea of using circle packings as an approximation of the Riemann mapping [18]. So far many important properties have been established for discrete analytic functions, such as the discrete maximum principle and Schwarz s lemma [6], the discrete uniformization theorem [15], and so forth. For a comprehensive introduction to the theory of discrete analytic functions, we refer to [17]. It is known that certain circle patterns with fixed regular combinatorics admit rich structure. For example, it has been pointed out that the circle patterns with square grid combinatorics introduced by Schramm [16] and the hexagonal circle patterns [5, 8, 9] are related to integrable systems. Some explicit examples of discrete analogues of analytic functions have been presented which are associated with Schramm s patterns: expðzþ, erf ðzþ, Airy function [16], z g, logðzþ [4]. Also, discrete analogues of z g and logðzþ associated with hexagonal circle patterns are discussed in [5, 8, 9]. Among those examples, it is remarkable that the discrete analogue of the power function z g associated with the circle patterns of Schramm type has

2 2 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda a close relationship with the sixth Painlevé equation (P VI ) [7]. It is desirable to construct a representation formula for the discrete power function in terms of the Painlevé transcendents as was mentioned in [7]. The discrete power function can be formulated as a solution to a system of di erence equations on the square lattice ðn; mþ A Z 2 with a certain initial condition. A correspondence between the dependent variable of this system and the Painlevé transcendents can be found in [14], but the formula seems somewhat indirect. Agafonov has constructed a formula for the radii of circles of the associated circle pattern at some special points on Z 2 in terms of the Gauss hypergeometric function [3]. In this paper, we aim to establish an explicit representation formula of the discrete power function itself in terms of the hypergeometric t function of P VI which is valid on Zþ 2 ¼fðn; mþ A Z 2 j n; m b 0g and for g A Cn2Z. Based on this formula, we generalize the domain of the discrete power function to a discrete analogue of the Riemann surface. On the other hand, the fact that the discrete power function is related to P VI has been used to establish the immersion property [4] and embeddedness [2] of the discrete power function with real exponent. Although we cannot expect such properties and thus the correspondence to a certain circle pattern for general complex exponent, we have found a special case of Re g ¼ 1 where the discrete power function is an immersion. Another purpose of this paper is to prove the immersion property of this case. This paper is organized as follows. In section 2, we give a brief review of the definition of the discrete power function and its relation to P VI. The explicit formula for the discrete power function is given in section 3. We discuss the extension of the domain of the discrete power function in section 4. In section 5, we show that the discrete power function for Re g ¼ 1 is an immersion. Section 6 is devoted to concluding remarks. 2. Discrete power function 2.1. Definition of the discrete power function For maps, a discrete analogue of conformality has been proposed by Bobenko and Pinkall in the framework of discrete di erential geometry [10]. Definition 2.1. A map f : Z 2! C; ðn; mþ 7! f n; m is called discrete conformal if the cross-ratio with respect to every elementary quadrilateral is equal to 1: ð2:1þ ð f n; m f nþ1; m Þð f nþ1; mþ1 f n; mþ1 Þ ð f nþ1; m f nþ1; mþ1 Þð f n; mþ1 f n; m Þ ¼ 1:

3 An Explicit Formula for the Discrete Power Function 3 The condition (2.1) is a discrete analogue of the Cauchy-Riemann relation. Actually, a smooth map f : D H C! C is conformal if and only if it satisfies ð2:2þ ð f ðx; yþ f ðx þ ; yþþð f ðx þ ; y þ Þ f ðx; y þ ÞÞ lim!0 ð f ðx þ ; yþ f ðx þ ; y þ ÞÞð f ðx; y þ Þ f ðx; yþþ ¼ 1 for all ðx; yþ A D. However, using Definition 2.1 alone, one cannot exclude maps whose behaviour is far from that of usual holomorphic maps. Because of this, an additional condition for a discrete conformal map has been considered [2, 4, 7, 11]. Definition 2.2. A discrete conformal map f n; m is called embedded if inner parts of di erent elementary quadrilaterals ð f n; m ; f nþ1; m ; f nþ1; mþ1 ; f n; mþ1 Þ do not intersect. An example of an embedded map is presented in Figure 1. This condition seems to require that f ¼ f n; m is a univalent function in the continuous limit, and is too strict to capture a wide class of discrete holomorphic functions. In fact, a relaxed requirement has been considered as follows [2, 4]. Definition 2.3. A discrete conformal map f n; m is called immersed, or an immersion, if inner parts of adjacent elementary quadrilaterals ð f n; m ; f nþ1; m ; f nþ1; mþ1 ; f n; mþ1 Þ are disjoint. See Figure 2 for an example of an immersed map. Let us give the definition of the discrete power function proposed by Bobenko [4, 7, 11]. Definition 2.4. Let f : Z 2 þ! C; ðn; mþ 7! f n; m be a discrete conformal map. If f n; m is the solution to the di erence equation Fig. 1. An example of the embedded discrete conformal map. Fig. 2. An example of the discrete conformal map that is not embedded but immersed.

4 4 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda ð2:3þ gf n; m ¼ 2n ð f nþ1; m f n; m Þð f n; m f n 1; m Þ f nþ1; m f n 1; m þ 2m ð f n; mþ1 f n; m Þð f n; m f n; m 1 Þ f n; mþ1 f n; m 1 with the initial conditions ð2:4þ f 0; 0 ¼ 0; f 1; 0 ¼ 1; f 0; 1 ¼ e gpi=2 for 0 < g < 2, then we call f a discrete power function. The di erence equation (2.3) is a discrete analogue of the di erential equation gf ¼ zðqf =qzþ for the power function f ðzþ ¼z g, which means that the parameter g corresponds to the exponent of the discrete power function. It is easy to get the explicit formula of the discrete power function for m ¼ 0 (or n ¼ 0). When m ¼ 0, (2.3) is reduced to a three-term recurrence relation. Solving it with the initial condition f 0; 0 ¼ 0, f 1; 0 ¼ 1, we have ð2:5þ 8 2l Y l 2k þ g ðn ¼ 2lÞ; >< 2l þ g 2k g k¼1 f n; 0 ¼ Y l 2k þ g >: ðn ¼ 2l þ 1Þ; 2k g k¼1 for n A Z þ. When m ¼ 1 (or n ¼ 1), Agafonov has shown that the discrete power function can be expressed in terms of the hypergeometric function [3]. One of the aims of this paper is to give an explicit formula for the discrete power function f n; m for arbitrary ðn; mþ A Zþ 2. In Definition 2.4, the domain of the discrete power function is restricted to the first quadrant Zþ 2, and the exponent g to the interval 0 < g < 2. Under this condition, it has been shown that the discrete power function is embedded [2]. For our purpose, we do not have to persist with such a restriction. In fact, the explicit formula we will give is applicable to the case g A Cn2Z. Regarding the domain, one can extend it to a discrete analogue of the Riemann surface Relationship to P VI In order to construct an explicit formula for the discrete power function f n; m, we will move to a more general setting. The cross-ratio condition (2.1) can be regarded as a special case of the discrete Schwarzian

5 An Explicit Formula for the Discrete Power Function 5 KdV equation ð f n; m f nþ1; m Þð f nþ1; mþ1 f n; mþ1 Þ ð f nþ1; m f nþ1; mþ1 Þð f n; mþ1 f n; m Þ ¼ p n ð2:6þ ; q m where p n and q m are arbitrary functions in the indicated variables. Some of the authors have constructed various special solutions to the above equation [12]. In particular, they have shown that an autonomous case ð2:7þ ð f n; m f nþ1; m Þð f nþ1; mþ1 f n; mþ1 Þ ð f nþ1; m f nþ1; mþ1 Þð f n; mþ1 f n; m Þ ¼ 1 t ; where t is independent of n and m, can be regarded as a part of the Bäcklund transformations of P VI, and given special solutions to (2.7) in terms of the t functions of P VI. We here give a brief account of the derivation of P VI according to [14]. The derivation is achieved by imposing a certain similarity condition on the discrete Schwarzian KdV equation (2.7) and the di erence equation (2.3) simultaneously. The discrete Schwarzian KdV equation (2.7) is automatically satisfied if there exists a function v n; m satisfying ð2:8þ f n; m f nþ1; m ¼ t 1=2 v n; m v nþ1; m ; f n; m f n; mþ1 ¼ v n; m v n; mþ1 : By eliminating the variable f n; m, we get for v n; m the following equation ð2:9þ t 1=2 v n; m v n; mþ1 þ v n; mþ1 v nþ1; mþ1 ¼ v n; m v nþ1; m þ t 1=2 v nþ1; m v nþ1; mþ1 ; which is equivalent to the lattice modified KdV equation. the di erence equation (2.3) is reduced to It can be shown that ð2:10þ n v nþ1; m v n 1; m v nþ1; m þ v n 1; m þ m v n; mþ1 v n; m 1 v n; mþ1 þ v n; m 1 ¼ m ð 1Þ mþn l with g ¼ 1 þ 2m, where l A C is an integration constant. In the following we take l ¼ m so that (2.10) is consistent when n ¼ m ¼ 0 and v 1; 0 þ v 1; v 0; 1 þ v 0; 1. Assume that the dependence of the variable v n; m ¼ v n; m ðtþ on the deformation parameter t is given by ð2:11þ 2t d dt log v n; m ¼ n v nþ1; m v n 1; m v nþ1; m þ v n 1; m þ w nþm ; where w nþm ¼ w nþm ðtþ is an arbitrary function satisfying w nþmþ2 ¼ w nþm. we have the following Proposition. Then Proposition 2.5. Let q ¼ q n; m ¼ q n; m ðtþ be the function defined by q n; m ¼ t 1=2 ðv nþ1; m =v n; mþ1 Þ. Then q satisfies P VI

6 6 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda ð2:12þ d 2 q dt 2 ¼ q þ 1 q 1 þ 1 dq 2 1 q t dt t þ 1 t 1 þ 1 dq q t dt " # qðq 1Þðq tþ þ 2t 2 ðt 1Þ 2 ky 2 t k2 0 q 2 þ t 1 k2 1 ðq 1Þ 2 þð1 tðt 1Þ y2 Þ ðq tþ 2 ; with ð2:13þ k 2 y ¼ 1 4 ðm n þ m nþ2 ; k 2 0 ¼ 1 4 ðm n m þ nþ2 ; k 2 1 ¼ 1 4 ðm þ n m n 1Þ2 ; y 2 ¼ 1 4 ðm þ n þ m þ n þ 1Þ2 ; where we denote n ¼ð 1Þ mþn m. In general, P VI contains four complex parameters denoted by k y, k 0, k 1 and y. Since n; m A Z þ, a special case of P VI appears in the above proposition, which corresponds to the case where P VI admits special solutions expressible in terms of the hypergeometric function. In fact, the special solutions to P VI of hypergeometric type are given as follows: by ð2:14þ Proposition 2.6 ([13]). Define the function t n 0ða; b; c; tþ ðc B Z; n 0 A Z þ Þ t n 0ða; b; c; tþ ¼ detðjða þ i 1; b þ j 1; c; tþþ 1ai; j a n 0 ðn0 > 0Þ; 1 ðn 0 ¼ 0Þ; with ð2:15þ GðaÞGðbÞ Gða c þ 1ÞGðb c þ 1Þ jða; b; c; tþ ¼c 0 Fða; b; c; tþþc 1 GðcÞ Gð2 cþ t 1 c Fða c þ 1; b c þ 1; 2 c; tþ: Here, F ða; b; c; tþ is the Gauss hypergeometric function, GðxÞ is the Gamma function, and c 0 and c 1 are arbitrary constants. Then ð2:16þ q ¼ t 1; 0 1; 1 t0; n 0 t 1; n 0 þ1 1; 1; 1 1; 0 n 0 t0; n 0 þ1 k; l; m with tn ¼ t 0 n 0ða þ k þ 1; b þ l þ 2; c þ m þ 1; tþ gives a family of hypergeometric solutions to P VI with the parameters ð2:17þ k y ¼ a þ n 0 ; k 0 ¼ b c þ 1 þ n 0 ; k 1 ¼ c a; y ¼ b: k; l; m We call t n 0ða; b; c; tþ or tn the hypergeometric t function of P VI. Note 0 that we have k y þ k 0 þ k 1 þ y 1 A 2Z þ in Proposition 2.6, which can be realized by choosing the parameters as

7 An Explicit Formula for the Discrete Power Function 7 ð2:18þ Gk y ¼ 1 ðm nþ; 2 Gk 0 ¼ 1 ðm nþ; 2 k 1 ¼ mþ 1 ðn þ m þ 1Þ; 2 y ¼ m þ 1 ðn þ m þ 1Þ 2 when n þ m is even, or as ð2:19þ Gk y ¼ m 1 ðm nþ; 2 Gk 0 ¼ m 1 ðm nþ; 2 k 1 ¼ 1 ðn þ m þ 1Þ; 2 y ¼ 1 ðn þ m þ 1Þ 2 when n þ m is odd. These choices concur with the expressions (2.17) by specializing the values of parameters a, b and c. 3. Explicit formulae 3.1. Explicit formulae for f n; m and v n; m We present the solution to the simultaneous system of the discrete Schwarzian KdV equation (2.7) and the di erence equation (2.3) under the initial conditions ð3:1þ f 0; 0 ¼ 0; f 1; 0 ¼ c 0 ; f 0; 1 ¼ c 1 t r ; where g ¼ 2r, and c 0 and c 1 are arbitrary constants. We set c 0 ¼ c 1 ¼ 1 and t ¼ e pi ð¼ 1Þ to obtain the explicit formula for the original discrete power function. Note that t n 0ðb; a; c; tþ ¼t n 0ða; b; c; tþ by the definition. Moreover, we interpret Fðk; b; c; tþ for k A Z >0 as Fðk; b; c; tþ ¼0 and Gð kþ for k A Z b0 as Gð kþ ¼ð 1Þ k =k!. Theorem 3.1. For ðn; mþ A Zþ 2, the function f n; m ¼ f n; m ðtþ can be expressed as follows. (1) Case where n a m (or n 0 ¼ n). When n þ m is even, we have ð3:2þ f n; m ¼ c 1 t r n N ðr þ 1Þ N 1 ð r þ 1Þ N t n ð N; r N þ 1; r; tþ t n ð N þ 1; r N þ 2; r þ 2; tþ ; where N ¼ðnþmÞ=2 and ðuþ j ¼ uðu þ 1Þ...ðu þ j 1Þ is the Pochhammer symbol. When n þ m is odd, we have ð3:3þ f n; m ¼ c 1 t r n ðr þ 1Þ N 1 t n ð N þ 1; r N þ 1; r; tþ ð r þ 1Þ N 1 t n ð N þ 2; r N þ 2; r þ 2; tþ ; where N ¼ðn þ m þ 1Þ=2.

8 8 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda (2) Case where n b m (or n 0 ¼ m). When n þ m is even, we have ð3:4þ f n; m ¼ c 0 N ðr þ 1Þ N 1 ð r þ 1Þ N t m ð N þ 2; r N þ 1; r þ 2; tþ t m ð N þ 1; r N þ 2; r þ 2; tþ ; where N ¼ ðn þ mþ=2. When n þ m is odd, we have ð3:5þ ðr þ 1Þ f n; m ¼ c N 1 t m ð N þ 2; r N þ 1; r þ 1; tþ 0 ð r þ 1Þ N 1 t m ð N þ 1; r N þ 2; r þ 1; tþ ; where N ¼ðn þ m þ 1Þ=2. Proposition 3.2. For ðn; mþ A Z 2 þ, the function v n; m ¼ v n; m ðtþ can be expressed as follows. (1) Case where n a m (or n 0 ¼ n). When n þ m is even, we have ð3:6þ v n; m ¼ t n=2 ðrþ N t n ð N þ 1; r N þ 1; r þ 1; tþ ð r þ 1Þ N t n ð N þ 1; r N þ 2; r þ 2; tþ ; where N ¼ ðn þ mþ=2. When n þ m is odd, we have ð3:7þ v n; m ¼ c 1 t r n=2 t nð N þ 1; r N þ 2; r þ 1; tþ t n ð N þ 2; r N þ 2; r þ 2; tþ ; where N ¼ðn þ m þ 1Þ=2. (2) Case where n b m (or n 0 ¼ m). When n þ m is even, we have ð3:8þ v n; m ¼ t m=2 ðrþ N t m ð N þ 1; r N þ 1; r þ 1; tþ ð r þ 1Þ N t m ð N þ 1; r N þ 2; r þ 2; tþ ; where N ¼ ðn þ mþ=2. When n þ m is odd, we have ð3:9þ v n; m ¼ c 0 t ðmþ1þ=2 t mð N þ 2; r N þ 2; r þ 2; tþ t m ð N þ 1; r N þ 2; r þ 1; tþ ; where N ¼ðn þ m þ 1Þ=2. Note that these expressions are applicable to the case where r A CnZ. A typical example of the discrete power function and its continuous counterpart are illustrated in Figure 3 and Figure 4, respectively. Figure 5 shows an example of the case suggesting multivalency of the map. The proof of the above theorem and proposition is given in the next subsection. Remark 3.3. Agafonov has shown that the generalized discrete power function f n; m, under the setting of c 0 ¼ c 1 ¼ 1, t ¼ e 2ia ð0 < a < pþ and 0 < r < 1, is embedded [3].

9 An Explicit Formula for the Discrete Power Function 9 Fig. 3. The discrete power function with g ¼ 1 þ i. Fig. 4. The ordinary power function z 1þi. Fig. 5. The discrete power function with g ¼ 0:25 þ 3:35i. Remark 3.4. As we mention above, some special solutions to (2.7) in terms of the t functions of P VI have been presented [12]. It is easy to show that these solutions also satisfy a di erence equation which is a deformation of (2.3) in the sense that the coe cients n and m of (2.3) are replaced by arbitrary complex numbers. For instance, a class of solutions presented in Theorem 6 of [12] satisfies

10 10 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda ð3:10þ ða 0 þ a 2 þ a 4 Þ f n; m ¼ðn a 2 Þ ð f nþ1; m f n; m Þð f n; m f n 1; m Þ f nþ1; m f n 1; m ða 1 þ a 2 þ a 4 mþ ð f n; mþ1 f n; m Þð f n; m f n; m 1 Þ ; f n; mþ1 f n; m 1 where a i are parameters of P VI introduced in Appendix A. Setting the parameters as ða 0 ; a 1 ; a 2 ; a 3 ; a 4 Þ¼ðr; 0; 0; r þ 1; 0Þ, we see that the above equation is reduced to (2.3) and that the solutions are given by the hypergeometric t functions under the initial conditions (3.1) Proof of the results In this subsection, we give the proof of Theorem 3.1 and Proposition 3.2. One can easily verify that f n; m satisfies the initial condition (3.1) by noticing t 0 ða; b; c; tþ ¼1. We then show that f n; m and v n; m given in Theorem 3.1 and Proposition 3.2 satisfy the relation (2.8), the di erence equation (2.3), the compatibility condition (2.9) and the similarity condition (2.11) by means of the various bilinear relations for the hypergeometric t function. Note in advance that we use the bilinear relations by specializing the parameters a, b and c as ð3:11þ a ¼ N; b ¼ r N; c ¼ rþ1; N ¼ n þ m ; 2 when n þ m is even, or ð3:12þ a ¼ r N þ 1; b ¼ N; c ¼ rþ1; N ¼ n þ m þ 1 ; 2 when n þ m is odd. We first verify the relation (2.8). Note that we have the following bilinear relations ð3:13þ ð3:14þ ð3:15þ ðc 1Þt ðc 1Þt 0; 1; 1 1; 1; 1 n t nþ1 ¼ðc b 1Þtt 1; 1; 1 0; 1; 1 n t 0; 1; 1 ða bþt ða bþtt n ¼ðc b 1Þt 0; 1; 0 m tm 0; 1; 0 mþ1 ðb a þ 1Þt 0; 1; 1 tm 0; 0; 0 1; 1; 1 m tm ¼ðb c þ 1Þt 1; 1; 1 ðb a þ 1Þt 1; 1; 1 ¼ at 1; 1; 0 m t 1; 1; 1 ¼ at 0; 1; 0 m 0; 0; 0 mþ1 tm ¼ðb c þ 1Þt 0; 1; 0 mþ1 0; 1; 0 1; 1; 2 nþ1 tn þ bt 0; 1; 0 1; 1; 2 n m bt 1; 1; 0 mþ1 t 1; 0; 1 tm 1; 0; 1 tm m bt tn þ bt 0; 1; 1 þðc aþt 0; 0; 0 n 0; 0; 0 n 0; 0; 0 0; 2; 1 m tm ; 0; 0; 0 0; 2; 1 m tmþ1 ; 1; 0; 0 m tm ; 1; 0; 0 m tmþ1 ; 0; 1; 1 þðc aþt 1; 2; 2 tnþ1 ; 1; 2; 2 tn ;

11 An Explicit Formula for the Discrete Power Function 11 for the hypergeometric t functions, the derivation of which is discussed in Appendix A. Let us consider the case where n 0 ¼ n. When n þ m is even, the relation (2.8) is reduced to ð3:16þ rt ½1; 1; 1Š ½0; 1; 1Š n tnþ1 ¼ Ntt ½1; 1; 2Š ½0; 1; 0Š nþ1 tn ðr þ NÞt ½1; 2; 2Š ½0; 0; 0Š n tnþ1 ; rt where we denote ½0; 1; 1Š ½1; 1; 1Š n tn ¼ Nt ½1; 1; 2Š ½0; 1; 0Š n tn ðr þ NÞt ½1; 2; 2Š ½0; 0; 0Š n tn ; ð3:17þ t ½i 1; i 2 ; i 3 Š n 0 ¼ t n 0ð N þ i 1; r N þ i 2 ; r þ i 3 ; tþ; for simplicity. We see that the relations (3.16) can be obtained from (3.13) with the parameters specialized as (3.11). In fact, the hypergeometric t functions can be rewritten as ð3:18þ 0; 1; 1 tn ½1; 1; 1Š ¼ t n ða þ 1; b þ 1; cþ ¼t n ð N þ 1; r N þ 1; r þ 1Þ ¼tn ; for instance. When n þ m is odd, (2.8) yields ð3:19þ rt ½1; 2; 1Š ½1; 1; 1Š n tnþ1 ¼ð r þ NÞtt ½1; 2; 2Š ½1; 1; 0Š nþ1 tn Nt ½2; 2; 2Š n ½0; 1; 0Š tnþ1 ; rt ½1; 1; 1Š ½1; 2; 1Š n tn ¼ð r þ NÞt ½1; 2; 2Š ½1; 1; 0Š n tn Nt ½2; 2; 2Š ½0; 1; 0Š n tn ; which is also obtained from (3.13) by specializing the parameters as (3.12). Note that the hypergeometric t functions can be rewritten as ð3:20þ 0; 1; 1 tn ¼ t n ða þ 1; b þ 1; cþ ¼t n ð r N þ 2; N þ 1; r þ 1Þ ½1; 2; 1Š ¼ t n ð N þ 1; r N þ 2; r þ 1Þ ¼tn ; this time. In the case where n 0 ¼ m, one can similarly verify the relation (2.8) by using the bilinear relations (3.14) and (3.15). Next, we prove that (2.3) is satisfied, which is rewritten by using (2.8) as ð3:21þ r f n; m v n; m ¼ We use the bilinear relations nt 1=2 v 1 nþ1; m þ v 1 n 1; m m þ vn; 1 mþ1 þ : v 1 n; m 1 ð3:22þ n 0 t 0; 0; 0 1; 1 n 0 t0; n n 0 þ1 0 ¼ðb c þ 1Þt0; 1; 0 0; 0; 1 tn 0 1 þ at 1 1; 1; 1 t 1; 0; 0 n 0 þ1 tn 0 1 ; ða þ b c þ n 0 þ 1Þt ¼ at 1; 1; 1 0; 0 n 0 t1; n 0; 0; 0 1; 1 n 0 t0; n 0 n 0 0 þðb c þ 1Þt0; 1; 0 0; 1 t0; n ; 0

12 12 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda and ð3:23þ 0; 0; 0 1; 1; 2 n tn t ¼ t 1 t 1; 1; 1 0; 0; 1 nþ1 tn 1 þ t 1; 1; 1 0; 0; 1 n tn ; 0; 0; 0 1; 1; 0 tm t t m ¼ t 0; 1; 1 1; 0; 1 m tm 0; 1; 0 1; 0; 0 m t ¼ t m t 0; 1; 0 1; 0; 0 mþ1 tm 1 ; 1; 1; 1 0; 0; 1 mþ1 tm 1 þ t 1; 1; 1 0; 0; 1 m tm ; for the proof. Their derivation is also shown in Appendix A. Let us consider the case where n 0 ¼ n. When n þ m is even, we have ð3:24þ nt ½1; 2; 2Š ½1; 1; 1Š n tn ¼ Nt ½1; 1; 2Š ½1; 2; 1Š nþ1 tn 1 þ Nt 1 t ½0; 1; 1Š ½2; 2; 2Š nþ1 tn 1 ; mt ½1; 2; 2Š ½1; 1; 1Š n tn ¼ Nt ½0; 1; 1Š ½2; 2; 2Š n tn þ Nt ½1; 1; 2Š ½1; 2; 1Š n tn ; from the bilinear relations (3.22) by specializing the parameters a, b and c as given in (3.11). These lead us to ð3:25þ v 1 nþ1; m þ v 1 n 1; m ¼ c 1 1 t rþðnþ1þ=2 n N ½1; 2; 2Š ½1; 1; 1Š n tn ½0; 1; 1Š ½1; 2; 1Š nþ1 tn 1 t t ; By using ð3:26þ v 1 n; mþ1 þ v 1 n; m 1 ¼ c 1 1 t rþn=2 m N ½1; 2; 2Š ½0; 1; 0Š n tn t ¼ t 1 t ½0; 1; 1Š ½1; 2; 1Š nþ1 tn 1 ½1; 2; 2Š ½1; 1; 1Š n tn ½0; 1; 1Š ½1; 2; 1Š n tn t t þ t ½0; 1; 1Š ½1; 2; 1Š n tn ; which is obtained from the first relation in (3.23), one can verify (3.21). n þ m is odd, we have the bilinear relations : When ð3:27þ nt ½2; 2; 2Š ½1; 2; 1Š n tn ¼ð r þ NÞt ½1; 2; 2Š ½2; 2; 1Š nþ1 tn 1 þðr þ N 1Þt 1 t ½1; 1; 1Š ½2; 3; 2Š nþ1 tn 1 ; mt ½2; 2; 2Š ½1; 2; 1Š n tn from (3.22) with (3.12), and ð3:28þ t ¼ðr þ N 1Þt ½2; 2; 2Š ½1; 1; 0Š n tn ¼ t 1 t ½1; 1; 1Š ½2; 3; 2Š n tn ½1; 1; 1Š ½2; 2; 1Š nþ1 tn 1 þð r þ NÞt þ t ½1; 1; 1Š ½2; 2; 1Š n tn ; ½1; 2; 2Š ½2; 2; 1Š n tn ; from the first relation in (3.23). These lead us to (3.21). We next consider the case where n 0 ¼ m. When n þ m is even, we get the bilinear relations ð3:29þ and ð3:30þ mt nt ½1; 2; 2Š ½1; 1; 1Š m tm ½1; 2; 2Š ½1; 1; 1Š m tm ¼ Nt ¼ Nt ½1; 2; 2Š ½2; 1;2Š tm tm ½1; 1; 2Š 2; 1Š mþ1 t½1; m 1 ½0; 1; 1Š ½2; 2; 2Š m tm ¼ t ½1; 1; 2Š ½2; 2; 2Š m tm þ Nt 1 t þ Nt t ½0; 1; 1Š 2; 2Š mþ1 t½2; m 1 ; ½1; 1; 2Š ½1; 2; 1Š m tm ; ½1; 1; 2Š 2; 2Š mþ1 t½2; m 1 ;

13 An Explicit Formula for the Discrete Power Function 13 from (3.22) and the second relation in (3.23), respectively. By using these relations, one can show (3.21) in a similar way to the case where n 0 ¼ n. When n þ m is odd, we use the bilinear relations ð3:31þ mt ½2; 2; 2Š ½1; 2; 1Š m tm ¼ð r þ NÞt ½1; 2; 2Š 2; 1Š mþ1 t½2; m 1 þðr þ N 1Þt 1 t ½1; 1; 1Š 3; 2Š mþ1 t½2; m 1 ; and ð3:32þ nt ½2; 2; 2Š ½1; 2; 1Š m tm ¼ðr þ N 1Þt ½1; 2; 1Š ½2; 1;1Š tm tm ¼ t ½1; 1; 1Š ½2; 3; 2Š m tm ½1; 1; 1Š 2; 1Š mþ1 t½2; m 1 þð r þ NÞt þ t ½1; 1; 1Š ½2; 2; 1Š m tm ; ½1; 2; 2Š m ½2; 2; 1Š tm ; which are obtained from (3.22) and the third relation in (3.23), respectively, to show (3.21). We next give the verification of the compatibility condition (2.9) by using the bilinear relations ð3:33þ ðc aþt 0; 1; 1 1; 0 n 0 t 1; n 0 þ1 bt 0; 1; 1 1; 0 n 0 t 1; 0; 0; 0 2; 1 n 0 t 1; n 0 þ1 0; 0; 0 2; 1 t 1; ¼ðt 1Þt 1; 1; 1 1; 0 n 0 t0; n 0 þ1 ; ðc aþtt n 0 þ1 btn 0 n 0 þ1 ¼ðt 1Þtn 0 n 0 þ1 : The derivation of these is discussed in Appendix A. We first consider the case where n 0 ¼ n. When n þ m is even, we get 0; 1; 0 1; 1 t 1; ð3:34þ ð r þ N þ 1Þt ½1; 1; 1Š ½0; 1; 2Š n tnþ1 þðr þ NÞt ½1; 2; 2Š ½0; 0; 1Š n tnþ1 ¼ðt 1Þt ½0; 1; 1Š n ½1; 1; 2Š tnþ1 ; ð r þ N þ 1Þtt ½1; 1; 1Š ½0; 1; 2Š n tnþ1 þðr þ NÞt ½1; 2; 2Š ½0; 0; 1Š n tnþ1 from the bilinear relations (3.33). Then we have ¼ðt 1Þt ½1; 1; 2Š ½0; 1; 1Š n tnþ1 ; ð3:35þ t 1=2 v n; m þ v nþ1; mþ1 ¼ t ðnþ1þ=2 ðrþ ðt 1Þ N t ½1; 1; 2Š n ð r þ 1Þ ½1; 2; 2Š Nþ1 t t n ½0; 1; 1Š tnþ1 ½0; 1; 2Š nþ1 ; v n; m þ t 1=2 v nþ1; mþ1 ¼ t n=2 ðrþ ðt 1Þ N t ½0; 1; 1Š n ð r þ 1Þ ½1; 2; 2Š Nþ1 t t n ½1; 1; 2Š tnþ1 ½0; 1; 2Š nþ1 ; from which we arrive at the compatibility condition (2.9). we have When n þ m is odd, ð3:36þ Nt ½1; 2; 1Š ½1; 1; 2Š n tnþ1 þ Nt ½2; 2; 2Š ½0; 1; 1Š n tnþ1 ¼ðt 1Þt ½1; 1; 1Š ½1; 2; 2Š n tnþ1 ; Ntt ½1; 2; 1Š ½1; 1; 2Š n tnþ1 þ Nt ½2; 2; 2Š ½0; 1; 1Š n tnþ1 ¼ðt 1Þt ½1; 2; 2Š ½1; 1; 1Š n tnþ1 ; from (3.33). Calculating t 1=2 v n; m þ v nþ1; mþ1 and v n; m þ t 1=2 v nþ1; mþ1 by means of these relations, we see that we have (2.9). In the case where n 0 ¼ m, one can verify the compatibility condition (2.9) in a similar manner.

14 14 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda Let us finally verify the similarity condition (2.11), which can be written as n w nþm t d dt log v nv nþ1; m ð3:37þ n; m ¼ : v nþ1; m þ v n 1; m Here, we take the factor w nþm as w nþm ¼ r½ð 1Þ nþm 1Š. The relevant bilinear relations for the hypergeometric t function are ð3:38þ 0; 0; 0 ðd þ nþtn 0; 1; 1 tn ¼ at 1 1; 1; 1 t 1; 0; 0 nþ1 tn 1 ; 0; 1; 1 0; 0; 0 ðd þ b c þ 1Þtm tm ¼ðb cþ1þt 0; 0; 0 ðd þ a þ mþtm 0; 1; 1 tm 1; 1; 1 ¼ at 1; 0; 0 m tm : 0; 1; 0 0; 0; 1 m tm ; The derivation of these is obtained in Appendix A. We first consider the case where n 0 ¼ n. When n þ m is even, it is easy to see that we have ð3:39þ v nþ1; m n ¼ Nt 1 t½0; 1; 1Š ½2; 2; 2Š nþ1 tn 1 ; v nþ1; m þ v ½1; 2; 2Š ½1; 1; 1Š n 1; m tn tn from the bilinear relation (3.24). We get ð3:40þ ½1; 2; 2Š ðd þ nþtn ½1; 1; 1Š tn ¼ Nt 1 t ½0; 1; 1Š ½2; 2; 2Š nþ1 tn 1 ; from the first relation in (3.38) with (3.11). From this we can obtain the similarity condition (3.37) as follows. When n þ m is odd, we have ð3:41þ ½2; 2; 2Š ðd þ nþtn ½1; 2; 1Š tn ¼ t 1 ðr þ N 1Þt ½1; 1; 1Š ½2; 3; 2Š nþ1 tn 1 ; from the first relation in (3.38). This relation together with the first relation in (3.27) leads us to (3.37). Next, we discuss the case where n 0 ¼ m. When n þ m is even, we have ð3:42þ ½1; 2; 2Š ðd þ NÞtm ½1; 1; 1Š tm ¼ Nt ½1; 1; 2Š ½1; 2; 1Š m tm ; from the second relation in (3.38). Then we arrive at (3.37) by virtue of the second relation in (3.29). When n þ m is odd, we get n m 1 ½1; 2; 1Š ½2; 2; 2Š ½1; 1; 1Š ½2; 3; 2Š ð3:43þ D þ r þ tm tm ¼ðrþN 1Þtm tm ; 2 from the third relation in (3.38). Then we derive the similarity condition (3.37) by using the second relation in (3.31). This completes the proof of Theorem 3.1 and Proposition Extension of the domain First, we extend the domain of the discrete power function to Z 2. To determine the values of f n; m in the second, third and fourth quadrants, we have

15 An Explicit Formula for the Discrete Power Function 15 to give the values of f 1; 0 and f 0; 1 as the initial conditions. Set the initial conditions as ð4:1þ f 1; 0 ¼ c 2 t 2r ; f 0; 1 ¼ c 3 t 3r ; This is natural because these condi- where c 2 and c 3 are arbitrary constants. tions reduce to ð4:2þ f 1; 0 ¼ 1; f 0; 1 ¼ e pir ; f 1; 0 ¼ e 2pir ; f 0; 1 ¼ e 3pir at the original setting. Due to the symmetry of equations (2.7) and (2.3), we immediately obtain the explicit formula of f n; m in the second and third quadrant. Corollary 4.1. Under the initial conditions f 0; 1 ¼ c 1 t r and (4.1), we have ð4:3þ f n; m ¼ f n; m j c0 7! c 2 t 2r; f n; m ¼ f n; m j c0 7! c 2 t 2r ; c 1 7! c 3 t 2r; for n; m A Z þ. Next, let us discuss the explicit formula in the fourth quadrant. Naively, we use the initial conditions f 0; 1 ¼ c 3 t 3r and f 1; 0 ¼ c 0 to get the formula f n; m ¼ f n; m j c1 7! c 3 t 2r. However, this setting makes the discrete power function f n; m become a single-valued function on Z 2. In order to allow f n; m to be multi-valued on Z 2, we introduce a discrete analogue of the Riemann surface by the following procedure. Prepare an infinite number of Z 2 -planes, cut the positive part of the real axis of each Z 2 -plane and glue them in a similar way to the continuous case. The next step is to write the initial conditions (3.1) and (4.1) in polar form as ð4:4þ f ð1; pk=2þ ¼c k t kr ðk ¼ 0; 1; 2; 3Þ; where the first component, 1, denotes the absolute value of n þ im and the second component, pk=2, is the argument. We must generalize the above initial conditions to those for arbitrary k A Z so that we obtain the explicit expression of f n; m for each quadrant of each Z 2 -plane. Let us illustrate a typical case. When 3p=2 a argðn þ imþ a 2p, we solve the equations (2.7) and (2.3) under the initial conditions ð4:5þ f ð1; 3p=2Þ ¼c 3 t 3r ; f ð1; 2pÞ ¼c 4 t 4r ; to obtain the formula ð4:6þ f n; m ¼ f n; m j c0 7! c 4 t 4r ; c 1 7! c 3 t 2r ðn; m A Z þþ: We present the discrete power function with g ¼ 5=2 whose domain is Z 2 and the discrete Riemann surface in Figure 6 and 7, respectively. Note that the

16 16 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda Fig. 6. The discrete power function with g ¼ 5=2 whose domain is Z 2. Fig. 7. The discrete power function with g ¼ 5=2 whose domain is the discrete Riemann surface. necessary and su cient condition for the discrete power function to reduce to a single-valued function on Z 2 is (c k ¼ c kþ4 and) e 4pir ¼ 1, which means that the exponent g is an integer. 5. Associated circle pattern of Schramm type Agafonov and Bobenko have shown that the discrete power function for real g is an immersion and thus defines a circle pattern of Schramm type. We have generalized the discrete power function to complex g. It is natural to ask whether there are other cases where the discrete power function is associated with circle patterns. We have the following result: Theorem 5.1. The mapping f : Zþ 2! C; ðn; mþ 7! f n; m satisfying (2.1) and (2.3) with the initial condition ð5:1þ f 0; 0 ¼ 0; f 1; 0 ¼ 1; f 0; 1 ¼ e pig=2 ; g A C is an immersion when Re g ¼ 1. In this section, we give the proof of Theorem 5.1 along with the discussion in [4]. We also use the explicit formulae given in previous sections Circle pattern Setting ð5:2þ g ¼ 1 þ id; d A R;

17 An Explicit Formula for the Discrete Power Function 17 we associate the discrete power function with circle patterns of Schramm type. The proof of Theorem 5.1 is then reduced to properties of the radii of those circles. Lemma 5.1. initial condition A discrete power function f n; m defined by (2.1) and (2.3) with ð5:3þ f 0; 0 ¼ 0; f 1; 0 ¼ 1; f 0; 1 ¼ c 1 e pig=2 ; c 1 > 0 for arbitrary g A Cn2Z has the equidistant property ð5:4þ f 2n; 0 f 2n 1; 0 ¼ f 2nþ1; 0 f 2n; 0 ; f 0; 2m f 0; 2m 1 ¼ f 0; 2mþ1 f 0; 2m ; for any n b 1; m b 1. Moreover, if and only if Re g ¼ 1, we have ð5:5þ j f nþ1; 0 f n; 0 j¼jf n; 0 f n 1; 0 j; j f 0; mþ1 f 0; m j¼jf 0; m f 0; m 1 j; for any n b 1; m b 1. Proof. By using the formulae in Theorem 3.1 (or Proposition 3.2), we have ð5:6þ f 2n; 0 f 2n 1; 0 ¼ f 2nþ1; 0 f 2n; 0 ¼ g 2n þ g Y n k¼1 2k þ g 2k g ; f 0; 2m f 0; 2m 1 ¼ f 0; 2mþ1 f 0; 2m ¼ c 1 e pig=2 g 2m þ g which proves (5.4). We also have Y m k¼1 2k þ g 2k g ; ð5:7þ f 2nþ2; 0 f 2nþ1; 0 ¼ Ynþ1 k¼1 2k 2 þ g ; f 2nþ1; 0 f 2n; 0 ¼ Yn 2k 2 þ g : 2k g 2k g k¼1 Putting g ¼ 1 þ id, we obtain ð5:8þ f 2nþ2; 0 f 2nþ1; 0 ¼ Ynþ1 2k 1 þ id 2k 1 id ; k¼1 f 2nþ1; 0 f 2n; 0 ¼ Yn k¼1 2k 1 þ id 2k 1 id ; which implies j f 2nþ2; 0 f 2nþ1; 0 j¼jf 2nþ1; 0 f 2n; 0 j¼1. Using the first equation of (5.4), we see that the first equation of (5.5) follows. The second equation of (5.5) can be proved in a similar manner. Suppose that (5.5) holds, then from (5.7) we have ð5:9þ f 2nþ2; 0 f 2nþ1; 0 f 2nþ1; 0 f ¼ 2n þ g 2n; 0 2n þ 2 g which leads us to Re g ¼ 1. r

18 18 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda Proposition 5.2. Let f n; m satisfy (2.1) and (2.3) in Zþ 2 with initial condition (5.3). Then all the elementary quadrilaterals ð f n; m ; f nþ1; m ; f nþ1; mþ1 ; f n; mþ1 Þ are of the kite form, namely, all edges at each vertex f n; m with n þ m ¼ 1 ðmod 2Þ are of the same length, ð5:10þ j f nþ1; m f n; m j¼jf n; mþ1 f n; m j¼jf n 1; m f n; m j¼jf n; m 1 f n; m j: Moreover, all angles between the neighbouring edges at the vertex f n; m with n þ m ¼ 0 ðmod 2Þ are equal to p=2. Proof. For three complex numbers z i (i ¼ 1; 2; 3), we introduce a notation ð5:11þ ½z 1 ; z 2 ; z 3 Š¼arg z 1 z 2 z 3 z 2 : We first consider the quadrilateral ð f 0; 0 ; f 1; 0 ; f 1; 1 ; f 0; 1 Þ. Notice that f 0; 1 ¼ ic 1 e pd=2 which implies that ½ f 0; 1 ; f 0; 0 ; f 1; 0 Š¼p=2. Then it follows from (2.1) that ½ f 1; 0 ; f 1; 1 ; f 0; 1 Š¼p=2, j f 1; 0 f 0; 0 j¼jf 1; 1 f 1; 0 j and j f 0; 1 f 0; 0 j¼ j f 1; 1 f 0; 1 j. We next consider the quadrilateral ð f 1; 0 ; f 2; 0 ; f 2; 1 ; f 1; 1 Þ where, from Lemma 5.1, we have j f 1; 1 f 1; 0 j¼jf 2; 0 f 1; 0 j. We see from (2.1) that ½ f 2; 1 ; f 1; 1 ; f 1; 0 Š¼½f 1; 0 ; f 2; 0 ; f 2; 1 Š¼p=2 and j f 2; 1 f 1; 1 j¼jf 2; 1 f 2; 0 j. From Lemma 5.1 and ½ f 1; 0 ; f 2; 0 ; f 2; 1 Š¼p=2, we see that ½ f 2; 1 ; f 2; 0 ; f 3; 0 Š¼p=2. Then a similar argument can be applied to the quadrilateral ð f 2; 0 ; f 3; 0 ; f 3; 1 ; f 2; 1 Þ and so forth. In this manner, Proposition 5.2 is proved inductively. r From Proposition 5.2, it follows that the circumscribed circles of the quadrilaterals ð f n 1; m ; f n; m 1 ; f nþ1; m ; f n; mþ1 Þ with n þ m ¼ 1 ðmod 2Þ form a circle pattern of Schramm type [4, 16], namely, the circles of neighbouring quadrilaterals intersect orthogonally and the circles of half-neighbouring quadrilaterals with a common vertex are tangent (See Figure 8). Conversely, for a given circle pattern of Schramm type, it is possible to construct a discrete conformal mapping f n; m as follows. Let fc n; m g; ðn; mþ A V ¼fðn; mþ A Zþ 2 j n þ m ¼ 1 ðmod 2Þg be a circle pattern of Schramm type on the complex plane. Define f : Zþ 2! C; ðn; mþ 7! f n; m in the following manner: (a) If n þ m ¼ 1 ðmod 2Þ, then f n; m is the center of C n; m. (b) If n þ m ¼ 0 ðmod 2Þ, then f n; m :¼ C n 1; m V C nþ1; m ¼ C n; mþ1 V C n; m 1. By construction, it follows that all elementary quadrilaterals ð f n; m ; f nþ1; m ; f nþ1; mþ1 ; f n; mþ1 Þ are of the kite form whose angles between the edges with di erent lengths are p=2. Therefore, (2.1) is satisfied automatically. In what follows, the function f n; m, defined by (a) and (b), is called a discrete conformal map corresponding to the circle pattern fc n; m g.

19 An Explicit Formula for the Discrete Power Function 19 Fig. 8. A schematic diagram of quadrilaterals. We now use the radii of corresponding circle patterns to characterize the necessary and su cient condition that the discrete power function is an immersion. Theorem 5.3. Let f n; m satisfying (2.1) and (2.3) with initial condition (5.3) be an immersion. Then R n; m defined by ð5:12þ R n; m ¼jf nþ1; m f n; m j¼jf n; mþ1 f n; m j¼jf n 1; m f n; m j¼jf n; m 1 f n; m j satisfies ð5:13þ n R nþ1; m R n 1; m R nþ1; m þ R n 1; m þ m R n; mþ1 R n; m 1 R n; mþ1 þ R n; m 1 ¼ 0; and ð5:14þ R nþ1; mþ2 ¼ ½ðm þ 1ÞR n; mþ1 þ dr nþ1; m ŠR n; mþ1 ðr nþ1; m þ R n 1; m ÞþnR nþ1; m ðrn; 2 mþ1 R nþ1; mr n 1; m Þ ½ðm þ 1ÞR nþ1; m dr n; mþ1 ŠðR nþ1; m þ R n 1; m Þ nðrn; 2 mþ1 R ; nþ1; mr n 1; m Þ for ðn; mþ A V. Conversely, let R : V! R þ satisfy (5.13) and (5.14). Then R n; m defines an immersed circle pattern of Schramm type. The corresponding discrete conformal map f n; m is an immersion and satisfies (2.3). Proof. The proof of Theorem 5.3 occupies the remainder of this subsection. Suppose that the discrete power function f n; m is immersed. For n þ m ¼ 0 ðmod 2Þ, we may parametrize the edges around the vertex f n; m as ð5:15þ f nþ1; m f n; m ¼ r 1 e ib ; f n; mþ1 f n; m ¼ ir 2 e ib ; f n 1; m f n; m ¼ r 3 e ib ; f n; m 1 f n; m ¼ ir 4 e ib ;

20 20 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda Fig. 9. Parametrization of edges. where r i > 0(i¼1; 2; 3; 4) are the radii of the corresponding circles, since all the angles around f n; m are p=2. The constraint (2.3) reads gf n; m ¼ e ib 2n r 1r 3 þ 2im r 2r 4 ð5:16þ : r 1 þ r 3 r 2 þ r 4 Lemma 5.4. For n þ m ¼ 0 ðmod 2Þ we have: ð5:17þ f nþ1; mþ1 f nþ1; m ¼ e ib r 1 r 1 ir 2 r 1 þ ir 2 ; f nþ1; m f nþ1; m 1 ¼ e ib r 1 r 1 þ ir 4 r 1 ir 4 : Proof. The kite form of the quadrilateral ð f n; m ; f nþ1; m ; f nþ1; mþ1 ; f n; mþ1 Þ implies f nþ1; mþ1 f nþ1; m ¼ ðf nþ1; m f n; m Þe 2i½ f n; mþ1; f nþ1; m ; f nþ1; mþ1 Š (See Figure 9). The first equation of (5.17) follows by noticing that tan½ f n; mþ1 ; f nþ1; m ; f nþ1; mþ1 Š ¼ r 2 =r 1. The second equation is derived by a similar consideration on the quadrilateral ð f n; m 1 ; f nþ1; m 1 ; f nþ1; m ; f n; m Þ. r Setting f nþ2; m f nþ1; m ¼ r 1 e iðbþsþ and substituting (5.15) (5.17) into (2.3) at the point ðn þ 1; mþ, one arrives at ð5:18þ 2n r 1r 3 r 1 þ r 3 þ 2im r 2r 4 r 2 þ r 4 þ gr 1 ¼ 2ðn þ 1Þr þ e is þ 2mr 1 r 1 ðr 2 r 4 Þþiðr1 2 þ r 2r 4 Þ : 2r 1 ðr 2 þ r 4 Þ The real part of (5.18) gives ð5:19þ n r 1 r 3 r 1 þ r 3 þ m r 2 r 4 r 2 þ r 4 ¼ 0; which coincides with (5.13). Now we parametrize the edges around the vertex f nþ1; mþ1 with n þ m ¼ 0 ðmod 2Þ as

21 An Explicit Formula for the Discrete Power Function 21 ð5:20þ f nþ2; mþ1 ¼ f nþ1; mþ1 þ R 1 e ib 0 ; f nþ1; mþ2 ¼ f nþ1; mþ1 þ ir 2 e ib 0 ; f n; mþ1 ¼ f nþ1; mþ1 r 2 e ib 0 ; f nþ1; m ¼ f nþ1; mþ1 ir 1 e ib 0 : From the first equation of (5.17) and noticing that all angles around the vertex f nþ1; mþ1 are p=2, we have the following relation between b 0 and b: ð5:21þ e ib 0 ¼ ie ib r 1 ir 2 r 1 þ ir 2 : One can express f nþ2; mþ1 in two ways as (See Figure 9) ð5:22þ f nþ2; mþ1 ¼ f nþ2; m þ ir 1 e iðbþsþ ¼ f nþ1; m þ r 1 e iðbþsþ þ ir 1 e iðbþsþ ; and ð5:23þ f nþ2; mþ1 ¼ f nþ1; mþ1 þ R 1 e ib 0 ¼ f nþ1; m e ib r 1 r 1 ir 2 r 1 þ ir 2 þ R 1 e ib 0 : The compatibility implies ð5:24þ e is ¼ R 1 þ ir 1 R 1 ir 1 r 1 ir 2 r 1 þ ir 2 : Then, from the imaginary part of (5.18), one obtains ð5:25þ m r 2r 4 r 2 1 r 2 þ r 4 þ dr 1 ¼ðn þ 1Þ r2 1 r 2R 1 r 2 þ R 1 ; or solving (5.25) with respect to R 1 ¼ R nþ2; mþ1 ð5:26þ R nþ2; mþ1 ¼ ½ðn þ 1ÞR nþ1; m dr n; mþ1 ŠR nþ1; m ðr n; mþ1 þ R n; m 1 ÞþmR n; mþ1 ðrnþ1; 2 m R n; mþ1r n; m 1 Þ ½ðn þ 1ÞR n; mþ1 þ dr nþ1; m ŠðR n; mþ1 þ R n; m 1 Þ mðrnþ1; 2 m R : n; mþ1r n; m 1 Þ We may rewrite (2.3) at ðn þ 1; m þ 1Þ in terms of r i (i ¼ 1; 2; 3; 4) R 1, R 2 as 2n r 1r 3 þ 2im r 2r 4 þ gr 1 1 r 1 ir 2 ð5:27þ r 1 þ r 3 r 2 þ r 4 r 1 þ ir 2 ¼ i r 1 ir 2 r 2 R 1 r 1 R 2 2ðn þ 1Þ þ 2iðm þ 1Þ : r 1 þ ir 2 r 2 þ R 1 r 1 þ R 2 Eliminating g from (5.18) and (5.27), we get ð5:28þ r 3 m r 2 n ðmþ1þ þðnþ1þ ¼ 0: r 2 þ r 4 r 1 þ r 3 r 1 þ R 2 r 2 þ R 1 R 2 r 2

22 22 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda We then eliminate R 1 using (5.25) to obtain ð5:29þ R 2 ¼ ½ðm þ 1Þr 2 þ dr 1 Šr 2 ðr 1 þ r 3 Þþnr 1 ðr2 2 r 1r 3 Þ ½ðm þ 1Þr 1 dr 2 Šðr 1 þ r 3 Þ nðr2 2 r ; 1r 3 Þ which coincides with (5.14). This proves the first part of Theorem 5.3. To prove the second part, we use the following lemma. Lemma 5.5. Let R : V! R þ satisfy (5.13) and (5.14). Then, R satisfies (5.26) and ð5:30þ ½ðn þ 1ÞðR 2 nþ1; m R n; m 1R nþ2; m 1 Þ þ dr nþ1; m ðr n; m 1 þ R nþ2; m 1 ÞŠðR n; mþ1 þ R n; m 1 Þ mðrnþ1; 2 m R n; mþ1r n; m 1 ÞðR n; m 1 þ R nþ2; m 1 Þ¼0: Proof. Substituting (5.13) at ðn; mþ and at ðn þ 1; m þ 1Þ into (5.14) to eliminate R n 1; m and R nþ1; mþ2, we get (5.26). Substituting (5.14) at ðn þ 1; m 1Þ into (5.26), we get (5.30) under the condition dr nþ1; m ðr nþ2; m 1 R n; mþ1 Þ þðnþmþ1þðr nþ2; m 1 R n; mþ1 þ Rnþ1; 2 mþ 0 0 which can be verified from the compatibility with (5.13) and (5.14). r Eliminating d from (5.26) and (5.30), we get ð5:31þ ðr 2 nþ1; m R n; mþ1r nþ2; mþ1 ÞðR n; m 1 þ R nþ2; m 1 Þ þðr 2 nþ1; m R n; m 1R nþ2; m 1 ÞðR n; mþ1 þ R nþ2; mþ1 Þ¼0: In [16], it was proven that, given R n; m satisfying (5.31), the circle pattern with radii of the circles R n; m is immersed. Thus, the corresponding discrete conformal map f n; m is an immersion. Let us finally show that the discrete conformal map f n; m satisfies (2.3). Putting m ¼ 0 in (5.13), we have R nþ1; 0 ¼ R n 1; 0. This means that j f nþ1; 0 f n; 0 j¼jf n; 0 f n 1; 0 j for any n b 1. By using the ambiguity of translation and scaling of the circle pattern, one can set f 0; 0 ¼ 0, f 1; 0 ¼ 1 without loss of generality, and set! ð5:32þ f 2Nþ1; 0 f 2N; 0 ¼ f 2N; 0 f 2N 1; 0 ¼ exp 2i XN y j ðn ¼ 1; 2;...Þ: Putting ðn; mþ ¼ð2N; 0Þ in (5.26) we have j¼1 ð5:33þ d ¼ð2N þ 1Þ R2 2Nþ1; 0 R 2N; 1R 2Nþ2; 1 R 2Nþ1; 0 ðr 2N; 1 þ R 2Nþ2; 1 Þ :

23 An Explicit Formula for the Discrete Power Function 23 (i) 0 a 2y a p=2 (ii) p=2 a 2y < p Fig. 10. Configuration of points in Lemma 5.6. A geometric consideration leads us the following lemma: ð5:34þ Lemma 5.6. We have d ¼ð2N þ 1Þ tan y Nþ1 ðn ¼ 0; 1;...Þ: Proof. First, we consider the case of 0 a 2y a p=2, see Figure 10 (i). From BC ¼ AD we obtain ð5:35þ ðr 2 þ R 1 Þ sin 2y 0 þðh þ R 1 Þ sin 2y ¼ 2r 1 ; which yields ð5:36þ ðr 2 h þ r1 2 Þ½ðr 2 þ R 1 Þh þðr 2 R 1 r1 2 ÞŠ ¼ 0: Since r 2 h þ r 2 1 > 0, we have h ¼ðr2 1 r 2R 1 Þ=ðr 2 þ R 1 Þ and ð5:37þ tan y ¼ r2 1 r 2R 1 r 1 ðr 2 þ R 1 Þ : Thus we get (5.34) from (5.33). points is shown in Figure 10 (ii). When p=2 a 2y < p, the configuration of The equality BC ¼ AD implies ð5:38þ ðr 2 þ R 1 Þ sin 2y 0 þðh þ R 1 Þ sinðp 2yÞ ¼2r 1 ; which gives the same result as the case of 0 a 2y a p=2. Let us investigate the case of p=2 a 2y a 0, see Figure 11 (i). Since DC ¼ AB we have ð5:39þ ðh þ R 1 Þ cos 2y þðr 2 þ R 1 Þ cosðp 2y 0 Þ¼h þ r 2 ; which leads us to ð5:40þ ðr 2 h þ r 2 1 Þ½ðr 2R 1 r 2 1 Þh r2 1 ðr 2 þ R 1 ÞŠ ¼ 0:

24 24 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda (i) p=2 a 2y a 0 (ii) p < 2y a p=2 Fig. 11. Configuration of points in Lemma 5.6. Then we have h ¼ r1 2ðr 2 þ R 1 Þ=ðr 2 R 1 r1 2 Þ, and thus (5.37). Figure 11 (ii) illustrates the case of p < 2y a p=2. We see from AB ¼ DC that ð5:41þ ðh þ R 1 Þ cosðp j2yjþ þ h þ r 2 ¼ðr 2 þ R 1 Þ cosðp 2y 0 Þ; which also leads us to (5.40), and thus (5.37). Lemma 5.6. Therefore we have proved r From (5.32), (5.34) and the initial condition f 0; 0 ¼ 0, f 1; 0 ¼ 1, we see by induction that the points f n; 0 satisfy ð5:42þ gf n; 0 ¼ 2n ð f nþ1; 0 f n; 0 Þð f n; 0 f n 1; 0 Þ f nþ1; 0 f n 1; 0 : Similarly, we see that f 0; m satisfy ð5:43þ gf 0; m ¼ 2m ð f 0; mþ1 f 0; m Þð f 0; m f 0; m 1 Þ f 0; mþ1 f 0; m 1 : Thus it is possible to determine f n; m in Zþ 2 by using (2.1). Since (2.1) is compatible with (2.3), f n; m satisfies (2.3) simultaneously. This proves the second part of Theorem 5.3. r 5.2. Positivity of radii of circles Theorem 5.3 claims that if R n; m satisfying (5.13) and (5.14) is positive, then the corresponding f n; m is an immersion. In order to establish Theorem 5.1, we have to prove the positivity of R n; m determined by (5.13) and (5.14) with the initial condition R 1; 0 ¼ 1, R 0; 1 ¼ z ð> 0Þ. First, we show that positivity of R m; n for ðn; mþ A V is reduced to that of R n; nþ1 for n A Z þ.

25 An Explicit Formula for the Discrete Power Function 25 Proposition 5.7. Let the solution R n; m of (5.13) for ðn; mþ A V and (5.14) for n ¼ m A Z þ with initial data ð5:44þ R 1; 0 ¼ 1; R 0; 1 ¼ z ð> 0Þ be positive for m ¼ n þ 1, namely, R n; nþ1 > 0 for any n A Z þ. Then R n; m is positive everywhere in V and satisfies (5.14) for ðn; mþ A V. Proof. Equation (5.14) for ðn; mþ ¼ð0; 0Þ with initial data (5.44) determines R 1; 2. We use (5.13) and (5.14) for n ¼ m inductively to get R n; nþ1 and R nþ1; n. As was mentioned before, we see that R 2nþ1; 0 ¼ 1 and R 0; 2mþ1 ¼ z for all n; m A Z þ by putting n ¼ 0 and m ¼ 0, respectively, in (5.13). With these data one can determine R n; m in V by using (5.13). When n b m, we use (5.13) in the form of ðn mþr n; mþ1 þðnþmþr n; m 1 ð5:45þ R nþ1; m ¼ R n 1; m : ðn þ mþr n; mþ1 þðn mþr n; m 1 For positive R n 1; m ; R n; mþ1 and R n; m 1,wegetR nþ1; m > 0. When m b n, one can show in a similar way that R n; mþ1 > 0 for given positive R n; m 1, R nþ1; m and R n 1; m by using (5.13). One can show by induction that we have (5.30) for m ¼ n b 1, and we get (5.14) for n ¼ m þ 2. Similarly, one can show by induction that we have (5.30) at ðn þ 2k; nþ for n; k b 1, and (5.14) for ðn þ 2k; nþ. Thus we obtain (5.14) for n b m. One can show in a similar way that we have (5.14) for n a m by using (5.30) at ðn; n þ 2kÞ as an auxiliary relation. r Due to Proposition 5.7, the discrete function Z g with Re g ¼ 1 is an immersion if and only if R n; nþ1 > 0 for all n A Z þ. We next reduce the positivity to the existence of unitary solution to a certain system of di erence equations. Proposition 5.8. The map f : Zþ 2! C satisfying (2.1) and (2.3) with the initial condition f 0; 0 ¼ 0, f 1; 0 ¼ 1, f 0; 1 ¼ iz ðz > 0Þ is an immersion if and only if the solution ðx n ; y n Þ to the system of equations ð5:46þ ðx n 1Þ y nþ1 1 þðx n þ 1Þ y n 1 x n þ y nþ1 x n y n 1 ¼ 0; with g 2 ¼ n þ 2 1 xnþ1 1 y 1 nþ1 n þ 1 1 þ xn 1 y nþ1 ; x 0 y 0 ¼ g g 2 ; ð5:47þ y 0 ¼ z þ i z i ; is of the form x n ¼ e 2ia n, y n ¼ e 2ij n, where an ; j n A ð0; p=2þ.

26 26 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda Proof. Let f n; m be an immersion. Define a n ; j n A ð0; p=2þ through ð5:48þ f n; nþ2 f n; nþ1 ¼ e 2ia n ð f nþ1; nþ1 f n; nþ1 Þ; Using Proposition 5.2, one obtains f nþ1; nþ1 f n; nþ1 ¼ e 2ij n ð fn; n f n; nþ1 Þ: ð5:49þ f nþ1; nþ1 f n; nþ1 ¼ R n; nþ1 e ið2j n p=2þb n Þ ; f n; nþ1 f n 1; nþ1 ¼ R n; nþ1 e iðb nþ2a n 1 p=2þ ; f n; nþ2 f n; nþ1 ¼ R n; nþ1 e ið2j n p=2þ2a n þb n Þ ; f n; nþ1 f n; n ¼ R n; nþ1 e iðb nþp=2þ ; where b n ¼ argð f nþ1; n f n; n Þ. Figure 12 is a schematic diagram illustrating the configuration of the relevant quadrilaterals. Now the constraint (2.3) for ðn; n þ 1Þ is equivalent to gf n; nþ1 ¼ 2iR n; nþ1 e ib n n þ 1 ð5:50þ n e 2ij n þ e 2ia þ : n 1 e 2iðj nþa nþ 1 Putting these expressions into the equality ð5:51þ f nþ1; nþ2 f n; nþ1 ¼ e ið2j n p=2þb n Þ ðr n; nþ1 þ ir nþ1; nþ2 Þ; together with R nþ1; nþ2 ¼ R n; nþ1 tan a n and e ib nþ1 ¼ e iðb n p=2þ2j Þ n, one obtains n þ 1 n þ 2 ð5:52þ 2i sin a n e 2ij þ nþ1 þ e 2ia n e 2iðj nþ1þa nþ1 Þ 1 n þ 2 cos a n 1 þ e þ n þ 1 2iðj n a n 1 Þ e 2ia ¼ ge ia n : n e 2ij n Fig. 12. A schematic diagram of quadrilaterals for Proposition 5.8.

27 An Explicit Formula for the Discrete Power Function 27 On the other hand, equation (5.14) for m ¼ n is reduced to ð5:53þ dr n; nþ1 ¼ðn þ 1Þ R nþ1; nr nþ1; nþ2 R 2 n; nþ1 R nþ1; n þ R nþ1; nþ2 n R2 n; nþ1 R nþ1; nr n 1; n R nþ1; n þ R n 1; n : By a similar geometric consideration to the proof of Lemma 5.6, this implies ð5:54þ d ¼ ðnþ1þcotða n þ j n Þ n tanða n 1 j n Þ; which can be transformed to ð5:55þ g 2 ¼ n þ 1 1 e n 2iða nþj n Þ 1 þ e : 2iðj n a n 1 Þ By using (5.55), we see that (5.52) yields the first equation of (5.46) with x n ¼ e 2ia n and y n ¼ e 2ij n. The second equations of (5.46) come from (5.55). This proves the necessity part. Now let us suppose that there is a solution ðx n ; y n Þ¼ðe 2ia n ; e 2ij n Þ of (5.46) with a n ; j n A ð0; p=2þ. This solution together with (5.48) and (2.1) determines a sequence of orthogonal circles with their centers on f n; nþ1, and thus the points ð f n; nþ1 ; f ng1; nþ1 ; f n; n ; f n; nþ2 Þ. Now (2.1) determines f n; m on Zþ 2. Since a n; j n A ð0; p=2þ, the inner parts of the quadrilaterals ð f n; nþ1 ; f nþ1; nþ1 ; f nþ1; nþ2 ; f n; nþ2 Þ and of the quadrilaterals ð f n; n ; f nþ1; n ; f nþ1; nþ1 ; f n; nþ1 Þ are disjoint, which means that we have positive solution R n; nþ1 and R nþ1; n of (5.13) and (5.14). Given R nþ1; n and R n; nþ1, (5.13) determines R n; m for all ðn; mþ A V. Due to Proposition 5.7, R n; m is positive, and satisfies (5.13) and (5.14). Theorem 5.3 implies that the discrete conformal map g n; m corresponding to the circle pattern fc n; m g determined by R n; m is an immersion and satisfies (2.3). Since g n; n ¼ f n; n and g n; ng1 ¼ f n; ng1, equation (2.1) implies g n; m ¼ f n; m. This proves Proposition 5.8. r Note that although (5.46) is a system of equations, a solution ðx n ; y n Þ of (5.46) is determined by its initial value y 0. The system of equation (5.46) can be written in the following recurrent form: ð5:56þ or ð5:57þ ð5:58þ y nþ1 ¼ 1 þ x2 n 2x n y n x 2 n y n þ y n 2x n ; x nþ1 ¼ y nþ1 ¼ Fðx n ; y n Þ; x nþ1 ¼ Cðn; x n ; y nþ1 Þ; Fðx; yþ ¼y x 1 y 1 þ xy 1 2 xy þ x 1 ; y 2 ð2n þ 2 þ gþx n þ gy nþ1 y nþ1 ½ðg 2Þx n þðg 2n 4Þy nþ1 Š ; Cðn; x; yþ ¼ 1 x ð2n þ 3 þ idþxy 1 þð1 þ idþ ð2n þ 3 idþx 1 y þð1 idþ ;

28 28 Hisashi Ando, Mike Hay, Kenji Kajiwara and Tetsu Masuda and x 0 ¼ ð1þidþ=fð1 idþy 0 g. It is easy to see that jfðx; yþj ¼ jcðn; x; yþj ¼ 1 when jxj ¼jyj ¼1 and that jx 0 j¼1 when jy 0 j¼1, which implies that this system possesses unitary solutions. Moreover, we have the following theorem as for the arguments of the unitary solutions: Theorem 5.9. There exists a unitary solution ðx n ; y n Þ to the system of equation (5.46) with x n ; y n A A I nfg1g, where ð5:59þ A I ¼fe 2ib j b A ½0; p=2šg: Proof. We first investigate the properties of the function Fðx; yþ and Cðn; x; yþ restricted to the torus T 2 ¼ S 1 S 1 ¼fðx; yþjx; y A C; jxj ¼jyj ¼1g. Property 1. The function Fðx; yþ is continuous on A I A I nfðg1;g1þg. The function Cðn; x; yþ is continuous on A I A I for any n A Z þ. (Continuity on the boundary of A I A I is understood to be one-sided.) The points of discontinuity must satisfy ð5:60þ x 2 y þ y 2x ¼ 0; ð2n þ 3 idþy þð1 idþx ¼ 0: The first identity holds only for ðx; yþ ¼ðG1;G1Þ. unitary x, y. The second never holds for Property 2. For ðx; yþ A A I A I nfðg1;g1þg, we have Fðx; yþ A A I. For ðx; yþ A A I A I, we have Cðn; x; yþ A A I U A II U A IV, where A II :¼ fe 2ib j b A ðp=2; pšg and A IV :¼fe 2ib j b A ½ p=2; 0Þg. Property 2 is verified as follows: using the transformation ð5:61þ where x n ¼ e 2ia n form ð5:62þ u n ¼ tan a n ; v n ¼ tan j n ; and y n ¼ e 2ij n, we see that the first equation of (5.46) takes the v nþ1 ¼ u 2 n v n: It is obvious that u 2 v A ½0; þyš when ðu; vþ A ½0; þyš½0; þyšnfð0; 0Þ; ðþy; þyþg. The second equation of (5.46) can be expressed as ð5:63þ a nþ1 ¼ o n a n þ p 2 ; e2io n ¼ ð2n þ 3 þ idþx n ynþ1 1 þð1þidþ ð2n þ 3 idþxn 1y nþ1 þð1 idþ : By using the variables u n, v nþ1, we have ð5:64þ tanðo n a n Þ¼Fðn; u n ; v nþ1 Þ;

29 An Explicit Formula for the Discrete Power Function 29 where ð5:65þ Fðn; u; vþ ¼ ½ðn þ 1Þ ðn þ 2Þv2 Šu ð2n þ 3Þv þ dð1 þ uvþ ðn þ 2Þ ðn þ 1Þv 2 þð2n þ 3Þuv þ dvð1 þ uvþ : Lemma It holds that o n a n þ p=2 A ½ p=2; pš for a n ; j nþ1 A ½0; p=2š. Proof. Let us investigate the function F ðn; u; vþ for u; v A ½0; þyš. It is easy to see that ð5:66þ qfðn; u; vþ qu > 0; qfðn; u; vþ qv < 0 on ½0; þyš 2 except for the points satisfying ðn þ 2Þ ðnþ1þv 2 þð2nþ3þuv þ dvð1 þ uvþ ¼0. Consider the values of Fðn; u; vþ on the boundary of ½0; þyš 2. It is easy to see that ð5:67þ We find that Fðn; 0; þyþ ¼0; Fðn; 0; 0Þ ¼ d n þ 2 ; Fðn; þy; þyþ ¼ n þ 2 ; Fðn; þy; 0Þ ¼þy: d dfðn; 0; vþ dv < 0 except for the point qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v ¼ v þ :¼ d þ d 2 þ 4ðn þ 1Þðn þ 2Þ ; 2ðn þ 1Þ and that F ðn; u; 0Þ ¼fðn þ 1Þu þ dg=ðn þ 2Þ is monotone increasing. Note that ð5:68þ ðn þ 2Þu Fðn; u; þyþ ¼ n þ 1 du ; Fðn; þy; vþ ¼ ðn þ 2Þv2 þ dv þ n þ 1 : vðdv þ 2n þ 3Þ When d > 0, we see that dfðn; u; þyþ du except for the point u ¼ ðn þ 1Þ=d and that dfðn; þy; vþ dv > 0 < 0