Twofold Translative Tilings with Convex Bodies

0.14 MB·

by  Kirati Sriamorn

Twofold Translative Tilings with Convex Bodies
6
1
0 Kirati Sriamorn
2

n January 19, 2016
a
J

7 Abstract
1 Let K be a convex body. It is known that, in general, if K is a k-

fold translative tile (for some positive integer k), then K may not be
]
G a (onefold) translative tile. However, in this paper I will show that for
every convexbody K,K is atwofold translativetile if and only if K is a
M
translative tile.
.
h
t Keywords Multiple tiling · Twofold tiling · Convex body · Polytope ·
a
m Translative tile · Lattice tile
[ Mathematics Subject Classification 52C20 · 52C22

1
v 1 Introduction
2
1
LetD be a connectedsubsetofRn, andletk be apositiveinteger. We saythat
3
4 a family of convex bodies {K1,K2,...} is a k-fold tiling of D, if each point of
0 D which dose not lie in the boundary of any K , belongs to exactly k convex
i
. bodies of the family.
1
0 LetK beann-dimensionalconvexbody,andletX beadiscretemultisubset
6 of Rn. Denote by K+X the family
1
: {K+x: x∈X},
v
Xi where K +x = {y+x : y ∈ K}. We say that the family K +X is a k-fold
translative tiling with K, if K +X is a k-fold tiling of Rn. In particular, if
r
a X = Λ is a lattice, then K +Λ is called a k-fold lattice tiling with K. We call
K a k-fold translative (lattice) tile if there exists a k-fold translative (lattice)
tiling with K. A onefold tiling (tile) is simply called a tiling (tile).
LetP beann-dimensionalcentrallysymmetricpolytopewithcentrallysym-
metricfacets. Abelt ofP isthe collectionofitsfacetswhichcontainatranslate
of a given subfacet ((n−2)-face) of P.
For the case of onefold tilings, Venkov [1] and McMullen [2] proved the
following result.
Theorem 1.1. A convex body K is a translative tile if and only if K is a
centrally symmetric polytope with centrally symmetric facets, such that each belt
of K contains four or six facets.
1
Furthermore, a consequence of the proof of this result is that, every convex
translativetile is also a lattice tile. In the caseof general k-foldtilings, Gravin,
Robins and Shiryaev [3] proved that
Theorem 1.2. If a convex body K is a k-fold translative tile, for some positive
integer k, then K is a centrally symmetric polytope with centrally symmetric
facets.
Moreover,theyalsoshowedthat, everyrationalpolytope P thatiscentrally
symmetric and has centrally symmetric facets must be a k-fold lattice tile, for
some positive integer k. This resultimplies that there exists a polytope P such
that P is a k-fold translative tile (for some k > 1), but P is not a translative
tile. For example, the octagonshownin Fig. 1 is a 7-fold lattice tile, but is not
a translative tile.
Fig. 1: The octagon that is a 7-fold lattice tile
In this paper, I will prove the following surprising result:
Theorem 1.3. A convex body K is a twofold translative tile if and only if K
is a onefold translative tile.
In order to prove this result, I will modify the method used in [2]. As an
immediate consequence of Theorem 1.3, we have
Corollary 1.4. A convex body K is a twofold translative tile if and only if K
is a twofold lattice tile.
2 Some Notations
LetK be ann-dimensionalconvexbody, andletX be a discrete multisubsetof
Rn which contains the origin. Let q be a point on the boundary ∂K of K. We
define
X (q)={x∈X : q ∈K+x},
K
◦
X (q)={x∈X (q): q ∈int(K+x)},
K K
and
∂X (q)={x∈X (q): q ∈∂(K+x)}.
K K
In addition, we define
∂X (q)={x∈∂X (q): int(K)∩(K+x)=∅},
K K
2
K
K+x1
q
K+x2
K+x3
◦
Fig. 2: X (q) = {0,x ,x ,x } , X (q) = {x }, ∂X (q) = {0,x ,x } and
K 1 2 3 K 1 K 2 3
∂X (q)={x }
K 2
(see Fig. 2 for an example).
Now suppose that P is an n-dimensional centrally symmetric convex poly-
tope with centrally symmetric facets. Let G be a translate of a subfacet of P.
Denote by B (G) the belt of P determined by G. Let q be a point that lies in
P
a facet in B (G). Let S(G,q) be the (n−2)-dimensional plane that contains
P
the point q and parallels to G. We define
∂˙X (G,q)={x∈∂X (q): P ∩(P +x)⊂S(G,q)}.
P P
Let F be a subset of ∂P containing the point q, we define
∂X (G,F,q)={x∈∂X (q): S(G,q)∩F ∩(P +x)6=F ∩(P +x)},
P P
(see Fig. 3).
G P +x3
F2
q P P +x2 q P
F1
P +x1
Fig. 3: ∂˙X (G,q)={x }, ∂X (G,F ,q)={x } and ∂X (G,F ,q)={x }
P 2 P 1 1 P 2 3
Let F (G,q) be the union of those facets in B (G) which contain q. It
P P
is easy to see that, S(G,q) divides F (G,q) into two parts. After choosing a
P
3
direction, we may define these two parts F+(G,q) and F−(G,p) as shown in
P P
Fig. 4. DenotebyAng (G,q)theanglefromF+(G,q)toF−(G,q),anddenote
P P P
byang (G,q) the measureofAng (G,q)in radian. Obviously,ifq liesin some
P P
subfacet that parallels to G, then ang (G,q) < π, otherwise, ang (G,q) = π.
P P
We will denote by E+(G,q) the subfacet which parallels to G and is contained
P
in F+(G,q), but is not containing q (Fig. 4). We can also define E−(G,q) in
P P
the similar way.
E+(G,q)
P
F+(G,q)
P
G q G
P − P
F (G,q)
P
E+(G,q)
q P
−
EP(G,q) F+(G,q)
P
−
F (G,q)
− P
E (G,q)
P
Fig. 4: F+(G,q), F−(G,q), E+(G,q) and E−(G,q)
P P P P
3 Some Lemmas
Forapositiverealnumberεandapointp,denotebyB (p)theclosedballwith
ε
center p and radius ε.
Lemma 3.1. Suppose that K and K′ are convex bodies. If there exist a point
p∈∂K∩∂K′andapositiverealnumberεsuchthatB (p)∩int(K)∩int(K′)=∅,
ε
then int(K)∩int(K′)=∅.
Proof. Since B (p)∩K andB (p)∩K′ areconvex,by applyingthe basicresult
ε ε
ofConvexandDiscreteGeometry,we knowthatthereis ahyperplaneH which
separates B (p)∩K and B (p)∩K′. Assume that p′ ∈ int(K)∩int(K′). By
ε ε
the convexity,the line segment L between the point p and the point p′ must lie
in K ∩K′. Therefore, L∩B (p) must be contained in the hyperplane H, and
ε
hence p′ ∈ H. On the other hand, there is a positive real number δ such that
B (p′)⊂int(K)∩int(K′). So B (p′)⊂H, this is impossible.
δ δ
Lemma 3.2. Let D be a connected subset of Rn, and let k be a positive integer.
Suppose that a family of convex bodies {K ,K ,...} is a k-fold tiling of D.
1 2
We have that, for every i ∈ {1,2,...} and every point q ∈ ∂K , if q is an
i
interior point of D, then there must be a j ∈ {1,2,...} such that q ∈ ∂K and
j
int(K )∩int(K )=∅.
i j
Proof. For i6=j, let
Aj ={p∈∂K : B (p)∩∂K ∩int(K )=∅, for some ε>0},
i i ε j i
4
and
Bj ={p∈∂K \Aj : B (p)∩∂K ⊂K , for some ε>0}.
i i i ε j i
Wenotethat,ifp∈∂K \(Aj∪Bj),thenpmustliein∂K andforallε>0,we
i i i j
haveB (p)∩∂K ∩int(K )6=∅and(B (p)∩∂K )\K 6=∅(Fig. 5). Obviously,
ε j i ε j i
when K ∩K =∅, we have Aj =∂K and Bj =∅. We note that for a fixed i,
i j i i i
there are finitely many j suchthat K ∩K 6=∅. Hence, there are finitely many
i j
j such that Aj ∪Bj 6=∂K . Let
i i i
Ci = \Aji ∪Bij.
j6=i
By considering (n−1)-dimensional Lebesgue measure, one can show that the
closureofC is ∂K . Therefore,to provethe lemma,itsuffices to showthatthe
i i
statement is true for all q ∈C .
i
p1 p2
K K
j j
p3
p4
K K
i i
p7
p6
p5
(a) Ki andKj (b) points p1,p2,p3,p4,p5,p6
Fig. 5: Aj =∂K \{p ,p ,p ,p ,p }, Bj = {p ,p ,p } and ∂K \(Aj ∪Bj)=
i i 1 2 3 4 5 i 2 3 4 i i i
{p ,p }
1 5
Supposethatq ∈C ∩int(D). DenotebyF(q)thecollectionofconvexbodies
i
K containing the point q. We note that F(q) must be finite. Let
j
F′(q)={K ,K ,...}\F(q).
1 2
It is not hard to see that, there exists a positive real number ε such that
0
B (q)∩K = ∅, for any K ∈ F′(q). Since q ∈ C ∩int(D), we may assume,
ε0 j j i
withoutlossofgenerality,thatB (q)⊂int(D)andforeachj,wehaveB (q)∩
ε0 ε0
∂K ∩int(K ) = ∅ or B (q)∩∂K ⊂ K . Furthermore, we may also assume
j i ε0 j i
that for all j, both B (q)∩K and B (q)\int(K ) are connected. For a unit
ε0 j ε0 j
vector u, we denote by R(q,u) the ray parallel to u and starting at q. One can
find a unit vector u that satisfies
(i) R(q,u)∩K ={q} and R(q,−u)∩int(K )6=∅,
i i
(ii) thereisapointq′ ∈R(q,u)∩B (q)suchthatq′ ∈/ ∂K forallj =1,2,....
ε0 j
5
Since {K ,K ,...} is a k-fold tiling of D and q′ ∈B (q)⊂int(D), there exist
1 2 ε0
exactly k convex bodies K ,...,K such that q′ ∈int(K ), j = 1,...,k. We
i1 ik ij
note that {K ,...,K }⊂F(q) and K ∈/ {K ,...,K }. Denote by F˜(q) the
i1 ik i i1 ik
collection of convex bodies K which contain the point q as an interior point.
j
If {K ,...,K }⊂ F˜(q), then it is easy to see that K ∩···∩K ∩K must
i1 ik i1 ik i
haveaninteriorpoint, whichis impossible, since {K ,K ,...} is a k-foldtiling.
1 2
Now we suppose that K ∈/ F˜(q), for some j ∈ {1,...,k}. It is clear that
ij0 0
q ∈∂K . Wewillshowthatint(K )∩int(K )=∅. Sinceq ∈C ,wehavethat
ij0 i ij0 i
q ∈Aj0 or q ∈Bj0. Recall that for each j, we have B (q)∩∂K ∩int(K )=∅
i i ε0 j i
orB (q)∩∂K ⊂K . Ifq ∈Bj0,then B (q)∩∂K ⊂K . Fromthis, onecan
ε0 j i i ε0 j0 i
deduce that B (q)∩K ⊂K which is impossible, since q′ is in B (q)∩K
ε0 ij0 i ε0 ij0
but is not in K . Therefore q ∈ Aj0. It is not hard to see that ∂K divides
i i ij0
B (q)into twoparts,whereone ofthem dosenotcontainanypointofint(K ),
ε0 i
we denote this part by B′. Since q′ ∈ int(K ), it is obvious that q′ and
ij0
B (q)∩K must be contained in the same part. Because q′ ∈R(q,u), by the
ε0 ij0
property(i)ofthevectoru,weseethatq′mustlieinB′. ThereforeB (q)∩K
ε0 ij0
is contained in B′, and hence B (q)∩K ∩int(K ) = ∅. By Lemma 3.1, we
ε0 ij0 i
obtain int(K )∩int(K )=∅. This completes the proof.
i ij0
Corollary 3.3. Let X be a discrete multisubset of Rn containing the origin,
and let k be a positive integer. Suppose that P is a centrally symmetric convex
polytope with centrally symmetric facets, and G is its subfacet. Let q be a point
on a facet in B (G). If P +X is a k-fold tiling, then ∂X (G,F+(G,q),q) and
P P P
∂X (G,F−(G,q),q) are not empty.
P P
WewilldenotebyX+(G,q)andX−(G,q)thesets∂X (G,F+(G,q),q)and
P P P P
∂X (G,F−(G,q),q), respectively. For example, in Fig. 3, we have X+(G,q)=
P P P
{x } and X−(G,q)={x }.
1 P 3
4 Proof of Main Theorem
Lemma 4.1. If a convex body K is a twofold translative tile, then K is a
centrally symmetric polytope with centrally symmetric facets, such that each belt
of K contains four or six facets.
Proof. ByTheorem1.2,weknowthatK isacentrallysymmetricpolytopewith
centrally symmetric facets.
Let G be an arbitrary subfacet of K. Recall that we denote by B (G) the
K
belt of K determined by G. Let B (G) have m pairs of opposite facets. We
K
will show that m ≤ 3. To do this, we shall suppose that m ≥ 4, and obtain a
contradiction. Denote by K(G)the union ofthe facets which arenot contained
in B (G).
K
SupposethatK+X isatwofoldtranslativetiling,whereX isamultisubset
of Rn. Without loss of generality,we may assume that 0∈X. It is not hardto
see that, we can choose a point q ∈G to lie in none of K(G)+x, where x∈X.
6
◦ ◦
First, we will show that X (q) = ∅. If there is a x ∈ X (q), then by
K K
Corollary3.3,thereexistx ∈X+(G,q)andx ∈X−(G,q). Obviously,wehave
1 K 2 K
x 6= x . Since m ≥ 4, it is easy to see that both ang (G,q)+ang (G,q)
1 2 K K+x1
and ang (G,q)+ang (G,q) are greater than (m−1)π −(m−2)π = π.
K K+x2
ThereforeAng (G,q)andAng (G,q)arenotoppositeangles,andhence
K+x1 K+x2
ang (G,q)+ang (G,q)+ang (G,q)isgreaterthan(m−1)π+(m−3)π=
K K+x1 K+x2
2π (Fig. 6). This can be deduced that (K +x)∩(K +x )∩(K +x ) has an
1 2
interior point which is impossible, since K+X is a twofold tiling.
K
K+x2 q
K+x1
Fig. 6: ang (G,q)+ang (G,q)+ang (G,q)>2π
K K+x1 K+x2
We assert that ∂˙X (G,q) = ∅. Suppose that x ∈ ∂˙X (G,q). Since K is
K K
centrally symmetric, it is not hard to see that Ang (G,q) and Ang (G,q)
K K+x
are opposite angles (Fig. 7). By Corollary 3.3, we can choose x ∈ X+(G,q)
1 K
andx ∈X−(G,q). Similar to the aboveargument,one obtains that (K+x)∩
2 K
(K+x )∩(K+x ) has an interior point which is a contradiction.
1 2
K
q
K+x
Fig. 7: Ang (G,q) and Ang (G,q)
K K+x
Now we will show that X must be a usual set (not a multiset). If not, then
wemayassumethat0hasmultiplicity2. ByCorollary3.3,onecanchoosex ∈
1
X+(G,q) and x ∈ X+ (G,q) (see Fig. 8). From the above discussion, we
K 2 K+x1
knowthatAng (G,q), Ang (G,q) andAng (G,q) cannotbe opposite
K K+x1 K+x2
angles, hence ang (G,q)+ang (G,q)+ang (G,q) is greater than 2π.
K K+x1 K+x2
7
This implies that K∩(K+x )has aninteriorpoint. We note thatx 6=0,and
2 2
hence we obtain a contradiction.
K+x2
K
q
K+x1
Fig. 8: x ∈X+(G,q) and x ∈X+ (G,q)
1 K 2 K+x1
We shall divide the remaining proof into the following two cases:
◦
(i) Casem≥6: SinceX (q)=∅,weobtainX (q)=∂X (q). Wenotethat
K K K
ang (G,q)<πandforeachx∈∂X (q)\{0},wehaveang (G,q)≤π.
K K K+x
Because K+X is a twofold tiling, so the cardinality of ∂X (q) must be
K
greaterthan4. Onthe otherhand, sincem≥6,we knowthat the sumof
five(distinct)non-oppositeanglesisgreaterthan(m−1)π−(m−5)π=4π.
Therefore, the cardinality of ∂X (q) cannot be greater than 4, this is a
K
contradiction.
(ii) Case m = 4 or 5: Similar to the above, we have that the cardinality of
∂X (q) is greater than 4. If there are two points x,x′ ∈ ∂X (q) such
K K
that q lies in the relative interior of a facet of K +x and also lies in the
relativeinteriorofafacetofK+x′,thenangK+x(G,q)=angK+x′(G,q)=
π, and hence ang (G,q) is greater than ang (G,q) +
Pz∈∂XK(q) K+z K+x
angK+x′(G,q)+(m−1)π−(m−3)π=4π,whichisimpossible. Therefore,
there is at most one point x ∈ ∂X (q) such that p lies in the relative
K
interior ofa facet of K+x. We choosex ∈X+(G,q), x ∈X+ (G,q)
1 K 2 K+x1
and x ∈ X+ (G,q). We note that ang +ang < 2π and
3 K+x2 K+x1 K+x2
ang (G,q)+ang (G,q)+ang (G,q) > 2π. Hence, one can
K+x1 K+x2 K+x3
prove that E+(G,q)∩int(K +x ) 6= ∅ (Fig. 9). Now we choose q′ ∈
K 3
E+(G,q)∩int(K+x )tolieinnoneofK(G)+x,wherex∈X. Obviously,
K 3
◦
x ∈X (q′). Ontheotherhand,byusingthesameargumentastheproof
3 K
◦ ◦
of X (q)=∅, one obtains X (q′)=∅. This is a contradiction.
K K
Above all, we obtain m≤3.
By Lemma 4.1 and Theorem 1.1, one obtain Theorem 1.3.
8
K
K+x2
q
E+(G,q)
K
K+x3
K+x1
Fig. 9: E+(G,q)∩int(K +x )6=∅
K 3
References
[1] Venkov,B.A.: OnaclassofEuclideanpolyhedra.Vestnik LeningradUniv.
Ser. Math. Fiz. Him. 9, 11–31 (1954).
[2] McMullen,P.: Convexbodieswhichtilespacebytranslation.Mathematika
27, 113–121(1980).
[3] Gravin, N., Robins, S., Shiryaev, D.: Translational tilings by a polytope,
with multiplicity. Combinatorica32,629–648(2012).doi: 10.1007/s00493-
012-2860-3
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