Prime Number Theorem And ...

12 Iwaniec-Kowalski

12.1 Blueprint for Iwaniec-Kowalski Chapter 1

Here we collect facts from Chapter 1 that are not already in Mathlib. We will try to upstream as much as possible.

Definition 12.1.1
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Additive arithmetic function: satisfies \(f(mn) = f(m) + f(n)\) for coprime \(m\), \(n\).

Definition 12.1.2
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Completely additive arithmetic function: satisfies \(f(mn) = f(m) + f(n)\) for all \(m, n \ne 0\).

A completely additive function is additive.

Proof
Theorem 12.1.2

If \(a\) and \(b\) are coprime, then any divisor \(d\) of \(ab\) can be uniquely expressed as a product of a divisor of \(a\) and a divisor of \(b\).

Proof

Let \(a\) and \(b\) be coprime natural numbers, and let \(d\) be a divisor of \(ab\). Since \(a\) and \(b\) are coprime, we can use the fact that the divisors of \(ab\) correspond to pairs of divisors of \(a\) and \(b\). Specifically, we can express \(d\) as \(d = d_a \cdot d_b\), where \(d_a\) divides \(a\) and \(d_b\) divides \(b\). The uniqueness of this decomposition follows from the coprimality of \(a\) and \(b\), which ensures that the divisors of \(a\) and \(b\) do not share any common factors. Therefore, there is a one-to-one correspondence between the divisors of \(ab\) and the pairs of divisors of \(a\) and \(b\), which guarantees the uniqueness of the decomposition.

Theorem 12.1.3

If \(f\) is a multiplicative arithmetic function, then for coprime, nonzero \(a\) and \(b\), we have that \(\sum _{d | ab} f(d) = (\sum _{d | a} f(d)) \cdot (\sum _{d | b} f(d))\).

Proof

Since \(f\) is multiplicative, we can express the sum over divisors of \(ab\) in terms of the sums over divisors of \(a\) and \(b\). The key idea is to use the fact that the divisors of \(ab\) can be expressed as products of divisors of \(a\) and divisors of \(b\), due to the coprimality condition. Specifically, each divisor \(d\) of \(ab\) can be uniquely written as \(d = d_a * d_b\), where \(d_a\) divides \(a\) and \(d_b\) divides \(b\). Therefore, we can rewrite the sum as a double sum over the divisors of \(a\) and \(b\), which factorizes into the product of the two separate sums.

Theorem 12.1.4

If \(g\) is a multiplicative arithmetic function, then for any \(n \neq 0\), \(\sum _{d | n} \mu (d) \cdot g(d) = \prod _{p | n} (1 - g(p))\).

Proof

Multiply out and collect terms.

Theorem 12.1.5
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The Dirichlet convolution of \(\zeta \) with itself is \(\tau \) (the divisor count function).

Proof

By definition of \(\zeta \), we have \(\zeta (n) = 1\) for all \(n \geq 1\). Thus, the Dirichlet convolution \((\zeta * \zeta )(n)\) counts the number of ways to write \(n\) as a product of two positive integers, which is exactly the number of divisors of \(n\), i.e., \(\tau (n)\).

Theorem 12.1.6

The L-series of \(\tau \) equals the square of the Riemann zeta function for \(\Re (s) {\gt} 1\).

Proof

From the previous theorem, we have that the Dirichlet convolution of \(\zeta \) with itself is \(\tau \). Taking L-series on both sides, we get \(L(\tau , s) = L(\zeta , s) \cdot L(\zeta , s)\). Since \(L(\zeta , s)\) is the Riemann zeta function \(\zeta (s)\), we conclude that \(L(\tau , s) = \zeta (s)^2\) for \(\Re (s) {\gt} 1\).

Definition 12.1.3
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\(d_k\) is the \(k\)-fold divisor function: the number of ways to write \(n\) as an ordered product of \(k\) natural numbers. Equivalently, the Dirichlet convolution of \(\zeta \) with itself \(k\) times.

Theorem 12.1.7
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\(d_0\) is the multiplicative identity (indicator at 1).

Proof

By definition, \(d_k\) is the \(k\)-fold Dirichlet convolution of \(\zeta \). When \(k = 0\), this corresponds to the empty convolution, which is defined to be the multiplicative identity in the algebra of arithmetic functions. The multiplicative identity is the function that takes the value \(1\) at \(n=1\) and \(0\) elsewhere, which can be expressed as \(\zeta ^0\).

Theorem 12.1.8
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\(d_1\) is \(\zeta \).

Proof

By definition, \(d_k\) is the \(k\)-fold Dirichlet convolution of \(\zeta \). When \(k = 1\), this means we are taking the convolution of \(\zeta \) with itself once, which simply gives us \(\zeta \). Therefore, \(d_1 = \zeta ^1 = \zeta \).

Theorem 12.1.9

\(d_2\) is the classical divisor count function \(\tau \).

Proof

By definition, \(d_k\) is the \(k\)-fold Dirichlet convolution of \(\zeta \). When \(k = 2\), this means we are taking the convolution of \(\zeta \) with itself twice, which gives us \(\zeta * \zeta \). From the earlier theorem, we know that \(\zeta * \zeta = \tau \), where \(\tau \) is the divisor count function. Therefore, \(d_2 = \zeta ^2 = \tau \).

Theorem 12.1.10
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Recurrence: \(d_{k+1} = d_k * \zeta \).

Proof

By definition, \(d_k\) is the \(k\)-fold Dirichlet convolution of \(\zeta \). Therefore, \(d_{k + 1}\) is the \((k + 1)\)-fold convolution of \(\zeta \), which can be expressed as the convolution of \(d_k\) (the \(k\)-fold convolution) with \(\zeta \). Thus, we have \(d_{k + 1} = d_k * \zeta \).

Theorem 12.1.11

The L-series for \(d_k\) is summable for \(\Re (s) {\gt} 1\).

Proof

Since \(d_k\) is defined as the \(k\)-fold Dirichlet convolution of \(\zeta \), and we know that the L-series of \(\zeta \) converges for \(\Re (s) {\gt} 1\), it follows that the L-series of \(d_k\) also converges for \(\Re (s) {\gt} 1\). This is because the convolution of functions with convergent L-series will also have a convergent L-series in the same region. Therefore, we can conclude that \(L(d_k, s)\) is summable for \(\Re (s) {\gt} 1\).

The \(L\)-series of \(d_k\) equals \(\zeta (s)^k\) for \(\Re (s) {\gt} 1\).

Proof

From the definition of \(d_k\) as the \(k\)-fold Dirichlet convolution of \(\zeta \), we can express \(d_k\) as \(\zeta ^k\). The L-series of a Dirichlet convolution corresponds to the product of the L-series of the individual functions. Since \(L(\zeta , s)\) is the Riemann zeta function \(\zeta (s)\), it follows that \(L(d_k, s) = L(\zeta ^k, s) = (L(\zeta , s))^k = \zeta (s)^k\) for \(\Re (s) {\gt} 1\) where the series converges.

Theorem 12.1.13

\(d_k\) is multiplicative for all \(k\).

Proof

The function \(d_k\) is defined as the \(k\)-fold Dirichlet convolution of \(\zeta \). Since \(\zeta \) is a multiplicative function, and the Dirichlet convolution of multiplicative functions is also multiplicative, it follows that \(d_k\) is multiplicative for all \(k\). This can be shown by induction on \(k\), using the fact that the convolution of a multiplicative function with another multiplicative function remains multiplicative.

Theorem 12.1.14

Explicit formula: \(d_k (p^a) = (a + k - 1).choose (k - 1)\) for prime \(p\) and \(k \geq 1\).

Proof

The function \(d_k\) counts the number of ways to write a natural number as an ordered product of \(k\) natural numbers. For a prime power \(p^a\), the number of ways to factor it into \(k\) factors corresponds to the number of non-negative integer solutions to the equation \(x_1 + x_2 + ... + x_k = a\), where each \(x_i\) represents the exponent of \(p\) in the factorization of the corresponding factor. This is a classic combinatorial problem, and the number of solutions is given by the formula \((a + k - 1).choose (k - 1)\), which counts the ways to distribute \(a\) indistinguishable items (the prime factors) into \(k\) distinguishable boxes (the factors).

Theorem 12.1.15
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(1.25) in Iwaniec-Kowalski: a formula for \(d_k\) for all \(n\).

Proof

The function \(d_k\) is multiplicative, so to compute \(d_k(n)\) for a general natural number \(n\), we can factor \(n\) into its prime power decomposition: \(n = p_1^{a_1} p_2^{a_2} ... p_m^{a_m}\). Since \(d_k\) is multiplicative, we have:

\[ d_k(n) = d_k(p_1^{a_1}) \cdot d_k(p_2^{a_2}) \cdot ... \cdot d_k(p_m^{a_m}) \]

Using the explicit formula for prime powers from the previous theorem, we can substitute to get:

\[ d_k(n) = \prod _{i=1}^{m} (a_i + k - 1).choose (k - 1) \]

This gives us a complete formula for \(d_k(n)\) in terms of the prime factorization of \(n\).

Definition 12.1.4
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Divisor power sum with complex exponent.

Theorem 12.1.16
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For natural exponents, \(\sigma ^R\) agrees with \(\sigma \).

Proof

The function \(\sigma ^R\) is defined as the sum of the \(s\)-th powers of the divisors of \(n\). When \(s\) is a natural number \(k\), this definition coincides with the classical divisor power sum function \(\sigma _k(n)\), which also sums the \(k\)-th powers of the divisors of \(n\). Therefore, for natural exponents, we have \(\sigma ^R_k(n) = \sigma _k(n)\) when we view \(\sigma _k(n)\) as a complex number. This can be shown by directly comparing the definitions and noting that both functions sum over the same set of divisors with the same exponentiation.

Definition 12.1.5
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Arithmetic function with complex parameter \(\nu \). Evaluates as \(n\mapsto n^{\nu }\) for \(n\neq 0\) and \(0\) at \(n=0\).

Theorem 12.1.17

\(\sigma ^R(\nu ) = \zeta * \text{pow}^R(\nu )\), where \(\zeta \) is the constant function \(1\).

Proof

The function \(\sigma ^R(\nu )\) is defined as the sum of the \(\nu \)-th powers of the divisors of \(n\). The function \(\text{pow}^R(\nu )\) is defined as \(n \mapsto n^\nu \) for \(n \neq 0\) and \(0\) for \(n = 0\). The Dirichlet convolution of \(\zeta \) (the constant function \(1\)) and \(\text{pow}^R(\nu )\) is exactly \(\sigma ^R(\nu )\), since for each divisor \(d\) of \(n\), we have \((\zeta * \text{pow}^R(\nu ))(n) = \sum _{d|n} 1 \cdot d^\nu = \sigma ^R(\nu )(n)\). Thus, we have \(\sigma ^R(\nu ) = \zeta * \text{pow}^R(\nu )\).

Theorem 12.1.18

\(L(\text{pow}^R(\nu ), s) = \zeta (s - \nu )\) for \(\Re (s - \nu ) {\gt} 1\).

Proof

The function \(\text{pow}^R(\nu )\) is defined as \(n \mapsto n^\nu \) for \(n \neq 0\) and \(0\) for \(n = 0\). The L-series of \(\text{pow}^R(\nu )\) at \(s\) is given by the sum \(\sum _{n=1}^{\infty } n^{\nu - s}\). This series converges to the Riemann zeta function \(\zeta (s - \nu )\) for \(\Re (s - \nu ) {\gt} 1\), since the zeta function is defined as \(\zeta (s) = \sum _{n=1}^{\infty } n^{-s}\) for \(\Re (s) {\gt} 1\). Therefore, we have \(L(\text{pow}^R(\nu ), s) = \zeta (s - \nu )\) under the condition that \(\Re (s - \nu ) {\gt} 1\).

Theorem 12.1.19

The abscissa of absolute convergence of \(L(\text{pow}^R(\nu ), s)\) is at most \(\Re (\nu ) + 1\).

Proof

We apply ?? which states that if there exists a constant \(C\) such that \(\| f(n)\| \leq C \cdot n^r\) for all \(n\) sufficiently large, then the abscissa of absolute convergence of \(L(f, s)\) is at most \(r + 1\). In our case, we can take \(f(n) = n^\nu \) and observe that \(\| n^\nu \| = n^{\Re (\nu )}\). Thus, we can choose \(C = 1\) and \(r = \Re (\nu )\), which gives us the desired result that the abscissa of absolute convergence of \(L(\text{pow}^R(\nu ), s)\) is at most \(\Re (\nu ) + 1\).

\(\zeta (s)\zeta (s - \nu ) = \sum _{n=1}^{\infty } \sigma _\nu (n) n^{-s}\) for \(\Re (s) {\gt} 1\) and \(\Re (s - \nu ) {\gt} 1\).

Proof

The divisor power sum function \(\sigma _\nu \) is the Dirichlet convolution of the constant function \(1\) (i.e., \(\zeta \)) and the power function \(n \mapsto n^\nu \). The L-series of a Dirichlet convolution is the product of the L-series of the individual functions. Since \(L(1, s) = \zeta (s)\) and \(L(n \mapsto n^\nu , s) = \zeta (s - \nu )\), we have \(L(\sigma _\nu , s) = \zeta (s) \cdot \zeta (s - \nu )\) for \(\Re (s) {\gt} 1\) and \(\Re (s - \nu ) {\gt} 1\).