Advanced Number Theory Note #11: Derivation of a Möbius function formula for Ramanujan’s sums

I used a Möbius function formula for Ramanujan’s sums in a previous note and I want to show where this formula comes from in the present note (mainly as a technical memo for myself). The approach will be to show how Ramanujan’s sum

c_k(n) = \sum_{\substack{m\hspace{1 mm} mod \hspace{1 mm}k \\ gcd(m, k)=1}} e^{2\pi imn/k}            (1)

can be derived from a sum of the form

s_k(n) = \sum_{d|gcd(n, k)}f(d)g\big(\frac{k}{d}\big)        (2)

which is like a Dirichlet convolution f \ast g except that the sum is only over a subset of the divisors d of k, namely those which also divide n.

Formula (1) can be obtained as a special case of (2), via the result

c_k(n) = \sum_{d|gcd(n, k)}d\mu\big(\frac{k}{d}\big)        (3)

where \mu is the Möbius function. In fact the Möbius function itself is a Ramanujan’s sum, and I showed in Advanced Number Theory Note #10 how to obtain it as a sum of primitive roots of unity like (1) above. It can also be seen directly from (3) by setting n = 1, in which case there is only one term in the sum and we get c_k(1) = \mu(k). In addition, when k|n we have gcd(n, k) = k and

c_k(n) = \sum_{d|k}d \mu(k/d) = \varphi(k)

where the second equality was derived in Advanced Number Theory Note #1.

Focusing on the sum s_k(n) in (2) above, note that n occurs only in gcd(n, k) which is a periodic function with period k, so we conclude that s_k(n) must also be a periodic function with period k, so that

s_k(n + k) = s_k(n)

for all n. Therefore s_k(n) must have a finite Fourier expansion, and the precise form of this is obtained in the following theorem.

Theorem 1. Let s_k(n) = \sum_{d|gcd(n, k)}f(d)g(k/d). Then s_k(n) has the finite Fourier expansion

s_k(n) = \sum_{m\hspace{1 mm} mod \hspace{1 mm}k} a_k(m)e^{2\pi imn/k}            (4)

where

a_k(m) = \sum_{d|gcd(m, k)}g(d)f\big(\frac{k}{d}\big)\frac{d}{k}         (5)

Proof: We can apply Theorem 4 in Advanced Number Theory Note #9, with the roles of the letters m and n interchanged, to get the following formula for the coefficients a_k(m):

a_k(m) = \frac{1}{k}\sum_{m\hspace{1 mm} mod \hspace{1 mm}k} s_k(n) e^{-2\pi imn/k}

= \frac{1}{k}\sum_{n=1}^k \sum_{\substack{d|n \\ d|k}}f(d)g\big(\frac{k}{d}\big) e^{-2\pi imn/k}

We can write n = cd for integer c (since d|n) and observe that, since it is required that d|n in the above sum, it must be the case that the index c runs from 1 to k/d. Then we have

a_k(m) = \frac{1}{k}\sum_{d|k}f(d)g\big(\frac{k}{d}\big) \sum_{c=1}^{k/d} e^{-2\pi icdm/k}

We can now replace d by k/d in the sum on the right without affecting the sum (because both will run over the divisors of k) to get

a_k(m) = \frac{1}{k}\sum_{d|k}f\big(\frac{k}{d}\big)g(d) \sum_{c=1}^d e^{-2\pi icm/d}

But by Theorem 1 of Advanced Number Theory Note #9, the sum on c is zero unless d|m in which case the sum has the value d. Therefore

a_k(m) = \frac{1}{k}\sum_{\substack{d|k \\ d|m}}f\big(\frac{k}{d}\big)g(d)d

which proves (5). QED

Now we apply the formula in Theorem 1 with specialised f and g to obtain the Möbius function formula for Ramanujan’s sums in (3) above (which is the main thing we wanted to derive in this note).

Theorem 2. We have

c_k(n) = \sum_{d|gcd(n, k)}d\mu\big(\frac{k}{d}\big)

Proof: Taking f(k) = k and g(k) = \mu(k) in formula (4) in Theorem 1, we get

\sum_{d|gcd(n, k)}d\mu\big(\frac{k}{d}\big) = \sum_{m\hspace{1 mm} mod \hspace{1 mm}k} a_k(m)e^{2\pi imn/k}

where using the formula for a_k(m) in Theorem 1 we get

a_k(m) = \frac{1}{k}\sum_{\substack{d|k \\ d|m}}f\big(\frac{k}{d}\big)g(d)d

= \frac{1}{k}\sum_{d|gcd(m, k)}\frac{k}{d}\mu(d)d

= \sum_{d|gcd(m, k)}\mu(d)

= \big[\frac{1}{gcd(m, k)}\big]

where the last equality is due to a property of the Möbius function proved in Advanced Number Theory Note #1. Therefore a_k(m) = 1 if gcd(m, k) = 1 and a_k(m) = 0 if gcd(m, k) > 1, so we conclude

\sum_{d|gcd(n, k)}d\mu\big(\frac{k}{d}\big) = \sum_{\substack{m\hspace{1 mm} mod \hspace{1 mm}k \\ gcd(m, k)=1}} e^{2\pi imn/k} = c_k(n). QED