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viazovska

sphere packing · dim 8 & 24 · 2016

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The 240 shortest vectors of the E₈ lattice, projected to 2D. In its Coxeter plane they fall into 8 rings of 30 — the most symmetric 2D shadow of the densest packing in dimension 8. Scrub the projection plane away from the Coxeter plane and the order dissolves into a generic blur; scrub back and it snaps into the flower.

projection plane
Coxeter
0 = Coxeter plane (8 rings of 30). Drag toward a generic 8D plane and the symmetry breaks.
display

The 2D case, to build intuition. Pack equal circles on a lattice; the densest is the hexagonal one (Thue 1910), filling π/√12 ≈ 90.7% of the plane. Tilt the lattice and watch the density fall away from the 60° optimum.

lattice angle · θ
60°

Best sphere-packing density by dimension. It plummets as dimension grows — yet the exact optimum is known in only five dimensions: 1, 2, 3, 8, and 24. Dimensions 8 (E₈) and 24 (Leech) are Viazovska's; they sit visibly above the trend of best-known records.

Count the E₈ vectors shell by shell — 240 of squared length 2, then 2160, 6720, 17520, … — and you are reading off the q-expansion of the modular form E₄. The lattice's theta series literally is E₄(τ): the coefficient of qⁿ is 240·σ₃(n), where σ₃(n) = Σd|n d³. That identity is the reason a magic function for E₈ can be built from modular forms at all — and E₄ is the very same weight-4 form that feeds the construction. Hover a shell.

θE₈(τ) = E₄(τ) = 1 + 240q + 2160q² + 6720q³ + 17520q⁴ + 30240q⁵ + …
shell 1: the 240 roots drawn in the E₈ projection tab.

The magic function must vanish to second order at every E₈ shell — radii √(2n). Viazovska gets every one of those double zeros from a single exact factor, sin²(π r²/2), drawn below (solid). It touches zero, tangentially, exactly at the shell radii (markers). The dashed curve is a schematic admissible f: it starts at f(0)=1, dives non-positive past √2, and kisses zero at each shell — the qualitative shape the modular-form construction produces. f̂ is its mirror: non-negative everywhere, same double zeros.

Maryna Viazovska · 2016 · Fields Medal 2022

How much of space can you fill with equal, non-overlapping balls? In the plane the answer is the honeycomb, π/√12 ≈ 90.7%, settled by Thue and Fejes Tóth. In three dimensions it's the greengrocer's orange stack, π/√18 ≈ 74.0% — the Kepler conjecture, not proven until Hales's 1998 computer-assisted tour de force.

In dimensions 8 and 24 the answer is exact — and the proof, once you have the magic function, is half a page. Building the magic function is the other twenty.

The Cohn–Elkies bound

In 2003 Cohn and Elkies turned packing into a single inequality. Call a function f : ℝⁿ → ℝ admissible if it and its Fourier transform decay fast enough (Schwartz class is plenty). Their theorem:

if   f(0) = f̂(0) > 0,   f̂(ξ) ≥ 0 ∀ξ,   and   f(x) ≤ 0 for |x| ≥ r,
then every packing in ℝⁿ has density   Δ ≤ ωn · (r/2)n,

where ωn = πn/2 / Γ(n/2 + 1) is the volume of the unit ball (so ω₈ = π⁴/24). The proof is one page: average f over the points of any packing and apply Poisson summation; the three sign conditions make every term you can't control land on the correct side.

Why E₈ saturates it

The E₈ lattice is { x ∈ ℤ⁸ ∪ (ℤ+½)⁸ : Σxᵢ ∈ 2ℤ }. It is unimodular (covolume 1), its minimal vectors have length √2, and its squared lengths are exactly the positive even integers, so its shell radii are √2, 2, √6, √8, √10, … = √(2n). Packing balls of radius √2⁄2:

Δ(E₈) = ω₈ · (√2⁄2)8 = (π⁴/24)·(1/16) = π⁴/384 ≈ 0.25367.

Take r = √2 in Cohn–Elkies and the upper bound is ω₈·(√2/2)⁸ = π⁴/384 — exactly E₈'s density. So if an admissible f exists with the sign conditions at r = √2, E₈ is optimal. Here is the half-page that shows the bound is met, given a "magic" f that also vanishes at every shell. By Poisson summation on the self-dual E₈ (E₈* = E₈):

Σx∈E₈ f(x) = Σξ∈E₈ f̂(ξ).

The left side is f(0) + Σx≠0 f(x) ≤ f(0), since every nonzero vector has |x| ≥ √2 and there f ≤ 0. The right side is f̂(0) + Σξ≠0 f̂(ξ) ≥ f̂(0), since f̂ ≥ 0. With f(0) = f̂(0) the two ends pinch together, forcing equality and making the LP bound tight at E₈. ∎

The magic function, precisely

So the whole problem collapses to building one radial function on ℝ⁸ with these (wildly over-determined) properties:

Double zeros at infinitely many radii, for both f and f̂ at once — that is what made it look hopeless for thirteen years after Cohn–Elkies.

Viazovska's construction

Step 1 — split by Fourier parity. On radial Schwartz functions in even dimension the Fourier transform satisfies F² = id, so its eigenvalues are ±1. The Gaussian is the archetype: F[e^{−π|x|²}] = e^{−π|ξ|²} (eigenvalue +1). Write f = a + b with Fa = +a and Fb = −b; then f̂ = a − b. Now the condition "f ≤ 0 outside √2" constrains a+b and "f̂ ≥ 0" constrains a−b, and the two eigenfunctions can be engineered separately.

Step 2 — the double-zero factor. The function sin²(π|x|²/2) vanishes to exactly second order precisely when |x|² ∈ 2ℤ, i.e. at every shell |x| = √(2n). Pulling this factor out builds all the required double zeros in for free. (The magic function tab plots it against the E₈ shells.)

Step 3 — manufacture the eigenfunctions from modular forms. Take the Gaussian gτ(x) = e^{π i τ |x|²} with Im τ > 0. In dimension 8 its Fourier transform is ĝτ = (τ/i)^{−4} · g−1/τ. So if you integrate gτ against a function ψ(τ) along a vertical contour,

b(x) = ∫0i∞ ψ(τ) · sin²(π|x|²/2) · e^{π i τ |x|²} dτ ,

the eigenfunction equation Fb = −b turns into a modular transformation law for ψ under τ ↦ −1/τ — the (τ/i)^{−4} from the Gaussian is absorbed exactly when ψ is (quasi)modular of weight 2 − n/2 = −2. The sign conditions f ≤ 0 and f̂ ≥ 0 become positivity statements about the q-expansion of ψ, which one checks term by term.

Step 4 — name the forms. ψ is a weakly holomorphic form (a pole allowed at the cusp, i.e. a leading q⁻¹) built from the classical generators

E₂ = 1 − 24Σσ₁(n)qⁿ,   E₄ = 1 + 240Σσ₃(n)qⁿ,   E₆ = 1 − 504Σσ₅(n)qⁿ,
Δ = (E₄³ − E₆²)/1728 = q∏(1−qⁿ)24,   q = e2πiτ.

The −1 eigenfunction's ingredient is an honest weight −2 weakly holomorphic modular form; the canonical one is

E₄E₆ / Δ = q⁻¹ − 240 − 141444 q − …   (weight 4 + 6 − 12 = −2).

The +1 eigenfunction uses a weight −2 weakly holomorphic quasimodular form — the same generators but now involving E₂, whose near-modularity supplies the missing parity. Viazovska's exact linear combinations (and the contour/normalization constants) are two short displayed formulas in arXiv:1603.04246; the structure above is the whole idea. Within weeks Cohn, Kumar, Miller, Radchenko, and Viazovska ran the identical playbook one weight up for dimension 24 and the Leech lattice (arXiv:1603.06518). Fields Medal, 2022.

The proof in one page

  1. Cohn–Elkies: an admissible f with f(0)=f̂(0), f̂ ≥ 0, f ≤ 0 outside r caps every packing at ωn(r/2)n. [~1 page, 2003]
  2. At r = √2 that cap equals π⁴/384 = Δ(E₈). So a magic f makes E₈ optimal.
  3. Build f = a + b from Fourier eigenfunctions; pull out sin²(π|x|²/2) for the double zeros at the shells √(2n).
  4. Realize a, b as contour integrals of weakly holomorphic (quasi)modular forms of weight −2; modularity ⇔ the ±1 eigenfunction property, and the q-expansions give the sign conditions. [the ~20 pages]
  5. Poisson summation on self-dual E₈ pinches the bound to equality. ∎

Why 8 and 24?

E₈ and the Leech lattice are extraordinarily symmetric and self-dual; their theta series are modular forms, which is exactly what lets a modular magic function lock onto their shells. Nearby dimensions have no such miraculous lattice and the method gives nothing — the Cohn–Elkies bound there is strictly above the best packing, and the gap is still open. The erdős-style moral runs in reverse: here a hidden structure makes the impossible-looking exactly solvable.

Connections in the pack