The walk to τ(χ)
τ(χ)
|τ(χ)|2
τ(χ)τ(χ̄)
Twisting by a: τa(χ) = ∑χ(n)ζan
| a | τa(χ) | χ̄(a)τ(χ) | match |
|---|
Lesson 14 · Class numbers and L-functions · 55–70 minutes
Gauss and Jacobi Sums
A Dirichlet character is multiplicative; the functions n ↦ ζan are additive. The Gauss sum τ(χ) = ∑χ(n)ζn is where the two meet. Its size is always exactly √p, it builds the Jacobi sum of Lesson 10 from simpler pieces, and it lets any character be rewritten as a combination of the additive characters ζan. That rewriting is the key step in evaluating L(1, χ) in Lesson 15.
Where this lesson is headed
τ(χ) = ∑n=1p−1 χ(n)ζn, |τ(χ)|2 = p, τ(χ)τ(ψ) = J(χ, ψ)τ(χψ), χ(n) = τ(χ̄)−1∑m χ̄(m)ζmn
In words: for every nontrivial character the Gauss sum has absolute value √p; Jacobi sums are ratios of Gauss sums (so again |J| = √p); and a character is its own "Fourier expansion" with coefficients χ̄(m)/τ(χ̄). For the quartic character, τ(χ4) = η0 + iη1 − η2 − iη3 is built from the periods of Lesson 12.
p: a prime in P*; ζ = cos(2π/p) + i sin(2π/p)
τ(χ) = ∑ χ(n)ζn: the Gauss sum of χ
τa(χ) = ∑ χ(n)ζan: the twisted Gauss sum
J(χ, ψ) = ∑x χ(x)ψ(1 − x)
(·/p): the Legendre symbol, χ42
01. A sum of p − 1 unit vectors with length √p
Start with p = 13 and χ4. Each term χ4(n)ζn is a point on the unit circle; the path adds them up. Twelve unit steps could reach distance 12, but the walk ends at distance exactly √13 ≈ 3.606 from 0, on the dashed circle. Try the other characters and primes: always √p.
Guided experiments
Do these in order. "Set this up" chooses p and the character.
Why √p is surprising
If the p − 1 unit vectors χ(n)ζn pointed in random directions, their sum would have length about √(p − 1) on average, with large fluctuations. The Gauss sum hits √p exactly, every time, for every nontrivial character. The cancellation is perfect in a precise sense, and it comes from the interplay of two kinds of structure: χ respects multiplication, ζn respects addition.
For the Legendre symbol, Gauss proved much more: τ((·/p)) = √p exactly (positive) when p ≡ 1 (mod 4), and i√p when p ≡ 3 (mod 4). It took him years to settle that sign.
02. Size, products and the Fourier expansion
Each step has the general argument on the left and your p and character from Stage 01 on the right.
1
The quartic Gauss sum from periods
Group the terms of τ(χ4) by coset, where χ4 = ij on Cj:
τ(χ4) = η0 + iη1 − η2 − iη3
and similarly τ((·/p)) = (η0 + η2) − (η1 + η3). So Gauss sums are the "Fourier coefficients" of the four periods.
2
Separating the twist
For p ∤ a, substitute m = an: as n runs over the units so does m, and χ(n) = χ̄(a)χ(m):
τa(χ) = ∑ χ(n)ζan = χ̄(a)∑ χ(m)ζm = χ̄(a)τ(χ)
For a ≡ 0, τ0(χ) = ∑χ(n) = 0 = χ̄(0)τ(χ). So the formula holds for every a.
3
|τ(χ)|2 = p
Add |τa(χ)|2 over all a mod p, two ways. By step 2 it is (p − 1)|τ(χ)|2 (the a = 0 term is 0, and |χ̄(a)| = 1 otherwise). Expanding the square instead:
∑a |τa|2 = ∑n,m χ(n)χ̄(m) ∑a ζa(n−m) = p ∑n |χ(n)|2 = p(p − 1)
since the inner sum is p when n = m and 0 otherwise. So |τ(χ)|2 = p.
4
τ(χ̄) and the product τ(χ)τ(χ̄)
Conjugating, τ(χ)‾ = ∑χ̄(n)ζ−n = τ−1(χ̄) = χ(−1)τ(χ̄) by step 2. Hence
τ(χ̄) = χ(−1)·τ(χ)‾, τ(χ)τ(χ̄) = χ(−1)|τ(χ)|2 = χ(−1)p
For the odd χ4: τ(χ4)τ(χ̄4) = −p.
5
The quadratic Gauss sum
For the Legendre symbol (real and, for p ≡ 1 mod 4, even), step 4 gives τ2 = p, so τ((·/p)) = ±√p. By step 1 it is a real number, (η0 + η2) − (η1 + η3), and together with η0 + ⋯ + η3 = −1 this gives η0 + η2 = (−1 ± √p)/2, the fact used in Lesson 12. Gauss's theorem that the sign is + is stated here; the computation confirms it for every prime shown.
6
Jacobi sums are ratios of Gauss sums
Let χ, ψ and χψ be nontrivial. Group τ(χ)τ(ψ) = ∑x,yχ(x)ψ(y)ζx+y by t = x + y. For t ≠ 0 put x = tu, y = t(1 − u): the inner sum is χψ(t)J(χ, ψ). For t = 0 it is ∑χ(x)ψ(−x) = ψ(−1)∑χψ(x) = 0. So
τ(χ)τ(ψ) = J(χ, ψ)·τ(χψ)
Taking absolute values, |J|2 = p·p/p = p: a second proof of Lesson 10's key fact. For χ = ψ = χ4: τ(χ4)2 = J·τ((·/p)) = J√p.
7
A character as a combination of waves
Apply step 2 to χ̄ with a = n: ∑m χ̄(m)ζmn = χ(n)τ(χ̄). Since |τ(χ̄)| = √p ≠ 0,
χ(n) = (1/τ(χ̄)) ∑m=1p−1 χ̄(m) ζmn
for every integer n (both sides are 0 when p | n). The multiplicative function χ is a combination of the additive waves n ↦ ζmn. Inserting this into L(1, χ) = ∑χ(n)/n turns the L-value into finitely many sums ∑ζmn/n, each of which has a closed form. That is Lesson 15.
03. The Gauss sum theorem
Assemble the argument, then test it by computer.
Theorem. Let χ be a nontrivial character mod p. Then ∑χ(n)ζan = χ̄(a)τ(χ) for all a, |τ(χ)|2 = p, and τ(χ)τ(χ̄) = χ(−1)p. If χ, ψ, χψ are nontrivial, τ(χ)τ(ψ) = J(χ, ψ)τ(χψ), so |J(χ, ψ)|2 = p. Finally χ(n) = τ(χ̄)−1∑mχ̄(m)ζmn for all n.
Complete proof
Separation. For p ∤ a, reindexing by m = an gives τa(χ) = χ̄(a)τ(χ); for p | a both sides vanish since ∑χ(n) = 0.
Size. ∑a mod p|τa|² = (p − 1)|τ|² by separation, and = ∑n,mχ(n)χ̄(m)∑aζa(n−m) = p(p − 1) by orthogonality of the additive characters. Hence |τ|² = p. Conjugation gives τ(χ)‾ = τ−1(χ̄) = χ(−1)τ(χ̄), so τ(χ)τ(χ̄) = χ(−1)p.
Jacobi. Writing τ(χ)τ(ψ) = ∑tζt∑x+y=tχ(x)ψ(y), the t = 0 part is ψ(−1)∑χψ = 0 and each t ≠ 0 part is χψ(t)J(χ, ψ), so τ(χ)τ(ψ) = J(χ, ψ)τ(χψ).
Expansion. Separation for χ̄ with a = n gives ∑mχ̄(m)ζmn = χ(n)τ(χ̄); divide by τ(χ̄) ≠ 0. ∎
Computer check
For all 42 primes of P* below 1000 and all nontrivial characters mod p: |τ(χ)|² = p and τ(χ)τ(χ̄) = χ(−1)p; for χ4: τ(χ4)² = J·τ((·/p)), τ((·/p)) = +√p, the period formula, and the Fourier expansion at every n.
Not run yet.
Exit problem
Use p = 29.