Mobius¨ inversion revisited 1 Arithmetic functions and Dirichlet series >0 Suppose f : N ! C is an arithmetic function. In that case, we may form X f(n) L(f; s) := ; ns n>0 we will usually ignore convergence issues and refer to this as a formal Dirichlet series. If f and g are arithmetic functions, then we may define the arithmetic function f + g by the formula f + g(n) = f(n) + g(n). In that case, observe that the formula: L(f + g; s) = L(f; s) + L(g; s) holds. While we can multiply arithmetic functions as well via (fg)(n) = f(n)g(n), the result is not as closely connected with L-functions. Instead, consider the product of L-functions: 1 1 X f(n) X g(n) L(f; s) · L(g; s) = ( )( ): ns ns n=1 n=1 If we multiply out and group terms, f(2) f(3) g(2) g(3) (f(1) + + + ··· )(g(1) + + + ··· ) 2s 3s 2s 3s f(1)g(2) + f(2)g(1) f(1)g(3) + g(1)f(3) f(1)g(4) + f(2)g(2) + g(1)f(4) =(f(1)g(1) + + + + ··· ) 2s 3s 4s Extrapolating from this, the coefficients of L(f; s) · L(g; s) take the form: X f(a)g(b): ab=n This defines a new rule for composing arithmetic functions. Definition 1.1. If f and g are two arithmetic functions, then their Dirichlet convolution is the arith- metic function defined by X n X (f ∗ g)(n) = f(d)g( ) = f(a)g(b): d djn ab=n 1 2 1.1 The ring of arithmetic functions under convolution Example 1.2. We have used this notation already. If f is an arithmetic function, then we defined X X F (n) := f(d) = f(d) · 1: djn djn If we write 1 for the function that takes the constant value 1 on every positive integer, then in terms of convolution, this is the formula F := f ∗ 1. Example 1.3. We introduced the Mobius function µ(n). We then proved that µ ∗ 1 = χ1; where χ1 was the characteristic function of 1, i.e., it took the value 1 if n = 1 and was zero otherwise. Note that, by definition, f ∗ g = g ∗ f, i.e., Dirichlet convolution is a commutative composition. 1.1 The ring of arithmetic functions under convolution Now that we have introduced a new binary operation on the set of arithmetic functions, it is natural to ask whether the new operation gives rise to a commutative (unital) ring. Thus, let R be the set of arithmetic functions. Addition of arithmetic functions as defined above with additive identity 0 satisfies the axioms for the addition in a ring. Let us consider (R; +; 0; ∗). In order to check that ∗ defines a commutative unital ring, we need a prospective unit for ∗, we need to check ∗ is associative, and we want to check that ∗ distributes over +. Proposition 1.4. The function χ1 is a multiplicative unit for ∗, i.e., for any arithmetic function f, f ∗ χ1 = f = χ1 ∗ f. Proof. Since ∗ is commutative, it suffices to check one of these statements. In that case, consider: X n f(d)χ ( ): 1 d djn n n Since χ1( d ) = 0 unless d = 1, i.e., n = d, this sum collapses to give f(n), which is what we wanted to show. Proposition 1.5. The operation ∗ is associative, i.e., if f; g and h are three arithmetic functions, then (f ∗ g) ∗ h = f ∗ (g ∗ h). Proof. We just compute: X f ∗ g = f(a)g(b): ab=n Then, X (f ∗ g) ∗ h = f(a)g(b)h(c): abc=n Computing in the other direction is similar and yields the result. 3 1.1 The ring of arithmetic functions under convolution Proposition 1.6. The operation ∗ distributes over +, i.e., given three arithmetic functions f; g and h, f ∗ (g + h) = f ∗ g + f ∗ h. Proof. Again, we just compute: X X X X f∗(g+h) = f(a)(g+h)(b) = f(a)(g(b)+h(b)) = f(a)g(b)+ f(a)h(b) = f∗g+f∗h; ab=n ab=n ab=n ab=n as claimed. We can now give a very simple proof of Mobius¨ inversion. Theorem 1.7. If f is an arithmetic function, and F = f ∗ 1, then F ∗ µ = f. Proof. Indeed, F ∗ µ = (f ∗ 1) ∗ µ by definition. Then, by associativity of ∗, (f ∗ 1) ∗ µ = f ∗ (1 ∗ µ): Now, 1 ∗ µ = χ1 by the formula we proved last time, so f ∗ (1 ∗ µ) = f ∗ χ1 = f; where the last equality holds because χ1 is a multiplicative unit for ∗. Remark 1.8. It’s worth pointing out that this proof is not so different than the first proof we gave in class (the one in your book), but I think it is much easier to remember. Theorem 1.9. If f and g are multiplicative arithmetic functions, then f ∗ g is also multiplicative. Proof. Suppose m and n are coprime integers, then X mn (f ∗ g)(mn) = f(d)g( ): d djmn Now, if djmn, then d = d1d2 with d1jm and d2jn. In other words, X mn X mn f(d)g( ) = f(d1d2)g( ): d d1d2 djmn d1jm;d2jn Since f and g are multiplicative, and d1 and d2 are automatically coprime, X mn X m n f(d1d2)g( ) = f(d1)f(d2)g( )g( ); d1d2 d1 d2 d1jm;d2jn d1jm;d2jn the latter expression can be rewritten as a double sum giving exactly what we want. Corollary 1.10. If F = f ∗ 1 is multiplicative, then so is f. Proof. If F = f ∗ 1, then by Mobius¨ inversion F ∗ µ = f. Since µ is multiplicative, it follows from the preceding result that F ∗ µ is also multipliative, which is what we wanted to prove..