Partial sums of L(1, χ) in the complex plane
Partial sum
L(1, χ)
error × m
Error after m·p terms
| m | terms | error | error × m |
|---|
Lesson 13 · Class numbers and L-functions · 60–75 minutes
Dirichlet L-Functions
Lesson 12 found that a class number equals N(β)/2. To see why, the course needs a bridge between algebra and analysis, and Dirichlet built it: attach to each character χ the infinite series L(s, χ) = χ(1)/1s + χ(2)/2s + χ(3)/3s + ⋯. This lesson shows that the series converges at s = 1 (slowly), factors over the primes, and is not zero at s = 1. As a first payoff you will prove that there are infinitely many primes p ≡ 5 (mod 8), so P* never runs out.
Where this lesson is headed
L(s, χ) = ∑n≥1 χ(n)n−s = ∏q prime (1 − χ(q)q−s)−1, L(1, χ) ≠ 0 for χ ≠ χ0
In words: the series encodes unique factorization through its product over primes, converges at s = 1 because the character's partial sums stay bounded, and does not vanish there. Combined with orthogonality, the nonvanishing makes the sum of 1/q over primes q in any class a mod m diverge. For m = 8 and a = 5, everything can be checked by hand.
χ: a character mod m (here m = p, 4 or 8)
L(s, χ) = ∑ χ(n)/ns, s real
A(x) = ∑n≤x χ(n): the partial sums of χ
χ−4, χ8, χ−8: the nontrivial characters mod 8, with values on 1, 3, 5, 7 of (+, −, +, −), (+, −, −, +), (+, +, −, −)
π(x; 8, a): number of primes q ≤ x with q ≡ a (mod 8)
01. A series that settles, slowly
Start with p = 13 and the character χ̄4. The spiral shows the partial sums of L(1, χ̄4) = χ̄4(1) + χ̄4(2)/2 + χ̄4(3)/3 + ⋯ closing in on about 1.0864 − 0.5814i. The error falls like 1/N: ten times more terms, ten times less error. On the right, the Euler product and the race of primes mod 8.
Guided experiments
Do these in order. "Set this up" chooses p and the character.
The characters mod 8 at s = 1
| χ | 1, 3, 5, 7 | L(1, χ) | closed form |
|---|
All three are positive. Stage 02 proves this by grouping each series in blocks of 8, which is all the nonvanishing that Dirichlet's theorem for p ≡ 5 (mod 8) needs.
02. From a series to infinitely many primes
Each step has the general argument on the left and your p and character from Stage 01 on the right.
1
Definition and absolute convergence for s > 1
|χ(n)/ns| ≤ 1/ns, and ∑1/ns converges for s > 1, so L(s, χ) converges absolutely there, with |L(s, χ)| ≤ ζ(s) = ∑1/ns ≤ 1 + 1/(s − 1).
2
The Euler product
For s > 1 each factor (1 − χ(q)q−s)−1 = 1 + χ(q)q−s + χ(q)2q−2s + ⋯ is a convergent geometric series. Multiplying the factors for q ≤ Q gives ∑χ(n)n−s over all n whose prime factors are ≤ Q, each exactly once, by unique factorization and χ(ab) = χ(a)χ(b). Letting Q → ∞:
L(s, χ) = ∏q (1 − χ(q)q−s)−1 (s > 1)
3
Convergence at s = 1 by summation by parts
For χ ≠ χ0 mod m, a full period sums to 0 (Lesson 06), so |A(x)| ≤ m for all x. Writing χ(n) = A(n) − A(n − 1):
∑M<n≤N χ(n)/n = A(N)/N − A(M)/(M + 1) + ∑M<n<N A(n)(1/n − 1/(n + 1))
The last sum is at most max|A| × (1/(M + 1) − 1/N) in size, so the tail beyond M is at most 2 max|A|/(M + 1). The series converges at s = 1 (in fact for all s > 0), with error after N terms at most 2 max|A|/N: the 1/N decay seen in Stage 01.
4
Logarithms: primes appear
Taking the logarithm of the Euler product, with −log(1 − z) = z + z2/2 + ⋯:
log L(s, χ) = ∑q χ(q)q−s + R(s, χ), |R(s, χ)| ≤ ∑q ∑k≥2 q−k ≤ ∑n≥2 1/(n(n − 1)) = 1
So log L(s, χ) differs from a sum over primes by a bounded amount.
5
Picking out one class with orthogonality
Multiply by χ̄(a) and average over the φ(m) characters mod m. By the dual orthogonality of Lesson 06, the character average of χ̄(a)χ(q) is 1 when q ≡ a and 0 otherwise:
(1/φ(m)) ∑χ χ̄(a) log L(s, χ) = ∑q ≡ a (mod m) q−s + (bounded)
As s → 1+, the χ0 term goes to +∞ (L(s, χ0) differs from ζ(s) by finitely many factors, and ζ(s) ≥ 1/(s − 1)). If every other L(1, χ) is finite and nonzero, their logarithms stay bounded, so ∑q≡a 1/q diverges and there are infinitely many such primes.
6
Nonvanishing mod 8, by grouping
For the three nontrivial characters mod 8, group the series in blocks n = 8k + 1, …, 8k + 7 and write f(x) = 1/x − 1/(x + 2), which is positive and decreasing:
χ−4 (+ − + −): the block is f(8k + 1) + f(8k + 5) > 0.
χ8 (+ − − +): the block is f(8k + 1) − f(8k + 5) > 0.
χ−8 (+ + − −): the block is [1/(8k + 1) − 1/(8k + 5)] + [1/(8k + 3) − 1/(8k + 7)] > 0.
So L(1, χ) > 0 for all three, and by step 5 with m = 8, a = 5, there are infinitely many primes q ≡ 5 (mod 8). P* is infinite.
7
Nonvanishing for non-real characters mod p
For s > 1, summing the logarithms over all characters mod p and using orthogonality:
log ∏χ L(s, χ) = (p − 1) ∑qk ≡ 1 (mod p) 1/(k qks) ≥ 0
so ∏χ L(s, χ) ≥ 1. If a non-real χ had L(1, χ) = 0, then L(1, χ̄) = 0 too (conjugate series). Near s = 1, L(s, χ0) grows at most like 1/(s − 1), while L(s, χ)L(s, χ̄) shrinks like (s − 1)2 (the series for nontrivial χ can be differentiated term by term, stated), and the other factors stay bounded. The product would tend to 0, contradicting ≥ 1. So L(1, χ) ≠ 0 for every non-real χ, in particular for χ4 and χ̄4. (For real χ mod p, nonvanishing needs a different argument; for odd real χ it follows from the class number formula of Lesson 15.)
03. There are infinitely many primes in P*
Assemble the argument, then test the pieces numerically.
Theorem. For a nontrivial character χ mod m, L(s, χ) converges for s > 0, equals its Euler product for s > 1, and the error after N terms at s = 1 is at most 2 maxx|A(x)|/N. The three nontrivial characters mod 8 have L(1, χ) > 0, and non-real characters mod p have L(1, χ) ≠ 0. Consequently ∑q ≡ 5 (mod 8) 1/q diverges, and P* is infinite.
Complete proof
Series and product. For s > 1 the series converges absolutely and, by unique factorization, equals ∏(1 − χ(q)q−s)−1. For χ ≠ χ0, |A(x)| ≤ m, and summation by parts bounds the tail beyond N by 2 max|A|/N, so the series converges at s = 1.
Primes in a class. log L(s, χ) = ∑qχ(q)q−s + R with |R| ≤ 1. Averaging χ̄(a) log L(s, χ) over the characters mod m isolates ∑q≡aq−s up to a bounded error. As s → 1+ the χ0 term diverges and the others stay bounded provided L(1, χ) ≠ 0.
Mod 8. Grouping in blocks of 8 with f(x) = 1/x − 1/(x + 2) decreasing shows L(1, χ−4), L(1, χ8), L(1, χ−8) are positive, so ∑q≡5 (8)1/q = ∞. Hence P* is infinite.
Non-real χ mod p. ∏χL(s, χ) = exp((p − 1)∑qk≡11/(kqks)) ≥ 1; a zero of L(s, χ) at s = 1 would bring a matching zero of L(s, χ̄), a double zero against at most a simple pole of L(s, χ0), forcing the product to 0. ∎
Computer check
For the primes of P* below 200: the Euler product at s = 2 matches the series; the 1/N error bound 2 max|A|/N holds for χ4, χ̄4 and (·/p) at N = p, 2p, …, 30p; ∏χL(1.5, χ) ≥ 1; and the three mod-8 values match π/4, log(1 + √2)/√2 and π/(2√2).
Not run yet.