The Sphere Packing Problem in Dimension 4

1. Introduction

1.1. Background and Statement of the Main Result

Given a positive integer n, the sphere packing problem asks for the densest way to fill ${\mathbb{R}}^n$ with non-overlapping open unit balls. Write ${\Delta }_n$ for the supremum of achievable densities; this is the optimal packing density in dimension n. The problem has a clean answer in low dimensions and in the two highest dimensions where it is currently settled.

In dimension 1 the answer is ${\Delta }_1=1$, trivially. In dimension 2 it is ${\Delta }_2=\pi /\left(2\sqrt{3}\right)$, the density of the hexagonal lattice, proved by Thue [1] and Fejes Tóth [2]. Dimension 3 is the Kepler conjecture, resolved by Hales [3] and subsequently given a formal computer-assisted proof in [4]. Dimensions 8 and 24 were settled in 2016 by Viazovska [5] and by Cohn, Kumar, Miller, Radchenko, and Viazovska [6] respectively. This paper treats dimension 4 by a local Voronoi-cell reduction centered on the regular 24-cell.

The root lattice $D_4$ scaled by $\sqrt{2}$ gives a packing of unit balls with density

$$\Delta ={\displaystyle \frac{vol{B}^4}{covol\left(\sqrt{2}{D}_4\right)}}={\displaystyle \frac{{\pi }^2/2}{8}}={\displaystyle \frac{{\pi }^2}{16}}\approx 0.6169,$$

since the volume of the four-dimensional unit ball is ${\pi }^2/2$. That this is optimal among all lattice packings has been known since Blichfeldt [9]. The question of whether any non-lattice, possibly non-periodic arrangement could improve on it remains the central point of the dimension-four problem. The best published unconditional upper bound on the centre density in ${\mathbb{R}}^4$, due to refinements of the Cohn–Elkies linear programming method, stands at $0.130587$ [10], still strictly above the conjectured $1/8$. The exact status of the present reduction is recorded in Remark 1.

Theorem 1 

(Reduction to the $D_4$ density). Assume the chamber-positivity statement of Proposition 6 holds for every row of the 176-row atlas. Then every saturated unit-ball packing in ${\mathbb{R}}^4$ has Voronoi cells of volume at least 8, the packing density is at most ${\pi }^2/16$, and equality forces the scaled lattice $\sqrt{2}{D}_4$ up to translation.

Remark 1 

(Status of the proof of Theorem 1). The proof given here is unconditional in the root-aligned case: Lemma 1 reduces the density bound to the local statement that every Voronoi cell of a saturated packing has volume at least 8, and Theorem 2 together with Corollary 1 proves this local statement whenever every shell-active neighbour of the centre lies in one of the 24 root directions of $D_4$. For the complementary cap-cutting case, Theorem 4 gives a corrected reduction: the volume defect is monotone in each cap-cutting offset, so the only configuration that can violate the bound is the corner configuration in which every active radius equals the packing minimum 2. The volume bound at that corner is the finite chamber-positivity statement of Section 7. The 176-row atlas is fully enumerated and checked for row count, cycle structure, orbit weights, and cross-file labels; the symbolic positivity identity (11) remains the finite algebraic computation needed to turn Theorem 1 into a complete proof.

Lemma 1 

(Saturation and local density). It is enough to prove the cell bound for saturated packings. More precisely, let $\mathcal{P}\subset {\mathbb{R}}^4$ be a packing with minimum mutual distance at least 2. If every Voronoi cell in every saturated extension of $\mathcal{P}$ has volume at least 8, then the upper density of $\mathcal{P}$ is at most ${\pi }^2/16$.

Proof. 

A packing can be enlarged, by the usual Zorn argument, to a saturated packing ${\mathcal{P}}^{\prime }\supseteq \mathcal{P}$: keep adding centres at distance at least 2 from all existing centres until no such point remains. This operation cannot decrease upper density. Thus it suffices to bound saturated packings.

Let $V_x$ be the Voronoi cell of $x\in {\mathcal{P}}^{\prime }$. The cells $V_x$ cover ${\mathbb{R}}^4$ up to common boundaries of measure zero. Suppose $vol\left(V_x\right)\ge 8$ for every x. For a large Euclidean ball $B_R$, let $N\left(R\right)$ be the number of centres in $B_R$. Saturation gives a uniform covering radius at most 2: every point of ${\mathbb{R}}^4$ is within distance 2 of some centre, so the portion of a Voronoi cell meeting $B_R$ lies in $B_{R+2}$ around the same centre. Consequently the boundary contribution in counting cells across $\partial B_R$ is $O\left(R^3\right)$. Summing cell volumes over centres whose cells meet $B_R$ gives

$$8N\left(R\right)\le vol\left(B_{R+2}\right)+O\left(R^3\right).$$

Dividing by $vol\left(B_R\right)$ and letting $R\to \infty$ gives

$$\overline{\rho }\left({\mathcal{P}}^{\prime }\right)\le {\displaystyle \frac{1}{8}},$$

where $\overline{\rho }$ denotes the centre density. Multiplying by $vol\left(B_1^4\right)={\pi }^2/2$ gives the packing-density bound ${\pi }^2/16$. Since $\mathcal{P}\subseteq {\mathcal{P}}^{\prime }$, the same upper bound applies to $\mathcal{P}$.    □

Remark 2 

(What remains after Lemma 1). All global density issues are now absorbed into a single local statement: for a saturated packing with a centre at the origin, its Voronoi cell must have volume at least 8. The rest of the paper proves exactly this local statement.

The closely related kissing-number problem in dimension 4 was resolved by Musin [7], who showed the answer is 24, matching the 24 nearest neighbours of $\sqrt{2}{D}_4$. Musin [8] subsequently studied the associated 24-cell conjecture, which concerns the local geometry around the Voronoi cell of $\sqrt{2}{D}_4$ that is also central to the present argument.

1.2. The Method

The proof reduces Theorem 1 to a purely local geometric statement: every Voronoi cell of every saturated unit-ball packing in ${\mathbb{R}}^4$ has volume at least 8. Given that statement, the density bound follows immediately, since the Voronoi cells tile ${\mathbb{R}}^4$ and the density contributed by a single ball of volume ${\pi }^2/2$ housed in a cell of volume at least 8 is at most $({\pi }^2/2)/8={\pi }^2/16$.

In dimensions 8 and 24 the analogous step was accomplished via a magic function: a radial Schwartz function f satisfying $\widehat{f}\ge 0$, $f\left(0\right)=\widehat{f}\left(0\right)$, and $f\left(x\right)\le 0$ for $\parallel x\parallel \ge 2$, which saturates the Cohn–Elkies linear programming bound [11]. In dimension 4 the linear programming bound does reproduce ${\pi }^2/16$ (Cohn–Zhao [10]), but no explicit magic function is known, and that route is not available to us.

Our approach works entirely in configuration space. The Voronoi cell of the origin in a packing captures the local density contribution of the ball centred there. A key simplification comes from the observation (Lemma 8 below) that only packing centres within a specific annular shell around the origin can possibly reduce the Voronoi volume below 8; centres farther away already generate half-spaces containing $P_{24}$ by a Cauchy–Schwarz estimate. The centres inside that shell fall into two classes: those pointing in one of the 24 root directions of $D_4$, and all others.

For the first class the argument is a direct containment: the support function of the regular 24-cell $P_{24}$ achieves its minimum value of exactly 1 on those 24 directions (Proposition 1), which forces every corresponding half-space to contain $P_{24}$ throughout the shell. Any collection of such neighbours therefore leaves the Voronoi cell at least as large as $P_{24}$, which has volume 8 (Theorem 2).

For the second class a neighbour outside the root directions cuts a cap off $P_{24}$, but simultaneously the Voronoi cell gains extra volume on the antipodal side, since the constraint from that neighbour does not restrict the cell below $P_{24}$ on the opposite half. We make this precise as follows. When a single non-root-aligned neighbour $y_1$ is present, we check directly that every point of $P_{24}$ at height $\langle z,u_1\rangle \le {\rho }_1=r_1/2$ belongs to the Voronoi cell, establishing a slice containment $P_{24}^{\left(t\right)}\subseteq V^{\left(t\right)}$ for all $t\le {\rho }_1$. By central symmetry of $P_{24}$, the antipodal copy $-C_1$ of the cap lies inside V as well, providing a volume gain that equals $vol\left(C_1\right)$. This gives $vol\left(V\right)\ge 8-vol\left(C_1\right)$ outright; to remove the deficit term we use, not convexity, but the elementary monotonicity of an intersection of half-spaces in each of its offsets separately (Theorem 4, Lemma 13): the volume is monotone non-decreasing in ${\rho }_1$ with the other offsets fixed, so its minimum over the feasible range of ${\rho }_1$ occurs at ${\rho }_1=1$, i.e., $r_1=2$, where the configuration falls into the finite chamber atlas of Section 7. Establishing $vol\left(V\right)\ge 8$ at that single corner for every chamber — the chamber-positivity statement of Proposition 6 — is therefore what closes the cap-cutting case; its present verification status is recorded in Remark 1.

1.3. Organisation

Section 2 introduces the lattice $D_4$, its Weyl group, the regular 24-cell $P_{24}$, and the support function of $P_{24}$. Section 3 analyses which directions are safe (never cut $P_{24}$) and proves the angular deficiency formula. Section 4 establishes the shell localisation. Section 5 handles all root-aligned configurations. Section 6 develops the Gram-variable volume formula. Section 7 records the Gram-spectral square decomposition, the chamber ideal, and the exact finite atlas. Section 8 completes the proof by combining the cap-cutting geometry with the chamber positivity theorem. Section 9 records some brief closing remarks. Appendix A gives the simplex-by-simplex volume computation for $P_{24}$. Appendix B records the 176-row chamber register and describes the accompanying auxiliary files. Appendix C prints the active-facet rows. Appendix D describes the auxiliary file layout. Appendix E records exact constants, orbit arithmetic, and numerical illustrations.

2. The ${\mathbf{D}}_{\mathbf{4}}$ Comparison Cell

2.1. The Lattice and Root System

Let $e_1,e_2,e_3,e_4$ denote the standard orthonormal basis of ${\mathbb{R}}^4$. The root lattice is

$$D_4=\left\{m=(m_1,m_2,m_3,m_4)\in {\mathbb{Z}}^4:m_1+m_2+m_3+m_4\equiv 0(mod2)\right\}.$$

It has rank 4 as a free abelian group. A basis is given by the four simple roots

$${\alpha }_1=e_1-e_2,{\alpha }_2=e_2-e_3,{\alpha }_3=e_3-e_4,{\alpha }_4=e_3+e_4.$$

The full root system of $D_4$ is

$$\mathcal{R}=\{\pm e_i\pm e_j:1\le i<j\le 4\},$$

which contains $\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right)·4=24$ vectors, each of squared norm ${\parallel \alpha \parallel }^2=2$. For the unit-ball packing problem, centres must be placed at mutual distance at least 2. The lattice $\sqrt{2}{D}_4$ achieves nearest-neighbour distance exactly 2, and we work with it throughout:

$$\Lambda =\sqrt{2}{D}_4.$$

Lemma 2 

(Covolume). $covol(\Lambda )=8$.

Proof. 

The Gram matrix of the simple roots ${\alpha }_1,\dots ,{\alpha }_4$ with respect to the standard inner product on ${\mathbb{R}}^4$ is the Cartan matrix of $D_4$:

$$C_{D_4}=\left(\begin{array}{cccc}2& -1& 0& 0\\ -1& 2& -1& -1\\ 0& -1& 2& 0\\ 0& -1& 0& 2\end{array}\right).$$

The off-diagonal pattern encodes the Dynkin diagram of $D_4$, in which node ${\alpha }_2$ is connected by single bonds to each of ${\alpha }_1$, ${\alpha }_3$, and ${\alpha }_4$. Expanding the determinant gives $det{C}_{D_4}=4$, so $covol\left(D_4\right)=\sqrt{4}=2$. Scaling by $\sqrt{2}$ multiplies every basis vector by $\sqrt{2}$, hence scales the covolume by ${\left(\sqrt{2}\right)}^4=4$, giving $covol(\Lambda )=4·2=8$.    □

Remark 3. 

The density of the packing by unit balls centred at Λ is $\Delta =({\pi }^2/2)/8={\pi }^2/16$.

2.2. The Weyl Group of $D_4$

The Weyl group $W=W\left(D_4\right)$ consists of all orthogonal transformations of ${\mathbb{R}}^4$ generated by the reflections $s_{\alpha }\left(x\right)=x-\langle x,\alpha \rangle \alpha$ in the hyperplanes ${\alpha }^{\perp }$, for $\alpha \in \mathcal{R}$ (where we use ${\parallel \alpha \parallel }^2=2$ to simplify the standard formula). Concretely, $W\left(D_4\right)$ consists of all coordinate permutations of ${\mathbb{R}}^4$ combined with sign changes of an even number of coordinates:

$$|W\left(D_4\right)|=2^3·4!=192.$$

This is not the full hyperoctahedral group $W\left(B_4\right)$, which would allow any number of sign changes and has order $2^4·24=384$; only the even-parity sign changes appear in $W\left(D_4\right)$.

The group $W\left(D_4\right)$ acts on ${\mathbb{R}}^4$ preserving $\mathcal{R}$, $D_4$, and $\Lambda$. It acts transitively on the 24 roots, with each root having a stabiliser of order $192/24=8$. This transitivity is used in Lemma 6 below.

2.3. The Regular 24-Cell

The Voronoi cell of the origin in $\Lambda$ is

$$V_0=\left\{z\in {\mathbb{R}}^4:\parallel z\parallel \le \parallel z-\lambda \parallel \mathrm{for}\mathrm{all}\lambda \in \Lambda \right\}.$$

Points at distance $\parallel v\parallel \ge 2\sqrt{2}$ from the origin generate half-spaces that already contain the ball of radius $\sqrt{2}$ enclosing $V_0$, so it suffices to intersect over the 24 minimal vectors $v\in \sqrt{2}\mathcal{R}$:

$$P_{24}:=V_0=\bigcap _{v\in \sqrt{2}\mathcal{R}}\left\{z\in {\mathbb{R}}^4:\langle z,v\rangle \le 2\right\}.$$

(1)

The resulting polytope $P_{24}$ is the regular 24-cell: a four-dimensional self-dual convex polytope with 24 vertices, 96 edges, 96 square two-faces, and 24 octahedral three-faces, with its 24 facets in bijection with the 24 minimal vectors.

Lemma 3 

(Vertices of $P_{24}$). The vertex set of $P_{24}$ is the following set of 24 points: the eight vectors $\pm \sqrt{2}{e}_i$ and the sixteen vectors $\frac{1}{\sqrt{2}}({\epsilon }_1,{\epsilon }_2,{\epsilon }_3,{\epsilon }_4)$ with ${\epsilon }_i\in \{\pm 1\}$.

Proof. 

Each vertex of $P_{24}$ is determined by exactly four linearly independent tight inequalities from (1). Take $v_1=\sqrt{2}(e_1+e_2)$, $v_2=\sqrt{2}(e_1+e_3)$, $v_3=\sqrt{2}(e_1+e_4)$, $v_4=\sqrt{2}(e_2+e_3)$. The four equations $\langle z,v_k\rangle =2$ give

$$z_1+z_2=\sqrt{2},z_1+z_3=\sqrt{2},z_1+z_4=\sqrt{2},z_2+z_3=\sqrt{2}.$$

From the first two equations $z_2=z_3$; from the first and third $z_2=z_4$; from the fourth $2z_2=\sqrt{2}$, so $z_2=1/\sqrt{2}$; then $z_1=\sqrt{2}-z_2=1/\sqrt{2}$. The vertex is $\frac{1}{\sqrt{2}}(1,1,1,1)$. The remaining 20 inequalities are satisfied: for example, $\langle z,\sqrt{2}(e_1-e_2)\rangle =\sqrt{2}(1/\sqrt{2}-1/\sqrt{2})=0\le 2$. Varying the signs of ${\epsilon }_i$ produces all 16 sign-vector vertices. For the coordinate vertices, the four equations associated with $\sqrt{2}(e_1\pm e_2)$ and $\sqrt{2}(e_1\pm e_3)$ (taking the appropriate signs) force $z_2=z_3=z_4=0$ and $z_1=\sqrt{2}$, giving $\sqrt{2}{e}_1$; permuting and changing signs gives all eight. A complete enumeration of active quadruples produces no further solutions.    □

Lemma 4 

(Volume of $P_{24}$). $vol\left(P_{24}\right)=8$.

Proof. 

By Lemma 2 the Voronoi cells of $\Lambda$ tile ${\mathbb{R}}^4$ with each cell having volume $covol(\Lambda )=8$. Since $P_{24}$ is the Voronoi cell of the origin, $vol\left(P_{24}\right)=8$. One can also compute directly. Decompose $P_{24}$ into 24 congruent pyramids, each with apex at the origin and base one of the 24 octahedral facets. For the facet $F_{\alpha }$ dual to $v=\sqrt{2}\alpha \in \sqrt{2}\mathcal{R}$, the supporting hyperplane $\langle z,v\rangle =2$ has unit normal $v/\parallel v\parallel =v/2$ and lies at distance 1 from the origin.

The following calculation shows that each facet is a regular octahedron of edge length $\sqrt{2}$. Consider the facet $F_{e_1+e_2}$ defined by $\langle z,e_1+e_2\rangle =\sqrt{2}$. A vertex from Lemma 3 lies on this facet if and only if its inner product with $e_1+e_2$ equals $\sqrt{2}$. Among the coordinate vertices: $\langle \sqrt{2}{e}_1,e_1+e_2\rangle =\sqrt{2}$ and $\langle \sqrt{2}{e}_2,e_1+e_2\rangle =\sqrt{2}$, while all other coordinate vertices give values 0 or $-\sqrt{2}$. Among the sign vertices $\frac{1}{\sqrt{2}}({\epsilon }_1,{\epsilon }_2,{\epsilon }_3,{\epsilon }_4)$, the condition $\langle z,e_1+e_2\rangle =\sqrt{2}$ becomes $({\epsilon }_1+{\epsilon }_2)/\sqrt{2}·\sqrt{2}={\epsilon }_1+{\epsilon }_2=2$, hence ${\epsilon }_1={\epsilon }_2=1$, giving four vertices $\frac{1}{\sqrt{2}}(1,1,\pm 1,\pm 1)$. The six vertices of $F_{e_1+e_2}$ are therefore $\sqrt{2}{e}_1$, $\sqrt{2}{e}_2$, and $\frac{1}{\sqrt{2}}(1,1,\pm 1,\pm 1)$. To confirm that adjacent vertices differ by vectors of squared norm 2, note for example $\parallel \sqrt{2}{e}_1-\frac{1}{\sqrt{2}}{(1,1,1,1)\parallel }^2={\parallel ({\displaystyle \frac{1}{\sqrt{2}}},-{\displaystyle \frac{1}{\sqrt{2}}},-{\displaystyle \frac{1}{\sqrt{2}}},-{\displaystyle \frac{1}{\sqrt{2}}})\parallel }^2=2$ and $\parallel \frac{1}{\sqrt{2}}(1,1,1,1)-\frac{1}{\sqrt{2}}{(1,1,-1,1)\parallel }^2={\parallel (0,0,\sqrt{2},0)\parallel }^2=2$. By the Weyl-group symmetry the same edge-length holds for all pairs of adjacent vertices in any facet, confirming edge length $\sqrt{2}$ throughout.

A regular octahedron of edge length a has volume $(\sqrt{2}/3)a^3$; with $a=\sqrt{2}$ each facet has volume $(\sqrt{2}/3){\left(\sqrt{2}\right)}^3=4/3$. The four-dimensional pyramid formula gives each pyramid volume $\frac{1}{4}·1·\frac{4}{3}=\frac{1}{3}$. Summing over 24 facets gives $vol\left(P_{24}\right)=24·\frac{1}{3}=8$. (An independent simplex-by-simplex check appears in Appendix A).    □

2.4. The Support Function of $P_{24}$

Recall that the support function of a convex body $K\subset {\mathbb{R}}^4$ is $h_K\left(y\right)={max}_{z\in K}\langle z,y\rangle$; it records how far K extends in direction y [12,13].

Lemma 5 

(Support function of $P_{24}$). For $y\in {\mathbb{R}}^4$ with sorted absolute coordinates $a_1\ge a_2\ge a_3\ge a_4\ge 0$,

$$h_{P_{24}}\left(y\right)=max\left\{\sqrt{2}{a}_1,\frac{1}{\sqrt{2}}(a_1+a_2+a_3+a_4)\right\}.$$

(2)

Proof. 

Since $P_{24}=conv\left(\mathrm{vertex}\mathrm{set}\right)$ by Lemma 3, the support function is the maximum of $\langle v,y\rangle$ over the 24 vertices. For the eight vertices $\pm \sqrt{2}{e}_i$, the maximum of $\pm \sqrt{2}{y}_i$ over signs and indices equals $\sqrt{2}{a}_1$. For the sixteen vertices $\frac{1}{\sqrt{2}}({\epsilon }_1,\dots ,{\epsilon }_4)$, choosing ${\epsilon }_i=sgn\left(y_i\right)$ maximises $\langle v,y\rangle =\frac{1}{\sqrt{2}}{\sum }_i{\epsilon }_i{y}_i=\frac{1}{\sqrt{2}}{\sum }_i\left|y_i\right|=\frac{1}{\sqrt{2}}(a_1+a_2+a_3+a_4)$. The overall maximum is the larger of these two values.    □

Remark 4. 

The two terms in (2) reflect the two types of vertex. The coordinate vertices $\pm \sqrt{2}{e}_i$ dominate when one coordinate of y is much larger than the others (an ${\ell }^{\infty }$-type contribution), while the sign vertices $\frac{1}{\sqrt{2}}(\pm 1,\pm 1,\pm 1,\pm 1)$ dominate when all four coordinates are comparable (an ${\ell }^1$-type contribution).

2.5. Shell-Rigidity

Lemma 6 

(Shell rigidity). Suppose a unit-ball packing has a centre at the origin and also has centres at all 24 points $\sqrt{2}\alpha$, $\alpha \in \mathcal{R}$. Then the Voronoi cell of the origin is exactly $P_{24}$.

Proof. 

The 24 half-spaces $\{\langle z,v\rangle \le 2\}$ for $v\in \sqrt{2}\mathcal{R}$ are precisely the Voronoi half-spaces generated by the 24 root neighbours, so their intersection is $P_{24}$ by (1). It suffices to show that every additional packing centre $y\notin \{\sqrt{2}\alpha :\alpha \in \mathcal{R}\}$ generates a Voronoi half-space $H_y{=\{\langle z,y\rangle \le \parallel y\parallel }^2/2\}$ that already contains $P_{24}$.

We may apply any element of the Weyl group $W\left(D_4\right)$ to y without changing the configuration, since $W\left(D_4\right)$ preserves $\Lambda$ and $\mathcal{R}$. Thus we may assume the sorted absolute coordinates of y satisfy $a_1\ge a_2\ge a_3\ge a_4\ge 0$. Choose signs ${\epsilon }_i=sgn\left(y_i\right)$ and set $v=\sqrt{2}({\epsilon }_1{e}_1+{\epsilon }_2{e}_2)\in \sqrt{2}\mathcal{R}$. The packing constraint ${\parallel y-v\parallel }^2\ge 4$ expands as

$${\parallel y\parallel }^2-2\langle y,v\rangle +{\parallel v\parallel }^2\ge 4.$$

Since ${\parallel v\parallel }^2=2$ and $\langle y,v\rangle =\sqrt{2}({\epsilon }_1{y}_1+{\epsilon }_2{y}_2)=\sqrt{2}(a_1+a_2)$, this gives

$${\parallel y\parallel }^2-2\sqrt{2}(a_1+a_2)+2\ge 4,$$

hence $\sqrt{2}(a_1+a_2)\le {(\parallel y\parallel }^2-2)/2$. Since ${\parallel y\parallel }^2\ge 4$ (the packing constraint $\parallel y\parallel \ge 2$), we have ${(\parallel y\parallel }^2{-2)/2\le \parallel y\parallel }^2/2$, so

$$\sqrt{2}(a_1+a_2)\le {\displaystyle \frac{{\parallel y\parallel }^2}{2}}.$$

From Lemma 5, $h_{P_{24}}\left(y\right)=max\{\sqrt{2}{a}_1,\frac{1}{\sqrt{2}}(a_1+a_2+a_3+a_4)\}$. Since $a_1\le a_1+a_2$ and $a_1+a_2+a_3+a_4\le 2(a_1+a_2)$, both terms of $h_{P_{24}}\left(y\right)$ are at most $\sqrt{2}(a_1+a_2)\le {\parallel y\parallel }^2/2$. Therefore $h_{P_{24}}\left(y\right)\le {\parallel y\parallel }^2/2$, which by the definition of the support function means $P_{24}\subseteq H_y$. Since y was an arbitrary additional packing centre, every Voronoi half-space contains $P_{24}$, so $V_0\supseteq P_{24}$. The containment $V_0\subseteq P_{24}$ holds by definition of $P_{24}$ as the intersection over the 24 root half-spaces. Hence $V_0=P_{24}$.    □

3. Safe Directions and the Angular Deficiency

3.1. The Inscribed Ball

Lemma 7 

($P_{24}$ contains the unit ball). $B^4:=\{z\in {\mathbb{R}}^4:\parallel z\parallel \le 1\}\subseteq P_{24}$.

Proof. 

From (1), $P_{24}={\bigcap }_{\alpha \in \mathcal{R}}\{z:\langle z,\alpha /\sqrt{2}\rangle \le 1\}$. Each $\alpha /\sqrt{2}$ is a unit vector. Cauchy–Schwarz gives $\langle z,\alpha /\sqrt{2}\rangle \le \parallel z\parallel \le 1$ for $z\in B^4$.    □

3.2. Safe Directions

Proposition 1 

(Safe directions). For every unit vector $u\in {\mathbb{R}}^4$, $h_{P_{24}}\left(u\right)\ge 1$. Equality holds if and only if $u=\alpha /\sqrt{2}$ for some $\alpha \in \mathcal{R}$, equivalently if and only if, among the sorted absolute coordinates $a_1\ge a_2\ge a_3\ge a_4\ge 0$ of u, exactly two equal $1/\sqrt{2}$ and the other two equal 0.

Proof. 

Since $\parallel u\parallel =1$, Lemma 7 gives $u\in B^4\subseteq P_{24}$, so

$$h_{P_{24}}\left(u\right)=\underset{z\in P_{24}}{max}\langle z,u\rangle \ge \langle u,u\rangle =1.$$

For the equality characterisation, write $a_i=\left|u_i\right|$ sorted in decreasing order, and $S=a_1+a_2+a_3+a_4$. By (2), $h_{P_{24}}\left(u\right)=1$ requires both $\sqrt{2}{a}_1\le 1$ and $S/\sqrt{2}\le 1$, i.e., $a_1\le 1/\sqrt{2}$ and $S\le \sqrt{2}$. Under these constraints, for each i we have $a_i\le 1/\sqrt{2}$, so $a_i^2\le a_i/\sqrt{2}$ (since $t^2\le t/\sqrt{2}$ for $t\in [0,1/\sqrt{2}]$, with equality only at $t=0$ or $t=1/\sqrt{2}$). Summing:

$$1=\sum _i{a}_i^2\le {\displaystyle \frac{S}{\sqrt{2}}}\le 1.$$

Both sides must be equal, forcing $a_i^2=a_i/\sqrt{2}$ for each i, hence $a_i\in \{0,1/\sqrt{2}\}$. With $\sum a_i^2=1$, exactly two coordinates equal $1/\sqrt{2}$ and the other two equal 0. A unit vector with that pattern has the form $(\pm e_i\pm e_j)/\sqrt{2}=\alpha /\sqrt{2}$ for some $\alpha \in \mathcal{R}$. The converse is immediate: if $a_1=a_2=1/\sqrt{2}$ and $a_3=a_4=0$, then $h_{P_{24}}\left(u\right)=max\{1,1\}=1$.    □

Remark 5 

(Maximum of $h_{P_{24}}$). By the same formula, $h_{P_{24}}\left(u\right)\le \sqrt{2}$ for every unit u. Equality holds in two families: the 8 coordinate directions $\pm e_i$ (where $\sqrt{2}{a}_1=\sqrt{2}$) and the 16 directions $\frac{1}{2}(\pm 1,\pm 1,\pm 1,\pm 1)$ (where $S/\sqrt{2}=\sqrt{2}$, with Cauchy–Schwarz being tight). So $h_{P_{24}}$ maps $S^3$ into $[1,\sqrt{2}]$, with minimum exactly on the 24 root directions and maximum on these further 24 directions, which are the normalised directions of the two types of vertex of $P_{24}$.

3.3. The Angular Deficiency

Definition 1 

(Angular deficiency). For $u\in S^3$, the angular deficiency is $\delta \left(u\right):=h_{P_{24}}\left(u\right)-1\in [0,\sqrt{2}-1]$. A direction u is called root-aligned if $\delta \left(u\right)=0$, equivalently if $u=\alpha /\sqrt{2}$ for some $\alpha \in \mathcal{R}$.

Proposition 2 

(Safe-neighbour criterion). Let $y=ru$ with $r>0$ and $u\in S^3$. The half-space $H_y{:=\{z:\langle z,y\rangle \le \parallel y\parallel }^2/2\}$ satisfies

$$P_{24}\subseteq H_y⟺r\ge 2h_{P_{24}}\left(u\right)=2(1+\delta \left(u\right)).$$

Proof. 

Since $r>0$, the condition $\langle z,y\rangle \le {\parallel y\parallel }^2/2$ is equivalent to $\langle z,u\rangle \le r/2$. Thus $P_{24}\subseteq H_y$ iff $h_{P_{24}}\left(u\right)\le r/2$.    □

For a neighbour in the annular shell $\mathcal{A}=\{2\le \parallel y\parallel <2\sqrt{2}\}$ (whose relevance is established in the next section), $r\in [2,2\sqrt{2})$. Proposition 1 gives $2h_{P_{24}}\left(u\right)=2(1+\delta \left(u\right))\ge 2$, with equality exactly for root-aligned u. Therefore:

• A root-aligned neighbour satisfies $2h_{P_{24}}\left(u\right)=2\le r$ for every r in the shell, so $P_{24}\subseteq H_y$ throughout the shell. A root-aligned neighbour never cuts $P_{24}$, at any distance within $\mathcal{A}$.
• A neighbour in a direction u with $\delta \left(u\right)>0$ satisfies $P_{24}\subseteq H_y$ only when $r\ge 2(1+\delta (u\left)\right)>2$. For $r<2(1+\delta (u\left)\right)$ the half-space $H_y$ cuts a non-empty cap off $P_{24}$.

3.4. Exact Support-Function Values Used Later

The formula (2) gives the following exact values on the directions that appear in the comparison and boundary calculations:

The first row is the only row with value $h_{P_{24}}\left(u\right)=1$. It is the root row. The second and third rows show the two ways in which the support reaches $\sqrt{2}$: at coordinate directions and at sign-vector directions. Thus the active shell $2\le r<2\sqrt{2}$ is exactly the interval in which a non-root direction can potentially cut the comparison cell.

4. Shell Localisation

Lemma 8 

(Shell localisation). If $\parallel y\parallel \ge 2\sqrt{2}$, then $P_{24}\subseteq H_y$, so the packing centre at y does not reduce the Voronoi cell of the origin below $P_{24}$.

Proof. 

We need $h_{P_{24}}\left(y\right)\le {\parallel y\parallel }^2/2$. By Cauchy–Schwarz, $a_1={max}_i|y_i|\le \parallel y\parallel$, giving $\sqrt{2}{a}_1\le \sqrt{2}\parallel y\parallel$. Also $a_1+a_2+a_3+a_4\le \sqrt{4}\parallel y\parallel =2\parallel y\parallel$ by Cauchy–Schwarz in ${\mathbb{R}}^4$, giving $\frac{1}{\sqrt{2}}(a_1+a_2+a_3+a_4)\le \sqrt{2}\parallel y\parallel$. Thus $h_{P_{24}}\left(y\right)\le \sqrt{2}\parallel y\parallel$. The inequality $\sqrt{2}\parallel y\parallel \le {\parallel y\parallel }^2/2$ reduces to $2\sqrt{2}\le \parallel y\parallel$, which holds by assumption.    □

It follows that only packing centres in the closed annular shell

$$\mathcal{A}=\left\{y\in {\mathbb{R}}^4:2\le \parallel y\parallel <2\sqrt{2}\right\}$$

(3)

can reduce the Voronoi volume below 8. The lower bound comes from the packing constraint $\parallel y\parallel \ge 2$; the upper bound is Lemma 8. For an active centre $y_i\in \mathcal{A}$ write $y_i=r_i{u}_i$ with $r_i\in [2,2\sqrt{2})$ and $u_i\in S^3$. The Gram inner products between direction vectors of active centres are

$$s_{ij}:=\langle u_i,u_j\rangle ,1\le i<j\le N,$$

(4)

and the packing constraint between two active centres reads

$$p_{ij}:=r_i^2+r_j^2-2r_i{r}_j{s}_{ij}-4\ge 0.$$

(5)

For two distinct roots $\alpha ,\beta \in \mathcal{R}$ with $\alpha \ne -\beta$, $\langle \alpha ,\beta \rangle \in \{0,\pm 1\}$, so $s_{ij}\le 1/2$. The 24 scaled roots are pairwise packing-compatible at distance 2, as they must be since they are exactly the nearest neighbours of the origin in $\sqrt{2}{D}_4$.

Proposition 3 

(Compact active parameter set). After shell localisation every active datum can be written as

$$(r_1,\dots ,r_N;s_{ij})\in {[2,2\sqrt{2}]}^N\times {[-1,1]}^{\left({\displaystyle \genfrac{}{}{0pt}{}{N}{2}}\right)},$$

subject to the Gram positive-semidefinite constraints and the packing inequalities $p_{ij}\ge 0$. On a chamber interior the Gram subdeterminants used to define vertices are strictly positive, and each boundary point is obtained by imposing at least one extra equality among determinant, ordering, shell, or packing factors.

Proof. 

The interval bounds for the radii are exactly (3); replacing the open upper endpoint by a closed endpoint only adds a boundary face, and Lemma 8 says that face cannot lower volume below the comparison cell. Since $u_i\in S^3$, every Gram coordinate satisfies $|s_{ij}|\le 1$, and the full Gram matrix $\left(\langle u_i,u_j\rangle \right)$ is positive semidefinite of rank at most 4. The packing condition between $r_i{u}_i$ and $r_j{u}_j$ is precisely (5). A chamber records which quadruples of supporting hyperplanes form vertices and which inequalities order the same rational functions. Inside such a chamber the relevant determinants are non-zero; after choosing the orientation they are positive. If one vanishes, or if an ordering, shell, or packing factor vanishes, the point lies on a lower-dimensional face recorded by the same chamber system.    □

Lemma 9 

(Pairwise angular bound from packing). For two active centres $y_i=r_i{u}_i$ and $y_j=r_j{u}_j$,

$$s_{ij}\le {\displaystyle \frac{r_i^2+r_j^2-4}{2r_i{r}_j}}.$$

In particular, if $r_i=r_j=2$, then $s_{ij}\le 1/2$.

Proof. 

The packing inequality $\parallel r_i{u}_i-r_j{u}_j{\parallel }^2\ge 4$ expands to

$$r_i^2+r_j^2-2r_i{r}_j{s}_{ij}\ge 4.$$

Solving for $s_{ij}$ gives the displayed inequality. Setting $r_i=r_j=2$ gives $s_{ij}\le (4+4-4)/8=1/2$.    □

5. Root-Aligned Configurations

5.1. The Root-Restricted Volume Theorem

For $\alpha \in \mathcal{R}$ write $u_{\alpha }:=\alpha /\sqrt{2}\in S^3$ and $H_{\alpha }\left(r\right):=\{z\in {\mathbb{R}}^4:\langle z,u_{\alpha }\rangle \le r/2\}$. Note that $H_{\alpha }\left(r\right)\subseteq H_{\alpha }\left(r^{\prime }\right)$ for $r\le r^{\prime }$, and ${\bigcap }_{\alpha \in \mathcal{R}}{H}_{\alpha }\left(2\right)=P_{24}$ by (1).

Theorem 2 

(Root-restricted volume). Let $S\subseteq \mathcal{R}$ be non-empty and ${\left(r_{\alpha }\right)}_{\alpha \in S}\in {[2,2\sqrt{2})}^S$. Set $V_S:={\bigcap }_{\alpha \in S}{H}_{\alpha }\left(r_{\alpha }\right)$. Then $V_S\supseteq P_{24}$, hence $vol\left(V_S\right)\ge 8$. Moreover $vol\left(V_S\right)=8$ if and only if $S=\mathcal{R}$ and $r_{\alpha }=2$ for every $\alpha \in S$.

Proof. 

By Proposition 1, $h_{P_{24}}\left(u_{\alpha }\right)=1$ for every $\alpha \in \mathcal{R}$, so Proposition 2 gives $P_{24}\subseteq H_{\alpha }\left(r\right)$ for every $r\ge 2$. In particular $P_{24}\subseteq H_{\alpha }\left(r_{\alpha }\right)$ for each $\alpha \in S$, hence $P_{24}\subseteq V_S$ and $vol\left(V_S\right)\ge vol\left(P_{24}\right)=8$. If $S=\mathcal{R}$ and every $r_{\alpha }=2$, then $V_S={\bigcap }_{\alpha \in \mathcal{R}}{H}_{\alpha }\left(2\right)=P_{24}$, so $vol\left(V_S\right)=8$. For the converse, suppose $(S,r)$ is not the full data $(S=\mathcal{R}$, all $r_{\alpha }=2)$. We show $vol\left(V_S\right)>8$.

Since $r_{\alpha }\ge 2$ for all $\alpha \in S$, we have $H_{\alpha }\left(2\right)\subseteq H_{\alpha }\left(r_{\alpha }\right)$, so $V_S\supseteq {\bigcap }_{\alpha \in S}{H}_{\alpha }\left(2\right)$.

Case 1: $S\ne \mathcal{R}$. Pick any ${\alpha }_0\in \mathcal{R}\setminus S$. Then $V_S\supseteq {\bigcap }_{\alpha \ne {\alpha }_0}{H}_{\alpha }\left(2\right)$.

Case 2: $S=\mathcal{R}$ but $r_{{\alpha }_0}>2$ for some ${\alpha }_0$. Then $H_{{\alpha }_0}\left(r_{{\alpha }_0}\right)⊋H_{{\alpha }_0}\left(2\right)$, so $V_S=H_{{\alpha }_0}\left(r_{{\alpha }_0}\right)\cap {\bigcap }_{\alpha \ne {\alpha }_0}{H}_{\alpha }\left(2\right)⊋{\bigcap }_{\alpha \in \mathcal{R}}{H}_{\alpha }\left(2\right)=P_{24}$ strictly along the ${\alpha }_0$ facet.

In either case the facet $F_{{\alpha }_0}:=P_{24}\cap \partial H_{{\alpha }_0}\left(2\right)$ is a regular octahedron of positive 3-dimensional volume $4/3$. Every interior point of $F_{{\alpha }_0}$ satisfies every inequality $\alpha \ne {\alpha }_0$ strictly. Moving such a point a small distance in the direction $u_{{\alpha }_0}$ (increasing $\langle z,u_{{\alpha }_0}\rangle$ slightly past 1, while keeping $\langle z,u_{\alpha }\rangle <r_{\alpha }/2$ for all other $\alpha$) produces a non-empty open set $W\subset V_S$ with $W\cap P_{24}=\varnothing$ and $vol\left(W\right)>0$. Hence $vol\left(V_S\right)\ge vol\left(P_{24}\right)+vol\left(W\right)>8$.    □

Remark 6. 

Theorem 2 is a purely geometric statement about polytopes defined by subsets of the 24 facet hyperplanes of $P_{24}$: among all such polytopes, $P_{24}$ is the unique volume minimiser.

5.2. Consequences for Packings

Corollary 1 

(Root-aligned lower bound). Let $\mathcal{P}$ be a unit-ball packing with a centre at the origin and Voronoi cell V. If every shell-active neighbour of the origin lies in a root direction, meaning every $y\in \mathcal{P}\cap \mathcal{A}$ has the form $y=r{u}_{\alpha }$ for some $\alpha \in \mathcal{R}$ and $r\in [2,2\sqrt{2})$, then $vol\left(V\right)\ge 8$.

Proof. 

Let A be the set of shell-active neighbours, and for each $\alpha \in \mathcal{R}$ let $r_{\alpha }$ be the smallest r such that $r{u}_{\alpha }\in A$ (if any; otherwise omit $\alpha$ from S). By Theorem 2, $V_S={\bigcap }_{\alpha \in S}{H}_{\alpha }\left(r_{\alpha }\right)\supseteq P_{24}$. Every half-space $H_y$ for $y\in A$ contains $V_S$ (repeated direction uses the most restrictive constraint), and every half-space from outside the shell $\mathcal{A}$ contains $P_{24}$ by Lemma 8. Hence $V\supseteq P_{24}$ and $vol\left(V\right)\ge 8$.    □

Corollary 2 

(General safe-neighbour bound). With V and $A=\mathcal{P}\cap \mathcal{A}$ as above, suppose every $y=ru\in A$ satisfies $r\ge 2h_{P_{24}}\left(u\right)=2(1+\delta \left(u\right))$. Then $vol\left(V\right)\ge 8$.

Proof. 

By Proposition 2, $P_{24}\subseteq H_y$ for each such $y\in A$. Together with Lemma 8 for neighbours outside $\mathcal{A}$, the same intersection argument gives $V\supseteq P_{24}$.    □

6. Voronoi Volume as a Function of Gram Data

6.1. Vertices via Cramer’s Rule

The Voronoi cell of the origin is $V=\{z\in {\mathbb{R}}^4:\langle z,u_i\rangle \le r_i/2\mathrm{for}\mathrm{all}i\in A\}$ where we label the active neighbours $y_i=r_i{u}_i$. Each vertex of V is determined by exactly four active constraints holding with equality. For a quadruple $I=(i_1,i_2,i_3,i_4)$ let $U_I$ be the $4\times 4$ matrix with rows $u_{i_k}^T$, and set

$$G_I:=U_I{U}_I^T,{\delta }_I:=det{G}_I.$$

(6)

When ${\delta }_I\ne 0$, the unique vertex where constraints $i_1,\dots ,i_4$ are tight is

$$z_I=U_I^{-1}\frac{1}{2}{r}_I={\displaystyle \frac{1}{2{\delta }_I}}{U}_I^{T}adj\left(G_I\right)r_I,$$

(7)

using $U_I^{-1}=U_I^T{G}_I^{-1}$ and $G_I^{-1}=adj\left(G_I\right)/{\delta }_I$.

Lemma 10 

(Vertex coordinate formula). Each coordinate ${\left(z_I\right)}_k$ is a rational function of the Gram entries $s_{i_p{i}_q}=\langle u_{i_p},u_{i_q}\rangle$ and the radial variables $r_{i_k}$, with denominator ${\delta }_I$. Writing $adj\left(G_I\right)=\left({\gamma }^{pq}\right)$,

$${\left(z_I\right)}_k={\displaystyle \frac{1}{2{\delta }_I}}{\displaystyle \sum _{p=1}^4}{\left(u_{i_p}\right)}_k{\displaystyle \sum _{q=1}^4}{\gamma }^{pq}{r}_{i_q},k=1,2,3,4.$$

The cofactors ${\gamma }^{pq}$ are signed $3\times 3$ minors of $G_I$, hence polynomials of degree 3 in the Gram entries.

Proof. 

Direct expansion of (7) using $adj{\left(G_I\right)}_{pq}={(-1)}^{p+q}{M}_{qp}$.    □

Lemma 11 

(Squared norm of a Cramer vertex). For every non-degenerate quadruple I, the squared norm of the vertex $z_I$ is

$$\parallel z_I{\parallel }^2={\displaystyle \frac{1}{4}}{r}_I^T{G}_I^{-1}{r}_I={\displaystyle \frac{1}{4{\delta }_I}}{r}_I^{T}adj\left(G_I\right)r_I.$$

Moreover the inequality that a fifth half-space $j\notin I$ contains this vertex is a polynomial inequality after multiplication by $2{\delta }_I$:

$$\langle z_I,u_j\rangle \le {\displaystyle \frac{r_j}{2}}⟺{\displaystyle \sum _{p,q=1}^4}{s}_{j{i}_p}adj{\left(G_I\right)}_{pq}{r}_{i_q}\le r_j{\delta }_I,$$

with the direction of the inequality unchanged on a chamber interior because ${\delta }_I>0$.

Proof. 

The first identity follows from $z_I={\displaystyle \frac{1}{2}}{U}_I^T{G}_I^{-1}{r}_I$:

$$\parallel z_I{\parallel }^2={\displaystyle \frac{1}{4}}{r}_I^T{G}_I^{-1}{U}_I{U}_I^T{G}_I^{-1}{r}_I={\displaystyle \frac{1}{4}}{r}_I^T{G}_I^{-1}{G}_I{G}_I^{-1}{r}_I={\displaystyle \frac{1}{4}}{r}_I^T{G}_I^{-1}{r}_I.$$

Replacing $G_I^{-1}$ by $adj\left(G_I\right)/{\delta }_I$ gives the second form. For the incidence inequality,

$$\langle z_I,u_j\rangle ={\displaystyle \frac{1}{2{\delta }_I}}{\displaystyle \sum _{p,q=1}^4}\langle u_j,u_{i_p}\rangle adj{\left(G_I\right)}_{pq}{r}_{i_q},$$

and multiplying by $2{\delta }_I$ gives the stated polynomial inequality.    □

Proposition 4 

(Algebraic form of all cell incidences). On each chamber, the statements “quadruple I gives a vertex”, “quadruple I does not give a vertex”, and “two adjacent vertices span an edge” are all expressible by polynomial equalities and inequalities in the variables $s_{ij}$, $r_i$, ${\delta }_I$, and the Cramer multipliers.

Proof. 

The equation defining a candidate vertex is the Cramer equation $U_I{z}_I=r_I/2$, written in the rational form (7). Lemma 11 turns every additional containment inequality $\langle z_I,u_j\rangle \le r_j/2$ into a polynomial inequality after multiplication by the positive chamber determinant ${\delta }_I$. A candidate is rejected exactly when one of these inequalities has the wrong sign. An edge between two vertices occurs when their active quadruples share three labels and the segment joining the two vertices satisfies all remaining half-space inequalities; again this is a finite list of polynomial inequalities after clearing positive determinants. Thus the chamber combinatorics is algebraic over $\mathbb{Q}\left(\sqrt{2}\right)$.    □

6.2. The Volume Formula

Triangulate V with apex at the origin. For each simplex $\sigma =(0,z_{I_1},z_{I_2},z_{I_3},z_{I_4})$,

$$vol\sigma ={\displaystyle \frac{1}{24}}det(z_{I_1},z_{I_2},z_{I_3},z_{I_4})={\displaystyle \frac{{\Theta }_{\sigma }}{384{\delta }_{I_1}{\delta }_{I_2}{\delta }_{I_3}{\delta }_{I_4}}},$$

(8)

where ${\Theta }_{\sigma }$ is the $4\times 4$ determinant of the matrix with columns $U_{I_k}^{T}adj\left(G_{I_k}\right)r_{I_k}$. Summing over the triangulation and writing $D_{\Omega }$ for the common denominator of all simplex volumes:

$$volV-8={\displaystyle \frac{F_{\Omega }}{D_{\Omega }}},$$

(9)

where $F_{\Omega }$ and $D_{\Omega }$ are explicit polynomials in the Gram variables $s_{ij}$ and radial distances $r_i$.

Proposition 5 

(Cleared numerator). Fix a chamber Ω with oriented simplex set ${\mathcal{T}}_{\Omega }$. Let

$$D_{\Omega }=384\prod _{I\in {\mathcal{Q}}_{\Omega }}{\delta }_I^{m_I}$$

be a common positive denominator, where ${\mathcal{Q}}_{\Omega }$ is the list of vertex quadruples and $m_I$ is the largest exponent with which ${\delta }_I$ occurs in the simplex sum. Then

$$F_{\Omega }=\sum _{\sigma =(I_1,I_2,I_3,I_4)\in {\mathcal{T}}_{\Omega }}{ϵ}_{\sigma }{\Theta }_{\sigma }\prod _{I\in {\mathcal{Q}}_{\Omega }}{\delta }_I^{m_I-{\#}_{\sigma }\left(I\right)}-8D_{\Omega },$$

where ${ϵ}_{\sigma }\in \{\pm 1\}$ is the orientation sign and ${\#}_{\sigma }\left(I\right)$ is the number of times I occurs among $I_1,I_2,I_3,I_4$. This polynomial is the numerator in (9).

Proof. 

Equation (8) gives each oriented simplex as

$${\displaystyle \frac{{ϵ}_{\sigma }{\Theta }_{\sigma }}{384{\delta }_{I_1}{\delta }_{I_2}{\delta }_{I_3}{\delta }_{I_4}}}.$$

Multiplying the whole simplex sum by the common denominator $D_{\Omega }$ gives exactly the displayed polynomial. Subtracting $8D_{\Omega }$ records $volV-8$ rather than $volV$. Since the chamber orientation fixes all signs ${ϵ}_{\sigma }$, no absolute value remains in the algebraic expression.    □

Remark 7 

(Finiteness of chamber types). The parameter space of active configurations decomposes into finitely many combinatorial chambers, since each chamber type is determined by specifying which quadruples $I=(i_1,i_2,i_3,i_4)$ of active constraints are simultaneously tight at vertices of V. The number of such quadruples is bounded by $\left({\displaystyle \genfrac{}{}{0pt}{}{N}{4}}\right)$ where N is the number of active neighbours; N is itself bounded since all active centres lie in the shell $\mathcal{A}$, and the packing constraint $p_{ij}\ge 0$ gives a global upper bound on N in terms of the kissing number bound in ${\mathbb{R}}^4$ (which is 24 by Musin [7]). Thus the total number of chamber types is finite, and the argument of Section 8 applies to each one.

6.3. The 176 Chamber Register

The finite chamber register used in the chamber part of the argument has 176 representative rows. It is organised into 22 cycles of 8 Weyl types. In each cycle the types are

$$A,B,C,D,E,F,G,H,$$

with denominator patterns and orbit sizes

Thus each cycle has orbit weight

$$192+96+64+48+192+96+64+48=800,$$

and the full register has

$$22·8=176$$

representative chamber rows, with total Weyl orbit weight

$$22·800=17600.$$

The expanded register is given in Appendix B and in the accompanying auxiliary files:

$$\begin{array}{c}\mathtt{merged}\_\mathtt{176}\_\mathtt{chamber}\_\mathtt{register}.\mathtt{csv},\mathtt{merged}\_\mathtt{176}\_\mathtt{chamber}\_\mathtt{register}.\mathtt{json},\\ \mathtt{doi}\_\mathtt{chamber}\_\mathtt{register}.\mathtt{csv},\mathtt{chambers}/\mathtt{C}\mathtt{001}.\mathtt{json},\dots ,\mathtt{chambers}/\mathtt{C}\mathtt{176}.\mathtt{json}.\end{array}$$

The printed register uses the DOI-version anchor and $\kappa$-columns together with the supplementary active-facet and normal-form columns, so the finite chamber indexing is the same in the main text and in the auxiliary files.

For a chamber $\Omega =C_m$, the rational volume identity has the form

$$volV-8={\displaystyle \frac{F_{\Omega }}{D_{\Omega }}},$$

where $D_{\Omega }$ is one of the four denominator patterns above and is positive on the chamber interior. The chamber check attached to the register is the normal-form identity

$${NF}_{{\mathfrak{i}}_{\Omega }}\left(F_{\Omega }-\sum _{\nu }{Q}_{\Omega ,\nu }^2{M}_{\Omega ,\nu }\right)=0,$$

with $M_{\Omega ,\nu }\ge 0$ on the chamber. Boundary chambers are obtained by deleting vanished determinant or ordering factors and applying the same reduced normal-form identity.

6.4. Denominator Factors and Chamber Inequalities

The four denominator patterns in the register are products of Gram determinants and ordering factors. The determinant factor is always of the form

$${\Delta }_{\Omega }=\prod _{I\in {\mathcal{Q}}_{\Omega }}{\delta }_I,$$

where ${\mathcal{Q}}_{\Omega }$ is the list of vertex-defining quadruples in the chamber. The chamber inequalities are recorded in the auxiliary row file as

$${\delta }_I>0,r_i-2\ge 0,2\sqrt{2}-r_i>0,p_{ij}=r_i^2+r_j^2-2r_i{r}_j{s}_{ij}-4\ge 0.$$

They are supplemented by the ordering factors which keep the same vertex list throughout the chamber. In the notation of the register,

$$\begin{array}{cc}\hfill {\mathsf{D}}_a& ={\Delta }_{\Omega }{L}_1{L}_2{L}_3,{\mathsf{D}}_b={\Delta }_{\Omega }{L}_1{L}_2{Q}_1,\hfill \\ \hfill {\mathsf{D}}_c& ={\Delta }_{\Omega }{L}_1{Q}_1{Q}_2,{\mathsf{D}}_d={\Delta }_{\Omega }{Q}_1{Q}_2{Q}_3,\hfill \end{array}$$

where the $L_j$ are linear ordering factors and the $Q_j$ are quadratic packing factors. On the chamber interior each listed factor is positive, so the sign of $volV-8$ is the sign of $F_{\Omega }$. This is the reason the register only needs the numerator reductions.

6.5. Normal-Form Reduction Used in Every Chamber

Let

$$R_{\Omega }=\mathbb{Q}[s_{ij},r_i,{\delta }_I,{\lambda }_I:1\le i<j\le N,I\in {\mathcal{Q}}_{\Omega }]$$

be the polynomial ring attached to a chamber. The chamber ideal ${\mathfrak{i}}_{\Omega }$ is generated by the Gram relations, the vertex equations coming from Cramer’s rule, and the incidence equations stating which four-tuples define vertices. For every $I=(i_1,i_2,i_3,i_4)$ in ${\mathcal{Q}}_{\Omega }$, the vertex equation is written without denominators as

$$2{\delta }_I{z}_I-U_I^{T}adj\left(G_I\right)r_I=0.$$

After triangulation, all simplex determinants are placed over the common positive denominator $D_{\Omega }$. The numerator is then reduced modulo ${\mathfrak{i}}_{\Omega }$. The row file records the reduced identity in the form

$$F_{\Omega }=\sum _a{A}_{\Omega ,a}^2{P}_{\Omega ,a}+\sum _b{B}_{\Omega ,b}^2{L}_{\Omega ,b}+\sum _c{C}_{\Omega ,c}^2{Q}_{\Omega ,c}+\sum _d{U}_{\Omega ,d}{G}_{\Omega ,d},$$

where $P_{\Omega ,a}$, $L_{\Omega ,b}$, and $Q_{\Omega ,c}$ are, respectively, determinant, ordering, and packing factors which are non-negative on the chamber, while $G_{\Omega ,d}\in {\mathfrak{i}}_{\Omega }$. Reducing the right hand side modulo the same ideal gives the normal form of $F_{\Omega }$. Thus the sign calculation is local, algebraic, and finite.

A useful check is the reference root chamber. In that chamber all active radii equal 2, every active normal is a normalised root, and each determinant is one of

$$1,{\displaystyle \frac{3}{4}},frac12,frac14,$$

according to the number of shared coordinates among the four root normals. The vertex formula then returns exactly the twenty-four vertices of $P_{24}$, and the simplex determinant calculation in Appendix A gives

$$96·{\displaystyle \frac{1}{12}}=8.$$

The chamber numerator therefore vanishes at the reference point, as it must, and the row identities record how the numerator moves away from this reference point inside each chamber.

6.6. Boundary Strata

The chamber interiors cover the non-degenerate part of the parameter space. Boundary strata arise when one of the determinant, ordering, or packing factors is zero. If a determinant factor ${\delta }_I$ is zero, the corresponding four constraints no longer define a vertex and the row is replaced by the lower-dimensional row obtained by deleting I from ${\mathcal{Q}}_{\Omega }$. If an ordering factor vanishes, two adjacent chambers share the same boundary value of the volume numerator. If a packing factor $p_{ij}$ vanishes, two active neighbours are at mutual distance exactly 2, and the boundary expression is obtained by adding $p_{ij}=0$ to the chamber ideal. In each case the normal form is the specialization of the same row identity, so no separate limiting argument is needed.

6.7. An Explicit Vertex Computation

Take ${\mathcal{A}}_1=\{e_1+e_2,e_1+e_3,e_1+e_4,e_2+e_3,e_2+e_4,e_3+e_4\}$ and the quadruple $I=({\alpha }_{12},{\alpha }_{13},{\alpha }_{23},{\alpha }_{34})$ where ${\alpha }_{ij}=e_i+e_j$. The unit vectors are $u_{ij}=\frac{1}{\sqrt{2}}({\delta }_{1i}+{\delta }_{1j},{\delta }_{2i}+{\delta }_{2j},{\delta }_{3i}+{\delta }_{3j},{\delta }_{4i}+{\delta }_{4j})$ (where ${\delta }_{ki}$ is the Kronecker symbol), giving

$$G_{1234}=\left(\begin{array}{cccc}1& 1/2& 1/2& 0\\ 1/2& 1& 1/2& 1/2\\ 1/2& 1/2& 1& 1/2\\ 0& 1/2& 1/2& 1\end{array}\right).$$

Off-diagonal entry $\langle u_{ij},u_{kl}\rangle$ equals $1/2$ if the index sets share an element, and 0 if they are disjoint. Expanding along the first row gives $det{G}_{1234}=\frac{1}{2}-\frac{1}{8}-\frac{1}{8}=\frac{1}{4}$. At the reference point $r_k=2$ for all k, the adjugate formula gives $adj\left(G_{1234}\right){(2,2,2,2)}^T={(\frac{1}{2},0,0,\frac{1}{2})}^T$, and then $z_{1234}{|}_{r=2}=2·(\frac{1}{2}{u}_{12}+\frac{1}{2}{u}_{34})=u_{12}+u_{34}=\frac{1}{\sqrt{2}}(1,1,1,1)$, which is indeed a vertex of $P_{24}$ from Lemma 3. This confirms that the formula (7) recovers the known geometry of $P_{24}$ at the $D_4$ reference point.

7. Gram-Spectral Chamber Positivity

This section records the algebraic layer used for the non-root-aligned part of the argument. The preceding sections reduce the geometric problem to a finite family of Voronoi cells with outward normals in the active shell. Here the volume defect on each cell type is written as a rational function in Gram and radial variables, and the numerator is represented as a sum of squares modulo the chamber equations.

7.1. Chambers and Their Polynomial Rings

Fix one combinatorial chamber $\Omega$ from the 176-row Weyl atlas. Its active normals are denoted

$$u_1,\dots ,u_N\in S^3,y_i=r_i{u}_i,2\le r_i<2\sqrt{2},$$

and its vertex-defining quadruples are collected in ${\mathcal{Q}}_{\Omega }$. The Gram variables are

$$s_{ij}=\langle u_i,u_j\rangle ,1\le i<j\le N,$$

with $s_{ii}=1$. For $I=(i_1,i_2,i_3,i_4)\in {\mathcal{Q}}_{\Omega }$, let

$$G_I={\left(s_{i_a{i}_b}\right)}_{a,b=1}^4,{\delta }_I=det{G}_I.$$

We work in the polynomial ring

$$R_{\Omega }=\mathbb{Q}\left(\sqrt{2}\right)[s_{ij},r_i,{\delta }_I,z_{I,k},{\lambda }_J],$$

where the $z_{I,k}$ are vertex coordinates and the ${\lambda }_J$ record the linear inequalities used to keep the same face incidence throughout the chamber. Clearing denominators in Cramer’s rule gives the vertex relations

$$2{\delta }_I{z}_I-U_I^{T}adj\left(G_I\right)r_I=0,I\in {\mathcal{Q}}_{\Omega }.$$

(10)

The chamber ideal ${\mathfrak{i}}_{\Omega }\subset R_{\Omega }$ is generated by (10), the Gram determinant identities, the incidence equations for the chosen triangulation, and the row equations fixing the denominator pattern of $\Omega$. In the atlas files this ideal is recorded through its normal-form template; the row-wise JSON file lists the active facets, the 15 vertex quadruples, and the denominator pattern for that row.

Lemma 12 

(Radical chamber ideal). For every row $\Omega =C001,\dots ,C176$, the ideal ${\mathfrak{i}}_{\Omega }$ is radical on the open chamber where all listed determinant and ordering factors are non-zero.

Proof. 

The generators split into three triangular blocks. First, the Gram block is the determinantal coordinate ring of the ordered normal frame after the chosen Weyl normalisation. On the open set where the listed ${\delta }_I$ do not vanish, this block is smooth. Second, the Cramer block (10) is linear in the vertex coordinates $z_{I,k}$ once the Gram block and the non-zero determinant factors are fixed; hence it adds a free graph over the smooth Gram chart. Third, the incidence and ordering equations are square-free linear factors or square-free packing factors in the local coordinates chosen by the row. Localising at their product gives a reduced coordinate ring. Intersecting back with the row ring gives the stated radicality on the chamber interior. Equivalently, the Groebner basis used in the row reduction has square-free leading monomial set after the same localisation.    □

7.2. The Numerator Identity

The simplex formula (8) gives

$$volV-8={\displaystyle \frac{F_{\Omega }}{D_{\Omega }}}.$$

The denominator is one of the four patterns ${\mathsf{D}}_a,{\mathsf{D}}_b,{\mathsf{D}}_c,{\mathsf{D}}_d$. It is a product of positive Gram determinants, positive ordering factors, and non-negative packing factors not used as square multipliers. Thus the sign of the volume defect on the chamber interior is the sign of $F_{\Omega }$.

The Gram-spectral square decomposition used row by row has the form

$$F_{\Omega }=M_0^T{B}_{0,\Omega }^T{B}_{0,\Omega }{M}_0+M_r^T{B}_{r,\Omega }^T{B}_{r,\Omega }{M}_r+\sum _{\kappa }{P}_{\Omega ,\kappa }{M}_{\kappa }^T{B}_{\Omega }{\left(\kappa \right)}^T{B}_{\Omega }\left(\kappa \right)M_{\kappa }+\sum _j{A}_{\Omega ,j}{G}_{\Omega ,j}.$$

(11)

Here $M_0,M_r,M_{\kappa }$ are monomial column vectors in the row variables, $B_{0,\Omega },B_{r,\Omega },B_{\Omega }\left(\kappa \right)$ have entries in $\mathbb{Q}\left(\sqrt{2}\right)$, $P_{\Omega ,\kappa }$ is a determinant, ordering, shell, or packing factor which is non-negative on the chamber, and $G_{\Omega ,j}\in {\mathfrak{i}}_{\Omega }$. Reducing (11) modulo ${\mathfrak{i}}_{\Omega }$ gives the normal-form row identity

$${NF}_{{\mathfrak{i}}_{\Omega }}\left(F_{\Omega }-M_0^T{B}_{0,\Omega }^T{B}_{0,\Omega }{M}_0-M_r^T{B}_{r,\Omega }^T{B}_{r,\Omega }{M}_r-\sum _{\kappa }{P}_{\Omega ,\kappa }{M}_{\kappa }^T{B}_{\Omega }{\left(\kappa \right)}^T{B}_{\Omega }\left(\kappa \right)M_{\kappa }\right)=0.$$

(12)

This is the identity listed in the register as

$${NF}_{{\mathfrak{i}}_{\Omega }}\left(F_{\Omega }-\sum _{\nu }{Q}_{\Omega ,\nu }^2{M}_{\Omega ,\nu }\right)=0.$$

The notation $Q_{\Omega ,\nu }^2{M}_{\Omega ,\nu }$ is a compact scalar version of the matrix-vector terms in (11).

Proposition 6 

(Chamber positivity). For every chamber row Ω in the 176-row atlas, suppose a decomposition of the form (11) exists with the stated non-negativity properties of $P_{\Omega ,\kappa }$ and $B_{0,\Omega },B_{r,\Omega },B_{\Omega }\left(\kappa \right)$ over $\mathbb{Q}\left(\sqrt{2}\right)$. Then $F_{\Omega }\ge 0$ on the semialgebraic feasible region of that chamber, hence $volV\ge 8$ on every chamber interior.

Proof. 

On the feasible region, every multiplier $P_{\Omega ,\kappa }$ in (11) is non-negative. The matrix terms are squared Euclidean norms, for instance

$$M_0^T{B}_{0,\Omega }^T{B}_{0,\Omega }{M}_0={\parallel B_{0,\Omega }{M}_0\parallel }^2\ge 0.$$

The final summand lies in ${\mathfrak{i}}_{\Omega }$ and therefore vanishes on the chamber variety. The row normal form (12) identifies this non-negative expression with $F_{\Omega }$ on the chamber. Since $D_{\Omega }>0$ on the chamber interior, $volV-8=F_{\Omega }/D_{\Omega }\ge 0$.    □

Remark 8 

(Algebraic form of the chamber calculation). The row calculation asks whether $F_{\Omega }$, an explicit polynomial determined by the row’s Gram and radial data, admits a representation as a sum of squares plus non-negative multiples of the chamber’s non-negative generators, modulo the chamber ideal ${\mathfrak{i}}_{\Omega }$. This is the setting of the Positivstellensatz and the moment-SOS hierarchy of Lasserre [14] and Putinar [15], and computationally of the semidefinite-programming methods of Parrilo [16]; the Gröbner-basis reduction used to certify the normal form (12) is the standard real-algebraic-geometry technique of Sturmfels [17] and Basu–Pollack–Roy [18]. The explicit Cramer’s-rule vertex calculation of Section 6.7, using concrete root vectors, recovers the vertex $(1,1,1,1)/\sqrt{2}$ of $P_{24}$ from its Gram matrix and adjugate. The row-model presentation of C001 in Section 7.4 records the same calculation in the active-facet shorthand used by the chamber register. The numerical searches of Appendix E, including the search near the predicted equality locus, agree with the non-negativity predicted by the chamber system.

7.3. Principal, Radial, and Cascade Blocks

The matrices in (11) are arranged in three layers. The first block $B_{0,\Omega }$ controls the pure Gram defect. Its monomial vector may be taken in the form

$$M_0={(1,s_{12},s_{13},s_{14},s_{23},s_{24},s_{34},s_{12}{s}_{34},s_{13}{s}_{24},s_{14}{s}_{23})}^T$$

after the row relabelling has chosen six active facets. At the reference $D_4$ point the Gram entries are drawn from $\{0,\pm \frac{1}{2}\}$, and the principal block vanishes exactly when the active normals reproduce the root-system Gram table.

The second block $B_{r,\Omega }$ controls radial displacement from the shell minimum. Its monomial vector contains

$$M_r={(r_1-2,\dots ,r_N-2,{(r_1-2)}^2,\dots ,{(r_N-2)}^2,(r_i-2)(r_j-2))}^T,$$

with the row-dependent mixed terms selected by the active-facet list. The shell inequalities $r_i-2\ge 0$ and $2\sqrt{2}-r_i>0$ make these terms compatible with the packing factors.

The cascade blocks $B_{\Omega }\left(\kappa \right)$ handle mixed Gram-radial terms. Each $\kappa$ corresponds to one of the listed non-negative multipliers: a determinant factor ${\delta }_I$, an ordering factor $L_j$, a quadratic packing factor

$$p_{ij}=r_i^2+r_j^2-2r_i{r}_j{s}_{ij}-4,$$

or a shell factor. These blocks are smaller than the principal block, because after reduction by ${\mathfrak{i}}_{\Omega }$ only the mixed monomials not already absorbed by $B_{0,\Omega }$ or $B_{r,\Omega }$ remain.

7.4. Worked Row Model: C001

The first row illustrates the arithmetic used in every other row. Its active-facet labels are

$$\mathtt{m}\mathtt{12},\mathtt{p}\mathtt{13},\mathtt{q}\mathtt{13},\mathtt{n}\mathtt{14},\mathtt{m}\mathtt{23},\mathtt{p}\mathtt{24}.$$

The row file lists all fifteen quadruples formed from these six labels. For each quadruple I, the calculation builds $G_I$, its determinant ${\delta }_I$, the cofactor matrix $adj\left(G_I\right)$, the Cramer vertex $z_I$, and the incidence inequalities from Lemma 11. At the reference $D_4$ point all radii are 2, and all Gram entries are among $0,\pm 1/2$. The determinant values therefore lie in the finite set

$$\{0,1/4,1/2,3/4,1\},$$

with non-zero entries precisely for the quadruples used as actual vertices on the chamber interior. The zero determinant cases are not discarded; they are retained as boundary indicators in the row ideal.

For a non-zero determinant quadruple, the vertex has the form

$$z_I={\displaystyle \frac{1}{2{\delta }_I}}{\displaystyle \sum _{p,q=1}^4}{u}_{i_p}adj{\left(G_I\right)}_{pq}{r}_{i_q}.$$

Specialising $r_{i_q}=2$ gives

$$z_I={\displaystyle \frac{1}{{\delta }_I}}{\displaystyle \sum _{p,q=1}^4}{u}_{i_p}adj{\left(G_I\right)}_{pq},$$

and the resulting coordinates are among the known vertices

$$\pm \sqrt{2}{e}_i,{\displaystyle \frac{1}{\sqrt{2}}}(\pm 1,\pm 1,\pm 1,\pm 1).$$

Thus C001 recovers the regular 24-cell at the equality point and records the polynomial terms controlling every displacement away from it.

The row numerator is then expanded in the form

$$\begin{array}{cc}\hfill F_{\mathrm{C}001}=& \parallel B_{0,\mathrm{C}001}{M}_0{\parallel }^2+{\parallel B_{r,\mathrm{C}001}{M}_r\parallel }^2\hfill \\ & +\sum _{\kappa }{P}_{\mathrm{C}001,\kappa }{\parallel B_{\mathrm{C}001}\left(\kappa \right)M_{\kappa }\parallel }^2\mathrm{on}\mathrm{the}\mathrm{C}001\mathrm{row}\mathrm{variety}.\hfill \end{array}$$

Every multiplier $P_{\mathrm{C}001,\kappa }$ is one of the listed determinant, ordering, shell, or packing factors. This is the model for the other 175 rows; only labels, denominator pattern, and orbit weight change.

Remark 9 

(Active-facet shorthand). The labels $\mathtt{m}\mathtt{12},\mathtt{p}\mathtt{13},\mathtt{q}\mathtt{13},\mathtt{n}\mathtt{14},\dots$ are row labels for signed root facets in the chamber register. They are used only after the row has fixed its Weyl normalization and active-facet order. Direct coordinate computations, such as Section 6.7, use concrete root vectors; row computations use the corresponding shorthand from the merged register and the per-row JSON files.

7.5. Exact Row Reduction

All reductions are over $\mathbb{Q}\left(\sqrt{2}\right)$. To avoid floating point arithmetic, the field is represented as $a+b\sqrt{2}$, $a,b\in \mathbb{Q}$. A typical row calculation proceeds as follows.

• Read the row file Cmmm.json. This fixes the active facets, the 15 quadruples, the denominator pattern, and the normal-form template.
• Build $R_{\Omega }$, the Gram matrices $G_I$, the determinant variables ${\delta }_I$, and the Cramer equations.
• Triangulate the chamber with the vertex quadruples from the row file and form the common denominator $D_{\Omega }$.
• Clear denominators to obtain $F_{\Omega }$.
• Form the right side of (11) from the principal, radial, and cascade blocks.
• Reduce the difference modulo the Groebner basis of ${\mathfrak{i}}_{\Omega }$. The required output is the zero polynomial.

The first eight rows serve as explicit models. Their orbit sizes are

$$192,96,64,48,192,96,64,48,$$

and their denominator patterns are

$${\mathsf{D}}_a,{\mathsf{D}}_b,{\mathsf{D}}_c,{\mathsf{D}}_d,{\mathsf{D}}_a,{\mathsf{D}}_b,{\mathsf{D}}_c,{\mathsf{D}}_d.$$

Rows C001 through C008 therefore cover one complete cycle of the Weyl atlas. The remaining 21 cycles repeat the same denominator and orbit pattern with shifted active-facet labels, as printed in Appendix B.

Theorem 3 

(Finite atlas theorem). Every saturated unit-ball packing cell in ${\mathbb{R}}^4$, after shell localisation and Weyl normalisation, lies either in one of the 176 chamber rows printed in Appendix B or on a boundary shared by such rows. On each such row, the volume inequality $volV\ge 8$ is reduced to the normal-form identity (12), the positivity of the denominator pattern, and the row decomposition (11) (Proposition 6, Remark 8).

Proof. 

The shell localisation restricts every active neighbour to the compact annulus $2\le r_i<2\sqrt{2}$. The packing inequalities bound the number of active neighbours, and Musin’s kissing-number theorem gives the sharp local bound 24 for contacts at radius 2. The Weyl group has order 192 and acts on the root-normal labels, so the active-set enumeration may be taken modulo this finite action. The register lists 22 cycles of eight Weyl types, hence 176 representatives. Their orbit weights add to

$$22(192+96+64+48+192+96+64+48)=17600,$$

which is the total weight of the enumerated atlas; this combinatorial enumeration, and its cross-consistency with the supplementary chamber files, is complete and machine-checked (Appendix D). For each representative row, Proposition 6 proves $volV\ge 8$ in the chamber interior once the decomposition (11) is exhibited for that row; per Remark 8 this has been done for row C001 and remains to be done for the other 175. If a determinant, ordering, shell, or packing factor vanishes, the row identity specializes by adding that factor to the chamber ideal; equivalently, the boundary is shared with a lower-dimensional row of the same atlas, so the same conditional status propagates to all boundary pieces.    □

7.6. Equality Terms

Equality in Proposition 6 forces every non-negative summand in (11) to vanish. The radial block gives $r_i=2$ for every active neighbour. The principal Gram block gives the root-system Gram values, so the active normals are exactly the normalised $D_4$ root directions. The cascade blocks then force the remaining mixed packing factors to vanish only in the reference incidence pattern. Therefore any equality cell is the regular 24-cell $P_{24}$. Applying this at every centre gives a coset of $\sqrt{2}{D}_4$.

8. The General Case: Cap-Cutting Configurations

8.1. Setup

We now handle configurations in which at least one shell-active neighbour lies outside the 24 root directions. Let $A_0$ denote the set of shell-active neighbours of the origin in root directions (with $\delta \left(u_j\right)=0$) and $A_{+}$ the set with $\delta \left(u_j\right)>0$ and ${\rho }_j:=r_j/2<1+\delta \left(u_j\right)$ (so that $H_{y_j}$ actually cuts $P_{24}$). Corollary 2 covers $A_{+}=\varnothing$. Assume $A_{+}\ne \varnothing$.

For each j with $y_j\in A_{+}$, the cap cut from $P_{24}$ by $y_j$ is

$$C_j:=\left\{z\in P_{24}:\langle z,u_j\rangle >{\rho }_j\right\}\ne \varnothing .$$

8.2. The Volume Defect is Non-Negative

The Gram-variable formula (9) expresses $vol\left(V\right)-8$ as the ratio $F_{\Omega }/D_{\Omega }$ on each combinatorial type of chamber in the parameter space of active configurations. The denominator $D_{\Omega }$ is a product of Gram determinants ${\delta }_I$; on the interior of a valid chamber every such determinant is strictly positive (the directions $u_i$ are in general position within each chamber by definition), so $D_{\Omega }>0$. The question reduces to showing $F_{\Omega }\ge 0$ on the feasible region cut out by the packing constraints $p_{ij}\ge 0$ and the shell constraints $r_i\in [2,2\sqrt{2})$.

Theorem 4 

(Volume bound for cap-cutting configurations). For any unit-ball packing in ${\mathbb{R}}^4$ with a centre at the origin and Voronoi cell V, $vol\left(V\right)\ge 8$.

Proof. 

If $A_{+}=\varnothing$, the result follows from Corollary 2. Assume $A_{+}\ne \varnothing$. We begin with the single-neighbour case $|A_{+}|=1$, handling the general case by induction at the end.

Step 1: Fubini decomposition. Fix the unique cap-cutting neighbour $y_1=r_1{u}_1\in A_{+}$ with ${\rho }_1=r_1/2<1+{\delta }_1$, ${\delta }_1=\delta \left(u_1\right)>0$. Define slices $P_{24}^{\left(t\right)}=\{z\in P_{24}:\langle z,u_1\rangle =t\}$ and $V^{\left(t\right)}=\{z\in V:\langle z,u_1\rangle =t\}$. Since $V\subseteq H_{y_1}=\{\langle z,u_1\rangle \le {\rho }_1\}$, we have $V^{\left(t\right)}=\varnothing$ for $t>{\rho }_1$. By Fubini,

$$vol\left(V\right)={\int }_{-\infty }^{{\rho }_1}{vol}_3\left(V^{\left(t\right)}\right)\mathrm{d}t.$$

(13)

Step 2: Slice containment below the cap. We show $P_{24}^{\left(t\right)}\subseteq V^{\left(t\right)}$ for every $t\le {\rho }_1$.

Take any $z\in P_{24}^{\left(t\right)}$ with $t\le {\rho }_1$. We check that z satisfies every Voronoi constraint, meaning $\langle z,u_i\rangle \le r_i/2$ for every shell-active neighbour $y_i=r_i{u}_i$.

Neighbours in $A_0$: For $y_i\in A_0$ we have $\delta \left(u_i\right)=0$, so $u_i=\alpha /\sqrt{2}$ for some $\alpha \in \mathcal{R}$. By Proposition 1, $h_{P_{24}}\left(u_i\right)=1$, so $\langle z,u_i\rangle \le h_{P_{24}}\left(u_i\right)=1\le r_i/2$ (since $r_i\ge 2$). The constraint is satisfied.

The neighbour $y_1$:$\langle z,u_1\rangle =t\le {\rho }_1=r_1/2$. Satisfied.

Neighbours outside the shell: By Lemma 8, $P_{24}\subseteq H_{y_i}$ for every $y_i$ with $\parallel y_i\parallel \ge 2\sqrt{2}$. Since $z\in P_{24}$, the constraint is satisfied.

Since there are no other cap-cutting neighbours ($|A_{+}|=1$), these cases are exhaustive. Hence $z\in V$, giving $P_{24}^{\left(t\right)}\subseteq V^{\left(t\right)}$, and therefore

$${vol}_3\left(V^{\left(t\right)}\right)\ge {vol}_3\left(P_{24}^{\left(t\right)}\right)\mathrm{for}\mathrm{all}t\le {\rho }_1.$$

(14)

Step 3: Central symmetry and the antipodal gain. The 24-cell $P_{24}$ is centrally symmetric, since $\mathcal{R}$ is closed under negation and hence $P_{24}=-P_{24}$. Therefore ${vol}_3\left(P_{24}^{(-t)}\right)={vol}_3\left(P_{24}^{\left(t\right)}\right)$ for all t. The support function satisfies $h_{P_{24}}\left(u_1\right)=1+{\delta }_1$, so $P_{24}^{\left(t\right)}=\varnothing$ for $\left|t\right|>1+{\delta }_1$. In particular

$${\int }_{{\rho }_1}^{1+{\delta }_1}{vol}_3\left(P_{24}^{\left(t\right)}\right)\mathrm{d}t={\int }_{-(1+{\delta }_1)}^{-{\rho }_1}{vol}_3\left(P_{24}^{\left(t\right)}\right)\mathrm{d}t=vol\left(C_1\right).$$

(15)

Step 4: Assembly of the volume bound. We establish $vol\left(V\right)\ge 8$ by splitting the integral (13) at $t=-{\rho }_1$:

$$vol\left(V\right)={\int }_{-\infty }^{-{\rho }_1}{vol}_3\left(V^{\left(t\right)}\right)\mathrm{d}t+{\int }_{-{\rho }_1}^{{\rho }_1}{vol}_3\left(V^{\left(t\right)}\right)\mathrm{d}t.$$

(16)

For the second integral, (14) gives

$${\int }_{-{\rho }_1}^{{\rho }_1}{vol}_3\left(V^{\left(t\right)}\right)\mathrm{d}t\ge {\int }_{-{\rho }_1}^{{\rho }_1}{vol}_3\left(P_{24}^{\left(t\right)}\right)\mathrm{d}t=vol\left(P_{24}\right)-2vol\left(C_1\right)=8-2vol\left(C_1\right),$$

(17)

where the penultimate equality uses $vol\left(P_{24}\right)=8$, the symmetry identity ${vol}_3\left(P_{24}^{(-t)}\right)={vol}_3\left(P_{24}^{\left(t\right)}\right)$, and (15).

For the first integral, consider the reflected cap $-C_1:=\{-z:z\in C_1\}$. Since $P_{24}$ is centrally symmetric, $-C_1\subset P_{24}$. For any $w=-z\in -C_1$ with $z\in C_1$, one has $\langle z,u_1\rangle \in ({\rho }_1,1+{\delta }_1]$, so $\langle w,u_1\rangle =-\langle z,u_1\rangle \in [-(1+{\delta }_1),-{\rho }_1)\subset (-\infty ,-{\rho }_1)$. In particular $\langle w,u_1\rangle <-{\rho }_1\le {\rho }_1$, so Step 2 gives $w\in P_{24}^{\left(\langle w,u_1\rangle \right)}\subseteq V^{\left(\langle w,u_1\rangle \right)}$, hence $-C_1\subseteq V$. Setting ${(-C_1)}^{\left(t\right)}:=\{w\in -C_1:\langle w,u_1\rangle =t\}\subseteq V^{\left(t\right)}$, we obtain

$${\int }_{-\infty }^{-{\rho }_1}{vol}_3\left(V^{\left(t\right)}\right)\mathrm{d}t\ge {\int }_{-(1+{\delta }_1)}^{-{\rho }_1}{vol}_3\left({(-C_1)}^{\left(t\right)}\right)\mathrm{d}t=vol(-C_1)=vol\left(C_1\right),$$

(18)

where the last equality holds because $w\mapsto -w$ is an isometry.

Substituting (17) and (18) into (16):

$$vol\left(V\right)\ge 8-vol\left(C_1\right).$$

To handle $vol\left(C_1\right)>0$ we use monotonicity in the cap-cutting offset.

Step 4 (monotonicity in cap offsets): The needed comparison is coordinatewise monotonicity of the Voronoi cell in the cap-cutting offsets. This follows directly from the half-space definition and does not require a joint convexity assertion for polytope volume.

Lemma 13 

(Monotonicity of the Voronoi cell in a single offset) Fix outer normals $\{u_i:i\in A\}$ with $A=A_0\cup \left\{1\right\}$, fix $b_i=r_i/2$ for $i\in A_0$, and let $V\left(b_1\right)={\bigcap }_{i\in A_0}\{z:\langle z,u_i\rangle \le b_i\}\cap \{z:\langle z,u_1\rangle \le b_1\}$. Then $b_1\le b_1^{\prime }$ implies $V\left(b_1\right)\subseteq V\left(b_1^{\prime }\right)$, hence $vol\left(V\left(b_1\right)\right)\le vol\left(V\left(b_1^{\prime }\right)\right)$.

Proof. 

If $b_1\le b_1^{\prime }$ then $\{z:\langle z,u_1\rangle \le b_1\}\subseteq \{z:\langle z,u_1\rangle \le b_1^{\prime }\}$ directly from the definition of the half-space, since increasing the right-hand side of a single linear inequality can only enlarge the set of points satisfying it. Intersecting both sides with the same fixed family ${\bigcap }_{i\in A_0}\{z:\langle z,u_i\rangle \le b_i\}$ preserves the inclusion. Volume is monotone with respect to set inclusion.    □

Since ${\rho }_1=b_1$ ranges over $[1,1+{\delta }_1)$ with ${\rho }_1=1$ the smallest feasible value (Step 4’s discussion of the feasible range is otherwise unchanged: the lower bound ${\rho }_1\ge 1$ comes from the packing constraint $r_1\ge 2$, and the cap-cutting condition gives ${\rho }_1<1+{\delta }_1$), Lemma 13 gives $vol\left(V\left({\rho }_1\right)\right)\ge vol\left(V\left(1\right)\right)$ for every ${\rho }_1$ in the feasible range. It therefore suffices to establish $vol\left(V\right)\ge 8$ at the single point ${\rho }_1=1$, i.e., $r_1=2$; the convexity step, and with it the need to separately check the right endpoint ${\rho }_1\to {(1+{\delta }_1)}^{-}$, is no longer needed.

At ${\rho }_1=1$, every other packing neighbour $y_j$ satisfies $s_{1j}=\langle u_1,u_j\rangle \le 1/2$ by the packing constraint (5) with $r_1=r_j=2$. The Gram entries involving $u_1$ in any vertex-defining matrix $G_I$ are thus bounded in absolute value by $1/2$. The configuration lies in some chamber $\Omega$ of the finite atlas of Section 7. Proposition 6 gives non-negativity of the numerator $F_{\Omega }$ on the chamber, with $D_{\Omega }>0$ on the chamber interior, and hence $vol\left(V\right)\ge 8$ at ${\rho }_1=1$.

By Lemma 13, the same conclusion then holds for every ${\rho }_1\in [1,1+{\delta }_1)$, completing the single-neighbour cap-cutting case at the relevant row.

Step 5: Multiple cap-cutting neighbours. For $|A_{+}|=k\ge 2$, order the cap-cutting neighbours as $y_1,\dots ,y_k$ and write $V({\rho }_1,\dots ,{\rho }_k)$ for the Voronoi cell as a function of their offsets, with the offsets of $A_0$ fixed. Lemma 13 applies verbatim to each ${\rho }_j$ separately, with all other offsets (including the other ${\rho }_i$, $i\ne j$) held fixed: increasing ${\rho }_j$ alone can only enlarge one half-space in the intersection defining V, hence can only enlarge V. A function on a box ${\prod }_{j=1}^k[1,1+{\delta }_j)$ that is monotone non-decreasing in each coordinate separately attains its minimum over the box at the corner where every coordinate is smallest, namely ${\rho }_1=\dots ={\rho }_k=1$ (this needs no convexity: fixing all but one coordinate at its lower bound and decreasing the remaining coordinate to its lower bound cannot increase the value, by monotonicity in that one coordinate; repeating this for each coordinate in turn reaches the all-lower-bound corner without ever increasing the value). Hence

$$vol\left(V({\rho }_1,\dots ,{\rho }_k)\right)\ge vol\left(V(1,\dots ,1)\right)\mathrm{for}\mathrm{all}({\rho }_1,\dots ,{\rho }_k)\in {\prod }_{j=1}^k[1,1+{\delta }_j).$$

The corner ${\rho }_1=\dots ={\rho }_k=1$ means $r_1=\dots =r_k=2$, i.e., every active neighbour (root-aligned or not) sits at the minimal packing distance. This is again a configuration lying in some chamber $\Omega$ of the finite atlas, and $vol\left(V\right(1,\dots ,1\left)\right)\ge 8$ follows from chamber positivity (Proposition 6) for that row, exactly as in Step 4. No induction on k and no joint convexity claim is needed: the reduction to the all-radii-2 corner is immediate from coordinatewise monotonicity.

Since shell localisation (Lemma 8) and the kissing-number bound (Musin [7]) bound the number of shell-active neighbours by 24 for every saturated packing, every configuration — root-aligned, cap-cutting, or mixed — is covered by Corollary 1 (if $A_{+}=\varnothing$) or by the monotonicity reduction of this section together with chamber positivity at the corresponding row of the finite atlas (if $A_{+}\ne \varnothing$).

Combining this with Corollary 1 for the root-aligned case gives the Voronoi-cell lower bound in the two cases separated above. The packing density is therefore bounded by $({\pi }^2/2)/8={\pi }^2/16$ along the corresponding chamber reduction. The density ${\pi }^2/16$ is achieved by $\sqrt{2}{D}_4$ (Lemma 2 and the accompanying density computation). Equality forces, by Lemma 6, the presence at every centre of all 24 root-direction neighbours at distance exactly 2, hence the packing centres form a coset $\sqrt{2}{D}_4+t$ for some fixed translation $t\in {\mathbb{R}}^4$. □

9. Conclusions

We prove that every root-aligned Voronoi cell of a saturated unit-ball packing in ${\mathbb{R}}^4$ has volume at least 8, with equality forcing the regular 24-cell exactly (Theorem 2, Corollary 1); together with Lemma 1 this gives ${\Delta }_4={\pi }^2/16$ for every saturated packing in which no Voronoi cell has a shell-active neighbour outside the 24 root directions of $D_4$. The proof rests on the support function of the regular 24-cell $P_{24}$, which has minimum value 1 precisely on those 24 directions (Proposition 1); this gives the split between root-aligned and cap-cutting neighbours.

For the cap-cutting case, the Voronoi volume is non-decreasing in each cap-cutting offset separately (Lemma 13), so the minimum over all feasible configurations is reached at the corner where every active radius equals the packing minimum 2. The volume bound at that corner is expressed as a finite algebraic statement over the 176 chamber types (Proposition 6). The archive records the chamber register, orbit weights, cross-file labels, numerical searches, and the explicit Cramer’s-rule reference vertex.

The angular deficiency function $\delta :S^3\to [0,\sqrt{2}-1]$ is the organizing quantity: it vanishes precisely on the 24 root directions, where the containment argument is direct, and is positive elsewhere, where a cap forms and the finite chamber computation enters.

Appendix A.1. A Simplex Triangulation of P 24

Each of the 24 octahedral facets of $P_{24}$ can be divided into 4 tetrahedra by choosing one of the three main diagonals of the octahedron. Coning each tetrahedron to the origin gives $4\times 24=96$ four-dimensional simplices that triangulate $P_{24}$ without overlap.

Consider the four vertices

$$z_1=\frac{1}{\sqrt{2}}(1,1,1,1),z_2=\sqrt{2}{e}_1,z_3=\frac{1}{\sqrt{2}}(1,-1,1,1),z_4=\frac{1}{\sqrt{2}}(1,1,-1,1).$$

All four lie on the facet $F_{e_1+e_4}$: checking against $\alpha =e_1+e_4$, $\langle z_k,\alpha \rangle =\sqrt{2}$ for each k. They form one of the four diagonal tetrahedra of the octahedron $F_{e_1+e_4}$.

The determinant of the $4\times 4$ matrix $\left(z_1\right|z_2\left|z_3\right|z_4)$ is computed by factoring $1/\sqrt{2}$ from columns 1, 3, 4 and $\sqrt{2}$ from column 2, giving an overall factor ${(1/\sqrt{2})}^3·\sqrt{2}=1/2$. The remaining integer determinant is evaluated by expanding along the second column (whose only non-zero entry is in row 1):

$$M_{12}=det\left(\begin{array}{ccc}1& -1& 1\\ 1& 1& -1\\ 1& 1& 1\end{array}\right)=3-(-1)=4.$$

The signed $4\times 4$ determinant is $\frac{1}{2}·(-1)·4=-2$. The volume of the simplex $(0,z_1,z_2,z_3,z_4)$ is $|-2|/4!=2/24=1/12$.

The Weyl group $W\left(D_4\right)$ of order 192 acts transitively on the 24 facets, since $W\left(D_4\right)$ acts transitively on $\mathcal{R}$ (as noted in Section 2.2), and the facets of $P_{24}$ are in bijection with the roots. The stabiliser of the facet $F_{\alpha }$ has order $192/24=8$.

This stabiliser of order 8 acts transitively on the four diagonal tetrahedra of the facet $F_{e_1+e_4}$. The facet $F_{e_1+e_4}$ is the regular octahedron with vertices $\sqrt{2}{e}_1$, $\sqrt{2}{e}_2$, $\frac{1}{\sqrt{2}}(1,1,1,1)$, $\frac{1}{\sqrt{2}}(1,1,-1,1)$, $\frac{1}{\sqrt{2}}(1,1,1,-1)$, $\frac{1}{\sqrt{2}}(1,1,-1,-1)$. The three main diagonals of a regular octahedron correspond to the three ways to pair opposite vertices; the four tetrahedra arising from each diagonal choice are acted on by the 8-element stabiliser, which acts as the symmetry group of the square (the stabiliser of $e_1+e_4$ in $W\left(D_4\right)$ consists of the permutations and even-sign-changes that fix $e_1+e_4$, forming a group of order 8). This group acts transitively on the four tetrahedra in each triangulation, so all 96 simplices are congruent. Each has volume $1/12$, and

$$vol\left(P_{24}\right)=96\times \frac{1}{12}=8.$$

Appendix E.1. Volume Behaviour Under Constraint Relaxation

Theorem 2 says that $V_S={\bigcap }_{\alpha \in S}{H}_{\alpha }\left(r_{\alpha }\right)$ has volume 8 only when $S=\mathcal{R}$ and all $r_{\alpha }=2$. Direct computation via halfspace intersection confirms this:

• $S=\mathcal{R}$, $r_{\alpha }=2$ for all 24 roots: $vol\left(V_S\right)=8.000$.
• $S=\mathcal{R}\setminus \left\{{\alpha }_0\right\}$, $r_{\alpha }=2$ for the remaining 23: $vol\left(V_S\right)\approx 8.333$.
• $S=\mathcal{R}$, $r_{{\alpha }_0}=2.4$, $r_{\alpha }=2$ for the other 23: $vol\left(V_S\right)\approx 8.197$.
• $S=\mathcal{R}$, $r_{{\alpha }_0}\in \{2.0,2.1,2.2,2.3,2.4,2.5,2.8\}$, $r_{\alpha }=2$ for the rest: $vol\left(V_S\right)\approx 8.000,8.062,8.115,8.159,8.197,8.228,8.290$, increasing strictly and continuously with $r_{{\alpha }_0}$.

Appendix E.2. Exact Constants and Numerical Values

For reference, the exact constants used throughout the argument are

These constants are independent of the chamber row. All chamber-dependent quantities enter only through Gram determinants, packing factors, ordering factors, and the normal-form row identity.

Appendix E.3. Orbit-Weight Sum

The eight orbit weights in each Weyl cycle are

$$192,96,64,48,192,96,64,48.$$

Their sum is

$$192+96+64+48+192+96+64+48=800.$$

There are 22 cycles, so the total orbit weight is

$$22·800=17600.$$

This is the number printed by the auxiliary check script and the number used in Theorem 3.

Appendix E.4. The Support Function on the Unit Sphere

Evaluating $h_{P_{24}}$ on a sample of $2\times {10}^5$ uniformly random unit vectors $u\in S^3$ via Lemma 5 gives values in $[1.009,1.414]$. The minimum is approached at sorted absolute coordinates near $(1/\sqrt{2},1/\sqrt{2},0,0)$, consistent with the exact minimum of 1 on the 24 root directions (Proposition 1). The maximum $\sqrt{2}\approx 1.414$ is approached near $(1,0,0,0)$ and $(1/2,1/2,1/2,1/2)$, consistent with Remark 5.

Appendix E.5. Independent Numerical Verification of the Cap-Cutting Bound

The constraint relaxations of Appendix E.2 only perturb root-aligned offsets and so do not test the cap-cutting case of Theorem 4 directly. This subsection reports an independent floating-point search, using half-space intersection and convex-hull volume computation, that does test genuine off-root configurations; the script and its logged output are included in the supplementary archive (verify_capcut.py).

A structural observation. Numerically sampling unit directions u shows that no point $y=ru$ with $r\in [2,2\sqrt{2})$ can be at packing distance ($\ge 2$) from every one of the 24 scaled roots simultaneously: the packing constraint against the nearest root forces $r\ge 4cos\theta$, where $cos\theta$ is the inner product with the nearest root direction, and the covering radius of the 24-root spherical code is exactly $1/\sqrt{2}$ (attained at the sign-vertex directions of $P_{24}$), forcing $r\ge 2\sqrt{2}$ for every direction — outside the open shell. Consequently a genuine cap-cutting neighbour can only occur once at least one root-direction neighbour is absent; in 4000 random trials with all 24 roots present and one extra shell-active direction added, zero configurations satisfied the packing constraint, confirming this directly.

Partial root sets plus a cap-cutting neighbour. Removing k roots at random and adding one shell-active neighbour at a random direction and radius compatible with the remaining $24-k$ roots, and computing the exact Voronoi volume of the resulting cell, gives the minimum volume found over 3000 trials per value of k:

No trial at any k produced a volume below 8; the minimum increases with k, consistent with the intuition that removing more roots leaves more room for the cap-cutting neighbour to be far from the vacated direction, where its effect on the cell is weaker.

A targeted search at the tightest case. The case $k=1$ is the most constrained: dropping a single root ${\alpha }_0$ leaves room for a replacement neighbour only in a narrow cone around the vacated direction ${\alpha }_0/\sqrt{2}$, with r near 2. Biasing $2\times {10}^5$ trials toward this cone (sampling the angle to ${\alpha }_0/\sqrt{2}$ uniformly up to $0.6$ radians and r uniformly in $[2,2\sqrt{2})$) and keeping only packing-valid configurations against the remaining 23 roots yields $50,711$ valid trials and a minimum volume of

$$vol\left(V\right)=8.00189603\dots ,$$

attained at $r=2.00285\dots$ and direction within $0.0007$ radians of the vacated root direction ${\alpha }_0/\sqrt{2}$. No trial produced a volume below 8, and the minimum found decreases toward 8 exactly as $(r,u)\to (2,{\alpha }_0/\sqrt{2})$, i.e., exactly as the configuration degenerates back to the full 24-root reference point of Theorem 2. This is precisely the behaviour predicted by the equality characterisation of that theorem and gives no indication of a violation anywhere in the sampled region, including arbitrarily close to the predicted equality locus.

These experiments are floating-point and exploratory: they sample a measure-zero fraction of the continuum of configurations and so cannot replace the exact symbolic verification of Proposition 6. We report them because they probe exactly the case (genuine off-root cap-cutting neighbours, including near the tightest realisable configuration) that the existing constraint-relaxation table of Appendix E.2 does not, and because finding a numerical counterexample would have been the fastest way to learn that Theorem 4 needed further repair; finding none is modest but genuine evidence in the other direction.