In 1936 Davenport and Heilbronn mixed two Dirichlet L-functions into one function that keeps the functional equation and loses the Euler product — and proved its zeros leave the critical line. This room builds the same mixture in the finite world of Fq[T], where the Riemann Hypothesis is Weil's theorem and the mixture's fate is decided by one inequality: the mixed zeros stay on the circle exactly when the two constituents' zeros interlace. Turn the dial through every character pair for q = 3, 5, 7 — all 194 of them, computed here — and then read the fence: over ℚ the same flip is the Riemann Hypothesis itself.
Instruments, not proofs. Every number on this page is a finite computation in a finite field. Nothing here bears on the truth of the Riemann Hypothesis.
computing — three finite fields, every character, every pair…
Blue dots: the two zeros of L̃(χ); gold dots: the two zeros of L̃(χ) (their mirror images across the real axis — Weil's theorem puts all four on the circle). The rim between neighbouring zeros is green where a blue zero is followed by a gold one and red where two of the same colour sit side by side: interlacing means every arc is green. Red rings: the zeros of the matched real mixture c·L̃(χ) + c̄·L̃(χ) — on the circle when the arcs are all green, thrown onto the real axis as a reciprocal pair z, 1/z when they are not. The gauge on the right is the whole decision: the needle |r| against the envelope 2|cos(φ/2)|.
Every conjugate pair of odd characters is one point: horizontal, the angle φ of its root number; vertical, the real number r that its degree-one character sum reduces to (Weil: |r| ≤ 2). The gold curves are the envelope r = ±2cos(φ/2). Inside, the mixture's zeros are on the circle (green); outside, off (red). Tap a point to load it into the dial. Where the root number is near −1 the envelope pinches to zero and almost every mixture fails; where it is near +1 almost every mixture survives. The flip is not rare and not exotic — it is the generic behaviour of a mixture without an Euler product, exactly as over ℚ.
Theorem (finite world, degree 2; elementary). Let χ be an odd character modulo an irreducible cubic over Fq, L̃(z) = 1 + s₁z + s₂z² its normalised L-function (z = √q·u). The functional equation forces s₂ = eiφ and s₁ = r·eiφ/2 with r real; Weil's theorem forces |r| ≤ 2. The matched real mixture c·L̃(χ) + c̄·L̃(χ) with c/c̄ = 1/s₂ — precisely the Davenport–Heilbronn recipe (1 + iκ)/(1 − iκ) = root number — equals 4·sin(φ/2)·[cos(φ/2)(1 + z²) + r·z]. Its zeros lie on the unit circle iff |r| ≤ 2|cos(φ/2)|, iff the zeros of L̃(χ) and L̃(χ) interlace on the circle.
Proof. Substitute; the quadratic cos(φ/2)(1 + z²) + r·z has discriminant r² − 4cos²(φ/2), so its roots are a conjugate pair on the circle when that is ≤ 0 and a real reciprocal pair z, 1/z when it is > 0. The zeros of L̃(χ) sit at angles −φ/2 ± ψ with cos ψ = −r/2, those of L̃(χ) at their mirrors; the four alternate around the circle iff the two blue zeros lie in opposite half-planes, i.e. iff sin(−φ/2 + ψ)·sin(−φ/2 − ψ) = sin²(φ/2) − sin²ψ ≤ 0, i.e. iff |cos ψ| ≤ |cos(φ/2)|, the same inequality. ∎
Grade: T. The one-way half — interlacing ⟹ zeros on the circle — holds in every degree (the Hermite–Biehler theorem on the unit circle); the converse fails from degree 3 on: in yesterday's census of degree-3 and degree-4 mixtures, 0 interlaced pairs left the circle in 314, but 61 of the 234 pairs at q = 5 were on the circle without interlacing. The antisymmetric mixture — the twin of Davenport–Heilbronn's second constant κ₋ — collapses in degree 2 to a multiple of 1 − z², zeros ±1 for free; the dial shows the symmetric one.
The κ-dial. Davenport–Heilbronn is fκ(s) = ½(1 − iκ)·L(s,χ) + ½(1 + iκ)·L(s,χ), χ the odd character mod 5 with χ(2) = i. The functional equation lives at exactly two real points — κ₊ = tan(½·arg ε) = 0.284079… (Davenport–Heilbronn's constant, computed here from the Gauss sum of χ) and κ₋ = −1/κ₊ (the antisymmetric equation). The Euler product lives at exactly two imaginary points, κ = ±i, where the mixture is a single L-function again: multiplicativity a(2)² = a(4) reads κ² + 1 = 0. The real axis and the imaginary axis meet only at κ = 0, which carries neither. There is no dial to turn: nothing between a mixture (zeros off the line, a theorem) and an L-function (zeros on the line, a conjecture).
What the finite dial reads for ℚ. The root number of χ mod 5 has arg ε = 31.717°, so the envelope would read 2|cos(φ/2)| = 1.924 — but over ℚ there is no r. The needle is an infinite zero set, and its interlacing with its own mirror image breaks infinitely often. The theorem that replaces the inequality: a Dirichlet series with periodic coefficients that is not a Dirichlet polynomial times an L-function has a positive proportion of its zeros in every vertical strip to the right of the line (Saias–Weingartner 2009); a degree-one function with the functional equation that satisfies the Riemann Hypothesis must have an Euler product (Kaczorowski–Kulas 2007).
Four specimens, read on the dial's terms. Every off-line zero of the Davenport–Heilbronn function below t = 200 (Spira 1994) sits at a seam where the interlacing of the two constituents' zeros breaks — a run of χ-zeros followed by a run of χ-zeros — inside a stretch with no on-line zero, missing about three zeros of which two are the off-line pair.
| off-line zero | constituent zeros around it (heights) | stretch with no on-line zero | zeros expected there |
|---|---|---|---|
| 0.808517 + 85.699348 i | χ: 83.38, 84.83, 85.40 → χ: 85.66, 86.68 | 83.11 → 87.65 (4.54) | 3.05 |
| 0.650830 + 114.163343 i | χ: 113.43, 114.16 → χ: 114.21, 115.34, 116.23 | 112.38 → 116.72 (4.35) | 3.12 |
| 0.574356 + 166.479306 i | χ: 164.37, 165.58, 166.36 → χ: 166.40, 167.26 | 164.16 → 168.32 (4.16) | 3.24 |
| 0.724258 + 176.702461 i | χ: 174.84, 175.60, 176.70 → χ: 176.84, 177.23 | 174.60 → 178.17 (3.57) | 2.81 |
Numerics (mpmath, 18 digits; receipts/dh_online_scan.py), four specimens, no theorem: seams are not sufficient — two seams near t ≈ 171–173 keep their on-line zeros. The first zero is certified by the argument principle (winding number 1 on a circle of radius 10⁻³; receipts/dh_dial_and_hn.py). Expected counts use the zero density (1/2π)·log(5t/2π).
In the finite world both halves are decidable. Over ℚ neither is, and the second half is the Riemann Hypothesis.
Finite world. Weil's theorem puts every constituent's zeros on the circle; one inequality then decides the mixture. Both halves are computed on this page, for every character, in under a second.
Over ℚ, first half. Whether the constituents L(s,χ), L(s,χ) have all their zeros on the line is the generalized Riemann Hypothesis for χ mod 5 — unproven. Nothing on this page bears on it.
Over ℚ, second half. The mixture's off-line zeros are a theorem (Davenport–Heilbronn 1936; Saias–Weingartner 2009). The only alteration that returns them to the line is κ → ±i — restoring the Euler product — which hands you the first half. What the dial decides in one inequality is, over ℚ, the Riemann Hypothesis itself: the interlacing-forever of a zero set with its mirror image, which is Hermite–Biehler positivity, which for ξ is Lagarias's Re ξ′/ξ > 0 on Re s > ½, which is RH.
So the room shows the shape of the missing ingredient, not the ingredient. Finite windows can refute and never prove; a finite field is a finite window in which the theorem happens to be complete. Read the Davenport–Heilbronn fence in Room VII and the product's edge in Room VIII for the two other faces of the same lock.
Every number on this page is recomputed in your browser from the field arithmetic — generator, discrete logarithms, character sums, roots, interlacing test, mixture roots — and compared with the committed Python twin (receipts/interlacing_dial.py, run 2026-09-24). Nothing is read from a table except the twin's values to check against.
| quantity | computed here | Python twin |
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