 \magnification=\magstep1
\parindent=0cm
\baselineskip=10pt
\parskip=12pt
\def\hb{\hfill\break}
\def\Isom{{\rm Isom}_+}
\def\ve{\vfill\eject}

\vskip 2in
\centerline{TILING PROBLEMS}
\vskip 2in
Tiling theory is a broad subject that draws upon geometry, topology, group theory, number
theory, and other branches of mathematics. Both the breadth and depth of the subject
were demonstrated in lectures and discussions at the RGI (Regional Geometry Institute) 
held at Smith 
College, Northampton, Massachusetts,  July 5-31, 1993. The following problems were collected at 
a problem session on tilings  in the last week of the RGI.
\vskip 2in
\rightline{compiled by Marjorie Senechal}
\bigskip
\rightline{Director, Regional Geometry Institute}
\bigskip
\rightline{August 12, 1993}
\ve

{\bf 1. Presented by Olaf Delgado.}

Consider a tiling $T$ of the Euclidean plane by closed topological disks
having crystallographic full symmetry group $G$.

$T$ is called {\it combinatorially convex} (or is said to have a convex
realization) if and only if there exists a homeomorphism $\phi$ from the Euclidean
plane to itself such that

  (a) $\phi$ takes each tile of $T$ to a convex set, i.e. $\phi$ maps $T$ to a tiling
     $T^{\prime}$ by convex polygons

  (b) $\phi$ preserves symmetries, i.e. for each $g \in G$, 
$ \phi \circ g = g \circ \phi$.

Note that (b) implies that $G$ is at least a subgroup of the full
symmetry group of $T^{\prime}$, so $T^{\prime}$ has all the symmetries $T$ has.

An obvious necessary condition for combinatorial convexity is the
{\it intersection condition:}   No pair of tiles may have a
        disconnected intersection.


\proclaim Conjecture (Dress).
        The intersection condition is also sufficient for
        combinatorial convexity.

\proclaim  Problem. Prove this conjecture!


Remarks:

  1) If one doesn't care about symmetries, this conjecture has been
proven by Thomassen in 1980, who gave necessary and sufficient
conditions for convex embeddability of finite and infinite planar
graphs.

  2) I have a proof for the special case where all tiles are
asymmetric , i.e. no non-trivial element of G fixes any tile. The
proof appears in my master's thesis (1990, in German) and has not been published
elsewhere.   I will send you a postscript file by email upon request.


Generalizations:
  1) In which cases is there a convex realization having G as its
full symmetry group?
  2) Can you always guarantee 'strict' convexity, i.e. can you make
each vertex of a tile an extremal point of it?
  3) How about other geometries (spherical/elliptic and hyperbolic)?
  4) How about tilings of space? (This is probably very hard!)

References:

C. Thomassen, ``Planarity and duality of finite
and infinite graphs'',Journal of Combinatorial Theory, Series B29 (1980), 244-271.

\ve

{\bf 2. Presented by Wlodek Kuperberg.}

1. Suppose a convex right prism in $E^3$ (the base is a convex polygon
   and the sides are rectangles) tiles space by congruent replicas, 
   where rotations and reflections are allowed. Does the base necessarily
   admit a plane tiling?

2. Characterize the convex polyhedra in $E^3$ that admit a tiling  consistiing
   of translates of the polyhedron and translates of its symmetric image
   about a point.

Each of the above problems can be generalized to higher dimensions, but
they seem to be difficult enough in $E^3$.



{\bf 3.  Presented by Ludwig Danzer}

\proclaim Problem 1. Does there exist a prototile $T$ in $E^n$ (n = 2,3) such that no tiling of 
$E^n$ by congruent copies of $T$ admits any (isometric) symmetry different from the identity?

Remark: in $H^2$ there are many such tiles;  compare also the 
SCD (Schmitt - Conway - Danzer) biprism in $E^3$.

\proclaim Problem 2. Does there exist a tiling $\cal P$ of $E^n, n = 2,3$ with
finite family $\cal F$ of prototiles $T_1,T_2,\ldots, T_k$ such that $\cal P$ does not
 contain any infinite set of translates of a tile?

Remark: Appearance in infinitely many orientation of every tile is a much weaker
condition; see the pinwheel or the SCD-biprism. 

\proclaim Problem 3. Given a Penrose tiling ${\cal P}$. Mark all points $x$ by the symbol
$n$, for which $(x + \tau^nB^2)\cap {\cal P}$
has precisely the symmetry group $D_5$, but $(x + \tau^{n+1}B^2)\cap {\cal P}$  has
not. Also draw all lines of local $D_1$ symmetry (respecting the arrows of the tiles). 

{\it What types of pattern will occur this way?} Probably just 3; but the same 
3 also correspond to tilings of other species.

\proclaim Main Question. Let us say that $S_1$ is equivalent to $S_2$ if the patterns of $S_1$
equal those of $S_2$ up to similarities. How many equivalence classes are there associated with
 $D_5$? {\it How many are there for the icosahedral group} (different markings for all 
subgroups?)?  How is this equivalence relation related to the one of Baake ($S_1$ is locally
 derivable from $S_2$ and vice versa?)
\ve

{\bf 4. Presented by Peter Cromwell}

This problem/curiosity arose when I was implementing the 3-dimensional point groups
(symmetries of polyhedra) on a computer. These groups can be presented in a form
particularly suited to computers: the group elements can be enumerated by a set of nested 
loops with each element appearing exactly once.

Suppose we have generators $g_1,\ldots,g_n$ and a sequence of integers $t_1,\ldots,t_n$. We
form the following words $$\prod_{j = 1}^n(g_j)^{r_j}$$ where each
$r_j$ runs from 0 to $t_j -1$, independently. This produces $\prod t_j$ words which
may correspond to elements in some group.

Here are a few examples of rotation groups and sequences which generate them:

quadratic $(D_2)$\quad 2,2\hb
tetrahedral\quad \quad 3,2,2\hb
octahedral \quad\quad 3,2,2,2\hb
icosahedral \quad\quad 3,2,2,5

Reflections can be included by prepending a 2 to the sequence.

Does the class of groups whose elements can be enumerated by a  nested loop presentation
have a familiar name?

\vskip .3in

{\bf 5. Presented by Jim Propp, Massachusetts Institute of Technology.}

An Aztec diamond of order $n$ is a region consisting of $2n(n+1)$ unit squares,
arranged in centered rows of lengths $2,4,6,\ldots,2n-2,2n,2n,2n-2,\ldots,6,4,2..$
 
1) For what values of $k$ can the Aztec diamond of order $2k^2-1$ with a
$2k$-by-$2k$ hole in the center be tiled by skew tetrominoes?
 
2) Can every dissection of a convex polygon by straight cuts be extended
to one in which the pieces are all triangles?  (A straight cut is required
to run all the way through the interior of the original region.  One extends
a dissection by adding additional cuts.)
 
3) Say two tilings are mutually $r$-accessible if there exists a sequence of
tilings, beginning with one and ending with the other, such that each tiling
in the sequence agrees with the next tiling outside of some disk of radius $r$.
For example, it is possible to convert any domino tiling of a simply-connected
region into any other by means of local moves that rotate a 2-by-2 block by 90
degrees; hence there exists an $r$ (namely $\sqrt{2}$) such that all domino
tilings of a simply-connected region are $r$-accessible from one another.
Is this result typical of what happens for tilings of simply-connected regions
by polyominoes?
 
 
\vskip .3in


{\bf 6. Presented by Bill Abikoff, University of Connecticut.}

Geometric Construction of co-compact discrete groups of hyperbolic motions.

Let $H^n$ denote hyperbolic $n$-space and $\Isom$ its group of orientation preserving motions. If
$G < \Isom$ is discrete and $w_0$ has trivial isotropy group ($\gamma(w_0) = w_0$ has no 
solutions for $\gamma \in G\setminus \{{\rm Id}\}$), then $$D(w_0) = \bigcap\{
 w|
d(w,w_0) < d(w,\gamma(w))\quad {\rm for\ all}\; \gamma\in G,\gamma\not={\rm Id}\}$$ 
is the Dirichlet region for $G$. Here $d$ is the hyperbolic
distance. The Dirichlet region for a hperbolic group is one example of a tile for a
$G$-invariant tessellation.

Question: How can one construct, using geometric methods, cocompact  groups $G$, or 
equivalently a compact polyhedron $P$ which satisfies 
the conditions of Poincar\`e's polyhedron theorem? $P$ will then define a 
fundamental region for $G$ or, equivalently, a tile for a $G$-invariant tessellation of $H^n$.

In high dimensions ($n > 3$), only arithmetic methods for constructing cofinite 
volume groups are known to this proposer. A method of constructing cocompact groups was 
suggested by Armand Borel to
Dennis Sullivan in about 1976. I don't know a printed reference for it.

There are geometric techniques for the case $n = 2$ (uniformization, variation of shape, etc)
and $n=3$ (for example, Dehn surgery on knot complements).

A good reference for Poincar\`e's polyhedron theorem is Maskit's {\it Kleinian Groups},
published by Springer-Verlag.


\ve

{\bf 7. Presented by Yang Wang}

 Assume $m \in Z, m > 1$. 
A set $T\subseteq R^2$ is called an $m$-{\it reptile} if \hb
$$ T \ {\rm is\  compact} ,\ T^o \neq \emptyset,\   T = \overline{T}^o,\
 {\rm and}$$
$$\sqrt m T = T_1 \cup T_2 \cup \cdots \cup T_m.  \eqno{(1)}$$
where each $T_i$ is congruent to $T$.

{\it Example}. The rectangle of sides 1 and $\sqrt2$ is a 2-reptile.

\proclaim Problem 1. Classify all 2-reptiles. Is there a disconnected 2-reptile?

Robert Ammann claims to have solved this problem, but so far nothing has been written up.

\proclaim Problem 2.  Is there a 2-reptile that's also an $n$-reptile?

{\it Examples}. A square is a 4-reptile, but it's also a 9-reptile. The rectangle of
sides 1 and $\sqrt3$ is a 3-reptile as well as a 4-reptile. 

\proclaim Conjecture. There exists an $m$-reptile which is also an $n$-reptile 
if and only if either one of $m,n$ is a square or $m/n = r^2$ for some $r\in Q$.

Let $T$ be an $m$-reptile. Then we can write (1) as $$\sqrt m T = g_1(T)\cup g_2(T)\cup\cdots\cup 
g_m(T)\eqno{(2)}$$ where $$g_i \in {\rm Isom}(R^2),\ g_i(\vec x) = A_i(\vec x) + \vec b_i$$ where 
$A_i$ is a rotation matrix and $\vec b_i \in R^2$. Notice that for any given 
$g_1,g_2,\ldots,g_m\in {\rm Isom}(R^2)$ 
there exists a {\it unique} compact $T$ satisfying (2). However, $T$ may not have 
positive Lebesgue measure.

\proclaim Problem 3. For what $g_1,g_2,\ldots,g_m$ does $T$ have positive Lebesgue measure?
If $\mu(T) > 0$, does it mean $T^o \neq \emptyset?$

When $A_1 = A_2 = ... = A_m$, there are criteria to decide whether $\mu(T) > 0$. 
But no criterion is known for the general case.

{\it References}. A comprehensive list of references can be found in the following preprints:\hb
J. C. Lagarias and Y. Wang, ``Integral self-affine tiles in $R^n$ I. Standard and nonstandard
digit set;\hb
J.C. Lagarias and Y. Wang, ``Self-affine tiles in $R^n$.

\vskip .3in
{\bf 8. Presented by M.D. Kovalyov}

This is a problem of Lazlo Fejes Toth.

Let us consider a densest packing $P$ of equal circular discs of radius 1 on the Euclidean
plane. Let us remove from $P$ some collection $C = \{D_1,D_2,\ldots,D_n\}$ of $n$ discs. 
Suppose that $\cal D$ is an $n$-point set consisting of centers $p_1,p_2,\ldots,p_n$ of removed
discs, and $H$ is a hole, that is a plane region, bounded by discs of $P$ touching discs
of $C$. If we pack in the hole $H$ only $n-1$ discs of radius 1, must all their centers belong to the set $\cal D$?

I. Barany and N. Dolbilin proved the following ``local'' result. Suppose that 
$P^{\prime}$ is a 
packing of $n-1$ discs of radius 1 in the hole $H$, and ${\cal D}^{\prime}$  is the 
$(n-1) $-point
set, consisting of their centers $q_1,\ldots,q_{n-1}$. If for every $q_i$ there exists a 
point $p_j$ such that $|q_i - p_j| < \varepsilon \sim 0.1$,  then 
${\cal D}^{\prime} \subset {\cal D}$ .

The packing is said to be stable if there are no other packings close to it. 
 If $\cal P^{\prime}$ is a stable packing in $H$ with the similar condition
$|q_i - p_j| < \varepsilon = \sqrt3-1$, then ${\cal D}^{\prime} \subset 
{\cal D}$. Can you improve the value of $\varepsilon$ or solve the initial problem?  

\vskip .3in
{\bf 9. Presented by Nikolai Dolbilin.}

  A stereohedron is the Dirichlet (Vorono\"\i\ ) cell of a point of an
orbit of an $n$-dimensional crystallographic group $G$. Stereohedra are thus convex and tile $R^n$ 
in a face-to-face manner.
Let  $f_{n-1}$ be  the number of $(n-1)$-dimensional facets of a $n$
dimensional stereohedron.
Delone (1960) showed that $$f_{n-1} \leq 2(2^n - 1) + (h-1)2^n,$$
where $h$ is the order of the factor group $G/T$ and $T$ is the translation subgroup of $G$.
When $n = 3$, $h$ can be as large as 48 (the order of the symmetry group of the cube). Delone's
theorem then says that $f_{n-1} \leq 390$. However, the maximum value known so far is 38; this
was found by Peter Engel in a computer search.
 
\proclaim Problem 1. Is Delone's theoretical upper bound the best possible?

\proclaim Problem 2. Suppose that a convex polytope $P$ tiles $n$-dimensional Euclidean space 
in a face-to-face manner. Does there exist an upper bound for $f_{n-1}$ for each $n$?

References:

B.N. Delone (1961), ``A proof of the fundamental theorem of stereohedra'', 
Reports of the Academy of Sciences of the U.S.S.R., 138, 1270-1272. (In Russian.)

P. Englel (1981), ``\"Uber Wirkungsbereichsteilungen mit kubischer Symmetrie'', Zeitschrift f\"ur 
Kristallographie, 157, 259-275.
\ve


A solution to the following problem was presented before the RGI ended!

{\bf 10. Presented by H.S.M. Coxeter, University of Toronto}.

Let the $n$ vertices of a regular simplex in real Euclidean $(n-1)$-space be orthogonally 
projected onto a plane. Let this real plane be regarded as the Argand plane of complex numbers. 
Let the $n$ points so derived from the regular simplex be $z_1,z_2,\ldots,z_n$. Then $$
\Bigl({z_1 + z_2 + \cdots + z_n\over n}\Bigr)^2 = {z_1^2 + z_2^2 + \cdots + z_n^2\over n}.
\eqno{(*)}$$
Conversely, if $n$ complex numbers (not all equal) satisfy this equation, 
the corresponding $n$ points in the plane can be derived by orthogonal projection 
from the $n$ vertices of a regular $(n-1)$-dimensional simplex.

As a trivial example, note that the equation with $n=3$ holds for the 3 vertices of 
any equilateral triangle in the Argand plane.
\vskip .2in

{\it Solution by Serge Tabachnikov, University of Arkansas.}

First observe that (*) is invariant under isometries of $R^2$. 
Choose orthogonal coordinates in $R^{n-1}$ so that the origin is the centroid of
 the simplex and $R^2$ is generated by the first two
coordinates. Let the vertices of the simplex be $$X_1,\ldots,X_n;\quad X_i = 
(x_i^1,\ldots,x_i^{n-1}).$$ Then (*) reads: $$ z_1^2 + \cdots + z_n^2 = 0$$ 
(because $z _1 + \cdots + z_n = 0$) or, since $z_i = x_i^1 + \sqrt{-1}x_i^2$, 
$$\sum(x_i^1)^2 = \sum(x_i^2)^2; \quad \sum x_i^1x_i^2 = 0.\eqno{(**)}$$
Embed $R^{n-1}$ in $R^n$ as an affine hyperplane $\{x^n = c/\sqrt n\}$, where $c$ is the 
length of the side of the simplex. Then the vectors $\vec x_i =
(x_i^1,x_i^2,\ldots,x_i^{n-1},c/\sqrt n)$
are pairwise orthogonal and have equal length. Hence the $n\times n$ matrix whose rows are
the $\vec x_i$ is, up to a constant factor, orthogonal. Therefore the first two columns are
 perpendicular vectors of equal length and (**) holds.

Conversely, if (**) holds and $\sum x_i^1 = \sum x_i^2 = 0$, then we place 
the three vectors
$$(x_1^1,\ldots,x_n^1),\ (x_1^2,\ldots,x_n^2),\ (c/\sqrt n,\ldots,c/\sqrt n)$$
where $c^2 =  \sum x_i^1 = \sum x_i^2$, into an orthogonal frame in $R^n$ consisting of
vectors of equal length. The vectors $X_i = (x_i^1,\ldots,x_i^{n-1})$
form a regular simplex in $R^{n-1}$.
\ve

{\bf 11. Presented by K. Robert Gutschera}

The  Auslander conjecture is a generalization of 
Bieberbach's Theorem on crystallographic groups, where we allow affine transformations of 
$R^n$ rather than just isometries.

We say that a subgroup $\Gamma \subset Aff(R^n) = GL(n,R) \times R^n$ is an
{\it affine crystallographic group} if $\Gamma$ acts properly discontinuously
on $R^n$ and $R^n/\Gamma$ is compact in the quotient topology. (Recall that
 a group acts properly discontinously on a space $X$ if the map
$\Gamma\times X \rightarrow X\times X,\ (\gamma,x) \mapsto (\gamma x,x)$ is proper
when we give $\Gamma$ the discrete topology; equivalently, if $K_1,K_2$
 are compact subsets of $X$, then $\{ \gamma\in \Gamma |\gamma K_1\cap K_2 \neq\emptyset\}$
is always finite.) For $\Gamma\subset {\rm Isom}(R^n) = O(n,R)\times R^n$, this is
equivalent to any of the usual definitions of a crystallographic group (e.g., that
$\Gamma$ acts on $R^n$ without accumulation points and with a compact fundamental
domain.){\it Note: here $\times$ means {\bf semidirect} product}.

The key portion of Bieberbach's Theorem (which answered a question posed by Hilbert
in his Problem 18) is that if $\Gamma \subset {\rm Isom}(R^n)$ is crystallographic, it
contains a lattice of translations of finite index in $\Gamma$ (namely $\Gamma\cap R^n$);
in other words $\Gamma$ must be {\it virtually Abelian}. (A group is ``virtually $P$''
if it has a subgroup of finite index with property $P$.)

Recall that a group is {\it polycyclic} if it has a finite composition series, each
factor of which is cyclic (this can be thought of as being the same thing as solvable --
the two properties are all but equivalent in this context.)

\proclaim Conjecture (Auslander/Milnor). An affine crystallographic group must be
virtually polycyclic.

An attempt to prove this was made by L. Auslander; it was later (1977) stated as 
an explicit conjecture by J. Milnor.  

It is known that:

-- If $\Gamma$ is virtually polycyclic, it can be realized as a subgroup of $Aff(R^n)$
acting properly discontinuously. Whether it can  be realized as acting with compact quotient as well is not known (in contrast to the Bieberbach case, where the
 compactness of
$R^n/\Gamma$ is irrelevant to the types of abstract groups $\Gamma$ that can arise).

-- If $n\leq 3$ the conjecture is true (Fried and Goldman, 1983).

-- If $\Gamma\subset O(1,n-1)\times R^n$ (the group of Lorentz {\rm Isom}etries of $R^n$)
the conjecture is true (Goldman and Kamishima, 1984).

-- If we relax the requirement that $R^n/\Gamma$ be compact, the conjecture is 
{\it false}. A counterexample (due to Margulis) with $\Gamma \subset O(1,2)\times R^3$ 
exists, where $\Gamma$ contains the free group on two generators (and thus cannot be
polycyclic).

{\it The geometric point of view.} Differential geometers generally use a different language
to discuss these questions. If we define a {\it space form} to be a manifold with a
flat connection which is geodesically complete (roughly speaking, any flat manifold that
doesn't have holes in it) then the Bieberbach Theorem becomes, ``If $M$ is a compact
Riemannian space form, then $\pi_1(M)$ is virtually Abelian'', and the Auslander
conjecture becomes ``If $M$ is a compact  space form, then $\pi_1(M)$ is 
virtually polycyclic''. All of the geometry and topology of a space form $M$ is carried by $\pi_1(M)$, so understanding the possible fundamental groups of space
 forms allows one
to understand them completely.

Some examples:

 $n=2$. Let $\Gamma$ be generated by $$\Bigl(\pmatrix{1&0\cr
                                                 0&1\cr},e_1\Bigr)\ \quad {\rm and}
\quad \Bigl(\pmatrix{1&1\cr
                     0&1\cr},e_2\Bigr).$$ Here $\Gamma \simeq Z^2$ but $\Gamma$
does not contain a lattice of translations.

 $n=3$. Let $$\Gamma = \{\Bigl(\pmatrix{1&t&0\cr
                                     0&1&0\cr
                                     0&0&1\cr},(r,s,t)\Bigr)|r,s,t\in Z\}.$$ 
Here one can show that in fact $\Gamma$ contains no subgroup {\rm Isom}orphic to $Z^3$.

{\it Other questions}: Since the geometers who work on this problem usually think in terms 
of $\pi_1(M)$ rather than in terms of tilings, answers to many rather basic 
tiling questions are unknown, or at the very least not stated in the literature. 
For example, if $\Gamma\subset Aff(R^n)$ acts freely and with discrete orbits, 
must $\Gamma$ also act properly discontinuously, or must $R^n/\Gamma$ be a manifold for some
other reason)? (This question is harder than it appears.) Is being a tiling in the usual
sense the same as proper discontinuity? It's not even immediately clear what the ``right''
way to construct the fundamental domain is.

References:

J. Milnor, ``On fundamental groups of complete affinely flat manifolds'', Adv. in
Math. 25 (1977), 178-187.

D. Fried and W. Goldman, ``Three-dimensional affine crystallographic groups'', Adv.
in Math. 47 (1983), 1-49 (this contains an extensive bibliography).

W. Goldman and Y. Kamishima, ``The fundamental group of a compact flat Lorentz space
form is virtually polycyclic'', J. Diff. Geom. 19 (1984), 233-240.

L. Auslander, ``Examples of locally affine spaces'', Annals of Math., 64, No. 2, (1956),
 255-259 (the second example given above, and an implicit assumption of the first 
``other question'' above may be found in this paper).

Exposition of the Bieberbach theorems may be found in 

L. Auslander, ``An account of the theory of crystallographic groups'', 
Proc. AMS 16 (1965), 1230-1236,\ \  and

J.A. Wolf, {\it Spaces of Constant Curvature}, McGraw-Hill, Newy York, 1967, and 
Publish or Perish, Boston, 1974, 98-107. (There is, however, some trouble in this
book with the definition of discontinuity.)

\ve

{\bf 12.  Presented by Goetz Gelbrich, Griefswald.}

In ``Crystallographic reptiles''$^{\rm 1}$
we investigated tilings of the form
$\Gamma A$ where $A\subset R^2$ is homeomorphic to a disk, and $\Gamma$ is a 
crystallographic subgroup of $Isom(R^2)$. In addition, we required that there
is an expanding affine map
$g:R^2 \rightarrow R^2$ such that $g\Gamma g^{-1}\subset\Gamma$
and $gA = \gamma_1A \cup\ldots\cup \gamma_kA$ for certain
$\gamma_1,\ldots,\gamma_k\in \Gamma$.
In this case $\{\gamma_1,\ldots,\gamma_k\}$ is a complete set of right coset
representatives of $g\Gamma g^{-1}\subset\Gamma$. 
 
Now relax the assumptions: the tiling should be of the form $\Gamma_0A$
where $\Gamma_0$
is a subset of $\Gamma$. Then some $\gamma_i\neq\gamma_j$ may represent the
same right coset of
$g\Gamma g^{-1}$. For the $\Gamma$-tilings we have a finiteness theorem and a
classification algorithm strongly using the algebraic structure. For a
systematic approach
to $\Gamma_0$-tilings, we should try to get some algebraic information.
 \medskip
Suppose given a $\Gamma_0$-tiling. (We can assume $1\in \Gamma_0$.) Define a
group $\tilde{\Gamma}$
as follows. The generators are the isomorphy classes of pairs
$(A_1,A_2)$ of tiles intersecting in a common edge. (Isomorphy means that one pair is mapped
onto another by some $\gamma\in\Gamma$.) For each isomorphy class of
vertices of the tiling we obtain a relation of the generators
(walking once around the 
vertex). If $\gamma_1A = A_1, \gamma_2A = A_2$, map 
$[(A_1,A_2)]$ to $\gamma_2^{-1}\gamma_1\in\Gamma$, obtaining a homomorphism
$\phi:\tilde{\Gamma}\rightarrow\Gamma$.
 
{\bf Question:}
Is there useful information contained in $Im(\phi)\subset\Gamma$?
In $Ker(\phi)$?
 
Consider also the case when $\Gamma$ is not a crystallographic group
but the smallest
group containing $\gamma_1\ldots,\gamma_k$ that is invariant under $g$ (i.e,
$g\Gamma g^{-1}\subset\Gamma$), for example the pinwheel tiling$^{\rm 2}$.

$^{\rm 1}$\ to appear in {\it Geometriae Dedicata},
see also the references there\hfill\break
$^{\rm 2}$\ see e.g.: C.\ Radin, {\it The pinwheel tilings of the plane},
preprint, Univ.\ of Texas, Austin (to appear)
 

\ve

{\bf 13. Presented by Marjorie Senechal}

A protoset (set of prototiles) is said to be aperiodic if every tiling admitted by the
protoset is nonperiodic. Famous examples include Penrose's two decorated rhombs,
Ammann's decorated square and rhomb, and Danzer's four tetrahedra. A less well known but 
very important example is the protoset of the pinwheel tiling.

Special techniques are used to prove that a protoset $\cal P$ is aperiodic. For example,
in some cases (including those mentioned above) it is possible to show that
in every tiling admitted by $\cal P$ the tiles can be grouped into larger copies of
the prototiles, in only one way. Let us call this the {\it substitution property}.  
 By a well-known result, all  tilings with the substitution property are
nonperiodic. 

\proclaim Problem 1. Do there exist aperiodic protosets whose tilings do {\it not} 
have the substitution  property?

\proclaim Problem 2. Do there exist protosets whose tilings have the substitution
property but which are not aperiodic?

At first it would seem that there are some very obvious counterexamples to Problem 2. For
example, Danzer has shown that the 4-reptile chair has the following property:

({\bf *}) one can
 build a periodic tiling with this prototile
 whose vertex atlas is a subset of the vertex atlas of its nonperiodic substitution tilings.

(A vertex configuration of a tiling $T$ is any vertex together with the tiles that meet there;
 the set of all vertex configurations of a tiling, up to some appropriate  isometry group,
 is  its {\it vertex atlas}.)

This property would
seem to imply that there are no edge or vertex decorations for the chairs that force 
nonperiodicity.

But in fact such decorations may exist anyway. For example, as Danzer has shown,
 the pinwheel triangle also has the property ({\bf *})! It is possible that
the Amman tiling does too: as
 De Bruijn first noted,  the set $\cal T$ of all tilings
whose vertex atlas coincides with that of the Ammann tilings is  larger than
the set ${\cal T}^{\prime}$ of Ammann tilings; indeed, it may even contain periodic tilings.


References: (none of these papers has been published as of August, 1993).

N.G. DeBruijn (1989), ``Remarks on Beenker patterns''.

C. Radin (1992), ``The pinwheel tilings of the plane'' (to appear in Annals of Mathematics).

Le Tu Quoc Thang (1993), ``Local structure of quasiperiodic tilings having 8-fold symmetry''. 
\vfill\eject

{\bf 14. Presented by K.Bezdek and T.Hausel}

Let T be a tiling of the d-dimensional Euclidean space by parallel
cubes where each tile is a unit cube except one tile that is a cube
having a non-integer edge-length greater than one. Prove that T is
uniquely determined up to congruence.

 \vfill\eject
 
\rightline{CONTRIBUTORS}\bigskip

Bill Abikoff\hb
Department of Mathematics\hb
University of Connecticut\hb
Storrs, CT \hb
abikoff@math.uconn.edu

Karoly Bezdek\hb
Department of Geometry
1088 Budapest
Hungary
kbezdek@ludens.elte.hu


H.S.M. Coxeter\hb
Department of Mathematics\hb
University of Toronto\hb
Toronto\hb
M5S 1A1, Canada

Peter Cromwell\hb
Department of Pure Mathematics\hb
University of Liverpool\hb
Liverpool L69 3BX, England\hb
spmr02@liverpool.ac.uk

Ludwig Danzer\hb
Fachbereich Mathematik\hb
Universitat Dortmund\hb
Postfach 50 05 00\hb
4600 Dortmund 50, Germany\hb
um018@unidozr.hrz.uni-dortmund.de

Olaf Delgado\hb
Universit\"at Bielefeld\hb
Fachbereich Mathematik\hb
Postfach 100131\hb
4800 Bielefeld, Germany\hb
delgado@mathematik.uni-bielefeld.de
 
Nikolai Dolbiliin\hb
Steklov Institute\hb
Ul. Vavilova 44, K. 2\hb
117333 Moscow, Russia\hb
nikolai@dolbilin.mian.su

Michail Kovalyov\hb
Department of Mathematics\hb
Moscow State University of Forest\hb
103473 Moscow, Russia\hb
chepanov@top.mian.su

Goetz Gelbrich\hb
Fachbereich Mathematik\hb
Sahnstr. 15a\hb
17489 Greifswald, Germany\hb
gelbrich@math-inf.uni-greifswald.d400.de
\ve
Robert Gutschera\hb
Department of Mathematics\hb
Wellesley College\hb
Wellesley, MA 02181\hb
krg@olaf.wellesley.edu

Daniel Huson\hb
Universit\"at Bielefeld\hb
Fachbereich Mathematik\hb
Postfach 100131\hb
4800 Bielefeld, Germany\hb
huson@mathematik.uni-bielefeld.de

Jim Propp\hb
Department of Mathematics\hb
Massachusetts Institute of Technology\hb
Cambridge, MA 02139\hb
propp@math.mit.edu

Marjorie Senechal\hb
Department of Mathematics\hb
Smith College\hb
Northampton, MA 01063\hb
senechal@minkowski.smith.edu

Serge Tabachnikov\hb
Department of Mathematics\hb
University of Arkansas\hb
Fayetteville, AR 72701\hb
serge@uafsysb.uark.edu

Yang Wang\hb
School of Mathematics\hb
Georgia Institute of Technology\hb
Atlanta, GA 30332\hb
wang@math.gatech.edu

\bye

The Ammann and pinwheel decorations have a global character. 
In the Ammann case, this
 can be explained in the context of the projection method; indeed the decorations as well as
the tilings can
be constructed by that method.
Although it is not obvious how  substitution tilings with nonconvex prototiles
might be obtained by projection, it is not impossible that ``long-range'' decorations
of the Ammann type exist for them.

\proclaim Problem 3. Which, if any, tilings with nonconvex prototiles can be obtained by
projection?
