de Bruijn’s pentagrids (2)

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 ${\mathbb{Z}}^5$ 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 ${\mathbb{Z}}^5$, which explains why Penrose tilings have a local $D_5$-symmetry.

The symmetry group of the standard $n$-dimensional hypercubic lattice
$$\mathbb{Z}{\overrightarrow{e}}_1+\cdots +\mathbb{Z}{\overrightarrow{e}}_n\subset {\mathbb{R}}^n$$
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 $(\overrightarrow{a},\pi )\in B_n$ is given by
$$T_{ij}=T(\overrightarrow{a},\pi {)}_{ij}=a_j{\delta }_{i,\pi (j)}$$
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 $S_{2n}$ via the morphism
$$\tau ((\overrightarrow{a},\pi ))(k)=\{\begin{array}{l}\pi (k)+n{\delta }_{-1,a_k}\text{if }1\le k\le n\\ \pi (k-n)+n(1-{\delta }_{-1,a_{k-n}})\text{if }n+1\le k\le 2n\end{array}$$
the image is generated by the permutations
$$\{\begin{array}{l}\alpha =(1,2)(n+1,n+2),\\ \beta =(1,2,\dots ,n)(n+1,n+2,\dots ,2n),\\ \gamma =(n,2n)\end{array}$$
and to a permutation $\sigma \in \tau (B_n)\subset S_{2n}$ we assign the signed permutation $n\times n$ matrix $T_{\sigma }=T({\tau }^{-1}(\pi ))$.

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 $D_5=⟨x,y|x^5=e=y^2,xyx=y⟩$ 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 ${\tau }^{-1}$.
$$\begin{array}{cc}\sigma & (\overrightarrow{a},\pi )\\ (2,5)(3,4)(7,10)(8,9)& ((1,1,1,1,1),(2,5)(3,4))\\ (1,5,4,3,2)(6,10,9,8,7)& ((1,1,1,1,1)(1,5,4,3,2))\\ (1,6)(2,5)(3,4)(7,10)(8,9)& ((-1,1,1,1,1),(2,5)(3,4))\\ (1,10,9,3,2)(4,8,7,6,5)& ((-1,1,1,-1,1),(1,5,4,3,2))\end{array}$$
giving the signed permutation matrices
$$\begin{array}{ccc}& x& y\\ A& \left[\begin{array}{ccccc}0& 1& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\\ 1& 0& 0& 0& 0\end{array}\right]& \left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0\end{array}\right]\\ B& \left[\begin{array}{ccccc}0& 1& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& -1& 0\\ 0& 0& 0& 0& 1\\ -1& 0& 0& 0& 0\end{array}\right]& \left[\begin{array}{ccccc}-1& 0& 0& 0& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0\end{array}\right]\end{array}$$
$D_5$ 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
$$A\simeq T\oplus W_1\oplus W_2\text{and}B\simeq V\oplus W_1\oplus W_2$$
Representation $A$ realises $D_5$ as a rotation symmetry group of the hypercube lattice ${\mathbb{Z}}^5$ in ${\mathbb{R}}^5$, and next we have to find a $D_5$-projection ${\mathbb{R}}^5=A\to W_1={\mathbb{R}}^2$.

As a complex representation $A{\downarrow }_{C_5}$ decomposes as a direct sum of $1$-dimensional representations
$$A{\downarrow }_{C_5}=V_1\oplus V_{\zeta }\oplus V_{{\zeta }^2}\oplus V_{{\zeta }^3}\oplus V_{{\zeta }^4}$$
where $\zeta =e^{2\pi i/5}$ and where the action of $x$ on $V_{{\zeta }^i}=\mathbb{C}{v}_i$ is given by $x.v_i={\zeta }^i{v}_i$. The $x$-eigenvectors in ${\mathbb{C}}^5$ are
$$\{\begin{array}{l}{v}_0=(1,1,1,1,1)\\ v_1=(1,\zeta ,{\zeta }^2,{\zeta }^3,{\zeta }^4)\\ v_2=(1,{\zeta }^2,{\zeta }^4,\zeta ,{\zeta }^3)\\ v_3=(1,{\zeta }^3,\zeta ,{\zeta }^4,{\zeta }^2)\\ v_4=(1,{\zeta }^4,{\zeta }^3,{\zeta }^2,\zeta )\end{array}$$
The action of $y$ on these vectors is given by $y.v_i=v_{5-i}$ because
$$x.(y.v_i)=(xy).v_i=(y{x}^{-1}).v_i=y.(x^{-1}.v_i)=y.({\zeta }^{-i}{v}_i)={\zeta }^{-1}(y.v_i)$$
and therefore $y.v_i$ is an $x$-eigenvector with eigenvalue ${\zeta }^{5-i}$. As a complex $D_5$-representation, the factors of $A$ are therefore
$$T=\mathbb{C}{v}_0,W_1=\mathbb{C}{v}_1+\mathbb{C}{v}_4,\text{and}{W}_2=\mathbb{C}{v}_2+\mathbb{C}{v}_3$$
But we want to consider $A$ as a real representation. As ${\zeta }^j=cos(\frac{2\pi j}{5})+isin(\frac{2\pi j}{5})=c_j+i{s}_j$ hebben we can take the vectors in ${\mathbb{R}}^5$
$$\{\begin{array}{l}\frac{1}{2}(v_1+v_4)=(1,c_1,c_2,c_3,c_4)=u_1\\ -\frac{1}{2}i(v_1-v_4)=(0,s_1,s_2,s_3,s_4)=u_2\\ \frac{1}{2}(v_2+v_3)=(1,c_2,c_4,c1,c3)=w_1\\ -\frac{1}{2}i(v_2-v_3)=(0,s_2,s_4,s_1,s_3)=w_2\end{array}$$
and $A$ decomposes as a real $D_5$-representation with
$$T=\mathbb{R}{v}_0,W_1=\mathbb{R}{u}_1+\mathbb{R}{u}_2,\text{and}{W}_2=\mathbb{R}{w}_1+\mathbb{R}{w}_2$$
and if we identify $\mathbb{C}$ with ${\mathbb{R}}^2$ via $z\leftrightarrow (Re(z),Im(z))$ we can describe the $D_5$-projection morphism ${\pi }_{W_1}:{\mathbb{R}}^5=A\to W_1={\mathbb{R}}^2$ via
$$(y_0,y_1,y_2,y_3,y_4)\mapsto y_0+y_1\zeta +y_2{\zeta }^2+y_3{\zeta }^3+y_4{\zeta }^4=\sum _{i=0}^4{y}_i(c_i,s_i)$$
Note also that $W_1$ is the orthogonal complement of $T\oplus W_2$, so is equal to the linear subspace in ${\mathbb{R}}^5$ determined by the three linear equations
$$\{\begin{array}{l}\sum _{i=0}^4{x}_i=0\\ \sum _{i=0}^4{c}_{2i}{x}_i=0\\ \sum _{i=0}^4{s}_{2i}{x}_i=0\end{array}$$

Okay, now take the Rhombic tiling corresponding to the regular pentagrid defined by ${\gamma }_0,\dots ,{\gamma }_4$ satisfying $\sum _{i=0}^4{\gamma }_i=0$. Let $\overrightarrow{k}=(k_0,\dots ,k_4)\in {\mathbb{Z}}^5$ and define the open hypercube $H_{\overrightarrow{k}}$ corresponding to $\overrightarrow{k}$ as the set of points
$$(x_0,\dots ,x_4)\in {\mathbb{R}}^5:\mathrm{\forall }0\le i\le 4:k_i–1<x_i<k_i$$ From the vector $\overrightarrow{\gamma }=({\gamma }_0,\dots ,{\gamma }_4)$ determining the Rhombic tiling we define the $2$-dimensional plane $P_{\overrightarrow{\gamma }}$ in ${\mathbb{R}}^5$ given by the equations $$\{\begin{array}{l}\sum _{i=0}^4{x}_i=0\\ \sum _{i=0}^4{c}_{2i}(x_i-{\gamma }_i)=0\\ \sum _{i=0}^4{s}_{2i}(x_i-{\gamma }_i)=0\end{array}$$ The point being that $P_{\overrightarrow{\gamma }}$ is the linear plane $W_1$ in ${\mathbb{R}}^5$ translated over the vector $\overrightarrow{\gamma }$, 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 $\overrightarrow{\gamma }=({\gamma }_0,\dots ,{\gamma }_4)$ are the points
$$\sum _{i=0}^4{k}_i(c_i,s_i)$$
with $\overrightarrow{k}=(k_0,\dots ,k_4)\in {\mathbb{Z}}^5$ such that $H_{\overrightarrow{k}}\cap P_{\overrightarrow{\gamma }}\ne \mathrm{\varnothing }$.

To see this, let $\overrightarrow{x}=(x_0,\dots ,x_4)\in P_{\overrightarrow{\gamma }}$, then $\overrightarrow{x}-\overrightarrow{\gamma }\in W_1$, but then there is a vector $\overrightarrow{y}\in {\mathbb{R}}^2$ such that
$$x_j–{\gamma }_j=\overrightarrow{y}.{\overrightarrow{v}}_j\mathrm{\forall }0\le j\le 4$$
But then, with $k_j=\lceil \overrightarrow{y}.{\overrightarrow{v}}_j+{\gamma }_j\rceil$ we have that $\overrightarrow{x}\in H_{\overrightarrow{k}}$ and we note that $V(\overrightarrow{y})=\sum _{i=0}^4{k}_i{\overrightarrow{v}}_i$ is a vetex of the Rhombic tiling associated to the regular pentagrid parameters $\overrightarrow{\gamma }=({\gamma }_0,\dots ,{\gamma }_4)$.

Here we used regularity of the pentagrid in order to have that $k_j=\overrightarrow{y}.{\overrightarrow{v}}_j+{\gamma }_j$ can happen for at most two $j$’s, so we can manage to vary $\overrightarrow{y}$ a little in order to have $\overrightarrow{x}$ in the open hypercube.

Here’s what we did so far: we have identified $D_5$ as a group of rotations in ${\mathbb{R}}^5$, preserving the hypercube-lattice ${\mathbb{Z}}^5$ in ${\mathbb{R}}^5$. If the $2$-plane $P_{\overrightarrow{\gamma }}$ is left stable under these rotations, then because rotations preserve distances, also the subset of lattice-points
$$S_{\overrightarrow{\gamma }}=\{(k_0,\dots ,k_4)|H_{\overrightarrow{k}}\cap P_{\overrightarrow{\gamma }}\ne \mathrm{\varnothing }\}\subset {\mathbb{Z}}^5$$
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 $\pi :A\to W_1$, 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 $P_{\overrightarrow{\gamma }}$ left stable under all rotations of $D_5$ is when $\overrightarrow{\gamma }=\overrightarrow{0}$, 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.