maths · two poles · zeta ball · the strut

The zeta strut

Put ½ at the centre: w = 1/(s − ½). The critical line becomes the imaginary axis, the strut through 0 and ∞; the functional equation s ↦ 1 − s becomes w ↦ −w, a sign flip through the centre; the pole s = 1 and the point s = 0 land at w = ±2, a dipole on the real axis. Then switch ζ for a function with the same symmetry and no Euler product, and watch zeros leave the strut.

±40

s-plane

critical line gold, real axis white, zeros as dots

painting

Riemann sphere, ½ in front, ∞ behind

the s-plane wrapped: the strut is the vertical great circle, the mirror the horizontal one, zeros march up the strut toward ∞

painting

Reading the strut: Z(t), the completed function made real

ξ(½ + it), the completed ζ: real on the line; its sign changes are the nontrivial zeros.

What the three paints show

ζ. In w the pole is the bright point at +2 and the trivial zeros −2, −4, … crowd in toward −2 from the other side: the dipole. Every nontrivial zero sits on the strut.

ξ. Pole and trivial zeros gone; ξ(s) = ξ(1 − s) makes the picture exactly symmetric under w ↦ −w. On the strut ξ is real and swings like cos; on the real axis it grows like cosh. The strip plots it divided by its own growth, Hardy's Z(t).

Davenport–Heilbronn. f = ((1 − iκ) L(s, χ) + (1 + iκ) L(s, χ̄))/2 with χ the odd character mod 5, χ(2) = i, and κ = (√(10 − 2√5) − 2)/(√5 − 1) ≈ 0.2841. It has the same completed symmetry Λ(s) = Λ(1 − s), so the strut and the mirror are still there, but no Euler product. Its first off-line zero is at s ≈ 0.8085 + 85.70i, marked in pink: in w it has left the strut. Raise the height past 86 to see it.

So the symmetry alone is not what pins zeros to the strut. The Euler product is the extra incidence, the way the yantra's 31 triple points were. Checked numerically to 30 digits in dh_check.py.

Key

critical line Re s = ½ (the strut, Im w)

real axis (the mirror, Re w)

zeros on the line

zeros off the line (Davenport–Heilbronn)

the dipole: pole s = 1 (w = 2) and s = 0 (w = −2)

Hue is the argument of the value, brightness its size. The sphere is the s-plane wrapped by stereographic projection with s = ½ at the front and ∞ at the back; the equator is |s − ½| = height/4.