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DIRICHLET CHARACTERS and the METHOD of HYPERBOLAS Contents 1. Dirichlet Convolution and Möbius Inversion 1 2. Dirichlet Charact

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DIRICHLET CHARACTERS and the METHOD of HYPERBOLAS Contents 1. Dirichlet Convolution and Möbius Inversion 1 2. Dirichlet Charact DIRICHLET CHARACTERS AND THE METHOD OF HYPERBOLAS E. CHERNYSH Contents 1. Dirichlet Convolution and M¨obiusInversion 1 2. Dirichlet Characters 3 3. Abel Summation 5 4. The Method of Hyperbolas 7 5. Topics and Solved Problems 8 1. Dirichlet Convolution and Mobius¨ Inversion Recall that an arithmetic function is a mapping α : N ! C. The family of such functions is denoted by A[N] and is a vector space over C when endowed with our usual notions of scalar multiplication and vector addition. Two crucial arithmetic functions are the δ and µ functions, the latter of which is called the M¨obius function. These are defined by ( 1 s = 1 δ(s) := (1) 0 else Now consider s 2 N. If s is not square-free, then we set µ(s) = 0. Else, write Qk k s = j=1 pj and let µ(s) := (−1) . This defines the M¨obiusfunction. Now, Definition. Given two arithmetic functions α; β we define α ∗ β, called the Dirichlet convolution, as follows: X X n X n (α ∗ β)(n) := α(r)β(s) = α (d) β = α β (d) (2) d d rs=n djn djn From these definitions it is clear that (α ∗ β)(n) is itself an arithmetic function and moreover we find that the operation ∗ is both commutative and associative in A[N]. We show now that it preserves multiplicativity of a mapping: Proposition 1. Let α; β 2 A[N] and suppose that α; β are weakly multiplicative. Then, so is (α ∗ β)(·). Date: May 14, 2017. 1 DIRICHLET CHARACTERS AND THE METHOD OF HYPERBOLAS 2 Proof. Let n; m 2 N be co-prime and note that any divisor d j nm may be written as rs where r j n and s j m. Therefore, X nm X nm (α ∗ β)(nm) = α (d) β = α (rs) β d d djnm rsjnm X n m = α(r)α(s)β β r s rjn;sjm X X n m = α(r)α(s)β β r s rjn sjm 0 1 0 1 X n X m = α(r)β · α(s)β @ r A @ s A rjn sjm where this last line is precisely the quantity (α ∗ β)(n) · (α ∗ β)(m). The above is in of itself a useful property but is not quite what we seek. Recall the definitions of the delta and M¨obiusfunctions δ; µ 2 A[N] which we have defined above. Obviously, any constant mapping γ : N ! C is an element of A[N] and the same can be said for the map 1 : N ! C which is defined by n 7! 1. What we now claim is the following: Lemma 1. One has (µ ∗ 1) ≡ δ. Proof. We may use the properties derived in the previous result as well as some of our introductory results. Obviously, 1 is strongly multiplicative. We claim now that µ is multiplicative. It is sufficient to show this for paqb where p; q are distinct primes. If one of a; b ≥ 2 then µ(paqb) = 0 and µ(pa)µ(qb) = 0 as well, and it is verified in this case. In the other-case we have the result by direct calculation on the number of such primes. By our previous proposition , it follows that (µ ∗ 1) is multiplicative. It is therefore sufficient to establish this for powers of primes (see the \behaviour" of the δ function!). We now proceed with the calculation where ` 2 N: ` X X (µ ∗ 1)(p`) = µ(d) = µ(pj) = 1 + µ(p) = 0 djp` j=0 If, however, ` = 0 (i.e. n = 1) then this simply yields 1. By the fundamental theorem of arithmetic, the proof is now complete. Equipped with this result we are now prepared to prove the rather useful M¨obius Inversion Formula. Loosely speaking, this is a discrete analogue to the problem of the inverse Laplace transform. Suppose we are given the following relation: X β(n) = α(d); α; β 2 A[N] and n 2 N djn DIRICHLET CHARACTERS AND THE METHOD OF HYPERBOLAS 3 How would one recover α? The answer lies in the previous lemma. Compute now the convolution with µ, then for each n we have β ∗ µ = (α ∗ 1) ∗ µ = α ∗ (µ ∗ 1) = α ∗ δ We must now compute α ∗ δ for some n. This is not difficult at all; write: X (α ∗ δ)(n) = α(r)δ(s) = α(n) rsjn Putting these facts together we obtain the following: P Theorem 1 (M¨obiusInversion Formula). Let α; β 2 A[N] be given by β(n) ≡ djn α(d). Then, for all such n: α(n) = (β ∗ µ)(n)(M) 2. Dirichlet Characters In this document be briefly introduce and explore the concept of a Dirichlet character modulo some integer n ≥ 1. Simply put, a Dirichlet character may be defined on the ∗ group Zn for any n as above. More-precisely: Definition. Given an integer n ≥ 1 a Dirichlet character modulo n is a group ho- ∗ ∗ momorphism: χ : Zn ! C , where the group operation is multiplication. ∗ Some notable properties are now in order. Note that χ 6≡ 0 on Zn. Indeed, this follows from the fact that a homomorphism fixes the identity, i.e. χ(1) = 1. It is clear that it makes sense to set χ(·) = 0 whenever the argument is an integer not co-prime to n. Hence, there is a natural extension to all of Z achieved by setting χ(z) := χ (z (mod n)) which is equivalent to a periodic extension of χ. Moreover, χ is a multiplicative map, since homomorphisms preserve the group operations. There is always the trivial Dirichlet character, called the principal character which is simply the constant-map χ0 ≡ 1. Our first useful property is the following one: ∗ ∗ Proposition 2. Let n 2 N and χ : Zn ! C a Dirichlet character modulo n. Then, for '(n) th ∗ all χ ≡ 1. Namely, χ(x) is an n root of unity for all x 2 Zn. Proof. Fix x (mod n), or rather, its equivalence class. Now, since χ is assumed to be a homomorphism one has of course χ(x)'(n) = χ x'(n) = χ(1) = 1. ∗ We note that it follows from the above that χ must always be a map from Zn to fjzj = 1g: the unit disc in C. Our next goal is to \turn" the set of all these characters into a group. We note that modulo any n taking two characters χ, the product: (χψ)(·) = χ(·) (·) allows us to discuss the \product characters". Namely, let us define by G the set of all characters modulo some n ≥ 1, which we fix. Under the product above we observe that G becomes a group in its own right. Indeed, it is clear that this operation commutes DIRICHLET CHARACTERS AND THE METHOD OF HYPERBOLAS 4 and it is only left to show that each character has an inverse character with respect to this operation. Fix χ, some character modulo n. If χ = χ0 there is nothing to show, as this is our identity element. Else, consider a character defined by: χ−1(·) := χ(·) −1 −1 clearly, χ·χ ≡ χ0 and it is only left to verify that χ is a Dirichlet character. Clearly, it fixes the identity element and preserves the group operations by simple properties of the complex conjugate and hence we know (G; ) is a group, as was asserted. What is of greater interest is the following: Theorem 2. Let q = pα for some integer α and a prime p. Then, where G denotes the group as above modulo q, we have #G = '(q). Moreover, G is a cyclic group. ∗ Proof. We may take a primitive root g for Zq. Now, since any character χ is a homo- ∗ morphism it will be completely determined on Zq by its value on g, i.e. χ(g). Now, we know from a previous proposition that χ(g) is a root of unity with '(q). That is, a χ(g) = exp 2πi · ; 0 ≤ a < '(q) '(q) showing that they are distinct for different values of a. We now show some orthogonality results, which are in practice quite useful. Theorem 3 (Orthogonality Theorems). Let n ≥ 1 be an integer. Then; (1) For each Dirichlet character modulo n one has: n ( X '(q) χ = χ0 χ(a) = (3) 0 else a=1 (2) For all a 2 N one has: ( X '(q) a ≡ 1 (mod n) χ(a) = (4) 0 else χ mod n (3) For any two Dirichlet characters χ, modulo n one has: n ( X '(q) χ = χ(a) (a) = (5) 0 else a=1 (4) For all a; b 2 N one has: ( X '(q) a ≡ b (mod n) and co-prime to n χ(a)χ(b) = (6) 0 else χ mod n Pn Proof. We begin by proving (1). In the case χ = χ0 the result is trivial since χ(a) = P a=1 ∗ χ(a) and there are precisely '(n) such elements. Otherwise, we still need only a2Zn DIRICHLET CHARACTERS AND THE METHOD OF HYPERBOLAS 5 P ∗ consider ∗ χ(a) but there is some b 2 such that χ(b) 6= 1. Since b is invertible a2Zn Zn ∗ ∗ modulo n, it is not difficult to see that bZn = Zn and therefore; X X X χ(b) χ(a) = χ(ab) = χ(α) ∗ ∗ ∗ a2Zn a2Zn α2Zn proving that this sum vanishes, since χ(b) 6= 1. To prove (2), we argue in a very similar fashion. Consider now the case a ≡ 1 (mod n), ∗ since the χ always fix the identity and we sum over a set of cardinality #Zn = '(n) the result follows.
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