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If gcd(a,b)=1, $1\leq b\leq a$, and $\mu(k)$ is the Möbius function, what is $$\sum_{k=0}^\infty\frac{\mu(ak+b)}{(ak+b)^s}$$ Can it be expressed in terms of other functions? Can I get it in the form of an Euler product?

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Try Dirichlet $L$-functions, and using Dirichlet characters to isolate the congruence. –  Eric Naslund Jan 5 '13 at 0:58
    
Is it always postive for s>1 eric? –  Ethan Jan 5 '13 at 0:59
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If $S(a,b;s)$ is your sum, then I would guess that the sign of $S(a,b;s)$ will be equal to the sign of $\mu(b)$. A numerical example: $S(1,3;2) \approx -0.142073$. –  Antonio Vargas Jan 5 '13 at 1:22
    
0<b<a+1 sorry i should have specified –  Ethan Jan 5 '13 at 1:28
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$S(5,3;2) \approx -0.118285$ –  Antonio Vargas Jan 5 '13 at 1:30

1 Answer 1

up vote 7 down vote accepted
+50

When $a>2$, this sum cannot be written as an Euler product $\prod_{p\text{ prime}} (\sum_{k\ge 0} a_{p^k} p^{-ks})$. We'll prove this by using the following fact: a Dirichlet series has an Euler product if and only if the corresponding arithmetic function is multiplicative ($f(nm)=f(n)f(m)$ if $(n,m)=1$) - prove this by expanding the Euler product!

First case: $a=b=1$. It is well-known that $\mu(n)$ is multiplicative (as an inverse of a multiplicative function), hence $$(*) \sum_{n \ge 1} \frac{\mu(n)}{n^s} = \prod_{p} (\sum_{k \ge 0} \mu(p^k) p^{-ks})=\prod_{p} (1-p^{-s})$$

Alternatively, since $\mu$ is the inverse of the constant function $\zeta = 1$ which has the representation $\sum_{n \ge 1} n^{-s} = \prod_{p} (1-p^{-s})^{-1}$, one can deduce $(*)$ by $(\sum_{n \ge 1} n^{-s}) (\sum_{n \ge 1} \mu(n) n^{-s})=1$.

Second case: $a=2,b=1$. The function $f(n)=\mu(n) \cdot 1_{2 \nmid n} = \begin{cases} 0 && 2|n \\ \mu(n) && 2 \nmid n \end{cases}$ is multiplicative, because both $\mu(n)$ and $1_{2 \nmid n} = \begin{cases} 0 && 2 |n \\ 1 && 2\nmid n \end{cases}$ are. Thus, we can expand the Dirichlet series of $f$ as follows: $$\sum_{n \ge 0} \frac{\mu(an+b)}{(an+b)^s} = \sum_{n \ge 0} \frac{\mu(2n+1)}{(2n+1)^s} = \sum_{m \ge 1} \frac{f(m)}{m^s}=$$ Here comes the Euler product: $$\prod_{p} (\sum_{k \ge 0} f(p^k) p^{-ks})= (\sum_{k \ge 0} f(2^k) 2^{-ks}) \prod_{p \neq 2} (\sum_{k \ge 0} \mu(p^k) p^{-ks}) = \prod_{p \neq 2} (1-p^{-s})$$

Third case: $a>2$. Sadly, $f(n)=\mu(n) \cdot 1_{n \equiv b \mod a}$ is no longer multiplicative. To show this, I'll assume Dirichlet's theorem on primes in arithmetic progressions. It states that if $n,m$ are coprime positive integers, then there are infinitely many primes satisfying $p \equiv m \mod n$.

Thus we can construct 2 distinct primes $p_1, q_1$ such that $p_1 \equiv 1 \mod a, p_2 \equiv b \mod a$. We have $f(p_1 q_1) = \mu(p_1 q_1) 1_{p_1 q_1 \equiv b \mod a} = 1$, and $f(p_1) f(q_1) = \mu(p_1)\mu(q_1) 1_{p_1 \equiv b \mod a}1_{q_1 \equiv b \mod a}= 1_{p_1 \equiv b \mod a, q_1 \equiv b \mod a}$.

Thus, $f(p_1 q_1)=f(p_1)f(q_1)$ implies $b \equiv 1 \mod a$. Now we'll construct 2 additional distinct primes $p_2, q_2$ such that $p_2, q_2 \equiv a-1 \mod a$. Since $p_2 q_2 \equiv 1 \mod a$, yet $p_2 \neq 1 \mod a$, we have $f(p_2)f(q_2) = 0$ yet $f(p_2 q_2)=1$.

It follows that $f$ isn't multiplicative.


Now we'll express our Dirichlet series as a linear combinations of Dirichlet series' with Euler product. In other words, we want to write $\mu(n) \cdot 1_{n \equiv b \mod a}$ as a linear combination of multiplicative functions.

We'll use characters. Characters modulo $m$ are functions from the positive integers to the complex numbers, satisfying 3 properties:

  1. They are periodic with period $m$: $\chi(a)=\chi(a+m)$.

  2. They are (completely) multiplicative: $\chi(ab) = \chi(a) \chi(b)$.

  3. If $(a,m) >1$, $\chi(a)=0$. Otherwise, $\chi(a) \neq 0$.

The theory of characters modulo $m$ is very well understood. For any $m$ you can construct distinct $\phi(m)$ characters. Given the factorization of $m$, it is a direct procedure to construct those characters. Read more here. There's a lot of interesting theory there, and explicit examples of characters.

Exercise 1: Show that $\chi(1)=1$ for any character, and that $\chi(a)$ is always a root of unity of order dividing $\phi(m)$.

Exercise 2: If $\chi$ is a character, $\overline{\chi}$ (complex conjugation) is also a character. $\overline{\chi}(a) = \chi(a^{-1})$, where $a^{-1}$ denotes a positive integer satisfying $a^{-1} \cdot a \equiv 1 \mod m$. Show that $\overline{\chi}(a) = \begin{cases} 0 && (a,m) > 1 \\ \frac{1}{\chi(a)} && (a,m)=1 \end{cases}$.

Exercise 3: The product of 2 characters is also a character.

Enough with the exercises. Suppose $\chi_{i}, 1 \le i \le \phi(m)$ are my distinct characters.

Theorem: Suppose $a$ is coprime to $m$.

$$\frac{1}{\phi(m)} \sum_{i=1}^{\phi(m)} \overline{\chi_i(a)}\chi_i(n) = 1_{n \equiv a \mod m}$$

The proof is actually not hard. If $(n,m)>1$ both sides are 0. Otherwise, by exercise 2, this identity is equivalent to

$$\sum_{i=1}^{\phi(m)} \chi_i(x) = \phi(m) 1_{x \equiv 1 \mod m}$$ where $x \equiv \frac{n}{a} \mod m$.

One direction is easy - if $x=1 \mod m$, all the characters evaluated on $x$ give the value 1.

The other direction is very nice - assume $x \neq 1 \mod m$. Let $S(x)=\sum_{i=1}^{\phi(m)} \chi_i(x)$. Use exercise 3 to show that $\{ \chi_i \}_{i} = \{ \chi_i \chi_j \}_{i}$, and deduce that $S(x) \chi_j(x) = S(x)$, for any $j$. Since the order of $x$ modulo $m$ is not $1$, one can find a character that doesn't give the value $1$ on $x$ (I can't elaborate without using more group theory), so for some $j$ we can cancel $\chi_j(x) - 1$ in the equality $S(x)(\chi_j(x) - 1)=0$ and get that $S(x)=0$. $\blacksquare$

With this theorem at hand, the decomposition is straightforward: we can write $\mu(n) \cdot 1_{n \equiv b \mod a}$ as $\frac{1}{\phi(a)} \sum_{i=1}^{\phi(a)} \overline{\chi_i(b)} \chi_i(n)\mu(n)$, and this shows: $$\sum_{n \ge 0} \frac{\mu(an+b)}{(an+b)^s} = \sum_{n \ge 1} \frac{\mu(n) 1_{n \equiv b \mod a}}{n^s}=$$ $$\frac{1}{\phi(a)} \sum_{i=1}^{\phi(a)} \overline{\chi_i(b)} (\sum_{n \ge 1} \frac{\mu(n) \chi_i(n)}{n^s})$$ And each summand has an Euler product, since $\mu(n) \chi_i(n)$ is multiplicative (a product of 2 multiplicative functions): $$\sum_{n \ge 1} \frac{\mu(n) \chi_i(n)}{n^s} = \prod_{p \text{ is prime}} (\sum_{k \ge 0} \mu(p^k)\chi_i(p^k)p^{-ks})=\prod_{p \text{ is prime}} (1-\chi_i(p)p^{-s})$$ You might recognize this as the inverse of the so-called L-function $L(s, \chi_i)$, defined by: $$L(s, \chi_i) = \sum_{n \ge 1} \frac{\chi_i(n)}{n^s} = \prod_{p \text{ is prime}} (1-\chi_i(p)p^{-s})^{-1}$$

Thus, we can write your function as $$\frac{1}{\phi(a)} \sum_{i=1}^{\phi(a)} \overline{\chi_i(b)} L(s,\chi_i)^{-1}$$

When $a=b=1$ or $a=2,b=1$ (the 2 cases discussed at the beginning), there is exactly one character, since $\phi(2)=\phi(1)=1$, and this sum becomes one element: $L(s, \chi)^{-1}$ where $\chi$ is the "trivial" character - the trivial character modulo $m$ is 1 when the value is coprime to $m$ and $0$ otherwise.

Hope this was readable. I can't help but notice that you ask many questions of this kind, reading chapter 16 of this classc will help you.

EDIT: In general, $$\sum_{k \ge 0} \frac{f(ak+b)}{(ak+b)^s} = \frac{1}{\phi(a)} \sum_{i=1}^{\phi(a)} \overline{\chi_i(b)} h(s,\chi_i)$$ where $h(s,\chi_i) = \sum_{n \ge 1} \frac{f(n)\chi_i(n)}{n^s}$, which has an Euler product expansion when $f$ is multiplicative. This show that you can express this kind of sums with at most $\phi(a)$ functions with Euler product expansions.

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I appreciate all the work you did –  Ethan Jan 13 '13 at 5:56
    
@Ethan No problem. My only question is whether this was clear for you? –  Ofir Jan 13 '13 at 6:04

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