The four coset sums
S0 + S2
S1 + S3
β = k + it
Lesson 08 · The quartic value β · 55–70 minutes
Coset Sums and B1,χ
Add up the members of each coset Cj from Lesson 07 and divide by p. For p = 13 the cosets {1, 3, 9}, {2, 5, 6}, {4, 10, 12}, {7, 8, 11} add to 13, 13, 26, 26: every total is a multiple of p. The four quotients Sj are whole numbers, and their differences form a Gaussian integer β = k + it, the character-weighted average of 1, …, p − 1 called B1,χ4. This number is the quartic value that the rest of the course studies.
Where this lesson is headed
β = B1,χ4 = (1/p) ∑n=1p−1 χ4(n)·n = (S0 − S2) + (S1 − S3)i = k + it
In words: each coset sum is a multiple of p, the four quotients Sj satisfy S0 + S2 = S1 + S3 = (p − 1)/4, and so β is a Gaussian integer with both coordinates odd. For even characters the same average would be 0, which is why the course uses the odd quartic character of P*.
p: a prime in P*
Sj = (1/p) ∑n ∈ Cj n, with n taken in 1 to p − 1
B1,χ = (1/p) ∑n=1p−1 χ(n)·n: the first generalized Bernoulli number
β = B1,χ4 = k + it
k, t: the coordinates of β
01. Add up the cosets
Start with p = 13. Each row lists one coset in increasing order with its total. The plot below builds B1,χ4 term by term: starting at 0, step by χ4(n)·n/p for n = 1, 2, …, p − 1. The walk wanders through non-integer points, but it always ends on a lattice point.
Guided experiments
Do these in order. "Set this up" chooses p.
The same average for other characters
B1,χ makes sense for any character. Compare the odd quartic character with the Legendre symbol, which is even for p ≡ 1 (mod 4).
| character | parity | ∑ χ(n)·n | B1,χ |
|---|
Even characters always give 0 (Stage 02, step 4). That is one reason the course needs an odd quartic character, and hence p ≡ 5 (mod 8).
02. Why the sums are whole, and what they add to
Each step has the general argument on the left and your p from Stage 01 on the right.
1
The fourth powers add up to a multiple of p
Let Σ = ∑h ∈ H h. Pick a ∈ H with a ≠ 1 (possible because |H| = (p − 1)/4 > 1, which needs p > 5). Multiplying by a permutes H, so
a·Σ ≡ Σ ⇒ (a − 1)Σ ≡ 0 ⇒ Σ ≡ 0 (mod p)
This is the orthogonality trick of Lesson 06 applied to the subgroup H. For p = 5, H = {1} and Σ = 1: the sums are not multiples of 5, which is the real reason P* starts at 13.
2
Every coset sum is a multiple of p
The members of Cj are the remainders of 2jh for h ∈ H. So
∑n ∈ Cj n ≡ 2j ∑h ∈ H h ≡ 0 (mod p)
and Sj = (1/p)∑Cj n is a whole number. (The actual integers n are between 1 and p − 1; only their total is controlled mod p.)
3
S0 + S2 = S1 + S3 = (p − 1)/4
C0 ∪ C2 is the set of squares (χ42 = (·/p) = 1 there). It is closed under n ↦ p − n, since −1 is a square, and n ≠ p − n. So its (p − 1)/2 members form (p − 1)/4 pairs, each adding to p:
p(S0 + S2) = p·(p − 1)/4
The non-squares C1 ∪ C3 pair up the same way.
4
B1,χ, and why even characters give 0
For a nontrivial χ, reindex n ↦ p − n:
∑ χ(n)n = ∑ χ(p − n)(p − n) = χ(−1)[p ∑ χ(n) − ∑ χ(n)n]
and ∑ χ(n) = 0, so ∑ χ(n)n = −χ(−1)∑ χ(n)n. If χ is even this forces B1,χ = 0. For odd χ nothing is forced; Lesson 15 shows B1,χ ≠ 0 for every odd χ, using L-functions.
5
β is a Gaussian integer
Group the sum defining B1,χ4 by coset, where χ4(n) = ij:
β = (1/p) ∑j ij ∑Cj n = S0 + iS1 − S2 − iS3
k = S0 − S2, t = S1 − S3
both whole numbers by step 2.
6
k and t are odd
By step 3, S2 = (p − 1)/4 − S0 and S3 = (p − 1)/4 − S1, so
k = 2S0 − (p − 1)/4, t = 2S1 − (p − 1)/4
For p ≡ 5 (mod 8), (p − 1)/4 is odd, and even minus odd is odd. In particular β ≠ 0, and N(β) = k2 + t2 is ≡ 2 (mod 4), twice an odd number. (Lesson 12 will meet that odd number as a class number.)
7
The conjugate character gives the conjugate value
χ̄4(n) is the conjugate of χ4(n), and the weights n are real, so
B1,χ̄4 = β̄ = k − it
Had we pinned χ̄4 instead (χ(2) = −i), we would get k − it: the pin decides the sign of t, and nothing else. Lesson 09 shows that trace and norm cannot see that sign.
03. The quartic value
Assemble the argument, then test it on every prime of P* below 1000.
Theorem. For p ∈ P*, each Sj is an integer, S0 + S2 = S1 + S3 = (p − 1)/4, and β = B1,χ4 = k + it with k = S0 − S2 = 2S0 − (p − 1)/4 and t = S1 − S3 = 2S1 − (p − 1)/4, both odd. Also B1,χ̄4 = β̄, and B1,χ = 0 for every even nontrivial character χ.
Complete proof
Integrality. Since p ∈ P*, |H| = (p − 1)/4 ≥ 3, so some a ∈ H has a ≠ 1; x ↦ ax permutes H, giving (a − 1)∑Hh ≡ 0 and, as a − 1 is a unit, ∑Hh ≡ 0. Then ∑Cjn ≡ 2j∑Hh ≡ 0, so Sj ∈ ℤ.
The relation. The squares C0 ∪ C2 are closed under n ↦ p − n (−1 is a square) with no fixed points, so they form (p − 1)/4 pairs of total p; hence S0 + S2 = (p − 1)/4, and likewise S1 + S3.
β. Grouping by coset, β = ∑ ijSj = (S0 − S2) + (S1 − S3)i, and substituting the relation gives k = 2S0 − (p − 1)/4, t = 2S1 − (p − 1)/4, odd because (p − 1)/4 is odd. Conjugating term by term gives B1,χ̄4 = β̄.
Even characters. Substituting n ↦ p − n and using ∑χ(n) = 0 gives ∑χ(n)n = −χ(−1)∑χ(n)n, which is 0 when χ(−1) = 1. ∎
Computer check
For each of the 42 primes of P* below 1000: computes the four coset sums and checks each is a multiple of p; checks the relation and that k, t are odd; compares β with the direct sum (1/p)∑χ4(n)n; and checks B1,χ = 0 for the Legendre symbol and every other even character.
Not run yet.
Exit problem
Use p = 37.