Advanced Number Theory Note #10: The Möbius function as a sum of primitive roots of unity

In this short note (intended mainly as a technical memo for myself) I want to give a careful and fully explicit proof of the fact that the Möbius function \mu(k) is the sum of the primitive kth roots of unity. This result is important because it can be regarded as a special case of other more general sums, such as Ramanujan’s sums and Gauss sums, which I will study closely in subsequent notes. The present note will therefore serve as a useful ‘base case’ for future reference.

The Möbius function \mu(k) and its basic properties were discussed in Advanced Number Theory Note #1. The Möbius inversion formula for Dirichlet convolutions was also discussed there, and plays a key role here. It says that

g(n) = \sum_{d \mid n} f(d)

if and only if

f(n) = \sum_{d \mid n} g(d) \mu(\frac{n}{d})

Also relevant is the result that every nth root of unity is a primitive dth root of unity for some divisor d of n. I constructed an explicit proof of this in an earlier note.

Let

f(d) = \sum_{\substack{1 \le a \le d \\ gcd(a,d)=1}} e^{2\pi ia/d}

and

g(k) = \sum_{1 \le m \le k} e^{2\pi im/k}

Since gcd(a, d) = 1, f(d) is the sum of all the primitive dth roots of unity, while g(k) is just the sum of all the kth roots of unity, primitive and non-primitive.

Since every kth root of unity is a primitive dth root of unity for some divisor d of k, it follows that adding up all the primitive dth roots of unity for each divisor d of k gives us all the kth roots of unity. Therefore we must have

g(k) = \sum_{d \mid k} f(d)                                       (1)

Example: For example, considering the 6th roots of unity we have

w = e^{2\pi i/6}     (primitive 6th root of unity)

w^ 2 = e^{2\pi i2/6} = e^{2\pi i/3}     (primitive 3rd root of unity)

w^3 = e^{2\pi i3/6} = e^{2\pi i/2}     (primitive 2nd root of unity)

w^4 = e^{2\pi i4/6} = e^{2\pi i2/3}     (primitive 3rd root of unity)

w^5 = e^{2\pi i5/6}     (primitive 6th root of unity)

w^6 = e^{2\pi i6/6} = 1     (primitive 1st root of unity)

The divisors of 6 are 1, 2, 3, and 6, and we see that all the primitive dth roots of unity for each of these divisors d of 6 are represented in the above list. Therefore adding up all the primitive dth roots of unity for each divisor d of 6 would give us all the 6th roots of unity, which is what (1) states. END

Now we can apply the Möbius inversion formula for Dirichlet convolutions to (1) to get

 

f(k) = \sum_{d \mid k}g(d)\mu(\frac{k}{d}) = \sum_{d \mid k}\mu(\frac{k}{d})\sum_{1 \le m \le d} e^{2\pi im/d}

 

But by Theorem 1 in Advanced Number Theory Note #9, with n = 1, the last sum on the right hand side is zero unless d = 1, so we get

f(k) = \mu(k)

which is what we set out to prove. QED