Riemann project / October 5, 2026 / Explanation v1
The surface we still
have to stay above.
For this theta subproblem, yes: we still need a lower bound at every height for the complete signed expression. We have an exact identity and justified boundary terms. We do not yet have the arithmetic estimate that forces its sign.
That would prove the first Laguerre inequality for the original function. That inequality alone would not prove RH.
Four coordinates. One sign boundary.
The four coordinates are height r, value Y, slope Y′, and curvature Y″. The gold surface marks equality in the required bound.
Illustrative coordinate space. No zeta values or original theta trajectory are plotted. Motion rotates the view; it does not evolve the function.
Drag horizontally or use the rotation slider. Dim sheets show selected heights from 0 to 2; the bright sheet follows the height slider. The display is cropped, not a global bound.
Move an illustrative state
These four numbers are free examples. They need not belong to any single function, let alone the original theta function. Try crossing the boundary by changing curvature.
Illustrative state: on the boundary
- Left side: Y′² − Y·Y″
- 0.0000
- Required lower bound
- 0.0000
- Difference: left − required
- 0.0000
Display arithmetic only. The sign of a chosen example says nothing about the sign of the actual function.
What the missing proof must do
The actual original function selects a very particular path through this four-coordinate space:
r ↦ (r, Y(r), Y′(r), Y″(r)).
We need to show that this path stays on the allowed side at every real height. That path is deliberately not drawn here: a finite plot would not supply the missing all-height proof.
The required lower bound, including its correction:
Y′² − Y·Y″ ≥ c(r)·Y²
c(r) = 2(1/4 − r²)/(r² + 1/4)²
For every real r, for the original theta coefficients.
“Above” means above this lower bound, not above a fixed screen direction. If Y is positive, lowering curvature helps the inequality; if Y is negative, the allowed direction reverses.
Below |r| = 1/2 the correction is positive. Above it, the correction is negative. Omitting it changes the question. At Y = 0 the complete expression is regular and becomes a nonnegative square; no division by Y is needed.
What we have — and what it does not give
The boundary repair is complete. The positive decaying primitive makes the pair moments integrable, and the integrations by parts retain their full boundaries. This is why the displayed expression is legitimate.
The sign estimate is absent. Positivity, decay and a good tail do not force this inequality. The existing NS101 control retains those features yet violates the corrected sign in its established parameter range.
The original arithmetic remains different. That control changes the original theta coefficients and does not obey the same nonlinear Jacobi initial-value problem. It therefore does not disprove the original target. A useful next method must exploit that difference to estimate the complete signed quantity.
This is a partial target. First-Laguerre positivity is only one condition in the broader real-zero problem. The NB cofinal gain and the Weil uniform-floor targets are separate open obligations. This page does not reduce the entire project to one final inequality.
Exact identity and how the 4D picture is drawn
In PR76’s fixed normalization, X(r) = xi(1/2 + ir)/2. Let Theta(x) be the sum of exp(−pi·n²x) over all integers n, let h(u) = exp(u/2)·Theta(exp(2u)), and let f(u) = 2 cosh(u/2) − h(u). Its Fourier transform is Y(r).
p = r² + 1/4
X = pY/4
16(X′² − X·X″)
= p²(Y′² − Y·Y″) + 2(r² − 1/4)Y².
The canvas draws the zero set of the right side in independent coordinates (r, y, v, w). For y ≠ 0, each height sheet has w = v²/y + 2(r² − 1/4)y/p². At y = v = 0 every w lies on the boundary; these lines are included separately. The curvature segment through the selected example marks its allowed and violating portions.
Display ranges are 0 ≤ r ≤ 2, |y| ≤ 1.2, |v| ≤ 1.4 and |w| ≤ 2.2. The renderer samples and clips the boundary, scales these four axes, rotates in four dimensions, then projects to three dimensions and onto the screen. Thus a sheet may end at the display box; it is not a mathematical endpoint.
The coordinate scales are only for viewing. No physical metric, symplectic structure or new geometric theorem is asserted. The sliders do not impose the derivative consistency of a genuine function. For example, the actual even function has Y′(0) = 0, and complete moments already give strict first-Laguerre positivity at r = 0. Free slider settings need not respect that.
Y is the meromorphic completed zeta factor with poles at r = ±i/2; it is not the entire function X. This model uses real r throughout.