Does every higher-dimensional tiling by translations of convex polytope tiles have an affine transformation taking it to a Voronoi diagram

Does every higher-dimensional tiling by translations of convex polytope tiles have an affine transformation taking it to a Voronoi diagram

Definitions and mathematical statement

A parallelohedron is a full-dimensional convex polytope $P\subset{\mathbb{R}}^d$ that tiles ${\mathbb{R}}^d$ by translations, with disjoint interiors of distinct tiles. For a full-rank lattice $\Lambda$, its Euclidean Dirichlet--Voronoi cell at zero is \[ V(\Lambda)=\{x\in{\mathbb{R}}^d:\|x\|\le\|x-v\| \text{ for every }v\in\Lambda\}. \] The conjecture asks whether every parallelohedron is of the form $A V(\Lambda)+b$ for some invertible linear map $A$, vector $b$, and lattice $\Lambda$. An equivalent viewpoint allows a positive-definite quadratic form in place of the Euclidean norm. The requested equivalence is affine, not necessarily an isometry or a similarity. Finding a translational tiling alone does not identify it as a Voronoi cell after a linear change of coordinates.

Formulation and concept references: [S1].

Short English statement

Can every convex polytope that tiles space by translation be transformed affinely into a lattice's nearest-point region?

Research attempt

Scope and literature check.

This attempt by GPT-6 Astra Ultra was prepared on 27 September 2026. The dimension is unrestricted and the desired transformation is affine. Garber's version 6 of [S1], dated 31 January 2025, proves the five-dimensional case; its Section 2 also explains the classical reduction from arbitrary translational tilings to a face-to-face lattice tiling. Neither that reduction nor the five-dimensional theorem is reproved here. The current arXiv version and journal entry list Garber as sole author, consistently with the inherited source entry; earlier versions and some repository records also list Magazinov.

The route below is to turn an already identified tiling lattice and its facet data into a positive-definite matrix problem. The local-to-global step can be proved exactly using volume. We then solve that matrix problem for parallelotopes and for every centrally symmetric convex hexagon in the plane, including an explicit formula. These are known affirmative families, rederived to test the method. The unresolved step is existence of a compatible positive-definite matrix for arbitrary parallelohedra in arbitrary dimension.

Lemma 477.1: changing the metric is exactly the affine freedom.

For a symmetric positive-definite matrix $Q$, write \[ \|x\|_Q^2=x^{\mathsf T}Qx,\qquad V_Q(\Lambda)=\{x:2x^{\mathsf T}Qt\le t^{\mathsf T}Qt \text{ for all }t\in\Lambda\}. \] Choose an invertible matrix $L$ with $Q=L^{\mathsf T}L$. Expanding $\|x-t\|_Q^2-\|x\|_Q^2$ shows \begin{equation}\label{wsref-0-61-34-37-37-3a-61-66-66-69-6e-65} L V_Q(\Lambda)=V(L\Lambda). \end{equation} Conversely, pulling the Euclidean norm back through an invertible linear map gives precisely such a positive-definite form. Translation first puts the central symmetry centre of the tile at zero. Thus constructing $Q$ for the centred polytope is sufficient for the catalogue conclusion.

Positive definiteness cannot be replaced by invertibility or by positive values on a few selected vectors. It makes the distance coercive and ensures that nearest lattice points exist. An indefinite quadratic form does not define a nearest-point cell in this sense.

A volume fact used in the local-to-global argument.

For every full-rank lattice $\Lambda$ and positive-definite $Q$, \begin{equation}\label{wsref-0-61-34-37-37-3a-76-6f-6c-75-6d-65} \operatorname{vol}V_Q(\Lambda)=\det\Lambda, \end{equation} where $\det\Lambda$ denotes its positive covolume. To justify this, coercivity and lattice discreteness give a nearest lattice point to every $x$. Points with two nearest lattice points lie in a countable union of affine bisector hyperplanes, hence a null set. The translates of $V_Q(\Lambda)$ therefore cover and have disjoint interiors. The cell is bounded: choose a bounded fundamental parallelepiped, move $x$ into it by a lattice translate, and use its finite $Q$-diameter as a uniform bound on distance to the lattice. If $x\in V_Q$, zero is nearest, so this bound also bounds $\|x\|_Q$.

Integrating $\sum_{t\in\Lambda}\mathbf1_{V_Q(\Lambda)}(x-t)=1$ almost everywhere over a fundamental parallelepiped proves (a477:volume). The same argument shows $\operatorname{vol}P=\det\Lambda$ whenever $P+\Lambda$ is a translational tiling with disjoint interiors. This proof does not assume the conjecture: it uses a metric Voronoi cell only after a positive-definite $Q$ has been specified.

Lemma 477.2: a finite facet certificate is sufficient.

Let $P$ be centred at zero and have a face-to-face lattice tiling $P+\Lambda$. For each facet $F$, let $t_F\in\Lambda$ be the centre of its neighbouring tile and let $a_F$ be an outward normal, so \begin{equation}\label{wsref-0-61-34-37-37-3a-66-61-63-65-74-73} P=\bigcap_F\left\{x:a_F\cdot x\le\tfrac12a_F\cdot t_F\right\}. \end{equation} The half-support value follows from central symmetry: the shared facet $P\cap(P+t_F)$ is preserved by $x\mapsto t_F-x$, so its centre is $t_F/2$. The support constant is positive because zero is in the interior.

Suppose there are a symmetric positive-definite matrix $Q$ and positive scalars $\lambda_F$ such that \begin{equation}\label{wsref-0-61-34-37-37-3a-63-65-72-74-69-66-69-63-61-74-65} Qt_F=\lambda_F a_F \qquad\text{for every facet }F. \end{equation} Then $P=V_Q(\Lambda)$.

Proof. The lattice inequalities for the particular vectors $t_F$ are exactly the inequalities in (a477:facets), after multiplication by $\lambda_F>0$. Since a Voronoi cell satisfies the inequalities for all lattice vectors, $V_Q(\Lambda)\subseteq P$. Both sets have volume $\det\Lambda$ by (a477:volume). A proper inclusion of compact full-dimensional convex sets has a positive-volume difference: a separating hyperplane and a small interior ball produce a region of positive volume in the larger set. Thus inclusion with equal volumes forces equality. This proves the lemma.

This argument closes a possible infinite-constraint gap. It would be incorrect simply to assume that checking neighbouring lattice vectors automatically checks all lattice vectors for an arbitrary norm. Here the missing implication follows from a specified positive-definite quadratic form, inclusion in the correct direction, and equal volumes.

If $P$ is already $V_Q(\Lambda)$ with its Voronoi lattice tiling, the shared facet with $P+t_F$ lies on its $Q$-bisector, so its outward normal is a positive multiple of $Qt_F$. Thus the certificate is also necessary for this fixed tiling. The classical tiling reduction cited in [S1] supplies the canonical face-to-face lattice when the initial tiling has offsets between rows or other non-face-to-face features. Our argument does not silently assume that the particular translate set initially handed to us is itself a lattice.

Attempt to solve the general matrix system.

The equations in (a477:certificate) are linear in the entries of $Q$ and the $\lambda_F$, supplemented by symmetry, positivity of the scalars, and $Q\succ0$. Multiplying $Q$ and all $\lambda_F$ by the same positive scalar changes no cell, so one may impose $\operatorname{tr}Q=1$. For exact rational facet data, a rational candidate can be checked using those linear equations and Sylvester's criterion, namely positivity of all leading principal minors of the symmetric matrix. That is a finite certificate for an individual tile.

Finding a nonzero solution to the linear equations is not enough. For example, prescribe in the plane \[ t_1=(1,0),\quad t_2=(0,1),\qquad a_1=(1,2),\quad a_2=(2,1). \] Both $a_i\cdot t_i$ are positive. If $Qt_i=\lambda_i a_i$ with $\lambda_i>0$, symmetry forces $\lambda_1=\lambda_2=\lambda$ and \[ Q=\lambda\begin{pmatrix}1&2\\2&1\end{pmatrix}. \] Its determinant is $-3\lambda^2$, so it is indefinite. These data do not arise from the asserted facet neighbours of a tile for $\mathbb Z^2$: the associated parallelogram \[ |x+2y|\le1/2,\qquad|2x+y|\le1/2 \] has area $1/3$, not covolume $1$. It is itself a parallelogram and does tile by a different lattice. Thus this is a counterexample to a local linear-algebra shortcut, not to Voronoi's conjecture. It identifies where actual tiling compatibility must enter any general proof.

Numerical feasibility has a related limitation. Eigenvalues that appear positive at floating point precision are not an exact certificate near the boundary of the positive cone. Conversely, failure of a search restricted to diagonal matrices or a prescribed symmetry class does not rule out an unrestricted $Q$. The conjecture requires an existence argument, not the success of a particular search parameterization.

Proposition 477.3: every parallelotope has an explicit metric.

Let $B$ be any invertible real $d\times d$ matrix and put \[ P=B[-1/2,1/2]^d,\qquad\Lambda=B\mathbb Z^d. \] For any positive diagonal matrix $D$, take \begin{equation}\label{wsref-0-61-34-37-37-3a-62-6f-78-6d-65-74-72-69-63} Q=B^{-\mathsf T}DB^{-1}. \end{equation} For $x=Bu$ and $t=Bn$, one has $\|x-t\|_Q^2=\sum_jD_{jj}(u_j-n_j)^2$. Each term is minimized independently by a nearest integer. Zero is a simultaneous minimizer exactly when $|u_j|\le1/2$ for every $j$. Therefore $V_Q(\Lambda)=P$, including boundary ties.

For a sheared square, $B=\left(\begin{smallmatrix}1&s\\0&1\end{smallmatrix}\right)$ and $D=I$ give \[ Q=\begin{pmatrix}1&-s\\-s&1+s^2\end{pmatrix}, \qquad\det Q=1. \] Its off-diagonal entries are part of the solution. Requiring the Euclidean metric for the original lattice, or requiring $Q$ to be diagonal in the original coordinates, would unnecessarily discard these valid examples.

Proposition 477.4: an explicit metric for every central hexagon.

Let $P$ be a strictly convex centrally symmetric hexagon, with vertices in counterclockwise order $a,b,c,-a,-b,-c$. Here strict convexity means that consecutive edges are not collinear; a degenerate six-vertex presentation reduces to a parallelogram. The centre is in the interior. The vectors $a,c$ are independent and their positive cone contains $b$. After an invertible linear change of coordinates, write \[ a=(1,0),\qquad c=(0,1),\qquad b=(u,v). \] Counterclockwise turning at $a,b,c$, respectively, gives \begin{equation}\label{wsref-0-61-34-37-37-3a-68-65-78-63-6f-6e-64-69-74-69-6f-6e-73} u>0,\quad v>0,\quad u+v>1,\quad |u-v|<1. \end{equation} For example, the three edge cross products are $1+v-u$, $u+v-1$, and $1+u-v$. Conversely, these inequalities give the indicated convex hexagon with zero in its interior.

Define \begin{equation}\label{wsref-0-61-34-37-37-3a-68-65-78-6d-65-74-72-69-63} h=\frac{1-u^2-v^2}{2uv},\qquad Q=\begin{pmatrix}1&h\\h&1\end{pmatrix}. \end{equation} The inequalities (a477:hexconditions) are exactly what is needed for $-1<h<1$: these are equivalent to $(u-v)^2<1<(u+v)^2$. Hence $Q\succ0$. Furthermore \[ a^{\mathsf T}Qa=b^{\mathsf T}Qb =c^{\mathsf T}Qc=1. \] The six vertices lie on a centred ellipse for this metric.

Set \[ t_1=a+b,\qquad t_3=c-a,\qquad \Lambda=\mathbb Zt_1+\mathbb Zt_3,\qquad t_2=b+c=t_1+t_3. \] The three positive-edge lines are the $Q$-bisectors of $0$ and $t_1,t_2,t_3$, respectively. For the edge from $a$ to $b$, \[ (b-a)^{\mathsf T}Q(a+b)=b^{\mathsf T}Qb-a^{\mathsf T}Qa=0, \quad a^{\mathsf T}Q(a+b)=\tfrac12(a+b)^{\mathsf T}Q(a+b). \] The edge from $b$ to $c$ is identical with the letters shifted. For the edge from $c$ to $-a$, use the sum $c-a=t_3$ and the equality of the $Q$-norms of $c$ and $-a$. The opposite edges correspond to $-t_1,-t_2,-t_3$. Since zero is in the interior and every right-hand side is positive, the halfspaces have the required orientation. It follows that $V_Q(\Lambda)\subseteq P$.

There is no assumption here that this candidate lattice already tiles by $P$. Instead, calculate both volumes. The shoelace formula gives \[ \operatorname{area}P =\det(a,b)+\det(b,c)+\det(a,c)=u+v+1, \] while \[ \det\Lambda=\det(t_1,t_3) =\det\begin{pmatrix}1+u&-1\\v&1\end{pmatrix}=u+v+1. \] Equation (a477:volume) now forces $V_Q(\Lambda)=P$. This simultaneously constructs the lattice tiling and proves the affine Voronoi property. There is no circular use of a presumed tiling by the lattice just constructed.

For $u=v=1$, the metric is $Q=\left(\begin{smallmatrix}1&-1/2\\-1/2&1\end{smallmatrix}\right)$, with determinant $3/4$, and the lattice basis is $(2,1),(-1,1)$. The common area is $3$. For general $u,v$, the smallest eigenvalue is $1-|h|$, so it can tend to zero as the hexagon approaches a degenerate boundary. Strict inequalities are needed for each nondegenerate tile; a semidefinite limit alone is not a certificate for it.

Combining this result with Proposition 477.3 covers the planar classification of parallelohedra, namely parallelograms and centrally symmetric hexagons, as recalled in [S1]. The classification is an attributed input. The metrics and the volume proof for those two families are supplied explicitly above.

A closure operation giving higher-dimensional examples.

If $P_i=V_{Q_i}(\Lambda_i)$ for $i=1,2$, then \begin{equation}\label{wsref-0-61-34-37-37-3a-70-72-6f-64-75-63-74} P_1\times P_2 =V_{\operatorname{diag}(Q_1,Q_2)}(\Lambda_1\times\Lambda_2). \end{equation} The squared distance is a sum of two independent squared distances. If each coordinate is nearest to zero, every lattice pair has at least as large a total distance. Conversely, testing lattice pairs with one coordinate zero enforces the two separate conditions. Thus the equality holds, and affine images of these products are also covered by Lemma 477.1. Products of the explicit planar examples and intervals give verified families in every dimension.

This does not exhaust higher-dimensional parallelohedra. In particular, a proof that every tile decomposes into these products has not been given and is not part of the cited tiling structure theorem. The five-dimensional theorem in [S1] uses substantially richer local combinatorics; its dimension-specific analysis is not replaced by the elementary product construction.

Verification and the exact general gap.

The rational checks accompanying this attempt test the explicit hexagon formula on many positive rational $u,v$ satisfying (a477:hexconditions): positive determinant, equal vertex norms, all three edge orthogonality relations, and equality of polygon area with lattice covolume. These are checks of algebraic identities, not a classification of tiles. The preceding symbolic proof establishes the family for all admissible real $u,v$.

The proposed general route has two distinct tasks. First obtain the facet-neighbour data of the canonical lattice tiling, using the known structure theory. Then prove that its linear compatibility equations have a solution meeting the positive-definite cone and all positive facet scalings. Lemma 477.2 would finish the argument once that second task were accomplished. Local positivity, isolated low-dimensional sections, and examples from products do not establish it for arbitrary facet systems. The indefinite-matrix diagnostic shows why positivity cannot simply be omitted from the task.

UNRESOLVED. We have a proved finite certificate for a specified tile and lattice, explicit solutions for parallelotopes and all planar central hexagons, and closure under products and affine maps. Existence of such a certificate for every parallelohedron in unrestricted dimension is not proved here. The known five-dimensional theorem is credited to [S1], and no new general solution is claimed.

Sources

• S1 A. Garber, Voronoi conjecture for five-dimensional parallelohedra, Inventiones Mathematicae $2025$; arXiv:1906.05193. https://arxiv.org/abs/1906.05193. Formulation source.
• E1 Alexey Garber, Voronoi conjecture for five-dimensional parallelohedra, arXiv:1906.05193v6, 31 January 2025, Sections 1--2 and Theorem 4.1; full-text version of [S1]. https://arxiv.org/html/1906.05193v6. Used for the scope of the known theorem and the face-to-face lattice reduction; the elementary matrix and planar arguments here are rederived.