Ramanujan sums as supercharacters

Abstract

The theory of supercharacters, recently developed by Diaconis-Isaacs and André, is used to derive the fundamental algebraic properties of Ramanujan sums. This machinery frequently yields one-line proofs of difficult identities and provides many novel formulas. In addition to exhibiting a new application of supercharacter theory, this article also serves as a blueprint for future work since some of the abstract results we develop are applicable in much greater generality.

Appendix: Computing the matrix M i (p α)

This appendix contains a detailed derivation of the description (4.11) for the matrix M _i (p ^α) given in Sect. 4.2.

The proof of Lemma 4.1 tells us that a _i,j,k is independent of the particular chosen representative z of K _k . Since $\mathbb{Z}/p^{\alpha }\mathbb{Z}$ is abelian, we also note that

$$ a_{i,j,k} = a_{j,i,k} $$

(A.1)

and

$$ a_{i,j,k} =0 \quad{\iff}\quad a_{i,k,j}=0. $$

(A.2)

Statement (A.2) requires some explanation. Observe that a _i,j,k =0 holds if and only if x+y=z has no solutions (x,y,z) in K _i ×K _j ×K _k . Since K _j =−K _j and K _k =−K _k by (4.9), it follows that the preceding happens if and only if x+z′=y′ has no solutions (x,z′,y′) in K _i ×K _k ×K _j . On the other hand, it is important to note that a _i,j,k =a _i,k,j does not hold in general since the fixed representative z of K _k used in (4.10) plays a distinguished role.

We first break down the evaluation of the a _i,j,k into five special cases, from which the structure of the matrix M _i =M _i (p ^α) can eventually be deduced.

Lemma A.1

For $G = \mathbb{Z}/p^{\alpha}\mathbb{Z}$ and $K_{i} = \{ xp^{\alpha -i+1} \in\mathbb{Z}/p^{\alpha}\mathbb{Z}: p \nmid x\}$, we have

• (a) if k>i and j≠k, then a _i,j,k =0,

• (b) if j=k>i, then a _i,j,k =ϕ(p ⁱ⁻¹),

• (c) if j=k=i, then a _i,j,k =p ⁱ⁻¹−2p ⁱ⁻²,

• (d) if j>k and i≠j, then a _i,j,k =0,

• (e) if i=j>k, then a _i,j,k =ϕ(p ⁱ⁻¹).

Proof

We first prove (a). Letting k>i and j≠k, we may assume that i≤j by (A.1). If x _i =xp ^α−i+1 and y _j =yp ^α−j+1 belong to K _i and K _j , respectively, then it follows that

$$ x_i + y_j = p^{\alpha-j+1} \bigl(xp^{j-i} + y \bigr) $$

(A.3)

belongs to K _j since xp ^j−i+y is not divisible by p (recall that p∤x and p∤y by the definition of K _i and K _j ). Since j≠k, it follows from (A.3) that x _i +y _j cannot belong to K _k , from which it follows that a _i,j,k =0.

Next we consider (b). Suppose that j=k>i and fix z _k =zp ^α−k+1 in K _k . Since j=k, the computation (A.3) tells us that for each x _i in K _i , there exists a unique y _j =z−xp ^j−i such that x _i +y _j =z _k . Thus, a _i,j,k =|K _i |=ϕ(p ⁱ⁻¹).

The proof of (c) is somewhat more involved. Let i=j=k and fix z _k =zp ^α−k+1 where p∤z. For any element x _i =xp ^n−k+1 of K _i =K _k , there exists a unique a ₀ in $\mathbb{Z}/p^{\alpha} \mathbb{Z}$ such that

$$ x_i + a_0 = z_k. $$

(A.4)

Since p ^α−k+1 divides both x _i and z _k , it must also divide a ₀, so that we can write a ₀=ap ^n−k+1 for some a. In light of (A.4), we now have x+a=z, so that a=z−x. Therefore a ₀ belongs to K _k if and only if p∤(z−x). We now note that if p|(z−x), then x _i would serve as a solution to x _i +y=z _k where y belongs to some K _ℓ with ℓ<i=k. By statement (b), it follows that

$$\begin{aligned} a_{i,j,k} &= |K_k| - \sum_{\ell=1}^{k-1} \phi \bigl(p^{\ell-1} \bigr) \\ &= \phi \bigl(p^k \bigr) - \sum_{\ell=2}^{k-1} \bigl(p^{\ell-1}-p^{\ell-2} \bigr) -1 \\ &= \bigl(p^{k-1} - p^{k-2} \bigr)- \bigl(p^{k-2} - p^{k-3} \bigr) - \cdots- (p-1) -1 \\ &= p^{k-1} - 2 p^{k-2}, \end{aligned}$$

as claimed.

Now we consider statement (d). Suppose that j>k and i≠j. In light of (A.1), we may assume that i<j. Maintaining the same notation and conventions as in the proof of statement (a), we again arrive at (A.3) and conclude that x _i +y _j belongs to K _j . Since j>k, we conclude that a _i,j,k =0.

Finally, let us prove (e). Suppose that i=j>k and let z _k =zp ^α−k+1 in K _k be given. For each x _i =xp ^α−i+1 in K _i , we have

$$ xp^{\alpha-i+1} + \bigl(zp^{i-k} - x \bigr)p^{\alpha-i+1} = \bigl(zp^{i-k} \bigr)p^{\alpha-i+1} = zp^{\alpha-k+1} = z_k. $$

(A.5)

Now p|(zp ^i−k) because i>k, so it follows that p∤(zp ^i−k−x) since p∤x. Therefore (zp ^i−k−x)p ^α−i+1 belongs to K _i . Looking at (A.5), we conclude that for each x _i in K _i , there exists a unique y _i in K _i such that x _i +y _i =z _k . Since i=j, we conclude that a _i,j,k =|K _i |=ϕ(p ⁱ⁻¹), as desired. □

Having proven the preceding lemma, it is now straightforward to see that M _i has the form (4.11). The complete reasoning is presented below.

• (1) If j,k<i, then (M _i ) _j,k =0. In other words, the upper-left (i−1)×(i−1) submatrix of M _i contains only zeros. Indeed, letting i′=j and j′=i>k (so that i′≠j′ since j<i), it follows that (M _i ) _j,k =a _i,j,k =a _j,i,k =a _i′,j′,k =0 by (A.1) and (d) of Lemma A.1.

• (2) If j<i=k, then (M _i ) _j,k =ϕ(p ^j−1), so that the first i−1 entries of the ith column of M _i are given by 1,ϕ(p),ϕ(p ²),…,ϕ(p ⁱ⁻²). As before, we set i′=j and j′=i and observe that (M _i ) _j,k =a _i,j,k =a _j,i,k =a _i′,j′,k =ϕ(p ^i′−1)=ϕ(p ^j−1) by (b) of Lemma A.1 since j′=k>i′.

• (3) If i=j>k, then (M _i ) _j,k =ϕ(p ⁱ⁻¹), so that the first (i−1) entries of the ith row of M _i are each ϕ(p ⁱ⁻¹). To see this, simply note that (M _i ) _j,k =a _i,j,k =ϕ(p ⁱ⁻¹) by (e) of Lemma A.1.

• (4) If i=j=k, then (M _i ) _j,k =a _i,i,i =p ⁱ⁻¹−2p ⁱ⁻² by (c) of Lemma A.1.

• (5) If j=k>i, then (M _i ) _j,k =ϕ(p ⁱ⁻¹). In other words, the final (n+1−i) entries along the main diagonal of M _i are ϕ(p ⁱ⁻¹). This follows immediately from (b) of Lemma A.1.

• (6) If j>i,k, then (M _i ) _j,k =0 follows from (d) of Lemma A.1. Therefore the final (n+1−i) rows of M _i have only zeros to the left of the main diagonal.

• (7) If i,j<k, then (M _i ) _j,k =0. In other words, the last (n+1−i) columns of M _i have only zeros above the main diagonal. This follows immediately from (a) of Lemma A.1.