# An isoperimetric problem for three-dimensional parallelohedra

Zsolt Lángi  Morphodynamics Research Group and Department of Geometry,
Budapest University of Technology, Egry József utca 1., Budapest 1111, Hungary
<EMAIL_ADDRESS>
###### Abstract.

The aim of this note is to investigate isoperimetric-type problems for
$3$-dimensional parallelohedra; that is, for convex polyhedra whose translates
tile the $3$-dimensional Euclidean space. Our main result states that among
$3$-dimensional parallelohedra with unit volume the one with minimal mean
width is the regular truncated octahedron.

###### Key words and phrases:

parallelohedra, zonotopes, discrete isoperimetric problems, tiling, Kepler’s
conjecture, honeycomb conjecture, Kelvin’s conjecture

###### 2010 Mathematics Subject Classification:

52B60, 52A40, 52C22

The author is supported by the National Research, Development and Innovation
Office, NKFI, K-134199, the János Bolyai Research Scholarship of the Hungarian
Academy of Sciences, and grants BME IE-VIZ TKP2020 and ÚNKP-20-5 New National
Excellence Program by the Ministry of Innovation and Technology.

##  1\. Introduction

Among the convex polyhedra in Euclidean $3$-space $\mathbb{R}^{3}$ most known
both inside and outside mathematics, we find the so-called _parallelohedra_ ;
that is, the convex polyhedra whose translates tile space. All parallelohedra
in $3$-space can be defined as the Minkowski sums of at most six segments with
prescribed linear dependencies between the generating segments, and thus, they
are a subclass of _zonotopes_. Parallelohedra in $\mathbb{R}^{3}$ are also
related to the Voronoi cells of lattices via Voronoi’s conjecture, stating
(and proved in $\mathbb{R}^{3}$, see [6, 10, 12]) that any parallelohedron is
an affine image of such a cell. Well-known examples of parallelohedra are the
cube, the regular rhombic dodecahedron, and the regular truncated octahedron,
which are the Voronoi cells generated by the integer, the face-centered cubic
and the body-centered cubic lattice, respectively.

Many papers investigate isoperimetric problems for zonotopes (see, e.g. [1, 2,
21]), with special attention on the relation between mean width and volume
(see [5, 9]). On the other hand, apart from the celebrated proof of Kepler’s
Conjecture by Hales [16], which, as a byproduct, implies that among
parallelohedra with a given inradius, the one with minimum volume is the
regular rhombic dodecahedron, there is no known isoperimetric-type result for
parallelohedra.

In our paper we offer a new representation of $3$-dimensional parallelohedra
that might be useful to investigate their geometric properties. We use this
representation to prove Theorem 1.

######  Theorem 1.

Among unit volume $3$-dimensional parallelohedra, regular truncated octahedra
have minimal mean width.

The structure of the paper is as follows. In Section 2 we introduce and
parametrize $3$-dimensional parallelohedra, and describe the tools and the
notation necessary to prove our result. In Section 3 we prove Theorem 1.
Finally, in Section 4, we collect additional remarks, and recall some old, and
raise some new problems.

##  2\. Preliminaries

###  2.1. Properties of three-dimensional parallelohedra

Minkowski [23] and Venkov [30] proved that $P$ is a convex $n$-dimensional
polytope for which there is a face-to-face tiling of $\mathbb{R}^{n}$ with
translates of $P$ if and only if

  * (i)

$P$ and all its facets are centrally symmetric, and

  * (ii)

the projection of $P$ along any of its $(n-2)$-dimensional faces is a
parallelogram or a centrally symmetric hexagon.

Dolbilin [7] showed that these properties are already necessary if we require
only the existence of a not necessarily face-to-face tiling with translates of
$P$. The polytopes satisfying properties (i-ii) are called _$n$ -dimensional
parallelohedra_.

The combinatorial classes of $3$-dimensional parallelohedra are well known
(cf. Figure 1). More specifically, any $3$-dimensional parallelohedron is
combinatorially isomorphic to one of the following:

  1. (1)

a cube,

  2. (2)

a hexagonal prism,

  3. (3)

Kepler’s rhombic dodecahedron (which we call a regular rhombic dodecahedron),

  4. (4)

an elongated rhombic dodecahedron,

  5. (5)

a regular truncated octahedron.

We call a parallelohedron combinatorially isomorphic to the polyhedron in (i)
a _type (i) parallelohedron_. It is also known that every parallelohedron $P$
in $\mathbb{R}^{3}$ is a zonotope. More specifically, a type (1)-(5)
parallelohedron can be attained as the Minkowski sum of $3,4,4,5,6$ segments,
respectively. A typical parallelohedron is of type (5), as every other
parallelohedron can be obtained by removing some of the generating vectors of
a type (5) parallelohedron.

Since every $3$-dimensional parallelohedron $P\subset\mathbb{R}^{3}$ is a
zonotope, every edge of $P$ is a translate of one of the segments generating
$P$. Furthermore, for any generating segment $S$, the faces of $P$ that
contain a translate of $S$ form a _zone_ , i.e. they can be arranged in a
sequence $F_{1},F_{2},\ldots,F_{k},F_{k+1}=F_{1}$ of faces of $P$ such that
for all values of $i$, $F_{i}\cap F_{i+1}$ is a translate of $S$. By property
(ii) in the list in the beginning of Section 2, this sequence contains $4$ or
$6$ faces. In these cases we say that $S$ generates a $4$-belt or a $6$-belt
in $P$, respectively. The numbers of $4$\- and $6$-belts of a type (i)
parallelohedron are $3$ and $0$, $3$ and $1$, $0$ and $4$, $1$ and $4$, and
$0$ and $6$ for $i=1,2,3,4,5$, respectively.

Figure 1. The five combinatorial types of $3$-dimensional parallelohedra. The
type (5) parallelohedron in the picture is the regular truncated octahedron
generated by the six segments connecting the midpoints of opposite edges of a
cube. The type (3) polyhedron is the regular rhombic dodecahedron generated by
the four diagonals of the same cube. The rest of the polyhedra in the picture
are obtained by removing some generating segments from the type (5)
polyhedron.

######  Remark 1.

Let $P=\sum_{i=1}^{m}[o,v_{i}]\subset\mathbb{R}^{3}$ be a parallelohedron, and
set $I=\\{1,2,\ldots,m\\}$. Then the volume, the surface area and the mean
width of $P$ is

(1) $\operatorname{vol}_{3}(P)=\sum_{\\{i,j,k\\}\subset I}|V_{ijk}|,$ (2)
$\operatorname{surf}(P)=2\sum_{\\{i,j\\}\subset I}|v_{i}\times v_{j}|,\hbox{
and}$ (3) $w(P)=\frac{1}{2}\sum_{i=1}^{m}|v_{i}|,$

respectively, where $V_{ijk}$ is the determinant of the matrix with column
vectors $v_{i},v_{j},v_{k}$.

###### Proof.

The formula in (1) is the well-known volume formula for zonotopes (cf. e.g.
[28]). The one in (2) can be proved directly for type (5) parallelohedra,
which implies its validity for all parallelohedra. Finally, Steiner’s formula
[27] yields that the second quermassintegral $W_{2}(P)$ of $P$ is
$\frac{1}{6}\sum_{ij}(\pi-\alpha_{ij})l_{ij}$, where $\alpha_{ij}$ and
$l_{ij}$ is the dihedral angle and the length of the edge between the $i$th
and $j$th faces of $P$. Thus, from the zone property it follows that
$W_{2}(P)=\frac{\pi}{3}\sum_{i=1}^{m}|v_{i}|$. On the other hand, for any
convex body $K\subset\mathbb{R}^{3}$ we have $w(K)=\frac{3}{2\pi}W_{2}(K)$,
which implies (3). ∎

If we choose the generating segments in the form $[o,p_{i}]$ for some vectors
$p_{i}\in\mathbb{R}^{3}$, then in case of a type (1) or type (4)
parallelohedron, the vectors $v_{i}$ are in general position, i.e. any three
of them are linearly independent, whereas in case of the other types there are
triples of vectors that are restricted to be co-planar. In particular, if $P$
is a type (5) parallelohedron, then each of the four pairs of hexagon faces
defines a linear dependence relation on the generating segments. More
specifically, if $v_{1},v_{2},v_{3},v_{4}$ are normal vectors of the four
pairs of hexagon faces of the parallelohedron, then the directions of the six
generating segments are the ones defined by the six cross-products
$v_{i}\times v_{j}$, $i\neq j$. Note that all triples of the $v_{i}$s are
linearly independent, as otherwise some of the generating segments of $P$ are
collinear. Thus, we have

(4) $P=\sum_{1\leq i<j\leq 4}[o,\beta_{ij}(v_{i}\times v_{j})]$

for some real numbers $\beta_{ij}$, which, without loss of generality, we may
assume to be non-negative. Every $3$-dimensional parallelohedron can be
obtained from a type (5) parallelohedron by removing some of the generating
segments, or equivalently, by setting some of the $\beta_{ij}$s to be zero. In
particular, $P$ is of type (5), or (4), or (3), or (2), or (1) if and only if

  * •

$\beta_{ij}>0$ for all $i\neq j$, or

  * •

exactly one of the $\beta_{ij}$s is zero, or

  * •

exactly two of the $\beta_{ij}$s is zero:
$\beta_{i_{1}j_{1}}=\beta_{i_{2},j_{2}}=0$, and they satisfy
$\\{i_{1},j_{1}\\}\cap\\{i_{2},j_{2}\\}=\emptyset$, or

  * •

exactly two of the $\beta_{ij}$s is zero:
$\beta_{i_{1}j_{1}}=\beta_{i_{2},j_{2}}=0$, and they satisfy
$\\{i_{1},j_{1}\\}\cap\\{i_{2},j_{2}\\}\neq\emptyset$, or

  * •

exactly three of the $\beta_{ij}$s are zero. and there is no
$s\in\\{1,2,3,4\\}$ such that the indices of all nonzero $\beta_{ij}$s contain
$s$, respectively.

It is easy to check that in the remaining cases $P$ is planar.

Thus, we can use this representation for all parallelohedra appearing in the
paper with the assumption that $\beta_{ij}\geq 0$ for all $1\leq i<j\leq 4$.

Since no three of the vectors $v_{i}$ are linearly independent, but clearly
all four of them are, up to multiplying by a constant they have a unique
nontrivial linear combination
$\lambda_{1}v_{i}+\lambda_{2}v_{2}+\lambda_{3}v_{3}+\lambda_{4}v_{4}=o$. Since
in the representation of $P$ we considered only the directions of the
$v_{i}$s, we may clearly assume that $v_{1}+v_{2}+v_{3}+v_{4}=o$. This implies
that the tetrahedron $\operatorname{conv}\\{v_{1},v_{2},v_{3},v_{4}\\}$ is
_centered_ , that is, its center of gravity is $o$. From the equation
$v_{1}+v_{2}+v_{3}+v_{4}=o$ it follows that the absolute values of the
determinants of any three of the $v_{i}$s are equal. We choose this common
value $1$, which yields that the volume of the tetrahedron is $\frac{2}{3}$.
Throughout the paper for any $i,j,k\in\\{1,2,3,4\\}$, we denote by $V_{ijk}$
the value of the determinant with $v_{i},v_{j},v_{k}$ as column vectors. We
choose the indices in such a way that $V_{123}=1$. Since
$v_{1}+v_{2}+v_{3}+v_{4}=o$, we have that for any
$\\{i,j,s,t\\}=\\{1,2,3,4\\}$, the plane containing $o,v_{i},v_{j}$ strictly
separates $v_{s}$ and $v_{t}$, which implies that $V_{ijs}=-V_{ijt}$.

In the proof we often use the function $f:\mathbb{R}^{6}\to\mathbb{R}$,

(5)
$f(\tau_{12},\tau_{13},\tau_{14},\tau_{23},\tau_{24},\tau_{34})=\tau_{12}\tau_{13}\tau_{23}+\tau_{12}\tau_{14}\tau_{24}+\tau_{13}\tau_{14}\tau_{34}+\tau_{23}\tau_{24}\tau_{34}+$
$+(\tau_{12}+\tau_{34})(\tau_{13}\tau_{24}+\tau_{14}\tau_{23})+(\tau_{13}+\tau_{24})(\tau_{12}\tau_{34}+\tau_{14}\tau_{23})+(\tau_{14}+\tau_{23})(\tau_{12}\tau_{34}+\tau_{13}\tau_{24}).$

We remark that $f$ is _not_ the third elementary symmetric function on six
variables, since after expansion it has only $16<20=\binom{6}{3}$ members.

An elementary computation yields that for any $i,j,k,l,s,t\in\\{1,2,3,4\\}$,
we have

$|v_{i}\times v_{j},v_{k}\times v_{l},v_{s}\times
v_{t}|=V_{ijl}V_{stk}-V_{ijk}V_{stl}.$

Using this formula and the properties of cross-products, we may express the
volume, surface area and mean width of $P$ in our representation.

######  Remark 2.

For the parallelohedron $P=\sum_{i\neq j}\beta_{ij}[o,v_{i}\times v_{j}]$
satisfying the conditions above, we have

(6)
$\operatorname{vol}_{3}(P)=f(\beta_{12},\beta_{13},\beta_{14},\beta_{23},\beta_{24},\beta_{34}),$
(7)
$\operatorname{surf}(P)=2\big{(}(\beta_{12}\beta_{13}+\beta_{12}\beta_{14}+\beta_{13}\beta_{14})|v_{1}|+\ldots(\beta_{14}\beta_{24}+\beta_{14}\beta_{34}+\beta_{24}\beta_{34})|v_{4}|+$
$+\beta_{12}\beta_{34}(|v_{1}+v_{2}|)+\ldots+\beta_{14}\beta_{23}|v_{1}+v_{4}|\big{)},$
(8) $w(P)=\frac{1}{2}\sum_{1\leq i<j\leq 4}\beta_{ij}|v_{i}\times v_{j}|,$

respectively.

###  2.2. Preliminary lemmas

The following lemma is used more than once in the paper.

######  Lemma 1.

Let $T=\operatorname{conv}\\{p_{1},p_{2},p_{3},p_{4}\\}$ be an arbitrary
centered tetrahedron with volume $V>0$ where the vertices are labelled in such
a way that the determinant with columns $p_{1},p_{2},p_{3}$ is positive. For
any $\\{i,j,s,t\\}=\\{1,2,3,4\\}$, let $\gamma_{ij}=-\langle
p_{s},p_{t}\rangle$ and $\zeta_{ij}=\gamma_{ij}|p_{i}\times p_{j}|^{2}$. Then

  1. (1)

$f(\gamma_{12},\gamma_{13},\gamma_{14},\gamma_{23},\gamma_{24},\gamma_{34})=\frac{9}{4}V^{2}$,

  2. (2)

$\sum_{1\leq i<j\leq 4}\zeta_{ij}=\frac{27}{4}V^{2}$,

where $f$ is the function defined in (5).

###### Proof.

Let $\chi_{ij}=\langle p_{i},p_{j}\rangle$ for all $i,j$. Consider the Gram
matrix $G$ defined by $p_{1},p_{2},p_{3}$, and observe that as $T$ is
centered, the volume of the parallelepiped spanned by them is $\frac{3}{2}V$.
Since the determinant of the Gram matrix of $n$ linearly independent vectors
in $\mathbb{R}^{n}$ is the square of the volume of the parallelotope spanned
by the vectors, it follows that $\det(G)=\frac{9}{4}V^{2}$. Furthermore, since
$T$ is centered, we have $\chi_{14}=-\chi_{11}-\chi_{12}-\chi_{13}$, and we
may obtain similar formulas for $\chi_{24}$ and $\chi_{34}$. Now, substituting
these formulas into
$f(\gamma_{12},\ldots,\gamma_{34})=f(-\chi_{34},\ldots,-\chi_{12})$, we obtain
that

$f(\gamma_{12},\ldots,\gamma_{34})=\chi_{11}\chi_{22}\chi_{33}+2\chi_{12}\chi_{13}\chi_{23}-\chi_{11}\chi_{2}^{2}-\chi_{22}\chi_{13}^{2}-\chi_{33}\chi_{12}^{2}=\det(G),$

which implies the first identity.

To obtain the second identity, observe that for any
$\\{i,j,s,t\\}=\\{1,2,3,4\\}$, we have $\zeta_{ij}=-\langle
p_{s},p_{t}\rangle|p_{i}\times
p_{j}|^{2}=-\chi_{st}\left(\chi_{ii}\chi_{jj}-\chi_{ij}^{2}\right)$. This
observation and an argument similar to the one in the previous paragraph
yields that $\sum_{1\leq i<j\leq 4}\zeta_{ij}=3\det(G)=\frac{27}{4}V^{2}$. ∎

######  Lemma 2.

Under the condition that $\sum_{1\leq i<j\leq 4}\tau_{ij}=C>0$ and
$\tau_{ij}\geq 0$ for all $1\leq i<j\leq 4$, we have

$f(\tau_{12},\tau_{13},\tau_{14},\tau_{23},\tau_{24},\tau_{34})\leq\frac{2C^{3}}{27},$

with equality if and only if $\tau_{ij}=\frac{C}{6}$ for all $1\leq i<j\leq
4$. Furthermore, if, in addition, $\tau_{34}=0$, then

$f(\tau_{12},\tau_{13},\tau_{14},\tau_{23},\tau_{24},0)\leq\frac{16C^{3}}{243},$

with equality if and only if
$2\tau_{12}=\tau_{13}=\ldots=\tau_{24}=\frac{2C}{9}$.

###### Proof.

Without loss of generality, we assume that $C=1$. Since the set of points in
$\mathbb{R}^{6}$ satisfying the conditions of Lemma 2 is compact, $f$ attains
its maximum on it. Assume that $f$ is maximal at some point
$\tau=(\tau_{12},\ldots,\tau_{34})$.

_Case 1_ , $\tau$ has only positive coordinates. By the Lagrange multiplier
method, the gradient of $f$ at $\tau$ is is parallel to the gradient of the
function $\sum_{i\neq j}\tau_{ij}-1$ defining the condition. In other words,
all partial derivatives of $f$ are equal. Let us assume that this common value
is some $t\in\mathbb{R}$. Since $f$ is a linear function of any of its
variables, $\partial_{\tau_{ij}}f$ is equal to the coefficient of $\tau_{ij}$
in $f$. On the other hand, any member of the sum in $f$ contributes to partial
derivatives with respect to three of the variables. Thus, we have that
$t=\sum_{i\neq j}t\tau_{ij}=\sum_{i\neq
j}(\partial_{\tau_{ij}}f)(\tau)\tau_{ij}=3f(\tau)$.

Computing the partial derivatives, we have that

$(\partial_{\tau_{12}}f)(\tau)=\tau_{13}\tau_{23}+\tau_{14}\tau_{24}+\tau_{13}\tau_{24}+\tau_{14}\tau_{23}+(\tau_{13}+\tau_{24}+\tau_{14}+\tau_{23})\tau_{34},$

and we obtain similar expressions for all partial derivatives. Solving the
system of equations $(\partial_{\tau_{1i}}f)(\tau)=3f(\tau)$, $i=2,3,4$ for
$\tau_{12},\tau_{13},\tau_{14}$, and substituting back the solutions into the
definition of $f$, we obtain that
$f(\tau)=\frac{4}{27}(\tau_{23}+\tau_{24}+\tau_{34})$. By a similar argument,
$f(\tau)=\frac{4}{27}(\tau_{12}+\tau_{13}+\tau_{23})=\frac{4}{27}(\tau_{12}+\tau_{14}+\tau_{24})=\frac{4}{27}(\tau_{13}+\tau_{14}+\tau_{34})$
also follows. Adding up, we obtain that in this case
$4f(\tau)=\frac{8}{27}\sum_{i\neq j}\tau_{ij}=\frac{8}{27}$, implying that
$f(\tau)=\frac{2}{27}$.

We show that in case of equality all coordinates of $\tau$ are equal. The
equations in the previous paragraph yield that
$\tau_{12}+\tau_{13}+\tau_{23}=\ldots=\tau_{23}+\tau_{24}+\tau_{34}=\frac{1}{2}$,
and also that
$\tau_{12}+\tau_{13}+\tau_{14}=\ldots=\tau_{14}+\tau_{24}+\tau_{34}=\frac{1}{2}$.
Comparing suitable equations it readily follows that $\tau_{12}=\tau_{34}$,
$\tau_{13}=\tau_{24}$ and $\tau_{14}=\tau_{23}$. Replacing $\tau_{34}$,
$\tau_{24}$ and $\tau_{23}$ by $\tau_{12},\tau_{13}$ and $\tau_{14}$,
respectively, in the equations
$(\partial_{\tau_{12}}f)(\tau)=(\partial_{\tau_{13}}f)(\tau)=(\partial_{\tau_{14}}f)(\tau)$,
we obtain $\tau_{12}^{2}=\tau_{13}^{2}=\tau_{14}^{2}$, which characterizes
equality in this case.

_Case 2_ , if exactly one coordinate of $\tau$ is zero. Without loss of
generality, we may assume that $\tau_{34}=0$, and let
$\tau^{\prime}\in\mathbb{R}^{5}$ be the vector obtained by removing the last
coordinate of $\tau$. Let us define the function
$g(\tau_{12},\ldots,\tau_{24})=f(\tau_{12},\ldots,\tau_{24},0)$. Similarly
like in Case 1, if $f(\tau)$ is a local maximum, we have that all five partial
derivatives of $g$ are equal at $\tau^{\prime}$. Again, from this it follows
that this common value is equal to $3g(\tau^{\prime})$. Factorizing the left-
hand sides of the equations $\partial_{\tau_{13}}-\partial_{\tau_{14}}=0$ and
$(\partial_{\tau_{23}}g)(\tau^{\prime})-(\partial_{\tau_{24}}g)(\tau^{\prime})=0$,
we obtain that $\tau_{13}=\tau_{14}$, $\tau_{23}=\tau_{24}$, respectively.
After substituting these equalities into the equation
$(\partial_{\tau_{13}}g)(\tau^{\prime})-(\partial_{\tau_{23}}g)(\tau^{\prime})=0$
and factorizing the left-hand side, we obtain that $\tau_{13}=\tau_{23}$.
Again, substituting back into
$(\partial_{\tau_{12}}g)(\tau^{\prime})-(\partial_{\tau_{13}}g)(\tau^{\prime})=0$
and factorizing its left-hand side yields $\tau_{13}=2\tau_{12}$. From this,
we have $\tau_{12}=\frac{1}{9}$ and $\tau_{ij}=\frac{2}{9}$ for any
$i\in\\{1,2\\}$ and $j\in\\{3,4\\}$, which implies that
$g(\tau^{\prime})=\frac{16}{243}$.

_Case 3_ , if at least two coordinates of $\tau$ are zero. Then, using an
argument similar to that in the previous cases, we obtain that the maximum of
$f(\tau)$ is $\frac{1}{16}$, $\frac{1}{27}$ and $0$ if exactly two, exactly
three or more than three coordinates of $\tau$ are zero, respectively. ∎

##  3\. Proof of Theorem 1

Recall from Subsection 2.1 that we represent $P$ in the form

$P=\sum_{1\leq i<j\leq 4}[o,\beta_{ij}v_{i}\times v_{j}]$

for some $v_{1},v_{2},v_{3},v_{4}\in\mathbb{R}^{3}$ satisfying
$\sum_{i=1}^{4}v_{i}=o$, and for some $\beta_{ij}\geq 0$. By our assumptions,
$|V_{ijk}|=1$ for any $\\{i,j,k\\}\subset\\{1,2,3,4\\}$, where $V_{ijk}$
denotes the determinant with $v_{i},v_{j},v_{k}$ as columns, and we assumed
that $V_{123}=1$, which implies, in particular, that $V_{124}=-1$. Then
$T=\operatorname{conv}\\{v_{1},v_{2},v_{3},v_{4}\\}$ is a centered
tetrahedron, with volume $\operatorname{vol}_{3}(T)=\frac{2}{3}$.

By Remark 2, we need to find the minimum value of $w(P)=\frac{1}{2}\sum_{1\leq
i<j\leq 4}\beta_{ij}|v_{i}\times v_{j}|$ under the conditions that
$T=\operatorname{conv}\\{v_{1},v_{2},v_{3},v_{4}\\}$ is a centered tetrahedron
with volume $\operatorname{vol}_{3}(T)=\frac{2}{3}$, and the value of
$f(\beta_{12},\ldots,\beta_{34})$ being fixed. Equivalently, we need to find
the maximum value of $f(\beta_{12},\ldots,\beta_{34})$ under the condition
that $w(P)=\frac{1}{2}\sum_{1\leq i<j\leq 4}\beta_{ij}|v_{i}\times v_{j}|$ is
fixed, where $T=\operatorname{conv}\\{v_{1},v_{2},v_{3},v_{4}\\}$ is an
arbitrarily chosen centered tetrahedron with volume
$\operatorname{vol}_{3}(T)=\frac{2}{3}$.

Our main goal is to reduce this optimization problem to the special case where
$T$ is a regular tetrahedron, and then apply Lemma 2. We do it in two steps.
In the first step we find an upper bound for $f$ depending only on $T$ (and
thus, we eliminate all $\beta_{ij}$s from the conditions). In the second step
we show that our upper bound from the first step can be bounded from above by
a value of $f$ under the additional condition that $T$ is regular (or in other
words, we eliminate the tetrahedron $T$ from the conditions at the price of
bringing back the parameters $\beta_{ij}$).

_Step 1_.
Our main tool in this step is Lemma 3, proved by Petty [25] and later
rediscovered by Giannopoulos and Papadimitrakis [14] for arbitrary convex
bodies (see (3.4) in [14]). Before stating it, we remark that any affine image
of a parallelohedron is also a parallelohedron.

######  Lemma 3.

Let $P\subset\mathbb{R}^{n}$ be a convex polytope with outer unit facet
normals $u_{1},\ldots,u_{k}$. Let $F_{i}$ denote the $(n-1)$-dimensional
volume of the $i$th facet of $P$. Then, up to congruence, there is a unique
volume preserving affine transformation $L$ such that
$\operatorname{surf}(L(P))$ is maximal in the affine class of $P$.
Furthermore, $P$ satisfies this property if and only if its surface area
measure is isotropic, that is, if

(9) $\sum_{i=1}^{k}\frac{nF_{i}}{\operatorname{surf}(P)}u_{i}\otimes
u_{i}=\operatorname{Id}$

where $\operatorname{Id}$ denotes the identity matrix.

Any convex polytope satisfying the conditions in (9) is said to be in _surface
isotropic position_. We note that the volume of the projection body of any
convex polyhedron is invariant under volume preserving linear transformations
(cf. [26]). On the other hand, from Cauchy’s projection formula and the
additivity of the support function (see. [13]) it follows that the projection
body of the polytope in Lemma 3 is the zonotope
$\sum_{i_{1}}^{k}[o,F_{i}u_{i}]$. Note that the solution to Minkowski’s
problem [27] states that some unit vectors
$u_{1},\ldots,u_{k}\in\mathbb{R}^{n}$ and positive numbers
$F_{1},\ldots,F_{k}$ are the outer unit normals and volumes of the facets of a
convex polytope if and only if the $u_{i}$s span $\mathbb{R}^{n}$, and
$\sum_{i=1}^{k}F_{i}u_{i}=o$. On the other hand, a translate of the
parallelohedron $P$ in our investigation can be written in the form
$\frac{1}{2}\sum_{1\leq i<j\leq 4}[o,\pm\beta_{ij}v_{i}\times v_{j}]$, which,
by the previous observation, is the projection body of a centrally symmetric
polytope with outer unit facet normals $\pm\frac{v_{i}\times
v_{j}}{|v_{i}\times v_{j}|}$, and volumes of the corresponding facets
$\frac{\beta_{ij}}{2}|v_{i}\times v_{j}|$, where $1\leq i<j\leq 4$. Since
$u\otimes u=(-u)\otimes(-u)$ for any $u\in\mathbb{R}^{n}$, any solution $P$ to
our optimization problem satisfies the conditions in (9) with the vectors
$\frac{v_{i}\times v_{j}}{|v_{i}\times v_{j}|}$ in place of the $u_{i}$s, and
the quantities $\beta_{ij}|v_{i}\times v_{j}|$ in place of the $F_{i}$s. Thus,
applying also the formula in Remark 2 for $w(P)$, it follows that

(10) $\sum_{1\leq i<j\leq 4}\frac{3\beta_{ij}}{2w(P)|v_{i}\times
v_{j}|}(v_{i}\times v_{j})\otimes(v_{i}\times v_{j})=\operatorname{Id},$

which in the following we assume to hold for $P$.

Recall that $x\otimes y=xy^{T}$ and $\langle x,y\rangle=x^{T}y$ for any column
vectors $x,y\in\mathbb{R}^{n}$. Hence, multiplying both sides in (10) by
$v_{3}^{T}$ from the left and $v_{4}$ from the right, it follows that

$\langle v_{3},v_{4}\rangle=\frac{3\beta_{12}}{2w(P)|v_{1}\times
v_{2}|}V_{123}V_{124}.$

Since $V_{123}=-V_{124}=1$, we can express $\beta_{12}$ as
$\beta_{12}=-\langle v_{3},v_{4}\rangle|v_{1}\times v_{2}|\frac{2w(P)}{3}$. By
symmetry, for any $\\{i,j,s,t\\}=\\{1,2,3,4\\}$ it follows that
$\beta_{ij}=-\langle v_{s},v_{t}\rangle|v_{i}\times v_{j}|\frac{2w(P)}{3}$,
where it may be worth noting that as $\beta_{ij}\geq 0$ holds for any value of
$i,j$, if $P$ satisfies (10), then $\langle v_{s},v_{t}\rangle\leq 0$ for all
$s,t$. Set $\zeta_{ij}=-\langle v_{s},v_{t}\rangle|v_{i}\times
v_{j}|=\frac{3\beta_{ij}}{2w(P)}$ for all $i,j$. Substituting back these
quantities into the formulas for $\operatorname{vol}_{3}(P)$ and $w(P)$ and
simplifying, we may rewrite our optimization problem in the following form:
find the maximum of
$\frac{27\operatorname{vol}_{3}(P)}{\left(2w(P)\right)^{3}}=f(\zeta_{12},\ldots,\zeta_{34})$,
where $\zeta_{ij}$ is defined as above, in the family of all centered
tetrahedra $T$ with $\operatorname{vol}_{3}(T)=\frac{2}{3}$, under the
condition that $\sum_{1\leq i<j\leq 4}\zeta_{ij}|v_{i}\times v_{j}|=3$. Here,
it is worth noting that the last condition is satisfied for any centered
tetrahedron $T$ with volume $\frac{2}{3}$ by Lemma 1, and thus, it is
redundant.

_Step 2_.
Consider the function

$f(\zeta_{12},\ldots,\zeta_{34}),\hbox{ where }\zeta_{ij}=-\langle
v_{s},v_{t}\rangle|v_{i}\times v_{j}|\hbox{ for }\\{i,j,s,t\\}=\\{1,2,3,4\\},$

and $v_{1},v_{2},v_{3},v_{4}$ are the vertices of a centered tetrahedron $T$
with volume $\frac{2}{3}$. Set $\gamma_{ij}=-\langle v_{s},v_{t}\rangle$ and
$\tau_{ij}=\gamma_{ij}|v_{i}\times v_{j}|^{2}$ for all
$\\{i,j,s,t\\}=\\{1,2,3,4\\}$. To give an upper bound on the value of $f$, we
apply the Cauchy-Schwartz Inequality, which states that for any nonnegative
real numbers $x_{i},y_{i}$, $i=1,2,\ldots,k$, we have
$\sum_{i=1}^{k}x_{i}y_{i}\leq\sqrt{\sum_{i=1}^{k}x_{i}^{2}}\sqrt{\sum_{i=1}^{k}y_{i}^{2}}$,
with equality if and only if the $x_{i}$s and the $y_{i}$s are proportional.
To do this, we write each member of $f$ as the product
$\zeta_{ij}\zeta_{kl}\zeta_{mn}=\sqrt{\gamma_{ij}\gamma_{kl}\gamma_{mn}}\sqrt{\tau_{ij}\tau_{kl}\tau_{mn}}$.
Thus, we obtain that

$f(\zeta_{12},\ldots,\zeta_{34})\leq\sqrt{f(\gamma_{12},\ldots,\gamma_{34})}\sqrt{f(\tau_{12},\ldots,\tau_{34})}=\sqrt{f(\tau_{12},\ldots,\tau_{34})},$

where we used the fact that by Lemma 1, $f(\gamma_{12},\ldots,\gamma_{34})=1$
for all centered tetrahedra with volume $\frac{2}{3}$. Furthermore, observe
that by Lemma 1 we have $\sum_{1\leq i\leq j\leq 4}\tau_{ij}=3$ for all such
tetrahedra. Hence, by Lemma 2, we have
$f(\zeta_{12},\ldots,\zeta_{34})\leq\sqrt{2}$.

Now, assume that $f(\zeta_{12},\ldots,\zeta_{34})=\sqrt{2}$. Then, by Lemma 2,
$\tau_{ij}=\frac{1}{2}$ for all values $i\neq j$. This, by the Cauchy-Schwartz
Inequality, implies that for some $t\in\mathbb{R}$, $\gamma_{ij}=t$ for all
values $i\neq j$. Since $T$ is centered, this implies that $\gamma_{ii}=3t$
for all $i$s. In other words, the Gram matrix of the vectors
$v_{1},\ldots,v_{4}$ is a scalar multiple of the matrix
$4\operatorname{Id}-E$, where $E$ is the matrix with all entries equal to $1$.
Since the Gram matrix of a vector system determines the vectors up to
orthogonal transformations, and it is easy to check that
$4\operatorname{Id}-E$ is the Gram matrix of the vertex set of a centered
regular tetrahedron of circumradius $\sqrt{3}$, the equality part in Theorem 1
follows.

##  4\. Remarks and open problems

First, in the next table we collect the results of our investigation for the
minimal values of the mean widths of the different types of parallelohedra.

Recall (cf. Section 2) that the different types of parallelohedra correspond
to the following polyhedra.

  * •

Type (1) parallelohedra: parallelopipeds.

  * •

Type (2) parallelohedra: hexagonal prisms.

  * •

Type (3) parallelohedra: rhombic dodecahedra.

  * •

Type (4) parallelohedra: elongated rhombic dodecahedra.

  * •

Type (5) parallelohedra: truncated octahedra.

Type |  Minimum of $w(P)$ |  Optimal parallelohedra
---|---|---
(1) |  $\frac{3}{2}$ |  cube
(2) |  $\frac{3^{7/6}}{2^{4/3}}\approx 1.43$ |  regular hexagon based right prism, with base and lateral edges of lengths $\frac{2^{2/3}}{3^{5/6}}$ and $\frac{3^{1/6}}{2^{1/3}}$, respectively
(3) |  $\frac{3^{1/2}}{2^{1/3}}\approx 1.37$ |  regular rhombic dodecahedron with edge length $\frac{\sqrt{3}}{2^{4/3}}$
(4) |  $\geq\frac{3^{4/3}}{2^{5/3}}\approx 1.36$ |  not known
(5) |  $\frac{3}{2^{7/6}}\approx 1.34$ |  regular truncated octahedron of edge length $\frac{1}{2^{7/6}}$
Table 1. Minima of the mean widths of different types of unit volume
parallelohedra

In the remaining part of this section we collect some old conjectures about
$3$-dimensional parallelohedra, and propose some new ones.

The origin of the Honeycomb Conjecture, stating that in a decomposition of the
Euclidean plane into regions of equal area, the regular hexagonal grid has the
least perimeter, can be traced back to ancient times [29]. This problem has
been in the focus of research throughout the second half of the 20th century
[8, 22], and was finally settled by Hales [15].

The most famous analogous conjecture for mosaics in $3$-dimensional Euclidean
space is due to Lord Kelvin [19], who in 1887 conjectured that in a tiling of
space with cells of unit volume, the mosaic with minimal surface area is
composed of slightly modified truncated octahedra. Even though this conjecture
was disproved by Weaire and Phelan in 1994 [31], who discovered a tiling of
space with two slighly curved ‘polyhedra’ of equal volume and with less total
surface area than in Lord Kelvin’s mosaic, the original problem of finding the
mosaics with equal volume cells and minimal surface area has been extensively
studied (see, e.g. [20, 11, 24]). On the other hand, in the author’s
knowledge, there is no subfamily of mosaics for which Kelvin’s problem is
solved. This is our motivation for the following conjecture, where we note
that a $3$-dimensional parallelohedron of unit volume has minimal surface area
if and only if the same holds for the translative, convex mosaic generated by
it.

######  Conjecture 1.

Among $3$-dimensional, unit volume parallelohedra $P$,
$\operatorname{surf}(P)$ is minimal if and only if $P$ is a regular truncated
octahedron.

The following conjecture, called Rhombic Dodecahedral Conjecture, can be found
in [1] (see also [2]). We note that this conjecture is the lattice variant of
the so-called Strong Dodecahedral Conjecture [3] proved by Hales in [17]. For
other variants of the Dodecahedral Conjecture, see also [4] or [18].

######  Conjecture 2 (Bezdek, 2000).

The surface area of any parallelohedron in $\mathbb{R}^{3}$ with unit inradius
is at least as large as $12\sqrt{2}\approx 16.97$, which is the surface area
of the regular rhombic dodecahedron of unit inradius.

As it was mentioned in the introduction, Voronoi cells of lattice packings of
congruent balls is an important subfamily of $3$-dimensional parallelohedra.
Regarding this subclass, Daróczy [5] gave an example of a packing whose
Voronoi cells have a smaller mean width than that of the regular rhombic
dodecahedron of the same inradius. This is the motivation behind our last
question. Here we note that the mean width of a cube, a regular truncated
octahedron and a regular rhombic dodecahedron of unit inradius is
$3>\sqrt{6}=\sqrt{6}$, respectively, which, in our opinion, makes the question
indeed intriguing.

######  Problem 1.

Find the minimum of the mean widths of $3$-dimensional parallelohedra of unit
inradius.

Acknowledgements
The author is indebted to A. Joós and G. Domokos for many fruitful discussions
of this problem, and to an unkown referee for helpful suggestions.

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