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1 2n C(n) ≡ n +1 n   since S (1, n)=2C(n). The fact that S(m, n) is always a whole number is a consequence of the recurrence relation 4S(m, n)= S(m +1, n)+ S(m, n + 1). (2) From the definition it is clear that S(n, m) = S(m, n). A further inspection of the formula shows that the zeroth row and zeroth column of the matrix of values S (m, n) agree with the diagonal, and these are the central binomial coefficients 2k S(k, 0) = S(0,k)= S(k,k)= . k   The table below lists super Catalan numbers for small values of m and n from 0 up to 10, with S (m, n) in row m and column n, with indexing starting at 0.

m n 0 12345678910 \ 0 1 2 6 20 70 252 924 3432 12870 48620 184756 1 2 2 4 10 28 84 264 858 2860 9724 33592 2 6 4 6 12 28 72 198 572 1716 5304 16796 3 20 10 12 20 40 90 220 572 1560 4420 12920 4 70 28 28 40 70 140 308 728 1820 4760 12920 5 252 84 72 90 140 252 504 1092 2520 6120 15504 6 924 264 198 220 308 504 924 1848 3960 8976 21318 7 3432 858 572 572 728 1092 1848 3432 6864 14586 32604 8 12870 2860 1716 1560 1820 2520 3960 6864 12870 25740 54340 9 48620 9724 5304 4420 4760 6120 8976 14586 25740 48620 97240 10 184756 33592 16796 12920 12920 15504 21318 32604 54340 97240 184 756

Table 1: S(m, n) for 0 m, n 10. ≤ ≤ Somewhat surprisingly, there is no combinatorial understanding for S (m, n) to date, unlike the Catalan numbers which have hundreds of combinatorial interpretations, compiled for example in [15]; see also sequence A000108 in the OEIS [12]. There are, however, some combinatorial or weighted interpretations of S (m, n) for some values of m and n which we will describe in the next section. In this paper we introduce also a closely related family of rational numbers, the circular super Catalan numbers given by S(m, n) Ω(m, n) . (3) ≡ 4m+n

2 We also give a table of the circular super Catalan numbers for small values of m and n below.

m n 012345 6 7 8 \ 1 3 5 35 63 231 429 6435 0 1 2 8 16 128 256 1024 2048 32 768 1 1 1 5 7 21 33 429 715 1 2 8 16 128 256 1024 2048 32 768 65 536 3 1 3 3 7 9 99 143 429 2 8 16 128 256 1024 2048 32 768 65 536 262 144 5 5 3 5 5 45 55 143 195 3 16 128 256 1024 2048 32 768 65 536 262 144 524 288 35 7 7 5 35 35 77 91 455 4 128 256 1024 2048 32 768 65 536 262 144 524 288 4194 304 63 21 9 45 35 63 63 273 315 5 256 1024 2048 32 768 65 536 262 144 524 288 4194 304 8388 608 231 33 99 55 77 63 231 231 495 6 1024 2048 32 768 65 536 262 144 524 288 4194 304 8388 608 33 554 432 429 429 143 143 91 273 231 429 429 7 2048 32 768 65 536 262 144 524 288 4194 304 8388 608 33 554 432 67 108 864 6435 715 429 195 455 315 495 429 6435 8 32 768 65 536 262 144 524 288 4194 304 8388 608 33 554 432 67 108 864 2147 483 648

Table 2: Ω(m, n) for 0 m, n 8. ≤ ≤ This second table is still symmetric, namely Ω(n, m) =Ω(m, n). The relation between the zeroth row, or the zeroth column, and the diagonal is now 1 1 1 2k Ω(k,k)= Ω(k, 0) = Ω(0,k)= . 4k 4k 42k k   A first indication of the possible interest of this second table is that the recurrence property of the super Catalan numbers becomes a Pascal-like property of the circular super Catalan numbers:

Ω(m, n) =Ω(m +1, n)+Ω(m, n + 1) so that each entry is the sum of the entry directly below it and to the right. It is not hard to see that these three conditions, together with the normalization condition that Ω(0, 0)= 1, determines the table.

Proposition 1 If a table of rational numbers c (m, n) for natural numbers m, n has the following properties:

1. c (0, 0)=1

2. c (m, n)= c (n, m) 1 1 3. c (m, m)= c (m, 0) = c (0, m) and 4m 4m 4. c (m, n)= c (m +1, n)+ c (m, n + 1)

then it must be that c (m, n)=Ω(m, n) for any natural numbers m and n.

3 We will see that the circular super Catalan numbers actually have a central algebraic and analytic interpretation in the context of integration or summation theory of polynomials over unit circles over finite fields of odd characteristic, in a manner that is to us quite remarkable, indeed almost unbelievable. Their role extends even to summations over suitably defined unit circles in relativistic geometries, and the curious relations between the Euclidean and relativistic formulas are an aspect of the three- fold symmetry of chromogeometry [16] which is an outgrowth of rational trigonometry [17]. This theory brings together Euclidean planar geometry with quadratic form x2 + y2 (blue) with two different forms of relativistic Einstein-Minkowski geometries with quadratic forms x2 y2 (red) and − xy (green). The relations and indeed symmetries between these three geometries play a crucial role in our derivations, as summations over unit circles in both Euclidean and relativistic geometries are closely related, and connections between them provide a powerful tool for explicit evaluations. In the rest of this Introduction we will explain the remarkable formulas in which the circular super Catalan numbers play the primary role and illustrate them with examples. We will also observe that the classical formulas for integration on the circle in the Euclidean plane can also be restated in terms of super Catalan numbers, so we obtain a bridge between characteristic zero and prime characteristic harmonic analysis. We will see that these formulas also suggest the study of a wide variety of additional (rational) numbers that play secondary roles.

1.1 Summing polynomials over unit circles over finite fields r We consider a finite field Fq where q = p is a power of an odd prime p, so that Fp Fq. The associated 2 ⊆ affine plane A consists of points [x, y] where x, y are in Fq, and we will in this paper identify this also with the associated vector space V2 of row vectors (x, y), so that linear transformations given by matrices act also on A2 on the right. Introduce three different metrical geometries: Euclidean (blue) and relativistic (red and green) with respective associated quadratic forms x2 + y2, x2 y2, and xy. The respective unit circles in − each geometry are then defined by

2 2 Sb (Fq)= Sb,q [x, y]: x, y Fq and x + y =1 ≡ ∈ 2 2 Sr (Fq)= Sr,q [x, y]: x, y Fq and x y =1 ≡  ∈ − Sg (Fq)= Sg,q [x, y]: x, y Fq and xy =1 ≡{ ∈ } where the subscripts b, r, and g naturally refer to blue, red, and green respectively; for brevity, we use the subscript c when we wish to refer to an undetermined one of these colors. We will also use Sc,q as a shorthand for Sc (Fq) when the underlying field is clear from context. The dichotomy of whether 1 is a square or not in Fq plays a role in the blue geometry and we − note that distinction through the Jacobi symbol, which is a generalization of the Legendre symbol. For any odd prime p, the Legendre symbol is defined as

1 1 if 1 is a square in Fp − − p ≡  1 if 1 is not a square in Fp    − − and for any prime power q = pr the Jacobi symbol is defined as 1 1 r − − . q ≡ p     4 This symbol comes into play in the following elementary count of the sizes of the unit circles.

Lemma 1 Let Sb,q, Sr,q, and Sg,q be the unit circles in the blue, red, and green geometries respectively over the field Fq. Then

1 Sg,q = q 1 Sr,q = q 1 and Sb,q = q − . | | − | | − | | − q   Note that while we naturally consider the blue Euclidean geometry to be primary, we list the results in the order green, red, and blue which we will see generally aligns with the natural increasing complexity of formulas. Now consider the linear space Pol2 (Fq) of polynomials in two variables α and β with coefficients in Fq, which we will introduce in terms of variants called polynumbers which have been used by Wildberger [18] and allow us more clearly to distinguish between polynomials as algebraic objects and as functions. Polynumbers can be evaluated as functions, but their definition is solely in terms of an array of coefficients, and so in the context of polynumbers, the monomials αkβl as k and l range over all natural numbers (including 0) are independent. We will use the convention that α0β0 = 1 represents the constant polynumber whose value as a function is identically 1 = 1q in Fq. For c one of b, r, or g, define the Fq-valued normalized summation functional

ψc,q : Pol (Fq) Fq 2 → to be the linear functional whose values on a monomial αkβl for k and l natural numbers are given by

k l k l ψg,q α β x y ≡ − [x,y]∈S
Xg,q k l k l ψr,q α β x y (4) ≡ − [x,y]∈S
Xr,q k l 1 k l ψb,q α β − x y . ≡ − q   [x,y]∈Sb,q
X

We will show that ψc,q satisfies the following three conditions which we state loosely here to avoid technicalities, namely:

(Normalization) For the polynumber 1 in Pol2 (Fq) we have ψc,q (1)=1q.

(Locality) For any polynomial f that as a function restricts to zero on the circle Sc,q we have ψc,q (f) = 0.

(Invariance) ψc,q is invariant under any linear transformation that preserves the bilinear form of the associated geometry c.

These conditions will be presented in a more precise manner in Section 4 where we describe the linear groups of symmetries of each of the geometries. We call any linear functional satisfying the three conditions above a circular integral functional. The perhaps curious coefficients involved in the definitions are explained by the following main theorem.

5 Theorem 1 The linear functionals ψg,q, ψr,q, and ψb,q given in (4) are the unique circular integral functionals in the green, red, and blue geometries respectively.

We prove the above theorem in two steps. First, we establish that ψc,q is indeed a circular integral functional in each geometry c. Next, we prove that if such a circular integral functional in each geometry exists, then it must be unique. In fact, not only are we able to prove the uniqueness of ψc,q, we are also able to give an explicit formula for ψc,q in each geometry, and curiously the derivation in the red and blue geometries stems from the result in the green geometry. Moreover, we can describe ψr,q and ψb,q in terms of the elements

S (m, n) Ω(m, n) mod p = mod p 4m+n (2m)!(2n)! = mod p 4m+nm!n!(m + n)!

m+n of Fp Fq which are well-defined since S (m, n) is always an integer and the term 4 in the denominator⊆ is never a multiple of p. Here is a simplified and restricted version of the formulas for ψc,q in the red and blue geometries. This is the heart of our understanding of the true significance of the super Catalan numbers, and their rational cousins the circular super Catalan numbers.

r Theorem 2 Consider the finite field Fq where q = p with p> 2.

1. In the red geometry, for 0 k + l < q 1, ≤ − n k l ( 1) Ω(m, n) mod p if k =2m and l =2n, ψr,q α β = −  0 otherwise.
 2. In the blue geometry, for 0 k + l < q 1, ≤ −

k l Ω(m, n) mod p if k =2m and l =2n, ψb,q α β =  0 otherwise.
 Here are some immediate consequences.

r Corollary 2.1 Consider the finite field Fq where q = p with p> 2 and any natural numbers k and l satisfying k + l < q 1. − k l k l 1. The normalized sums ψr,q α β and ψb,q α β lie in the prime field Fp and are independent of the power of p.

2. Suppose l =2n. We then have

k l n k l ψr,q α β =( 1) ψb,q α β . −

k l
k l
3. If k and l are both even, then ψr,q α β and ψb,q α β are never 0.

6 However to us it is even more remarkable that the single rational number Ω(m, n) determines the 2m 2n 2m 2n values ψr,q (α β ) and ψb,q (α β ) for all prime powers q>k + l + 1 in such a direct and uniform fashion. Here is an example to illustrate these formulas.

Example 1 Consider the Euclidean (blue) geometry in the case q = p = 13 so that we can identify F as 0, 1, 2,..., 12 . Here, 1=12=52 is a square, so −1 = 1. From Theorem 2, if 13 { } − q 0 k + l < q 1=12 then   ≤ − 2m 2n ψb,13 α β = Ω(m, n)mod13. The (blue) unit circle is

Sb, = [1, 0] [12, 0] [0, 1] [0, 12] [2, 6] [2, 7] [11, 6] [11, 7] [6, 2] [7, 2] [6, 11] [7, 11] 13 { } where there is no need for commas as the points are not naturally ordered. By manual inspection, working over F13 and omitting the zero terms,

4 1 4 ψb, α = − x 13 − q   [x,y]∈Sb,13
X = x4 − [x,yX]∈Sb,13 = 14 + 124 +24 +24 + 114 + 114 +64 +74 +64 +74 − =2.
Using the formula of Theorem 2, we may verify that 3 ψ α4 = Ω(2, 0)mod13 = mod13 = 2. b,11 8
Similarly

2 6 1 2 6 ψb, α β = − x y 13 − q   [x,y]∈Sb,13
X = x2y6 − [x,yX]∈Sb,13 = 2266 +2276 + 11266 + 11276 +6226 +7226 +62116 +72116 − =4
and again using the formula we verify that 5 ψ α2β6 = Ω(1, 3)mod13 = mod13 = 4. b,13 128

2m 2n For the example above, the values ψb, (α β ) are recorded for 0 m,n < 10 in the following 13 ≤ table. Note that they are indeed all non-zero, at least when m + n < 1 (q 1) = 6 according to 2 −

7 Corollary 2.1. 2m 2n ψb, (α β ): m n 0123456789 13 \ 0 1726273726 1 7594849594 2 2959484959 3 6495948495 4 2849594849 5 7484959484 6 3948495948 7 7594849594 8 2959484959 9 6495948495

2m 2n Table 3: The values of ψb, (α β ) for 0 m,n < 10. 11 ≤ Correspondingly, here are the values of the circular super Catalan numbers in the same range, reduced mod 13.

Ω(m, n) mod 13 : m n 0123 4 56789 \ 0 1726 2 71000 1 7594 8 61000 2 2959 2 51000 3 649 7 10 4 100 0 4 28210 6 3 100 7 5 765 4 3 2 10612 6 1111 1 11777 7 0000 0 07000 8 0000 0 67000 9 000 0 7 12700 0

Table 4: The values of Ω(m, n)mod13 for 0 m,n < 10. ≤ So while we see that there is indeed agreement between the circular super Catalan numbers and the 2m 2n q−1 summations ψb,11 (α β ), this only holds in the range we have stipulated, namely m + n< 2 =6 which we call the upper triangular range. In fact we cannot in general expect the circular super Catalan numbers beyond the upper tri- angular range to determine the summation values, as the numerators of the super Catalan numbers

8 generally contain large factorials which will become 0 modulo any finite prime p beyond some point. However for fixed m and n all but finitely many prime powers q will satisfy the upper triangular range condition. The complete formulas for the red and blue geometries which go beyond this constraint are more involved but still quite attractive, and they point towards an additional family of integers that extend the super Catalan numbers. These formulas are given in terms of sums of particular coefficients of the rational polynumbers 1+ α k 1 α l fk,l = fk,l (α) − ≡ 2 2     which we will call circular polynumbers, suitably reduced mod p. We use the convention that for a polynumber f in α, the expression αk f denotes the coefficient of αk in f, and allow ourselves to write   l αk1 f + αk2 f + + αkl f = αki f. ··· i=1 ! X The key point that connects  these  circular polynumbers  and the super Catalan numbers is the following result of Georgiadis, Munemasa, and Tanaka in [8] restated in terms of the circular super Catalan numbers and the polynumbers fk,l rather than Krawtchouk matrices. Theorem 3 For any natural number m and n, n m+n Ω(m, n)=( 1) α f m, n. − 2 2 We now state the complete summation formulas for ψc,q in each geometry; these will make separate appearances as Theorems 12, 15, and 21 respectively. The formulas involve geometric constants that depend on the sizes of the unit circles in each geometry, and the symbol x represents the usual floor function, namely the greatest integer less than or equal to x. ⌊ ⌋ r Theorem 4 Consider the finite field Fq where q = p with p> 2. 1. In the green geometry,

k l 1q if Sg,q (k l) , ψg,q α β = | | | −  0 otherwise.
 1  k+l 2. In the red geometry, with w = Sr,q and R = , 2 | | |Sr,q | j k m+n+dw α fk,l mod p if k =2m, l =2n, ψ (αkβl)=    r,q  |d|≤R  X     0  otherwise.    1 k+l 3. In the blue geometry, with w = 2 Sb,q and R = , | | Sb,q | |

m+n+dw n α ( 1) fk,l mod p if k =2m, l =2n, ψ (αkβl)=    − b,q  |d|≤R  X     0 otherwise.   9 k l k l So we see that our main formulas to evaluate ψr,q(α β ) and ψb,q(α β ) involve an equally spaced ladder of coefficients of fk,l or fk,l symmetrically placed around the central coefficient which is the circular super Catalan number− Ω (m, n) when k =2m and l =2n, and is zero otherwise. 8 2 In the following examples, we will compute the value of ψb,q evaluated at the monomial α β in two different fields F13 and F7. Since k and l are both even, m = 4 and n = 1. Both computations require us to find certain coefficients of the polynumber 8 2 n 1 1+ α 1 α ( 1) fk,l =( 1) f , = − − − 8 2 − 2 2     which, upon expansion, is equal to 1 3 13 1 7 7 7 1 13 3 1 α α2 α3 + α4 + α5 + α6 α7 α8 α9 α10. −1024 − 512 − 1024 − 128 512 256 512 − 128 − 1024 − 512 − 1024 Note that the expression is palindromic around the central value of α5 and that the coefficient of that term is 7 = Ω(4, 1). 256

Example 2 Continuing our previous example where q = p = 13, we deduce that Sb, = 12 so | 13| 1 k+l 10 w = 2 Sb,13 =6 and R = = 12 =0. The formula of ψb,q in Theorem 4 above then reads | | Sb,13 | |  

8 6 5+6d ψb, (α β )= α ( f , )mod13 13   − 8 2 |d|≤0 X   5 = α ( f , )mod13 − 8 2 7 =  mod13 256 =8. Here we see that the condition 0 k + l < q 1 reduces the sum to just the central term. ≤ − Example 3 Now, consider the corresponding formula when q = p = 7. In this case, 1 is not a − 1 k+l 2+8 square so according to Lemma 1, w = 2 Sb,7 =4 and R = = 8 =1. It follows that | | Sb,7 | |  

8 2 5+4d ψb, (α β )= α ( f , )mod7 7   − 8 2 |d|≤1 X   1 5 9 = α + α + α ( f , )mod7 − 8 2 3 7 3 1 =   +   
mod7 = mod7= 1. −512 256 − 512 64   It can be verified that

Sb, = [1, 0] [6, 0] [0, 1] [0, 1] [2, 2] [5, 5] [2, 5] [5, 2] 7 { } and we can check directly by working modulo 7 that

2 8 2 8 2 8 2 8 2 8 ψb,7(α β )=2 2 +5 5 +2 5 +5 2 =1.

10 1.2 Connection to characteristic zero integration theory The theory of integrating polynomials on a sphere, even in higher dimensions, dates back at least to the paper of Baker [2]. Later, Folland [7] offered a shorter way of computing the integral of a polynomial over a sphere, and both Baker’s and Folland’s formulas utilized the gamma function Γ. Here we restate the main theorem in Folland’s paper without proof involving the usual rotationally invariant measure dµ on the (d 1)-dimensional unit sphere in Rd coming from the usual Riemannian structure on Rd. −

d 2 2 2 Theorem 5 Let Sd = [x , x ,...,xd] R : x + x + + x =1 . We have that 1 2 ∈ 1 2 ··· d

 a1 a2 ad 2Γ(b1)Γ(b2) Γ(b d) x1 x2 xd dµ = ··· ··· Γ(b + b + + bd) ZSd 1 2 ··· 1 if all the natural numbers ai are even with bi = 2 (ai + 1) , and is zero otherwise. In the next of our papers, entitled Circular Super Catalan numbers and classical Polynomial Integration over Spheres, we will show that algebraic reformulations of the classical formulas of Baker and Folland for integrating polynomials over spheres can be obtained from computations with spherical coordinates, and that the results obtained can be expressed without reference to transcendental functions, such as the gamma function Γ, and also without irrational numbers, such as π. In the two-dimensional case we will see that the integrals we obtain over the Euclidean circle can be algebraically expressed using the circular super Catalan numbers, and yield formulas which are essentially identical to the ones obtained in this paper in the finite field case. Indeed, by taking d = 2 and a1 =2m, a2 =2n in Theorem 5, we can make the following simple manipulations: 2Γ(b )Γ(b ) 2Γ m + 1 Γ n + 1 1 2 = 2 2 Γ(b1 + b2) Γ(m +
n + 1)
2 m 1 m 3 1 √π n 1 n 3 1 √π = − 2 × − 2 ×···× 2 − 2 × − 2 ×···× 2

(m +
n)!


2π ((2m 1) (2m 3) 1)((2n 1) (2n 3) 1) = − × − ×···× − × − ×···× 2m+n (m + n)! 2π (2m)!(2n)! = 4m+nm!n!(m + n)! =2πΩ(m, n) so if the measure is normalized, that is, if

dµ =1 ZS2 then we have that

Ω(m, n) if a1 =2m, a2 =2n, xa1 xa2 dµ = 1 2  S2 0 otherwise Z  which is very akin to the formula of ψ presented in Theorem 2. Note that the value of the integral b,q  is always a rational number. This is analogous to the finite field situation since Q is a prime subfield of R.

11 In the last of our series of three papers, entitled Super Catalan numbers and Algebraic Integration over Spheres, we initiate a new algebraic theory of integration on spheres which avoids classical measure theory entirely, does not rely on limit processes or real numbers, and allows us to realize the circular super Catalan numbers and additional variants and generalizations as encoding fundamental integral formulas for polynomials over circles and spheres in general characteristic zero fields. While there is something combinatorial about the connection between the super Catalan numbers and their circular variants, with sums over the unit circle in finite fields, the broader picture that we obtain shows that their real significance is algebraic and indeed analytic. We are obtaining here formulas which ought to be at the heart of modular Fourier analysis over general fields, where we study not complex-number-valued functions on a circle or a sphere, but rather functions taking values in the field over which the circle or sphere is defined. The kind of analysis that emerges here has the pleasant property of being completely specifiable without recourse to any infinite procedures, neither in assumptions on the underlying number system, nor on the transcendental nature of functions. Everything is completely algebraic and concrete, and importantly can be checked by a computer without relying on approximations.

1.3 Recent work on the super Catalan numbers Before we get to the main discussion of this paper, we want to mention some recent developments in the research on the super Catalan numbers. These numbers were perhaps first brought back to the public’s attention by Gessel in his paper [10]. First we note from the definition of the super Catalan numbers in (1) that S (m, n) is always even for every m and n except when m = n = 0, so sometimes the numbers 1 T (m, n) S(m, n) ≡ 2 are studied instead. As mentioned in the introduction, there is no general combinatorial interpre- tation of S (m, n) to date. There are, however, combinatorial interpretations of T (m, n) for some specific m. A description of T (2, n) in terms of blossom trees was discovered by Schaeffer [14], and another in terms of cubic trees by Pippenger and Schleich [13]. Gessel [11] gave an explanation of T (2, n) in terms of pairs of Dyck paths with restricted heights, in which an interpretation of T (3, n) also appeared. A meaning for T (m, m+s)for0 s 3 in terms of restricted lattice paths was found by Chen and Wang [5]. A weighted interpretation≤ of ≤S(m, n) in terms of 2-Motzkin paths was discovered by Allen and Gheorghiciuc [1]. Their description involved a notion of positive and negative 2-Motzkin paths. Most recently, a combinatorial interpretation for T (3, n) and T (4, n) was given by Gheorghiciuc and Orelowitz [9] as the numbers of 3-tuples and 4-tuples of Dyck paths satisfying certain conditions. One (weighted) interpretation of the super Catalan numbers that will be prominent in this paper is as certain values of Krawtchouk polynomials and this was given by Georgiadis, Munemasa, and Tanaka in [8]. This will be discussed in more detail in Section 7.

2 Polynumbers and polynomials

To begin the discussion, we start by clarifying the distinction between polynomials over a field F as formal expressions and as functions. In characteristic zero this distinction can often be finessed, but in the finite field situation it pays to be rather precise and keep the notions carefully separate, without relying on a prior understanding of ”variables”.

12 To make this distinction explicit we introduce a more fundamental computational object called a polynumber. Evaluation associated to a polynumber gives a polynomial function, but different polynumbers may give rise to the same function. Specifically we will see that if two polynumbers f and g satisfy the equality f(t) = g(t) for all t Fq, then they are not necessarily equal as ∈ polynumbers. Since we are dealing here with finite mathematics and finite arithmetic, we propose to dispense with the usual reliance on ”infinite sets”. Throughout this paper, N will denote the type of natural numbers which for us includes the number 0, Z the type of integers, Q the type of rational numbers, and Fq will denote the finite field of size q where q is a power of an odd prime p. We aim for this to be in line with what a computer would do: a computer wants to recognize when an expression that it meets is of a given type, without having to rely on prior databases. In some places, we will indicate the multiplicative identity in F by 1q if there is possibility of confusion. We also use the convention that 00 = 1. A polynumber f over a field F is a (finite) list of numbers in F written vertically with a bar above it and another one to the left, as in

f0

f 1 f = . . .

f k

We call the numbers fi the entries of f. Two polynumbers f and g are equal precisely when one can be obtained from the other by adding or removing zeroes from the bottom. For example,

3 3 0

0 = 1 .

1 − − 2

2 0

The zero polynumber and the identity polynumber are respectively

0 0 and 1 1 . ≡ ≡

If k is the largest natural number such that fk = 0 then k = deg(f) is the degree of the polynumber f. 6 Addition of polynumbers is defined entrywise, adding zeroes at the bottom to make two polynumbers have the same number of entries if necessary. The scalar product of a polynumber f by a number λ F is also entrywise. ∈

13 Given two polynumbers

f0 g0

f g 1 1 f = . and g = . . .

f g k l

of degree k and l respectively, the product of f and g is another polynumber

h0

h 1 h = . .

h k+l

so that for 0 j k + l, ≤ ≤ j

hj = figj−i. i=0 X This is the familiar Cauchy product of sequences. We denote the algebra of polynumbers over F by Pol (F). Define the special polynumber 0 α . ≡ 1

A polynomial in α is a polynumber written as a linear combination of powers of the polynumber α, where for k N, αk will be a list of k zeroes followed by a single 1, with α0 = 1. For example, if ∈

2

1 2 f = 0

3

−

is a polynumber over Q, then

0 0 0 0 1 0 1 1 3 f =2 1 + +0 0 3 =2α + α 3α . 2 1 − 0 2 −

1 1

14 0 k k If f = f (α) = f α + f α + + fkα Pol (F), then fk is the coefficient of α in f and we 0 1 ··· ∈ denote this by αk f . When there are multiple coefficients to extract from f, we will write

  l αki f αk1 f + + αkl f. ≡ ··· i=1 ! X       We also define the evaluation of f at t F to be ∈ k f(t) f + f t + + fkt F. ≡ 0 1 ··· ∈ An element t F is said to be a zero of a polynumber f Pol (F) precisely when f (t) = 0. We ∈ ∈ say that f Pol (F) is irreducible over F precisely when f cannot be written as the product of two polynumbers∈ of degree strictly less than that of f. In a similar fashion we can define a 2-polynumber over F to be a two-dimensional array of numbers in F of the form

f00 f01 f0n ··· f f f n 10 11 ··· 1 f . . . . . ≡ . . .. .

f f f m0 m1 mn ···

Again we adopt the convention that adding further zeroes to the right and down leaves the 2- polynumber unchanged. Addition, subtraction, and scalar multiplication of such 2-polynumbers are pointwise, forming a vector space over F. Multiplication of 2-polynumbers is defined by the two-dimensional version of the Cauchy product, so that for example

5 9 2 1 2 5 1 − − = 17 0 9 . × 3 1 2 4 − 6 10 4

−

Let Pol2 (F) be the algebra of 2-polynumbers over F with multiplicative identity 1, not to be confused with the multiplicative identity of Pol (F) if the context is clear. After we define the polynumber β 0 1 ≡

any 2-polynumber can be written as a linear combination of powers of α and β, for example

2 1 1 0 − 3 0 0 4 2 2 2 3 2 3 f =2+3α α + β +7α β β +5α β +4αβ ≡ 17 0 0 − −

− 0 0 5 0

15 and then the algebra operations agree with the familiar operations on polynomials. So the 2- polynumber multiplication above corresponds to the more usual computation

(1+3α +2β αβ)(5+2α β +4αβ)=5+17α +9β +6α2 2β2 + 10α2β +9αβ2 4α2β2. − − − − Note however that here and throughout α and β are not variables; they are just particular polynum- bers. Evaluation is also obvious, for example if f is as above and t, u F then ∈ f (t, u)=2+3t t2 + u +7t2u u2 +5t3u2 +4tu3. − − This evaluation can also be extended in the usual way to give compositions of polynumbers, for example if g and h are themselves polynumbers, then

f (g, h)=2+3g g2 + h +7g2h h2 +5g3h2 +4gh3. − − If GL(2, F) denotes the group of 2 2 invertible matrices over the field F, then GL (2, F) acts on × the space Pol2 (F) as follows: for a polynumber f and

g11 g12 g = GL(2, F)   ∈ g21 g22   let f g be the polynumber defined by · (f g)(α, β) f ((α, β) g)= f (g α + g β,g α + g β) . · ≡ 11 21 12 22 This action respects evaluation, that is for any t, u F, ∈ (f g)(t, u)= f ((t, u) g)= f (g t + g u,g t + g u) . · 11 21 12 22 We will also use the convention that diag (a, b) refers to the diagonal matrix with entries a and b.

3 Unit circles in chromogeometry

r Throughout this paper, we will be dealing with the finite field Fq of size q = p for some prime p > 2 and a natural number r 1. Such a field is unique up to automorphism. The multiplicative ∗ ≥ subgroup of Fq, denoted by Fq = Fq 0 , is cyclic of order q 1, and so for any x Fq, we have the Fermat property −{ } − ∈ xq = x. For an odd prime p, recall that the Legendre symbol is defined as

1 if a is a quadratic residue modulo p and p does not divide a, a  1 if a is not a quadratic residue modulo p, p ≡     −  0 if p divides a.  

16 For the specific case a = 1, note that p never divides a and thus, rewriting the definition in the − language of finite fields,

1 1 if 1 is a square in Fp, − = − p  1 if 1 is not a square in Fp.    − − For a more general powers of primeq = pr, we resort to a generalization of the Legendre symbol known as the Jacobi symbol, which is defined as

1 1 r − − . q ≡ p     Note that both the Legendre and Jacobi symbol is integer-valued. This is consistent with the characterisation of finite fields Fq in which 1 is a square number, stated in the proposition below. −

Proposition 2 Let q = pr for some odd prime p and a natural number r 1. We have that 1 is ≥ − not a square in Fq precisely when p 1mod4 and r is odd. ≡ −

Proof. If p 1 mod 4 then it is known that 1 is a square in Fp, hence it is also a square in Fq ≡ − since Fp is a subfield of Fq. If p 3 mod 4 then 1 is not a square in Fp and hence we can extend Fp quadratically to ≡ − Fp (i)= x1q + yi : x, y Fp { ∈ } 2 where the formal symbol i satisfies i = 1. It is not hard to see that Fp (i) F 2 so 1 is a square − ≃ p − in F 2 . Now we use the fact that Fps is a subfield of Fpr precisely when s r. If r is even, then F 2 p | p is a subfield of Fpr and thus 1 will be a square in Fq = Fpr . − Now for the sake of contradiction assume that 1 is a square in Fpr where p 3 mod 4 and − 2 ≡ r is odd. Then there exists an element ı Fpr such that ı = 1. Consider K = 1q, ı = ∈ − h i a1q + bı : a, b Fp . It is easy to check that K is a subfield of Fpr isomorphic to Fp2 but this {implies that 2 divides∈ } r which is impossible. Hence our assumption is false and thus 1 is not a − square in Fpr if p 1 mod 4 and r is odd. ≡ − With that in mind, we can also restate the Jacobi symbol in terms of the finite field Fq as

1 1 if 1 is a square in Fq, − = − q  1 if 1 is not a square in Fq.    − −  Remark 5.1 Proposition 2 above is equivalent to the following: −1 = 1 precisely when q q − ≡ 1mod4.   − 2 Now consider V to be the vector space of row vectors v =(x, y) with entries in Fq. For convenience here we will be identifying the natural affine space A2 with its associated vector space, that is blurring the distinction between the vector (x, y) and the affine point [x, y]. We will be interested in three distinct metrical geometries that can be imposed on V2, and hence on A2, that interact in a somewhat symmetrical fashion; this is the phenomenon of chromogeometry, see [16] and [17].

17 Given an invertible symmetric matrix M with entries from Fq, define an associated symmetric bilinear form on V2 by u, v = uMvT h i for u, v V2. The associated quadratic form will be denoted by ∈ Q (v) v, v ≡h i and this element of Fq will be called the quadrance of v. The unit circle is then the set 2 S = S (M, Fq) v V : Q (v)=1 . ≡ ∈ Define H to be the subgroup of G GL(2, Fq) that preserves the bilinear form, that is, those h in G such that for any u and v in V2 we≡ have uh, vh = u, v . h i h i An equivalent condition is that h is M-orthogonal, namely hMhT = M.

The restriction of the action of G then gives an action of H on Pol2 (Fq) by (f h)(α, β)= f ((α, β) h) (5) · for any f Pol (Fq) and h H. ∈ 2 ∈ Define the following 2 2 matrices, with entries in Fq as follows: × 1 0 1 0 1 0 1 I J K . ≡  0 1  ≡  0 1  ≡ 2  1 0  −  1     We briefly note that the entry 2 in K is well-defined since the characteristic of Fq is p > 2. In this 2 paper, we will be considering primarily three different geometries on the affine plane A over Fq, given by the bilinear forms u, v = uMvT where M is either I, J or K: h i 2 2 1. The case M = I gives the blue geometry with quadrance Qb (v)= Qb (x, y)= x + y .

2 2 2. The case M = J gives the red geometry with quadrance Qr (v)= Qr (x, y)= x y . − 3. The case M = K gives the green geometry with quadrance Qg (v) = Qg (x, y) = xy. We note that this definition of green geometry differs from the one in [16] by a factor of 2. Chromogeometry asserts that these three geometries are intimately connected, giving a three-fold symmetry in metrical planar geometry. In each geometry there is a notion of unit circle. The blue unit circle Sb (Fq) (respectively the red unit circle Sr (Fq) and the green unit circle Sg (Fq)) is the set of all [x, y] V2 such that x2 + y2 = 1 (respectively x2 y2 = 1 and xy = 1). For brevity, we ∈ − shall use the notation Sb,q, Sr,q, and Sg,q to indicate the blue, red, and green unit circle respectively when the underlying field is clear from context, and Sc,q to indicate an arbitrary unit circle in the three geometries where c is one of b, r, or g. We will use the elongated name Sc (Fq) only if we need to emphasize the underlying field over which the unit circle is defined. Although it is natural to consider the blue (Euclidean) unit circle to be primary, we shall see that it is more beneficial to list the geometries in ascending complexity: green, red, and blue. This will be the order of presentation of the results in the three geometries adopted in this paper. We first establish a simple observation about the size of the unit circles through basic counting argument.

18 Lemma 2 Let Sb,q, Sr,q, and Sg,q be the unit circles in the blue, red, and green geometries respectively over the field Fq. Then

1 Sg,q = q 1 Sr,q = q 1 Sb,q = q − . | | − | | − | | − q   −1 ∗ ∗ Proof. We can parametrize any element in Sg,q as simply [t, t ] where t F . Since F = q 1, ∈ q q − the conclusion follows. Now, every element in Sr,q except [ 1, 0] can be expressed as −

1+ t2 2t , 1 t2 1 t2  − −  for some t Fq 1 . It is then obvious that Sr,q =1+(q 2) = q 1. ∈ − {± } | | − − Finally, every element in Sb,q except [ 1, 0] admits a parametrization − 1 t2 2t − , 1+ t2 1+ t2   2 for some t Fq such that 1 + t = 0. It thus follows that if 1 is not a square, then the number ∈ 6 − of such points is q + 1. Otherwise, there are exactly two solutions to the equation t2 = 1, so the − number of points on Sb,q is 1 + (q 2) = q 1. − − 4 Circular integral functionals

We now introduce the central object of this paper: a linear functional on the vector space Pol2 (Fq) over Fq which generalizes normalized integration over the unit circle in the usual characteristic zero situation. This will then be an algebraic integral on the unit circle valid for a general finite field Fq of odd characteristic. Note that here we consider functions which take values in the underlying field Fq, not the usual complex numbers C. We assume we have chosen a geometry on V2 which is one of blue (b), red (r) or green (g) and we refer to this color as c. There is then the unit circle Sc,q and the group Hc of invertible linear 2 transformations of V which preserve the bilinear form, and so Hc acts on Sc,q as a group of bijections. As above we can consider Hc as acting on Pol2 (F) via (5). A linear functional φ : Pol (Fq) Fq 2 → is called a circular integral functional precisely when the following three conditions are met:

(Normalization) φ satisfies φ (1)=1q.

(Locality) If f Pol (Fq) satisfies f (x, y) = 0 for all [x, y] in Sc,q then φ (f) = 0. In other words, ∈ 2 φ (f) is zero for any polynumber f that evaluates identically to 0 on Sc,q.

(Invariance) φ is Hc-invariant, i.e. φ (f h)= φ (f) for any f Pol (F) and h Hc. · ∈ 2 ∈ The Locality and Invariance conditions above are motivated by properties of the classical (nor- malized) integral functional over the unit circle. We shall see later that these three conditions are sufficient to assert the existence and uniqueness of such a circular integral functional φ in each of

19 the green, red, and blue geometries. Note that φ is completely determined by the values φ αkβl for natural numbers k and l by linearity.
Below we list some secondary properties that can be derived from the Locality and Invariance conditions. The purpose is two-fold: to illustrate how powerful the two conditions are, and also to obtain useful relations that appear in our analysis later. First we state two properties derived from the Locality property. Theorem 6 (Periodicity property) In all three geometries, if k q, then for any l 0 we have ≥ ≥ φ αkβl = φ αk−q+1βl and if l q then for any k 0 we have

≥ ≥ φ αkβl = φ αkβl−q+1 . (6) k l k−q+1 l Proof. Consider the polynumber α β α
β for a natural
number k q. For [x, y] in Sc,q we have − ≥ αkβl αk−q+1βl (x, y)= xk xk−q+1 yl − − = xk−q (xq x) yl
−
=0 by the Fermat condition xq x = 0. It follows by Locality and the linearity of φ that − φ αkβl = φ αk−q+1βl . The second identity φ αkβl = φ αkβl−q+1
is treated similarly.
Theorem 7 (Pascal-like property)
Define
I (m, n)= φ (α2mβ2n) for any natural numbers m, n. 1. In the red geometry, I (m, n)= I (m +1, n) I (m, n + 1) . (7) − 2. In the blue geometry, I (m, n)= I (m +1, n)+ I (m, n + 1) . (8) Proof. Define two polynumbers f = α2m+2β2n α2mβ2n+2 α2mβ2n 1 − − and f = α2m+2β2n + α2mβ2n+2 α2mβ2n. 2 − If we restrict f1 to the point [x, y] on the red unit circle Sr,q, we get f (x, y)= x2m+2y2n x2my2n+2 x2my2n = x2my2n x2 y2 1 =0 1 − − − − 2 2 since x y = 1 for all [x, y] Sr,q. By applying the Locality property,
− ∈ 0= φ (f1) = φ α2m+2β2n α2mβ2n+2 α2mβ2n − − = φ α2m+2β2n φ α2mβ2n+2 φ α2mβ2n − −
= I (m +1, n) I (m, n + 1) I (m, n)
− −

and the conclusion follows. The identity (8) is proved analogously by restricting f2 to the blue unit 2 2 circle Sb,q and applying Locality using the defining property that x + y = 1 for all [x, y] Sb,q. Now we derive some important identities from the Invariance property. ∈

20 Theorem 8 (Recurrence equation in the red geometry) In the red geometry, we have that for any u Fq such that u = 1 and any natural number k, ∈ 6 ± k k k s 1 u2 φ αk = 1+ u2 (2u)k−s φ αsβk−s (9) − s s=0  

X

Proof. One can verify that if u Fq such that u = 1 then ∈ 6 ±

1 1+ u2 2u h = 2 Hr 1 u  2u (1 + u2)  ∈ − −   since 1 0 1 0 h hT = .  0 1   0 1  − −     By the Invariance property, it follows that, for all u Fq such that u = 1, ∈ 6 ± φ αk = φ αk h · k
1+
u2 2u = φ α + β 1 u2 1 u2  − −  ! k 1 k 2 s k−s s k−s = k φ 1+ u (2u) α β (1 u2) s ! − s=0   k X
1 k s = 1+ u2 (2u)k−s φ αsβk−s (1 u2)k s s=0   − X

where the last equation follows by the linearity of φ. The conclusion then follows by multiplying both sides by the non-zero number (1 u2)k. − Theorem 9 (Recurrence equation in the blue geometry) In the blue geometry, we have that 2 for any u Fq such that u = 1 and any natural number k, ∈ 6 − k k k s 1+ u2 φ αk = 1 u2 (2u)k−s φ αsβk−s . (10) s − s=0  

X

Proof. The proof is identical to the proof in Theorem 8 above, where if u Fq such that u = 1 ∈ 6 ± then we use the Invariance property with

1 1+ u2 2u h = 2 Hb. 1+ u  2u 1 u2  ∈ − −   If 1 is not a square then there is no restriction on u, while if 1 is a square, then h is defined for − − all u Fq except the two elements in the field whose square is 1. ∈ −

21 We will soon see that the geometry of the three unit circles are very much connected to each other and this is a key feature of the derivation. We restate from (4) the three normalized summation functionals in each geometry: k l k l ψg,q α β = x y (11) − [x,y]∈S
Xg,q k l k l ψr,q α β = x y (12) − [x,y]∈S
Xr,q k l 1 k l ψb,q α β = − x y . (13) − q   [x,y]∈Sb,q
X We plan on establishing that these are the unique circular integral functionals in respectively the green, red, and blue geometries, which explains the reason for the curious negative signs involved. We then aim to give explicit formulas for these functionals, and these turn out to involve the circular super Catalan numbers in the red and blue geometries. Somewhat remarkably, the explicit formula for the normalized summation functional in the green geometry is quite easy to find, and this is the key to obtaining the formula in the red geometry, which then gives us the formula in the blue geometry. The entire derivation rests strongly on the three-fold symmetry of chromogeometry. The uniqueness of ψc,q is established in two steps. First, we prove the existence of circular integral functionals in each geometry, given in the following theorem. Next, we prove that if such a circular integral functional exists, it is uniquely determined. This is the heart of Theorems 11, 14, 17, and 19. This will then show that ψc,q is the unique circular integral functional in each geometry.

Theorem 10 The three normalized summation functionals ψc,q given in (11), (12), and (13) are circular integral functionals in respectively the green, red, and blue geometries.

Proof. We need show that ψc,q satisfies the Normalization, Locality, and Invariance conditions in each geometry c.

1. First, we prove that ψc,q satisfies the Normalization condition by Lemma 2. In the green geometry,

ψg,q (1)= 1q = Sg,q 1q = (q 1)1q =1q − −| | − − [x,yX]∈Sg,q since q1q = 0. The proof that ψr,q (1)=1q is done similarly. In the blue geometry, we have that 1 1 1 1 ψb,q (1)= − 1q = − Sb,q 1q = − q − 1q =1q − q − q | | − q − q   [x,yX]∈Sb,q        2 −1 since q = 1. This proves Normalization.   2. To prove Locality, suppose that f (x, y) = 0 for any [x, y] Sc,q. By linearity it follows that for any of the colors c, ∈

ψc,q (f)= f (x, y)= 0=0. ± ± [x,yX]∈Sc,q [x,yX]∈Sc,q

22 3. Finally, note that every h Hc induces a bijection on Sc,q since it is an invertible linear ∈ transformation. Take any h Hc and let [u, v] = [x, y] h for any [x, y] Sc,q. By working in the green geometry, ∈ ∈

k l k l k l k l ψg,q α β h = u v = x y = ψg,q α β · − − [u,v]∈S [x,y]∈S
Xg,q Xg,q
for any h. By linearity, it follows that ψg,q (f h)= ψg,q (f) for any f Pol2 (Fq). The red and blue geometry case are treated similarly. This· proves Invariance. ∈

Since ψc,q satisfies Normalization, Locality, and Invariance, ψc,q is a circular integral functional in each geometry c. The existence-implies-uniqueness proof in each geometry and an explicit formula for ψc,q will be presented in ascending order of complexity in three separate sections. We shall start with the green geometry before delving further to the red and finally blue geometry. Before we get to the main results, we illustrate some examples.

5 Examples of summing polynumbers over unit circles

Example 4 Let q = 17. Here, F = 0, 1, 2,..., 15, 16 and the unit circles are 17 { } [1, 1] [2, 9] [3, 6] [4, 13] [5, 7] [6, 3] [7, 5] [8, 15] [9, 2] Sg,17 =  [10, 12] [11, 14] [12, 10] [13, 4] [14, 11] [15, 8] [16, 16]     [1, 0] [16, 0] [6, 1] [6, 16] [11, 1] [11, 16] [3, 5] [3, 12]  Sr,17 =  [14, 5] [14, 12] [4, 7] [4, 10] [13, 7] [13, 10] [0, 4] [0, 13]     [1, 0] [16, 0] [0, 1] [0, 16] [4, 6] [4, 11] [13, 6] [13, 11] [6,4] Sb,17 =  [6, 13] [11, 4] [11, 13] [3, 3] [3, 14] [14, 3] [14, 14]   

 1 2  so that Sg, = Sr, = Sb, = 16. Note that − =1 since 1=16=4 . | 17| | 17| | 17| 17 −   It can be verified that, working in F17 over the blue geometry, we have

6 4 1 6 4 ψb, α β = − x y 17 − 17   [x,y]∈Sb,17
X 1604 + 16604 +0614 +06164 +4664 +46114 + 13664 + 136114+ = −  6644 +66134 + 11644 + 116134 +3634 +36144 + 14634 + 146144  =3.   Similarly one can directly calculate that

6 4 6 4 ψr, α β = x y =3 17 − [x,y]∈S 17
Xr, 6 4 6 4 ψg, α β = x y =0. 17 − [x,y]∈S 17
Xg, 23 2 Example 5 Let q =25=5 . We may identify F25 as F5 (ω) = a + bω : a, b F5 where ω is a 2 2 { ∈ } formal symbol satisfying ω =2 since the polynomial 2 α Pol (F5) is irreducible over F5. With this identification and the fact that − ∈

1 1 2 − = − =12 =1 25 5     one may find that the red unit circle is

[1, 0] [4, 0] [0, 2] [0, 3] [ω, 1] [ω, 4] [4ω, 1] [4ω, 4]  [2, 2ω] [2, 3ω] [3, 2ω] [3, 3ω] [2ω,ω] [2ω, 4ω] [3ω,ω] [3ω, 4ω]    Sr,25 =    [1 + ω, 2+3ω] [1 + ω, 3+2ω] [4 + 4ω, 2+3ω] [4 + 4ω, 3+2ω]    [1 + 4ω, 2+2ω] [1 + 4ω, 3+3ω] [4 + ω, 2+2ω] [4 + ω, 3+3ω]       and compute that   2 2 2 2 ψr, α β = x y =3. 25 − [x,y]∈S 25
Xr, 3 Example 6 Let q =27=3 . We can identify F27 as F (ω)= a + bω + cω2 : a, b, c F 3 ∈ 3 3 3 where ω is a formal symbol satisfying 1+ ω ω = 0 since the polynomial 1+ α α Pol (F3) is 3 − − ∈ 1 1 irreducible over F . Now, − = − = 1, and the blue unit circle is 3 27 3 −     [1, 0] [2, 0] [0, 1] [0, 2] [ω2, 2ω + ω2][ω2,ω +2ω2] [2ω2, 2ω + ω2]  [2ω2,ω +2ω2] [2ω + ω2,ω2][ω +2ω2,ω2] [2ω + ω2, 2ω2][ω +2ω2, 2ω2]       [2 + ω2, 1+ ω + ω2] [2 + ω2, 2+2ω +2ω2] [1 + 2ω2, 2+2ω +2ω2]       [1 + 2ω2, 1+ ω + ω2] [1 + ω + ω2, 2+ ω2] [2 + 2ω +2ω2, 2+ ω2]    Sb,27 =    [2+2ω +2ω2, 1+2ω2] [1 + ω + ω2, 1+2ω2][ω + ω2, 1+2ω + ω2]    [ω + ω2, 2+ ω +2ω2] [2ω +2ω2, 1+2ω + ω2] [2ω +2ω2, 2+ ω +2ω2]      [1 + 2ω + ω2,ω + ω2] [2 + ω +2ω2,ω + ω2]       ω ω2, ω ω2 ω ω2, ω ω2   [1 + 2 + 2 +2 ] [2 + +2 2 +2 ]      and thus we may compute  4 6 4 6 ψb,27 α β = x y =0. [x,y]∈S
Xb,27 6 The green geometry

The green geometry is the easiest geometry among all three to work with because of the simplicity of the green unit circle parametrization.

24 Theorem 11 (Existence implies uniqueness in green geometry) If a circular integral func- tional φg exists in the green geometry, then it must be unique.

−1 Proof. Assuming φg exists, then by the Invariance condition, using h = diag (u,u ) Hg, ∈ k l k l k −1 l k−l k l φg α β = φg α β h = φg (uα) u β = u φg α β · ∗

 k−l

k l for all u F . If q 1 does not divide k l, then u = 1 for some u =0, and thus φg α β = 0. ∈ q − − 6 6 It remains to show that φ αkβl is uniquely determined for all k and l such that (q 1) (k l). g
Define two polynumbers − | −
f (α, β) αk+1βl+1 αkβl = αkβl (αβ 1) 1 ≡ − − and f (α, β) αm(q−1) 1 2 ≡ − for any natural numbers k,l,m. For any [x, y] Sg,q, ∈ k l f1 (x, y)= x y (xy 1)=0 − m f (x, y)= xm(q−1) 1= xq−1 1=1m 1=0 2 − − − hence by Locality and the linearity of φg, we have that

k+1 l+1 k l φg α β = φg α β (14) and

m(q−1) φg α = φg (1)=1q. (15) Now, without loss of generality, assume k
l and write k = l + m (q 1) for some m N. We have that ≥ − ∈

k l l+m(q−1) l φg α β = φg α β m(q−1)
= φg α (by
repeated use of equation (14)) =1q (by (15).
0 t The case k

k l 1q if Sg,q (k l) , ψg,q α β = | | | − (16)  0 otherwise.
  25 7 The Georgiadis-Munemasa-Tanaka formula and circular polynumbers

We now introduce, for natural numbers k and l, the rational circular polynumbers

1+ α k 1 α l fk,l − (17) ≡ 2 2     and connect them to circular super Catalan numbers via a key formula of Georgiadis, Munemasa, and Tanaka [8]. These polynumbers will be crucial in our full formulas for ψc,q in the red and blue geometries. We will see quite a transition from the explicit rather simple formula in the green geometry to those in the other two geometries. Here is a table of these polynumbers for the first few values of k and l, organized by the sum k + l:

fk,l k + l =3 fk,l k + l =2 1 2 3 fk,l k + l =1 f0,3 (1 3α +3α α ) 1 2 8 fk,l k + l =0 f0,2 (1 2α + α ) − − 1 4 1 2 3 f0,1 (1 α) − f1,2 (1 α α + α ) 2 1 2 8 f0,0 1 − f1,1 (1 α ) − − 1 4 1 2 3 f1,0 (1 + α) − f2,1 (1 + α α α ) 2 1 2 8 f2,0 (1+2α + α ) − − 4 1 2 3 f3,0 8 (1+3α +3α + α )

fk,l k + l =5 fk,l k + l =4 1 2 3 4 5 f0,5 (1 5α + 10α 10α +5α α ) 1 2 3 4 32 f0,4 (1 4α +6α 4α + α ) − − − 16 1 2 3 4 5 − − f1,4 (1 3α +2α +2α 3α + α ) 1 3 4 32 f1,3 (1 2α +2α α ) − − 16 1 2 3 4 5 − − f2,3 (1 α 2α +2α + α α ) 1 2 4 32 f2,2 (1 2α + α ) − − − 16 1 2 3 4 5 − f3,2 (1 + α 2α 2α + α + α ) 1 3 4 32 f3,1 (1+2α 2α α ) − − 16 1 2 3 3 5 − − f4,1 (1+3α +2α 2α 3α α ) 1 2 3 4 32 f4,0 (1+4α +6α +4α + α ) − − − 16 1 2 3 4 5 f5,0 32 (1+5α + 10α + 10α +5α + α )

Table 5: The polynumbers fk,l for 0 k + l 5. ≤ ≤ The coefficients of these circular polynumbers are very closely connected to the entries of Krawtchouk (d) (d) matrices K , which are defined to be the (d + 1) (d + 1) matrices whose entries are Ki,j i × ≡ [α ] fd−j,j (see [3], [6]). So for example the Krawtchouk matrices of order zero up to three are

11 1 1 1 1 1 1 1  3 1 1 3  K(0) = (1) K(1) = K(2) =  2 0 2  K(3) = − − .     1 1 −  3 1 1 3  −  1 1 1   − −         −   1 1 1 1     − −    26 From the table above we can immediately observe two symmetry/anti-symmetry phenomena in the coefficients of the circular polynumbers. Define a polynumber

0 1 d−1 d f = f α + f α + + fd α + fdα 0 1 ··· −1 of degree d to be palindromic (sometimes called self-reciprocal) precisely when fk = fd−k for all 0 k d, and define it to be anti-palindromic precisely when fk = fd−k for all 0 k d. It is well-known≤ ≤ that the product of two palindromic polynumbers or two a−nti-palindromic≤ polynumbers≤ is palindromic, while the product of a palindromic and an anti-palindromic polynumber is anti- palindromic.

Lemma 3 (Palindromic property of fk,l) If 0 i, j k + l and i + j = k + l then ≤ ≤ i l j α fk,l =( 1) α fk,l. −

Proof. We note that fk,0 is a palindromic  polynumber for all k. Moreover, f0.l is palindromic if l is even and anti-polindromic if l is odd. Since fk,l = fk,0f0,l, the result then follows.

Lemma 4 (Symmetry property of fk,l) If 0 i k + l then ≤ ≤ i i i α fk,l =( 1) α fl,k. − Proof. We simply note that since    

1+ α l 1 α k fl,k = − 2 2     we have 1 α l 1+ α k fl,k ( α)= − = fk,l (α) . − 2 2     Thus, extracting the coefficient of αi from both sides,

i i i i i i i α fk,l = α fl,k ( α)= ( α) fl,k = ( 1) α fl,k =( 1) α fl,k − − − − h i h i as claimed.       Now, if the degree of a palindromic or an anti-palindromic polynumber f = f (α) is an even number, say 2d, then the term involving αd is called the central term. In particular, the central m+n term of the circular polynumber f2m,2n is one that involves α . In the applications to finite fields, for natural numbers k and l we want to reduce the coefficients of the circular polynumbers fk,l modulo an odd prime p > 2 so we can regard them as elements of k l Fp Fq. Note that this reduction makes sense since the coefficients of (1 + α) (1 α) are always integers⊆ and the denominators involved in the coefficients are all powers of 2. − Georgiadis, Munemasa, and Tanaka observed that the super Catalan numbers S(m, n) can be represented as a particular entry of some Krawtchouk matrix. Their derivation was combinatorial and involved a set of lattice paths from (0, 0) to (m + n, m + n) consisting of up- and right-steps, and the height of such a path at certain steps. We will restate their result rather in terms of circular super Catalan numbers and the central coefficients of the circular polynumbers fk,l and give an independent algebraic proof. This allows us to naturally identify the remaining coefficients around the central one as secondary but still important variants of circular super Catalan numbers.

27 Theorem 13 (Georgiadis, Munemasa, and Tanaka) For any natural number m and n,

n m+n Ω(m, n)=( 1) α f m, n. − 2 2 Proof. We calculate that  

2m 2n n m+n n m+n 1+ α 1 α ( 1) α f m, n =( 1) α − − 2 2 − 2 2       ( 1)n   = − αm+n (1 + α)2m (1 α)2n 4m+n − n ( 1)   t 2m 2n = − ( 1) . 4m+n − s t 0≤s≤2m    0≤X,t≤2n s+t=m+n

k By using the convention that = 0 if s< 0 or s>k, we may rearrange this sum as s   n m+n n m+n ( 1) m+n−s 2m 2n ( 1) α f m, n = − ( 1) − 2 2 4m+n − s m + n s s=0   −    Xm n ( 1)m + (2m)! (2n)! = − ( 1)s 4m+n − s!(2m s)! (m + n s)!( m + n + s)! s=0 − − − X m n ( 1)m (2m)!(2n)! + (m + n)! (m + n)! = − ( 1)s 4m+n (m + n)!(m + n)! − s!(m + n s)! (2m s)!( m + n + s)! s=0 − − − Xm n ( 1)m (2m)!(2n)! + m + n m + n = − ( 1)s . (18) 4m+n (m + n)!(m + n)! − s 2m s s=0 X   −  The summation on the right-hand side of (18) can be realized as

m n α2m (1 + α)m+n (1 α)m+n = α2m 1 α2 + − − so we get    

m n m+n ( 1) (2m)!(2n)! 2m 2 m+n ( 1) α f m, n = − α 1 α − 2 2 4m+n (m + n)!(m + n)! − m n   ( 1)m (2m)!(2n)!   +
m + n = − α2m ( 1)r α2r 4m+n (m + n)!(m + n)! − r r=0 ! X   (2m)!(2n)!  m + n = 4m+n (m + n)!(m + n)! m   (2m)!(2n)! = 4m+nm!n!(m + n)! =Ω(m, n) .

This finishes the proof.

28 8 The circular integral functional in the red geometry

We start this section by proving the existence implies uniqueness theorem in the red geometry over r the finite field Fq where q = p for an odd prime p and a natural number r 1. ≥ 8.1 Existence implies uniqueness in the red geometry Theorem 14 (Existence implies uniqueness in red geometry) In the red geometry, if a cir- cular integral functional φr exists, then it is unique.

Proof. Assume there is such a circular integral functional φr. By the linearity of φr, it is enough k l to prove the uniqueness of φr α β for any natural numbers k and l. Now, using Invariance, by choosing the diagonal matrices h diag (1, 1) and h diag ( 1, 1) in Hr respectively, we obtain 1 ≡
− 2 ≡ − k l k l k l l k l φr α β = φr α β h = φr α ( β) =( 1) φr α β · 1 − − k l k l  k l k k l φr α β
= φr α β h
= φr ( α) β =( 1) φr α β
· 2 − −

 
which shows that k l φr α β = 0 if either k or l is odd. (19) We will call this the double-parity property.
Thus we just need to deal with the case when k and 2m 2n 2m l are both even. Let Ir (m, n) φr (α β ). We first claim that Ir (m, 0) = φr (α ) is uniquely determined for any m N. By≡ Theorem 6, we have that ∈ k k−q+1 φr α = φr α

2m for any k q, so it suffices to show the uniqueness of φr (α ) when 2m q 1. We divide our analysis into≥ two main cases: ≤ − 2 2m−2 Case 1. Suppose that 2m < q 1. Assume the previous values φr (1), φr (α ), ...,φr (α ) − are known. By equation (9) in Theorem 8, for any u Fq for which u = 1 we have ∈ 6 ± 2m 2 2m 2m 2m 2 s 2m−s s 2m−s 1 u φr α = 1+ u (2u) φr α β . − s s=0  

X

By the double-parity property of φr, we can make a change of index s 2s to obtain 7→ m 2 2m 2m 2m 2 2s 2m−2s 2s 2m−2s 1 u φr α = 1+ u (2u) φr α β − 2s s=0  

Xm

2m 2 2s 2m−2s 2s 2 m−s = 1+ u (2u) φr α α 1 (by Locality) 2s − s=0     Xm
m−s
2m 2 2s 2m−2s 2s m−s−t m s 2t = 1+ u (2u) φr α ( 1) − α 2s − t s=0   t=0   ! Xm m−s
X m−s−t 2m m s 2 2s 2m−2s 2s+2t = ( 1) − 1+ u (2u) φr α (20) − 2s t s=0 t=0    X X

29 for all u Fq, u = 1. Breaking down the right-hand side of (20) by separating the terms containing 2m ∈ 6 ± φr (α ), we get

m 2 2m 2m 2m 2 2s 2m−2s 2m 1 u φr α = 1+ u (2u) φr α + − 2s s=0  

m−X1 m−s−1

m−s−t 2m m s 2 2s 2m−2s 2s+2t ( 1) − 1+ u (2u) φr α . (21) − 2s t s=0 t=0    X X

From our assumption, the double-summation on the right-hand side of (21) is uniquely determined. 2m The coefficient of φr (α ) in the first summation can be simplified to

m 2m 2s 1 2m 2m 1+ u2 (2u)2m−2s = 1+2u + u2 + 1 2u + u2 2s 2 − s=0     X
1

= (1 + u)4m + (1 u)4m 2 − by using the identity

k k k k k k (x + y)k +(x y)k = xsyk−s + ( 1)s xsyk−s =2 xsyk−s (22) − s − s s s=0   s=0   s=0   X X sXeven 2m valid for x, y Fq and k N. Thus by moving all the terms containing φr (α ) to the left-hand side, we can rewrite∈ (21) as∈

1 2m 2m 2 2m (1 + u) (1 u) φr α − 2 − − m−1 m−s−1

m−s−t 2m m s 2 2s 2m−2s 2s+2t = ( 1) − 1+ u (2u) φr α − 2s t s=0 t=0    X X

since 2m 1 1 2 1 u2 (1 + u)4m + (1 u)4m = (1 + u)2m (1 u)2m . − − 2 − −2 − − Now, define a polynumber
f Pol (Fq) where

∈ f (1 + α)2m (1 α)2m ≡ − − 2m 2m 3 2m 2m−1 =2 1qα + 1qα + + 1qα . 1 3 ··· 2m 1      −   We claim that f is non-zero. Suppose f is the zero polynumber; then all the coefficients are divisible by p. It follows, by a slightly tweaked version of (22)

k k k k k k (x + y)k (x y)k = xsyk−s ( 1)s xsyk−s =2 xsyk−s − − s − − s s s=0   s=0   s=0   X X sXodd that the sum of all the coefficients 2m 2m 2m 2 + + + = (1+1)2m (1 1)2m =22m 1 3 ··· 2m 1 − −      −  30 is also divisible by p, which is impossible since p > 2. We have established that f is a non-zero polynumber of degree at most 2m 1 < q 2. If all u Fq, u = 1 are zeroes of f, then − − ∈ 6 ± deg (f) = q 2 > 2m 1 deg (f), a contradiction. Hence there is some u Fq, u = 1 such − 2m − ≥ 2m 2m ∈ 6 ± that f (u)=(1+ u) (1 u) = 0, showing that φr (α ) can be determined uniquely using the − − 6 previous known values. 2 Case 2. Now suppose that 2m = q 1. Again, assume the previous values φr (1), φr (α ), . . . , 2m−2 − φr (α ) have been determined. Now if [x, y] Sr,q we must have that x + y = 0 because otherwise 2 2 ∈ 2m 6 we would have x y =0 = 1. By Locality, the polynumber (α + β) 1 restricts to zero on Sr,q − 6 − since if [x, y] Sr,q then ∈ (α + β)2m 1 (x, y)=(x + y)2m 1=(x + y)q−1 1=1 1=0 − − − − by the Fermat property. Thus,
m 2 2m 1 = φ (1)= φ (α + β)2m = φ αsβ2m−s . q r r s r s=0  
X
By the double-parity property of φr, we can make a change of index s 2s, so that 7→ m 2m 1 = φ α2sβ2m−2s q 2s r s=0   Xm
2m 2s 2 m−s = φr α α 1 (by Locality) 2s − s=0     Xm m−s
2m 2s m−s−t m s 2t = φr α ( 1) − α 2s − t s=0   t=0   ! Xm m−s X m−s−t 2m m s 2s+2t = ( 1) − φr α . (23) − 2s t s=0 t=0    X X
2m Now, we can break down (23) into two parts by separating φr (α ) from the rest: m m−1 m−s−1 2m 2m m−s−t 2m m s 2s+2t 1q = φr α + ( 1) − φr α . (24) 2s − 2s t s=0 s=0 t=0 X   X X   
2 2m−2
Since we assumed that φr (1), φr (α ), . . . , φr (α ) are known, all the terms in the double sum- mations in (24) are determined. Now, by using the identity (22), m 2m 1 = (1+1)2m =22m−1 =2q−2 2s 2 s=0 X   2m is not a multiple of p. We thus conclude that φr (α ) is also uniquely determined in this case. 2m The two cases above, together with the periodicity result from Theorem 6 show that φr (α ) is uniquely determined for any m N, proving our claim. By Theorem 7, φr satisfies the Pascal-like property ∈ Ir (m, n)= Ir (m +1, n) Ir (m, n + 1) − 2m for any m, n N. This means that if we know the values of φr (α ) for all m d N, then ∈ 2m 2n ≤ ∈ we can successively determine all the values of φr (α β ) for all natural numbers m and n with m + n = d by repeated application of the Pascal-like property. Since d is arbitrary, this establishes 2m 2n the uniqueness of φr (α β ) for any m and n and concludes the proof.

31 8.2 The formula for ψr,q

Theorem 14 above asserts that ψr,q is the unique circular integral functional in the red geometry by Theorem 10 and 14. We now derive a formula for ψr,q using the result we have established in the green geometry. Define the vector space

Fq (j)= z = x1q + yj : x, y Fq { ∈ } over Fq with ordered basis 1q, j and make it into a commutative associative algebra over Fq by { } 2 declaring that the formal symbol j satisfies j =1q. We will write x1q + yj more simply as x + yj and identify it also with the affine point [x, y]. The distributive law implies that the product of two elements z1 = x1 + y1j and z2 = x2 + y2j in Fq (j) is

z z (x x + y y )+(x y + x y ) j. 1 2 ≡ 1 2 1 2 1 2 2 1

We will refer to an element z Fq (j) as a red complex number; in other literature, it is also ∈ sometimes called a split-complex number. Given z = x + yj, the conjugate of z is defined to be z x yj and we may check that for z and z in Fq (j) we have ≡ − 1 2

z1 z2 = (z1z2).

2 2 2 2 The quadrance of z is Q (z) zz =(x y )1q +0j which we identity with the element x y in ≡ − − Fq. So we may regard Sr (Fq) = Sr,q as the collection of z Fq (j) such that Q (z)=1q. Note that this implies z−1 = z so if z = x + yj then ∈ z + z−1 z z−1 x = and y = − . (25) 2 2j

With this point of view, the group structure of Sr,q is evident: if z1and z2 are elements of Sr,q, then by associativity and commutativity

Q (z1z2)=(z1z2) (z1z2)=(z1z2) z1 z2

= z1 z1 z2 z2

= Q (z1) Q (z2)=1q so z z Sr,q. Moreover, Sr,q is a group of order q 1 by Lemma 2. 1 2 ∈ − There are two non-trivial idempotent elements in Fq (j), namely 1+ j 1 j e = and e = − 2 2 that satisfy e2 = e and e2 = e.

Clearly e, e also forms an ordered basis for Fq (j) and { } z = x + yj =(x + y) e +(x y) e. − In addition (x1e + y1e)(x2e + y2e)= x1x2e + y1y2e

32 so that the multiplication becomes pointwise in this basis. To get a bridge from the red geometry to the green geometry, observe that

[x, y] Sr (Fq) implies [x + y, x y] Sg (Fq) . ∈ − ∈

The map [x, y] [x + y, x y] from Sr (Fq) to Sg (Fq) is a bijection and in fact it is a group isomor- 7→ − phism. This important fact will help us establish the following theorem. Recall that ψr,q : Pol (Fq) 2 → Fq defined in (12) is by Theorem 10 and Theorem 14 the unique circular integral functional in the red geometry.

Theorem 15 In the red geometry over Fq define two geometric constants

1 1 k + l w = Sr,q = (q 1) and R = . 2 | | 2 − Sr,q | | We have that

m+n+dw α fk,l mod p if k =2m, l =2n, ψ (αkβl)=    (26) r,q  |d|≤R  X     0  otherwise.   u+v u−v Proof. For any [x, y] Sr,q, let u = x + y and v = x y so by inverting, x = and y = since ∈ − 2 2 q is an odd prime power and so 2 is invertible. Thus

k l k l k l u + v u v 1 k l ψr,q α β = x y = − = (u + v) (u v) . − − 2 2 −2k+l − [x,y]∈Sr,q [u,v]∈Sg,q     [u,v]∈Sg,q
X X X k l m n This means we can express ψr,q α β in terms of values of ψg,q (α β ) for some pairs (m, n). Re- verting back to x and y,

k l 1 k l ψr,q α β = (x + y) (x y) −2k+l − [x,y]∈S
Xg,q k l 1 k l = xsyk−s ( 1)l−t xtyl−t −2k+l s − t s=0 ! t=0 ! [x,yX]∈Sg,q X   X   k l 1 k l = ( 1)l−t xs+tyk+l−(s+t) −2k+l − s t s=0 t=0 X X    [x,yX]∈Sg,q l k l ( 1) t k l s+t k+l−(s+t) = − ( 1) ψg,q α β . (27) 2k+l − s t s=0 t=0    X X
Now that we have established the connection between the formula for ψr,q and ψg,q, we are ready to simplify equation (27). By Theorem 12,

s+t k+l−(s+t) ψg,q α β =1q
33 precisely when Sg,q (s + t) (k + l (s + t)) | | | − − or equivalently, 2(s + t)= k + l + d (q 1) (28) − for some d Z. If k and l have different parity, equation (28) has no solution for s and t since the ∈ s+t k+l−(s+t) left-hand side is even but the right-hand side is odd, so we conclude that ψg,q α β = 0 in this case which leads to ψ αkβl = 0. Let us therefore assume that k + l = 2N for some integer r,q
N.
Write R = k+l = k+l so R (q 1) k + l< (R +1)(q 1). If d R + 1, |Sr,q| q−1 − ≤ − ≥ j k j k k + l + d (q 1) k + l +(R +1)(q 1) > 2(k + l) 2(s + t) − ≥ − ≥ and if d (R + 1), ≤ − 2(s + t)= k + l + d (q 1) k + l (R +1)(q 1) < 0. − ≤ − − In both cases, equality is never achieved. Together with the observation that the equation 2(s + t)= k+l+d (q 1) has solutions when d R, we conclude that 2 (s + t)= k+l+d (q 1) has solutions for s and t−precisely when d R.| | ≤ − 1 | | ≤ By letting w = Sr,q , each d such that d R corresponds to the equation 2 | | | | ≤ k + l d (q 1) s + t = + − = N + dw. 2 2 We then can simplify the right-hand side of equation (27) to

l k k l ( 1) N+dw−s k l ψr,q α β = − ( 1) 1q. (29) 22N − s N + dw s |d|≤R s=0   − 
X X For each d, k k l ( 1)N+dw−s = αN+dw (1 + α)k (1 α)l − s N + dw s − s=0   −  X   so equation (29) reduces to

l k l ( 1) N+dw k l ψr,q α β = − α (1 + α) (1 α) 1q. (30) 22N − |d|≤R
X   Using Lemma 4,

αN+dw (1 + α)k (1 α)l =( 1)k αN−dw (1 + α)k (1 α)l − − − for all d R so if k is odd, (30) simplifies to   | | ≤ k l ψr,q α β =0.

34 We may therefore write k =2m and l =2n. In this case, N = m + n and (30) becomes

k l 1 m+n+dw 2m 2n ψr,q α β = α (1 + α) (1 α) 1q 22m+2n − |d|≤R
X   = αm+n+dw f mod p   2m,2n |d|≤R X     as desired. The above theorem gives us an exact formula for the unique circular integral functional ψr,q in terms of a sum of coefficients of fk,l.

Theorem 16 The special case 0 k + l < q 1 simplifies ψr,q to ≤ − k l n ψr,q α β =( 1) Ω(m, n) mod p. − Proof. To see this, note that from Theorem
15 that if 0 k + l < q 1 then R =0 and so ≤ − k l m+n ψr,q α β = α f2m,2n mod p n m+n n =( 1) α ( 1) f m, n mod p
−  − 2 2 =( 1)n Ω(m, n) mod p −   by Theorem 13.

9 The blue geometry

Moving from the red to the blue geometry, there is a dichotomy between whether 1 is a square or − not in Fq which is captured by the Jacobi symbol. Indeed, if 1 is a square then we will see that the − blue geometry behaves very similarly to the red geometry. However, if 1 is not a square in Fq, the − analysis is somewhat more complicated: we will then work in the quadratic field extension Fq (i) of Fq defined by Fq (i)= z = x1q + yi : x, y Fq { ∈ } 2 where the formal symbol i satisfies i = 1q. We shall write x1q + yi more simply as x + yi. Observe − 2 that Fq (i) F 2 because the polynumber 1 + α is irreducible in Pol (Fq). We can think of Fq (i) as ≃ q a vector space over Fq with the standard basis 1q, i and after we define the multiplication of two { } elements z1 = x1 + y1i and z2 = x2 + y2i as

z z (x x y y )+(x y + x y ) i 1 2 ≡ 1 2 − 1 2 1 2 2 1 the vector space Fq (i) is now an algebra over Fq. Note that this is analogous to the construction of the usual field of complex numbers C, and we could name Fq (i) as the blue complex numbers over Fq.

35 9.1 Existence implies uniqueness in the blue geometry when 1 is a − square

Throughout this subsection, suppose that 1 is a square in Fq, and let ı be an element of Fq such − that ı2 + 1 = 0.

Theorem 17 (Existence implies uniqueness in blue geometry) Let Fq be a field in which 1 − is a square. In the blue geometry, if a circular integral functional φb exists, then it is unique.

Proof. The flow of the proof is similar to that in the red geometry. First, it can be verified that the two matrices h1 diag (1q, 1q) and h2 diag ( 1q, 1q) belong to Hb. By applying Invariance to h1 ≡ − ≡ − k l and h2, we get the double-parity property again: φb α β = 0 if k or l is odd. By Theorem 7, we also get the Pascal-like property in the blue geometry

Ib (m, n)= Ib (m +1, n)+ Ib (m, n + 1)

2m 2n where Ib (m, n) φb (α β ) . Finally, by the periodicity condition in Theorem 6, it remains to show ≡ 2m the uniqueness of Ib (m, 0) = φb (α ) when 2m q 1. We divide our analysis into two main cases ≤ − again: 2 2m−2 Case 1. Suppose 2m < q 1. Assume φb (1), φb (α ), . . . , φb (α ) have been determined. By − Theorem 8, we have

2m 2 2m 2m 2m 2 s 2m−s s 2m−s 1+ u φb α = 1 u (2u) φb α β s − s=0  

X

for all u Fq, u = ı. Using the double-parity property, we can make a change of index s 2s and thus the∈ above equation6 ± simplifies to 7→

m 2 2m 2m 2m 2 2s 2m−2s 2s 2m−2s 1+ u φb α = 1 u (2u) φb α β 2s − s=0  

Xm

2m 2 2s 2m−2s 2s 2 m−s = 1 u (2u) φb α 1 α (by Locality) 2s − − s=0     Xm
m−s
2m 2 2s 2m−2s 2s t m s 2t = 1 u (2u) φb α ( 1) − α 2s − − t s=0   t=0   ! Xm m−s
X t 2m m s 2 2s 2m−2s 2s+2t = ( 1) − 1 u (2u) φb α . (31) − 2s t − s=0 t=0    X X

2m Separating φb (α ) from the rest in the right-hand side of (31),

m 2 2m 2m m−s 2m 2 2s 2m−2s 2m 1+ u φb α = ( 1) 1 u (2u) φb α + − 2s − s=0  

m−X1 m−1−s

t 2m m s 2 2s 2m−2s 2s+2t ( 1) − 1 u (2u) φb α . (32) − 2s t − s=0 t=0    X X

36 The double-summation on the right-hand side of (32) is therefore determined. Utilizing ı2 = 1, the 2m − coefficient of φb (α ) on the right-hand side of (32) can be written as m 2m 2s 1 1+(uı)2 (2uı)2m−2s = (1 + uı)4m + (1 uı)4m . 2s 2 − s=0   X

2m Collecting the terms containing φb (α ) to the left-hand side and doing some algebraic manipulation, 2m the coefficient of φb (α ) then becomes

m 1 1 2 1+ u2 2 (1 + uı)4m + (1 uı)4m = (1 + uı)2m (1 uı)2m − 2 − −2 − − and thus we can rewrite
(32) as

1 2m 2m 2 2m (1 + uı) (1 uı) φb α − 2 − − m−1 m−1−s

t 2m m s 2 2s 2m−2s 2s+2t = ( 1) − 1 u (2u) φb α − 2s t − s=0 t=0    X X

for all u Fq such that u = ı. We proceed with the same trick as in the red geometry case. Define a polynumber∈ 6 ± 2m 2m f (1 + αı) (1 αı) Pol (Fq) ≡ − − ∈ which we can show to be non-zero of degree at most 2m 1 < q 2. If every element u Fq where − − ∈ u = ı is a zero, then deg (f) q 2 > 2m 1 deg (f), a contradiction. Hence there exists some u 6of ±f where u = ı which is≥ not− a zero of f−and≥ thus f (u) = 0. This is enough to guarantee that 2m 6 ± 6 φb (α ) is uniquely determined. Case 2. Suppose 2m = q 1. If [x, y] Sb,q then x + ıy = 0 because if it were zero, then − ∈ 6 x2 + y2 = x2 +( ıx)2 =0 = 1. − 6 2m The polynumber (α + ıβ) 1 thus restricts to zero on Sb,q so by Locality, − 2m 1q = φb (α + ıβ) . (33)

2 2m−2 Assume φb (1), φb (α ), . . . , φb (α ) are all known. By expanding
the right-hand side of equation (33) and using the double-parity property to make a change of index s 2s, we have that 7→ m 2m 1 = φ α2s (ıβ)2m−2s q 2s b s=0   Xm
m−s 2m 2s 2m−2s = ( 1) φb α β − 2s s=0   Xm
m−s 2m 2s 2 m−s = ( 1) φb α 1 α (by Locality) − 2s − s=0     Xm m−s
m−s 2m 2s t 2t = ( 1) φb α ( 1) α − 2s − s=0   t=0 ! Xm m−s X m−s+t 2m 2s+2t = ( 1) φb α . (34) − 2s s=0 t=0   X X
37 2m By separating φb (α ) from the rest in (34),

m−1 m−1−s m m−s+t 2m 2s+2t 2m 2m 1q = ( 1) φb α + φb α . − 2s 2s s=0 t=0   s=0   X X
X
Since m 2m 1 = (1+1)2m =22m−1 =2q−2 2s 2 s=0 X   2m 2m is not a multiple of p, it follows that the coefficient of φb (α ) is not zero, thus proving that φb (α ) uniquely determined in this case. By similar argument in the proof of Theorem 14 employing the Pascal-like property in the blue 2m 2n geometry from Theorem 7, we conclude that φb (α β ) is uniquely determined for any natural numbers m and n, finishing the proof.

9.2 The formula for ψb,q when 1 is a square − Recall that from Theorem 10,

k l 1 k l ψb,q α β = − x y − q   [x,y]∈Sb,q
X is a circular integral functional in the blue geometry. By Theorem 17, ψb,q is then the unique circular integral functional in the blue geometry. We wish to find an explicit formula for ψb,q. Define a map

̺ : Sb (Fq) Sr (Fq) −→ [x, y] [x, yı] 7→ 2 where ı Fq satisfies ı = 1. It is easy to verify that this map is a bijection. This bijection therefore ∈ − bridges the red and blue geometry when 1 is a square in Fq. From this, not only can we find a k l − formula for ψb,q α β easily, it also shows that the red unit circle and the blue unit circle when 1 is a square are in a way similar. −

Theorem 18 In the blue geometry over Fq when 1 is a square, define two geometric constants − 1 1 k + l w = Sb,q = (q 1) and R = . 2 | | 2 − Sb,q | | We have that

m+n+dw n α ( 1) fk,l mod p if k =2m, l =2n, ψ αkβl =    − b,q  |d|≤R  X  
  0 otherwise.  Proof. We have that  k l k l l k l l k l l k l ψb,q α β = x y = ( ı) x (yı) = ( ı) x y =( ı) ψr,q α β . − − − − − − [x,y]∈S [x,yı]∈S [x,y]∈S
Xb,q X r,q Xr,q
k l The result follows from the formula of ψr,q α β in Theorem 15.
38 Corollary 18.1 If 1 is a square in Fq, then −

k l k l ψb,q α β if l 2mod4, ψr,q α β = 6≡  k l ψb,q α β
if l 2mod4.
 − ≡ 
1 9.3 The case − = 1 q −

It remains for us to derive  the similar results when 1 is not a square in Fq, an assumption we will make from this point henceforth. We shall see a slightly− more involved analysis in this setup as we need to work in the quadratic field extension Fq (i) defined at the beginning of this section. Again, we start by establishing the uniqueness of the circular integral functional in this geometry provided it exists.

Theorem 19 Let Fq be a field in which 1 is not a square. In the blue geometry, if a circular − integral functional φb exists, then it is unique.

Proof. The initial setup of the proof is identical to the case 1 is a square above. We can establish − the double-parity property and the same Pascal-like property

Ib (m, n)= Ib (m +1, n)+ Ib (m, n + 1)

2m so it is sufficient to prove the uniqueness of I (m, 0) = φb (α ). By using equation (10) in Theorem 8 and the double-parity property, we can again make a change of index s 2s to get 7→ m 2 2m 2m 2m 2 2s 2m−2s 2s 2m−2s 1+ u φb α = 1 u (2u) φb α β 2s − s=0  

Xm

2m 2 2s 2m−2s 2s 2 m−s = 1 u (2u) φb α 1 α (by Locality) 2s − − s=0   X

 2m which then can be expanded and the terms containing φb (α ) can be extracted to give us

m−1 m−1−s 2 2m 2m t 2m m s 2 2s 2m−2s 2s+2t 1+ u φb α = ( 1) − 1 u (2u) φb α − 2s t − s=0 t=0   

Xm X

m−s 2m 2 2s 2m−2s 2m + ( 1) 1 u (2u) φb α . (35) − 2s − s=0   X

We note that equation (35) is essentially equation (32), but it is now true for all u Fq. Similar to the existence-implies-uniqueness proof in Theorem 14 and Theorem 17,∈ it is sufficient 2m 2 to show the uniqueness of φb (α ) for 2m q 1. Assume all the previous terms φb (1), φb (α ), . . . , 2m−2 ≤ − φb (α ) are determined. It follows that all the terms in the double-summation on the right-hand side of (35) are determined. Now, here is where the proof diverges from the proof in the case where 1 is a square in Fq. We can think of equation (35) as an equation in Fq (i) by replacing some 1 − 2 − in our field Fq with i .

39 We thus have

m−1 m−1−s 2 2m 2m t 2m m s 2 2s 2m−2s 2s+2t 1 (ui) φb α = ( 1) − 1 u (2u) φb α − − 2s t − s=0 t=0   

Xm X

m s 2m 2s + i2 − 1+(ui)2 (2u)2m−2s φ α2m 2s b s=0   Xm−1m−
1−s

t 2m m s 2 2s 2m−2s 2s+2t = ( 1) − 1 u (2u) φb α − 2s t − s=0 t=0    Xm X

2m 2s + 1+(ui)2 (2ui)2m−2s φ α2m 2s b s=0   Xm−1 m−1−s

t 2m m s 2 2s 2m−2s 2s+2t = ( 1) − 1 u (2u) φb α − 2s t − s=0 t=0    X X

1 4m 4m 2m + (1 + ui) + (1 ui) φb α (36) 2 −

where we have used again the identity given in equation (22). After moving the single-summation on the right-hand side of (36) to the left-hand side, equation (36) now becomes

1 2m 2m 2 2m (1 + ui) (1 ui) φb α − 2 − − m−1 m−1−s

t 2m m s 2 2s 2m−2s 2s+2t = ( 1) − 1 u (2u) φb α − 2s t − s=0 t=0    X X

for all u Fq where the left-hand side comes from observing that ∈ 2m 1 1 2 1 (ui)2 (1 + ui)4m + (1 ui)4m = (1 + ui)2m (1 ui)2m . − − 2 − −2 − −

∗
2m 2m
Now, we claim that there exists some u Fq such that (1 + ui) (1 ui) = 0. First, we note that ∈ − − 6

m 2 2m (1 + ui)2m (1 ui)2m = (1 ( 1)s) (ui)s − − − − s s=0   sXodd 2m 2m 2m 2m =2i u u3 + u5 +( 1)m−1 u2m−1 . 1 − 3 5 −··· − 2m 1        −   Employing the result from the proof in Case 1 of Theorem 14 that not all of the integers

2m 2m 2m 2m , , , , 1 3 ··· 2m 3 2m 1      −   −  are divisible by p, this implies that

2m 2m 3 2m 5 m−1 2m 2m−1 f 1qα 1qα + 1qα +( 1) 1qα ≡ 1 − 3 5 −··· − 2m 1        −  40 ∗ is a non-zero polynumber in Pol (Fq) of degree at most 2m 1 < q 1. If every element u F is a − − ∈ q zero of f, then deg (f) q 1 > 2m 1 deg (f) ≥ − − ≥ ∗ which is again a contradiction. Thus there exists some u Fq such that f (u) = 0 and therefore 2m 2m 2m ∈ 6 (1 + ui) (1 ui) = 0. It follows that φb (α ) is uniquely determined when 2m q 1. The − − 2m 2n6 ≤ − uniqueness of φb (α β ) then follows by applying Periodicity condition from Theorem 6 and the Pascal-like property in the blue geometry repeatedly.

9.4 The formula for ψb,q when 1 is not a square − We have already established that ψb,q is the unique circular integral functional in the blue geometry 1 when − = 1. We proceed to find an explicit formula for ψb,q in this setting, using analysis q −   in Fq (i). An element z = x + yi Fq (i) will be referred to as a blue complex number. Given z = x + yi, the conjugate of z is∈ defined to be z x yi, and the quadrance of z is defined 2 2 ≡ − 2 2 Q (z) zz =(x + y )+0i which we identify with the element x + y in Fq. ≡ We can now regard the blue unit circle Sb (Fq) as the set of blue complex numbers z such that −1 Q (z) = 1. Note that this implies z = z. The group structure is evident: if z1 = x1 + y1i and z2 = x2 + y2i are two elements on Sb (Fq), then Q (z1z2) = Q (z1) Q (z2)=1so z1z2 Sb (Fq). This ∗ ∈ is in fact a cyclic group, being a subgroup of the cyclic group Fq (i) . Now, define a map

̺ : Sb (Fq) Sr (Fq (i)) −→ [x, y] [x1q +0i, 0+ yi] . 7→ 2 Here Sb (Fq) is a group of order q + 1 and Sr (Fq (i)) is a group of order q 1 since the underlying 2 − field Fq (i) has q elements. It can be verified easily that given [x, y] Sb (Fq), its map ̺ ([x, y]) = ∈ [x1q +0i, 0+ yi] is an element of Sr (Fq (i)) since

2 2 2 2 2 2 (x1q +0i) (0 + yi) = x 1q + y 1q = x + y 1q =1q − where the last equation follows since [x, y] Sb (Fq). Moreover, it can be
checked that ̺ is an injective ∈ homomorphism and ̺ (Sb (Fq)) has a group structure.

Lemma 5 ̺ (Sb (Fq)) is a cyclic subgroup of Sr (Fq (i)) of order q +1.

Proof. Take any z , z ̺ (Sb (Fq)), thus there exist [x ,y ] , [x ,y ] Sb (Fq) such that z = 1 2 ∈ 1 1 2 2 ∈ 1 ̺ ([x1,y1]) and z2 = ̺ ([x2,y2]). We can identify z1 and z2 as two red complex numbers z1 = x1+(y1i) j and z2 = x2 +(y2i) j in Fq(i)(j). Since

z1z2 =(x1 + yij)(x2 + y2ij) 2 = x1x2 + y1y2i +(x1y2i + x2y1i) j =(x x y y )+(x y + x y ) ij 1 2 − 1 2
1 2 2 1 and (x x y y )2 +(x y + x y )2 = x2 + y2 x2 + y2 =1 1 2 − 1 2 1 2 2 1 1 1 2 2 it follows that z1z2 ̺ (Sb (Fq)). Moreover, since ̺ is one-to-one,
̺ (Sb (Fq
)) = Sb (Fq) = q + 1 and ∈ | | | | ̺ (Sb (Fq)) is cyclic since it is a subgroup of a cyclic group.

41 With that in mind, we see that

k l k l l k l l k l ψb,q α β = x y =( i) x (yi) =( i) x y . (37) − − [x,y]∈S (F ) [x,yi]∈̺(S (F )) [x,y]∈̺(S (F ))
Xb q Xb q Xb q

The key difference from the previous case is now the summation is taken over ̺ (Sb (Fq)) instead of k l the whole group Sr (Fq (i)). This slight deviation in deriving the formula of ψb,q α β means that we cannot easily express ψ in terms of ψ but as we shall see in the following theorem, the bijection b,q r,q
̺ still acts as a bridge between the red and the blue geometry.

Theorem 20 In the blue geometry over Fq when 1 is not a square, define two geometric constants − 1 1 k + l w = Sb,q = (q + 1) and R = . 2 | | 2 Sb,q | | We have that

m+n+dw n α ( 1) fk,l mod p if k =2m, l =2n, ψ αkβl =    − b,q  |d|≤R  X  
  0 otherwise.   Proof. Since Sr (Fq (i)) is a cyclic group, let ζ be a generator for Sr (Fq (i)) so that it has order q2−1 2 +1 q−1 q 1. Since ̺ (Sb (Fq)) is cyclic of order q + 1 by Lemma 5, it is generated by ζ q = ζ . In other words,− λ(q−1) ̺ (Sb,q)= ζ : 0 λ q . ≤ ≤ It follows from (37) that 

k l l k l ψb,q α β =( i) x y −
[x,y]∈X̺(Sb,q) z + z−1 k z z−1 l =( i)l − (by identity (25)) − 2 2j z ̺ S     ∈X( b,q) l q ( i) k l = − ζλ(q−1) + ζ−λ(q−1) ζλ(q−1) ζ−λ(q−1) 2k+ljl − λ=0 Xq k
l
( i)l k l = − ζλ(2s−k)(q−1) ( 1)l−t ζλ(2t−l)(q−1) 2k+ljl s − t λ=0 s=0   ! t=0   ! Xk lX q X il k l = ( 1)t ζλ(2s+2t−k−l)(q−1). (38) 2k+ljl − s t s=0 t=0 λ X X    X=0 Now, we note that

q 1q if 2 (s + t) (k + l) is divisible by q + 1, ζλ(2s+2t−k−l)(q−1) = −  λ 0 otherwise X=0   42 which follows from basic geometric series. Similar to the proof in Theorem 15, if k and l have different parity then there is no solution to the equation

2(s + t)=(k + l)+ d (q + 1)

k+l k+l for some d Z. Next, by letting R = = q+1 we claim that 2 (s + t)=(k + l)+ d (q + 1) ∈ Sb,q | | has solutions for s and t precisely when d R.j Indeed,k if d R + 1, then | | ≤ ≥ k + l + d (q + 1) k + l +(R +1)(q + 1) > 2(k + l) 2(s + t) ≥ ≥ and if d (R + 1) then ≤ − k + l + d (q + 1) k + l (R +1)(q + 1) < 0 2(s + t) . ≤ − ≤ In both cases, equality is never attained. The claim follows by noting that the equation 2 (s + t)= k + l + d (q + 1) has solutions when d R. 1 | | ≤ Let k + l =2N and w = Sb,q . We can reduce the right-hand side of (38) to 2 | | l k l i N+dw−s k l i N+dw k l ( 1) 1q = α (1 + α) (1 α) 1q. 22N jl − s N + dw s 22N jl − |d|≤R s=0   −  |d|≤R X X X   The same analysis pertains as in the red geometry: the sum vanishes if k and l are both odd, and if k =2m and l =2n then the above sum evaluates to

n m+n+dw m+n+dw n ( 1) α f m, n mod p = α ( 1) f m, n mod p. − 2 2   − 2 2 |d|≤R |d|≤R X   X     This completes the proof.

9.5 A combined formula Recall from Lemma 2 that 1 Sb,q = q − . | | − q   We can combine the two formulas for ψb,q in Theorems 18 and 20 into a single formula where the Jacobi symbol is implicit.

Theorem 21 (General formula of ψb,q) In the blue geometry over a general finite field Fq with q odd, define the geometric constants 1 1 1 k + l w Sb,q = q − and R = . ≡ 2 | | 2 − q Sb,q    | | We have that

m+n+dw n α ( 1) fk,l mod p if k =2m, l =2n, ψ (αkβl)=    − (39) b,q  |d|≤R  X     0 otherwise.   43 Not only does the above formula capture the number-theoretic aspect of the size of the finite field Fq, it is also strikingly similar to the formula of ψr,q. This shows that there is an intricate connection between the red and the blue geometry, both of which stem from the result in the green geometry. Moreover, if the degree of the input monomial αkβl is restricted, the values of the circular integral functional ψb,q can be described solely in terms of the circular super Catalan numbers Ω(m, n).

Theorem 22 If k + l < q 1 then − k l ψb,q(α β )=Ω(m, n) mod p.

Proof. By Theorem 21, R = k+l = 0 and thus Sb,q | | k l m+n n ψb,q(α β )= α ( 1) f m, n =Ω(m, n) mod p − 2 2 by Theorem 13.  

10 Using periodicity to establish simpler formulas for ψr,q and ψb,q

Up to this point we have established the uniqueness of the circular integral functionals ψc,q defined in (11), (12), and (13) in each geometry, and given explicit formulas involving summing certain coefficients of the circular polynumbers fk,l in the red and blue situations. This section aims to give alternate explicit formulas for the circular integral functionals ψr,q and ψb,q utilizing the Periodicity condition presented in Theorem 6. Although this alternate formula is perhaps more practical from a computational point of view, we consider it as secondary. There are at least two good reasons for that: the symmetry in our formula for ψr,q and ψb,q in Theorem 15 and 21 is more apparent. The problem of summing polynomial functions over unit circles over finite fields in the non-trivial case becomes the problem of summing an equally spaced ladder of coefficients of the circular polynomial, up to a sign, symmetrically placed around the central coefficient which is the circular super Catalan number. Moreover, the periodicity condition was derived from the Normalization, Locality, and Invariance conditions. k l k−q+1 l By Theorem 6, if k q, then for any l 0 we have ψc,q α β = ψc,q α β . And if l q, then for any k 0 we have≥ ψ αkβl = ψ ≥ αkβl−q+1 . We may combine the two equations in≥ the c,q c,q

form ≥
k+rq l+sq
k+r l+s ψc,q α β = ψc,q α β for all natural numbers k,l,r,s.

To illustrate this, consider again the case q = 13 in Table 3 to observe a periodicity phenomenon k l with k = 2m and l = 2n. All the information required to determine the values ψb,13 α β is then

44 contained in the following block 0 m, n 6, as summing any odd powers of α or β will yield 0. ≤ ≤ 2m 2n ψb, (α β ): m n 0123456 13 \ 0 1726273 1 7594849 2 2959484 3 6495948 4 2849594 5 7484959 6 3948495

2m 2n Table 6: All the information to determine ψb,13 (α β ). k l k l In general, all the information to determine all the values of ψr,q α β and ψb,q α β is contained 2m 2n 2m 2n 1 in the block of values of ψr,q (α β ) and ψb,q (α β ) where 0 m, n (q 1), which we will ≤
≤ 2 −
call the principal block of ψc,q. We divide the presentation of the alternate formula in two: a simplified formula for ψr,q and ψb,q where 1 is a square, which is connected by Theorem 20, and a − simplified formula for ψb,q where 1 is not a square. Here, we let δi,j be the delta Kronecker function − 1 if i = j, δi,j =  0 if i = j.  6 Theorem 23 If w = 1 (q 1) and 0 m, n w, then 2 − ≤ ≤ 2m 2n n m+n+w 2 ψr,q α β = ( 1) Ω(m, n) + 2[α ]f m, n + δm n,q mod p − 2 2 4m+n + −1   −1
and if q =1,   2m 2n m+n+w n 2 n ψb,q α β = Ω(m, n) + 2[α ]( 1) f m, n + ( 1) δm n,q mod p. − 2 2 4m+n − + −1  
2m 2n n k l Proof. We will only prove the red geometry case since ψb,q (α β )=( 1) ψr,q α β by Theorem − k+l 18. If m + n

45 The result then follows. We also give a simplified formula of ψb,q where 1 is not a square which we will state without proof. −

Theorem 24 In the blue geometry where 1 is not a square, let w = 1 (q + 1) . If 0 m, n w, − 2 ≤ ≤ we have that 2m 2n m+n+w n ψb,q α β = Ω(m, n) + 2[α ]( 1) f m, n mod p. − 2 2

11 Supporting evidence and examples

Finally, we will revisit our earlier examples to illustrate the use of the various formulas we have established. We will do each example twice, using the main formula of ψc,q, and then also the simplified formula involving the periodicity condition.

6 4 Example 7 Let q = 17. From Example 4, we know that ψr,17 (α β )=3. Now, using the result in Theorem 15, 6 4 2 3 ψr, α β =( 1) Ω(3, 2)mod17 = mod17 = 3. 17 − 256
k+l 64 Moreover, if k = 40 and l = 24, then R = q−1 = 16 =4, and therefore j k   40 24 32+8d ψr,17 α β = [α ](f40,24)mod17 |Xd|≤4
0 8 16 24 32 40 48 56 64 = [α ] + [α ] + [α ] + [α ] + [α ] + [α ] + [α ] + [α ] + [α ] f40,24.

Through inspection,
1 [α0]f = [α64]f = 40,24 40,24 264 90744 [α8]f = [α56]f = 40,24 40,24 264 16 48 15156452 [α ]f , = [α ]f , = 40 24 40 24 − 264 24 40 264053432 [α ]f , = [α ]f , = 40 24 40 24 − 264 1650 887 238 [α32]f = 40,24 264 and thus

40 24 1 90744 15156452 264053432 1650 887 238 ψr, α β = 2 + + mod17 17 264 264 − 264 − 264 264    
33345 = mod17 249 =4.

46 Using the periodicity condition,

40 24 6+2×17 7+1×17 ψr,17 α β = ψr,17 α β 8 8
= ψr,17 α β
4 0 =( 1) Ω(4, 4)+2 α f , mod17 −
8 8 35 1 = + mod17  215 215   =4.

2 2 2 Example 8 Let q =25=5 . It has been computed in Example 5 that ψr,25 (α β )=3. Now, with our formula, 2 2 1 1 ψr, α β =( 1) Ω(1, 1)mod5 = mod5= 3. 25 − −8
k+l 54 Moreover, if k = 30 and l = 24 then R = q−1 = 24 =2, and therefore j k 30 24 27+12d   ψr,25 α β = [α ](f30,24 (α))mod5 |Xd|≤2
3 15 27 39 51 = [α ] + [α ] + [α ] + [α ] + [α ] f30,24 mod5.

It can be computed that

3 51 124 [α ]f , = [α ]f , = 30 24 30 24 − 254 360 272 [α15]f = [α39]f = 30,24 30,24 254 24129392 [α27]f = 30,24 254 thus it follows that

30 24 124 360 272 24129392 ψr, α β = 2 + mod5 25 − 254 − 254 254  
3106 211 = mod5 251 =2.

Now, with the periodicity condition,

30 24 5+1×25 24 ψr,25 α β = ψr,25 α β 6 24
= ψr,25 α β
12 3 = ( 1) Ω(3, 12)+2 α f , mod5 −
6 24 7429 177 = mod5
67108864 − 227   14 681 = mod5 134217728 =2.

47 3 1 2 Example 9 Let q =27=3 . From Example 6, − = 1 and ψb, (α )=2 which agrees with 3 − 27   1 k+l 52 Theorem 20. Moreover, if k = l = 26, then w = 2 (q +1) = 14 and R = q+1 = 28 = 1. Therefore, j k  

26 26 26+14d 13 ψb, α β = [α ]( 1) f , mod3 27 − 26 26 |Xd|≤1
12 26 40 = [α ] + [α ] + [α ] f , mod3. − 26 26 Since
230 230 [α12]f = [α40](f (α)) = 26,26 26,26 252 26 10400600 [α ]f , = 26 26 − 252 we have that

26 26 10400600 460 460 ψb, α β = mod3 27 252 − 252  
9940140 = mod3 252 =0.

Note that we do not need to employ the periodicity condition since m, n w. ≤ References

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