Lamp at ½
+The zeta ball, projected from the other pole. Put the lamp at the touch point ½ and the screen at the top, tangent at ∞: the chart is u = ¼ / (s − ½). The frame does not move — the poles still sit at ±½, the third strut at ±i/2, the circle |u| = ½ is where both charts agree — only the two ends of the ½–∞ strut have swapped. The old lamp, ∞, is now the centre; the old touch point, ½, is now at infinity. In this chart the functional equation is u ↦ −u and the whole dial is a rotation about the centre. And every zero, trivial and critical, falls into the origin: the trivial ones along the real axis from one side, the critical ones along the strut from both.
+ + + +u = ¼ / (s − ½), centred on ∞
+window ±0.60 · strut gold, mirror white, poles teal, zeros white
+hover for u and s
+Cross-section through the real axis
+ +The ball of diameter ½ on the s-plane. Old lamp at the top, ∞, casting onto the floor: the poles land on 0 and 1. New lamp at the touch point ½, casting onto the ceiling: the same poles land directly above, at u = ∓½, and ∞ itself lands at the centre. The trivial zeros, bunched near the top of the ball, fall onto the ceiling at −1/(8n+2): −0.1, −0.056, −0.038, … into the centre from the left.
+Key
+the strut: critical line Re s = ½, now the imaginary axis
+the mirror: real axis of s, still the real axis
+the ball's poles s = 0, 1 at u = ∓½ (ζ's pole is the right one)
+nontrivial zeros at u = ∓i/(4t)
+trivial zeros at u = −1/(8n+2)
+poles of χ at u = +1/(8n+2), the mirror images of its zeros
+the circle |u| = ½: the ball's horizontal equator, the one circle both charts draw in the same place
+the strip's edges Re s = 0 and Re s = 1: two circles of diameter ½ kissing at the centre. The critical strip is everything outside them, pinched to nothing at the lamp
+Hue is the argument of the value, brightness its size. The dial turns the ball about the ½–∞ strut; the chart's own crosshair stays put.
+Same frame, other lamp
+Stereographic projection from the top of a ball of diameter d onto the floor, and from the bottom onto the ceiling, differ by z ↦ d²/z̄. With d = ½ that is u = ¼/(s − ½), read from inside the ball so the mirror comes out the right way round. Points on the circle |s − ½| = ½ — through 0, 1 and ½ ± i/2 — are the ball's horizontal equator and land in the same place under both lamps. Everything else trades near for far: the small disc around ½ becomes the whole outside of the lens, and the whole far field of s becomes the small disc around the centre.
+So this is the chart centred on the point the s-plane does not have. The real origin ½ has gone to infinity and the imaginary origin ∞ sits at 0.
+The dial is a rotation
+s ↦ 1 − s is u ↦ −u, a point reflection through the centre. Turning the ball by θ about the ½–∞ strut is u ↦ e−iθu, a rotation of the plane. The quarter turn is the Wick rotation: it lays the strut, with every nontrivial zero on it, along the real axis. No Möbius map, no moving circles. Switch the paint to ξ and the half turn leaves the picture exactly as it was; with ζ it does not, and χ is the difference.
+The zeros fall in
+Nontrivial zeros ½ ± it sit at u = ∓i/(4t): on the strut, approaching the centre from above and below. Trivial zeros −2n sit at u = −1/(8n + 2): on the mirror, approaching from the left only — the positive real axis carries ζ(2), ζ(3), … → 1 and no zeros at all. One-sided along the real line, two-sided along the imaginary one: the real line reaches the lamp as a ray, the imaginary line as a circle through it. Both sequences converge to a point of this chart. In the s-chart that point is missing, which is the only reason they looked like they ran away.
+The strip itself is the outside of two kissing circles, the images of Re s = 0 and Re s = 1. Zoom in and the strip narrows like a horn, its two edges tangent to the strut at the centre: in this chart the strip has no width at the lamp. That is a fact about the chart, not about the zeros — Davenport–Heilbronn's off-line zeros would fall into the same cusp — but it is why, seen from ∞, on the line and near the line are the same thing.
+χ, what the half turn leaves over
+χ(s) = ζ(s)/ζ(1 − s) = 2sπs−1 sin(πs/2) Γ(1 − s): the gamma factor by itself. Every nontrivial zero cancels out of it. What remains lives on the mirror — zeros at u = −½, −1/(8n+2) on the left, poles at u = +½, +1/(8n+2) on the right, each the point reflection of the other, since χ(s)χ(1 − s) = 1. On the strut |χ| = 1 exactly: pure phase, winding as the Riemann–Siegel θ(t). ζ on the line is that phase times a real amplitude, Hardy's Z. Fold χ in and the picture becomes ξ, symmetric under the half turn.
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