Tag: Penrose

an einStein

Published March 28, 2023 by lieven

On March 20th, David Smith, Joseph Myers, Craig Kaplan and Chaim Goodman-Strauss announced on the arXiv that they’d found an ein-Stein (a stone), that is, one piece to tile the entire plane, in uncountably many different ways, all of them non-periodic (that is, the pattern does not even allow a translation symmetry).

This einStein, called the ‘hat’ (some prefer ‘t-shirt’), has a very simple form : you take the most symmetric of all plane tessellations, $* 632$ in Conway’s notation, and glue sixteen copies of its orbifold (or if you so prefer, eight ‘kites’) to form the gray region below:

(all images copied from the aperiodic monotile paper)

Surprisingly, you do not even need to impose gluing conditions (unlike in the two-piece aperiodic kite and dart Penrose tilings), but you’ll need flipped hats to fill up the gaps left.

A few years ago, I wrote some posts on Penrose tilings, including details on inflation and deflation, aperiodicity, uncountability, Conway worms, and more:

To prove that hats tile the plane, and do so aperiodically, the authors do not apply inflation and deflation directly on the hats, but rather on associated tilings by ‘meta-tiles’ (rough outlines of blocks of hats). To understand these meta-tiles it is best to look at a large patch of hats:

Here, the dark-blue hats are the ‘flipped’ ones, and the thickened outline around the central one gives the boundary of the ’empire’ of a flipped hat, that is, the collection of all forced tiles around it. So, around each flipped hat we find such an empire, possibly with different orientation. Also note that most of the white hats (there are also isolated white hats at the centers of triangles of dark-blue hats) make up ‘lines’ similar to the Conway worms in the case of the Penrose tilings. We can break up these ‘worms’ into ‘propeller-blades’ (gray) and ‘parallelograms’ (white). This gives us four types of blocks, the ‘meta-tiles’:

The empire of a flipped hat consists of an H-block (for Hexagon) made of one dark-blue (flipped) and three light-blue (ordinary) hats, one P-block (for Parallelogram), one F-block (for Fylfot, a propellor blade), and one T-block (for Triangle) for the remaining hat.

The H,T and P blocks have rotational symmetries, whereas the underlying block of hats does not. So we mark the intended orientation of the hats by an arrow, pointing to the side having two or three hat-pieces sticking out.

Any hat-tiling gives us a tiling with the meta-tile pieces H,T,P and F. Conversely, not every tiling by meta-tiles has an underlying hat-tiling, so we have to impose gluing conditions on the H,T,P and F-pieces. We can do this by using the boundary of the underlying hat-block, cutting away and adding hat-parts. Then, any H,T,P and F-tiling satisfying these gluing conditions will come from an underlying hat-tiling.

The idea is now to devise ‘inflation’- and ‘deflation’-rules for the H,T,P and F-pieces. For ‘inflation’ start from a tiling satisfying the gluing (and orientation) conditions, and look for the central points of the propellors (the thick red points in the middle picture).

These points will determine the shape of the larger H,T,P and F-pieces, together with their orientations. The authors provide an applet to see these inflations in action.

Choose your meta-tile (H,T,P or F), then click on ‘Build Supertiles’ a number of times to get larger and larger tilings, and finally unmark the ‘Draw Supertiles’ button to get a hat-tiling.

For ‘deflation’ we can cut up H,T,P and F-pieces into smaller ones as in the pictures below:

Clearly, the hard part is to verify that these ‘inflated’ and ‘deflated’ tilings still satisfy the gluing conditions, so that they will have an underlying hat-tiling with larger (resp. smaller) hats.

This calls for a lengthy case-by-case analysis which is the core-part of the paper and depends on computer-verification.

Once this is verified, aperiodicity follows as in the case of Penrose tilings. Suppose a tiling is preserved under translation by a vector $\vec{v}$ . As ‘inflation’ and ‘deflation’ only depend on the direct vicinity of a tile, translation by $\vec{v}$ is also a symmetry of the inflated tiling. Now, iterate this process until the diameter of the large tiles becomes larger than the length of $\vec{v}$ to obtain a contradiction.

Siobhan Roberts wrote a fine article Elusive ‘Einstein’ Solves a Longstanding Math Problem for the NY-times on this einStein.

It would be nice to try this strategy on other symmetric tilings: break the symmetry by gluing together a small number of its orbifolds in such a way that this extended tile (possibly with its reversed image) tile the plane, and find out whether you discovered a new einStein!

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Imagination and the Impossible

Published May 24, 2021 by lieven

Two more sources I’d like to draw from for this fall’s maths for designers-course:

1. Geometry and the Imagination

A fantastic collection of handouts for a two week summer workshop entitled ’Geometry and the Imagination’, led by John Conway, Peter Doyle, Jane Gilman and Bill Thurston at the Geometry Center in Minneapolis, June 1991, based on a course ‘Geometry and the Imagination’ they taught twice before at Princeton.

Among the goodies a long list of exercises in imagining (always useful to budding architects) and how to compute curvature by peeling potatoes and other vegetables…

The course really shines in giving a unified elegant classification of the 17 wallpaper groups, the 7 frieze groups and the 14 families of spherical groups by using Thurston’s concept of orbifolds.

If you think this will be too complicated, have a look at the proof that the orbifold Euler characteristic of any symmetry pattern in the plane with bounded fundamental domain is zero :

Take a large region in the plane that is topologically a disk (i.e. without holes). Its Euler characteristic is $1$ . This is approximately equal to $N$ times the orbifold Euler characteristic for some large $N$ , so the orbifold Euler characteristic must be $0$ .

This then leads to the Orbifold Shop where they sell orbifold parts:

a handle for 2 Euros,
a mirror for 1 Euro,
a cross-cap for 1 Euro,
an order $n$ cone point for $(n - 1) / n$ Euro,
an order $n$ corner reflector for $(n - 1) / 2 n$ Euro, if you have the required mirrors to install this piece.

Here’s a standard brick wall, with its fundamental domain and corresponding orbifold made from a mirror piece (1 Euro), two order $2$ corner reflectors (each worth $.25$ Euro), and one order $2$ cone point (worth $.5$ Euro). That is, this orbifold will cost you exactly $2$ Euros.

If you spend exactly $2$ Euros at the Orbifold Shop (and there are $17$ different ways to do this), you will have an orbifold coming from a symmetry pattern in the plane with bounded fundamental domain, that is, one of the $17$ wallpaper patterns.

For the mathematicians among you desiring more details, please read The orbifold notation for two-dimensional groups by Conway and Daniel Huson, from which the above picture was taken.

2. On the Cohomology of Impossible Figures by Roger Penrose

The aspiring architect should be warned that some constructions are simply not possible in 3D, even when they look convincing on paper, such as Escher’s Waterfall.

M.C. Escher, Waterfall – Photo Credit

In his paper, Penrose gives a unified approach to debunk such drawings by using cohomology groups.

Clearly I have no desire to introduce cohomology, but it may still be possible to get the underlying idea across. Let’s take the Penrose triangle (all pictures below taken from Penrose’s paper)

The idea is to break up such a picture in several parts, each of which we do know to construct in 3D (that is, we take a particular cover of our figure). We can slice up the Penrose triangle in three parts, and if you ever played with Lego you’ll know how to construct each one of them.

Next, position the constructed pieces in space as in the picture and decide which of the two ends is closer to you. In $Q_{1}$ it is clear that point $A_{12}$ is closer to you than $A_{13}$ , so we write $A_{12} < A_{13}$ .

Similarly, looking at

Q_{2}

and

Q_{3}

we see that

A_{23} < A_{21}

and that

A_{31} < A_{32}

Next, if we try to reassemble our figure we must glue

A_{12}

A_{21}

, that is

A_{12} = A_{21}

, and similarly

A_{23} = A_{32}

and

A_{31} = A_{13}

. But, then we get

A_{13} = A_{31} < A_{32} = A_{23} < A_{21} = A_{12} < A_{13}

which is clearly absurd.

Once again, if you have suggestions for more material to be included, please let me know. One Comment

de Bruijn’s pentagrids (2)

Published March 17, 2021 by lieven

Last time we’ve seen that de Bruijn’s pentagrids determined the vertices of Penrose’s P3-aperiodic tilings.

These vertices can also be obtained by projecting a window of the standard hypercubic lattice $Z^{5}$ by the cut-and-project-method.

We’ll bring in representation theory by forcing this projection to be compatible with a $D_{5}$ -subgroup of the symmetries of $Z^{5}$ , which explains why Penrose tilings have a local $D_{5}$ -symmetry.

The symmetry group of the standard $n$ -dimensional hypercubic lattice
$Z {\vec{e}}_{1} + \dots + Z {\vec{e}}_{n} \subset R^{n}$
is the hyperoctahedral group of all signed $n \times n$ permutation matrices
$B_{n} = C_{2}^{n} ⋊ S_{n}$
in which all $n$ -permutations $S_{n}$ act on the group $C_{2}^{n} = {1, - 1}^{n}$ of all signs. The signed permutation $n \times n$ matrix corresponding to an element $(\vec{a}, π) \in B_{n}$ is given by
$T_{i j} = T (\vec{a}, π)_{i j} = a_{j} δ_{i, π (j)}$
The represenation theory of $B_{n}$ was worked out in 1930 by the British mathematician and clergyman Alfred Young

We want to do explicit calculations in

B_{n}

using a computational system such as GAP, so it is best to describe

B_{n}

as a permutation subgroup of

S_{2 n}

via the morphism

τ ((\vec{a}, π)) (k) = {\begin{cases} π (k) + n δ_{- 1, a_{k}} if 1 \leq k \leq n \\ π (k - n) + n (1 - δ_{- 1, a_{k - n}}) if n + 1 \leq k \leq 2 n \end{cases}

the image is generated by the permutations

{\begin{cases} α = (1, 2) (n + 1, n + 2), \\ β = (1, 2, \dots, n) (n + 1, n + 2, \dots, 2 n), \\ γ = (n, 2 n) \end{cases}

and to a permutation

σ \in τ (B_{n}) \subset S_{2 n}

we assign the signed permutation

n \times n

matrix

T_{σ} = T (τ^{- 1} (π))

We use GAP to set up $B_{5}$ from these generators and determine all its conjugacy classes of subgroups. It turns out that $B_{5}$ has no less than $953$ different conjugacy classes of subgroups.

gap> B5:=Group((1,2)(6,7),(1,2,3,4,5)(6,7,8,9,10),(5,10));
Group([ (1,2)(6,7), (1,2,3,4,5)(6,7,8,9,10), (5,10) ])
gap> Size(B5);
3840
gap> C:=ConjugacyClassesSubgroups(B5);;
gap> Length(C);
953

But we are only interested in the subgroups isomorphic to $D_{5}$ . So, first we make a sublist of all conjugacy classes of subgroups of order $10$ , and then we go through this list one-by-one and look for an explicit isomorphism between $D_{5} = ⟨ x, y | x^{5} = e = y^{2}, x y x = y ⟩$ and a representative of the class (or get a ‘fail’ is this subgroup is not isomorphic to $D_{5}$ ).

gap> C10:=Filtered(C,x->Size(Representative(x))=10);;
gap> Length(C10);
3
gap> s10:=List(C10,Representative);
[ Group([ (2,5)(3,4)(7,10)(8,9), (1,5,4,3,2)(6,10,9,8,7) ]),
Group([ (1,6)(2,5)(3,4)(7,10)(8,9), (1,10,9,3,2)(4,8,7,6,5) ]),
Group([ (1,6)(2,7)(3,8)(4,9)(5,10), (1,2,8,4,10)(3,9,5,6,7) ]) ]
gap> D:=DihedralGroup(10); gap> IsomorphismGroups(D,s10[1]);
[ f1, f2 ] -> [ (2,5)(3,4)(7,10)(8,9), (1,5,4,3,2)(6,10,9,8,7) ]
gap> IsomorphismGroups(D,s10[2]);
[ f1, f2 ] -> [ (1,6)(2,5)(3,4)(7,10)(8,9), (1,10,9,3,2)(4,8,7,6,5) ]
gap> IsomorphismGroups(D,s10[3]);
fail
gap> IsCyclic(s10[3]);
true

Of the three (conjugacy classes of) subgroups of order $10$ , two are isomorphic to $D_{5}$ , and the third one to $C_{10}$ . Next, we have to transform the generating permutations into signed $5 \times 5$ permutation matrices using the bijection $τ^{- 1}$ .
$\begin{array}{cc} σ & (\vec{a}, π) \\ (2, 5) (3, 4) (7, 10) (8, 9) & ((1, 1, 1, 1, 1), (2, 5) (3, 4)) \\ (1, 5, 4, 3, 2) (6, 10, 9, 8, 7) & ((1, 1, 1, 1, 1) (1, 5, 4, 3, 2)) \\ (1, 6) (2, 5) (3, 4) (7, 10) (8, 9) & ((- 1, 1, 1, 1, 1), (2, 5) (3, 4)) \\ (1, 10, 9, 3, 2) (4, 8, 7, 6, 5) & ((- 1, 1, 1, - 1, 1), (1, 5, 4, 3, 2)) \end{array}$
giving the signed permutation matrices
$\begin{array}{ccc} x & y \\ A & [\begin{matrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \end{matrix}] & [\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{matrix}] \\ B & [\begin{matrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 & 0 \end{matrix}] & [\begin{matrix} - 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{matrix}] \end{array}$
$D_{5}$ has $4$ conjugacy classes with representatives $e, y, x$ and $x^{2}$ . the
character table of $D_{5}$ is
$\begin{array}{ccccc} (1) & (2) & (2) & (5) \\ 1_{a} & 5_{1} & 5_{2} & 2_{a} \\ D_{5} & e & x & x^{2} & y \\ T & 1 & 1 & 1 & 1 \\ V & 1 & 1 & 1 & - 1 \\ W_{1} & 2 & \frac{- 1 + \sqrt{5}}{2} & \frac{- 1 - \sqrt{5}}{2} & 0 \\ W_{2} & 2 & \frac{- 1 - \sqrt{5}}{2} & \frac{- 1 + \sqrt{5}}{2} & 0 \end{array}$
Using the signed permutation matrices it is easy to determine the characters of the $5$ -dimensional representations $A$ and $B$
$\begin{array}{ccccc} D_{5} & e & x & x^{2} & y \\ A & 5 & 0 & 0 & 1 \\ B & 5 & 0 & 0 & - 1 \end{array}$
decomosing into $D_{5}$ -irreducibles as
$A ≃ T \oplus W_{1} \oplus W_{2} and B ≃ V \oplus W_{1} \oplus W_{2}$
Representation $A$ realises $D_{5}$ as a rotation symmetry group of the hypercube lattice $Z^{5}$ in $R^{5}$ , and next we have to find a $D_{5}$ -projection $R^{5} = A \to W_{1} = R^{2}$ .

As a complex representation $A ↓_{C_{5}}$ decomposes as a direct sum of $1$ -dimensional representations
$A ↓_{C_{5}} = V_{1} \oplus V_{ζ} \oplus V_{ζ^{2}} \oplus V_{ζ^{3}} \oplus V_{ζ^{4}}$
where $ζ = e^{2 π i / 5}$ and where the action of $x$ on $V_{ζ^{i}} = C v_{i}$ is given by $x . v_{i} = ζ^{i} v_{i}$ . The $x$ -eigenvectors in $C^{5}$ are
${\begin{cases} v_{0} = (1, 1, 1, 1, 1) \\ v_{1} = (1, ζ, ζ^{2}, ζ^{3}, ζ^{4}) \\ v_{2} = (1, ζ^{2}, ζ^{4}, ζ, ζ^{3}) \\ v_{3} = (1, ζ^{3}, ζ, ζ^{4}, ζ^{2}) \\ v_{4} = (1, ζ^{4}, ζ^{3}, ζ^{2}, ζ) \end{cases}$
The action of $y$ on these vectors is given by $y . v_{i} = v_{5 - i}$ because
$x . (y . v_{i}) = (x y) . v_{i} = (y x^{- 1}) . v_{i} = y . (x^{- 1} . v_{i}) = y . (ζ^{- i} v_{i}) = ζ^{- 1} (y . v_{i})$
and therefore $y . v_{i}$ is an $x$ -eigenvector with eigenvalue $ζ^{5 - i}$ . As a complex $D_{5}$ -representation, the factors of $A$ are therefore
$T = C v_{0}, W_{1} = C v_{1} + C v_{4}, and W_{2} = C v_{2} + C v_{3}$
But we want to consider $A$ as a real representation. As $ζ^{j} = c o s (\frac{2 π j}{5}) + i s i n (\frac{2 π j}{5}) = c_{j} + i s_{j}$ hebben we can take the vectors in $R^{5}$
${\begin{cases} \frac{1}{2} (v_{1} + v_{4}) = (1, c_{1}, c_{2}, c_{3}, c_{4}) = u_{1} \\ - \frac{1}{2} i (v_{1} - v_{4}) = (0, s_{1}, s_{2}, s_{3}, s_{4}) = u_{2} \\ \frac{1}{2} (v_{2} + v_{3}) = (1, c_{2}, c_{4}, c 1, c 3) = w_{1} \\ - \frac{1}{2} i (v_{2} - v_{3}) = (0, s_{2}, s_{4}, s_{1}, s_{3}) = w_{2} \end{cases}$
and $A$ decomposes as a real $D_{5}$ -representation with
$T = R v_{0}, W_{1} = R u_{1} + R u_{2}, and W_{2} = R w_{1} + R w_{2}$
and if we identify $C$ with $R^{2}$ via $z \leftrightarrow (R e (z), I m (z))$ we can describe the $D_{5}$ -projection morphism $π_{W_{1}} : R^{5} = A \to W_{1} = R^{2}$ via
$(y_{0}, y_{1}, y_{2}, y_{3}, y_{4}) \mapsto y_{0} + y_{1} ζ + y_{2} ζ^{2} + y_{3} ζ^{3} + y_{4} ζ^{4} = \sum_{i = 0}^{4} y_{i} (c_{i}, s_{i})$
Note also that $W_{1}$ is the orthogonal complement of $T \oplus W_{2}$ , so is equal to the linear subspace in $R^{5}$ determined by the three linear equations
${\begin{cases} \sum_{i = 0}^{4} x_{i} = 0 \\ \sum_{i = 0}^{4} c_{2 i} x_{i} = 0 \\ \sum_{i = 0}^{4} s_{2 i} x_{i} = 0 \end{cases}$

Okay, now take the Rhombic tiling corresponding to the regular pentagrid defined by

γ_{0}, \dots, γ_{4}

satisfying

\sum_{i = 0}^{4} γ_{i} = 0

. Let

\vec{k} = (k_{0}, \dots, k_{4}) \in Z^{5}

and define the open hypercube

H_{\vec{k}}

corresponding to

\vec{k}

as the set of points

(x_{0}, \dots, x_{4}) \in R^{5} : \forall 0 \leq i \leq 4 : k_{i} - 1 < x_{i} < k_{i}

From the vector

\vec{γ} = (γ_{0}, \dots, γ_{4})

determining the Rhombic tiling we define the

2

-dimensional plane

P_{\vec{γ}}

R^{5}

given by the equations

{\begin{cases} \sum_{i = 0}^{4} x_{i} = 0 \\ \sum_{i = 0}^{4} c_{2 i} (x_{i} - γ_{i}) = 0 \\ \sum_{i = 0}^{4} s_{2 i} (x_{i} - γ_{i}) = 0 \end{cases}

The point being that

P_{\vec{γ}}

is the linear plane

W_{1}

R^{5}

translated over the vector

\vec{γ}

, so it is parallel to

W_{1}

. Here's the punchline:

de Bruijn’s theorem: The vertices of the Rhombic tiling produced by the regular pentagrid with parameters

\vec{γ} = (γ_{0}, \dots, γ_{4})

are the points

\sum_{i = 0}^{4} k_{i} (c_{i}, s_{i})

with

\vec{k} = (k_{0}, \dots, k_{4}) \in Z^{5}

such that

H_{\vec{k}} \cap P_{\vec{γ}} \neq \emptyset

To see this, let

\vec{x} = (x_{0}, \dots, x_{4}) \in P_{\vec{γ}}

, then

\vec{x} - \vec{γ} \in W_{1}

, but then there is a vector

\vec{y} \in R^{2}

such that

x_{j} - γ_{j} = \vec{y} . {\vec{v}}_{j} \forall 0 \leq j \leq 4

But then, with

k_{j} = ⌈ \vec{y} . {\vec{v}}_{j} + γ_{j} ⌉

we have that

\vec{x} \in H_{\vec{k}}

and we note that

V (\vec{y}) = \sum_{i = 0}^{4} k_{i} {\vec{v}}_{i}

is a vetex of the Rhombic tiling associated to the regular pentagrid parameters

\vec{γ} = (γ_{0}, \dots, γ_{4})

Here we used regularity of the pentagrid in order to have that

k_{j} = \vec{y} . {\vec{v}}_{j} + γ_{j}

can happen for at most two

j

’s, so we can manage to vary

\vec{y}

a little in order to have

\vec{x}

in the open hypercube.

Here’s what we did so far: we have identified

D_{5}

as a group of rotations in

R^{5}

, preserving the hypercube-lattice

Z^{5}

R^{5}

. If the

2

-plane

P_{\vec{γ}}

is left stable under these rotations, then because rotations preserve distances, also the subset of lattice-points

S_{\vec{γ}} = {(k_{0}, \dots, k_{4}) | H_{\vec{k}} \cap P_{\vec{γ}} \neq \emptyset} \subset Z^{5}

is left stable under the

D_{5}

-action. But, because the map

(k_{0}, \dots, k_{4}) \mapsto \sum_{i = 0}^{4} k_{i} (c_{i}, s_{i})

is the

D_{5}

-projection map

π : A \to W_{1}

, the vertices of the associated Rhombic tiling must be stable under the

D_{5}

-action on

W_{1}

, meaning that the Rhombic tiling should have a global

D_{5}

-symmetry.

Sadly, the only plane

P_{\vec{γ}}

left stable under all rotations of

D_{5}

is when

\vec{γ} = \vec{0}

, which is an exceptionally singular pentagrid. If we project this situation we do indeed get an image with global

D_{5}

-symmetry

but it is not a Rhombic tiling. What’s going on?

Because this post is already dragging on for far too long (TL;DR), we’ll save the investigation of projections of singular pentagrids, how they differ from the regular situation, and how they determine multiple Rhombic tilings, for another time. Comments closed