Skip to content →

Category: groups

Witt and his Niemeier lattices

Sunday, January 28th 1940, Hamburg

Ernst Witt wants to get two papers out of his system because he knows he’ll have to enter the Wehrmacht in February.

The first one, “Spiegelungsgruppen und Aufzählung halbeinfacher Liescher Ringe”, contains his own treatment of the root systems of semisimple Lie algebras and their reflexion groups, following up on previous work by Killing, Cartan, Weyl, van der Waerden and Coxeter.



(Photo: Natascha Artin, Nikolausberg 1933): From left to right: Ernst Witt; Paul Bernays; Helene Weyl; Hermann Weyl; Joachim Weyl, Emil Artin; Emmy Noether; Ernst Knauf; unknown woman; Chiuntze Tsen; Erna Bannow (later became wife of Ernst Witt)

Important for our story is that this paper contains the result stating that integral lattices generated by norm 2 elements are direct sums of root systems of the simply laced Dynkin diagrans $A_n, D_n$ and $E_6,E_7$ or $E_8$ (Witt uses a slightly different notation).



In each case, Witt knows of course the number of roots and the determinant of the Gram matrix
\[
\begin{array}{c|cc}
& \# \text{roots} & \text{determinant} \\
\hline
A_n & n^2+n & n+1 \\
D_n & 2n^2-2n & 4 \\
E_6 & 72 & 3 \\
E_7 & 126 & 2 \\
E_8 & 240 & 1
\end{array}
\]
The second paper “Eine Identität zwischen Modulformen zweiten Grades” proves that there are just two positive definite even unimodular lattices (those in which every squared length is even, and which have one point per unit volume, that is, have determinant one) in dimension sixteen, $E_8 \oplus E_8$ and $D_{16}^+$. Previously, Louis Mordell showed that the only unimodular even lattice in dimension $8$ is $E_8$.

The connection with modular forms is via their theta series, listing the number of lattice points of each squared length
\[
\theta_L(q) = \sum_{m=0}^{\infty} \#\{ \lambda \in L : (\lambda,\lambda)=m \} q^{m} \]
which is a modular form of weight $n/2$ ($n$ being the dimension which must be divisible by $8$) in case $L$ is a positive definitive even unimodular lattice.

The algebra of all modular forms is generated by the Eisenstein series $E_2$ and $E_3$ of weights $4$ and $6$, so in dimension $8$ we have just one possible theta series
\[
\theta_L(q) = E_2^2 = 1+480 q^2+ 61920 q^4+ 1050240 q^6+ \dots \]

It is interesting to read Witt’s proof of his main result (Satz 3) in which he explains how he constructed the possible even unimodular lattices in dimension $16$.

He knows that the sublattice of $L$ generated by the $480$ norm two elements must be a direct sum of root lattices. His knowledge of the number of roots in each case tells him there are only two possibilities
\[
E_8 \oplus E_8 \qquad \text{and} \qquad D_{16} \]
The determinant of the Gram matrix of $E_8 \oplus E_8$ is one, so this one is already unimodular. The remaining possibility
\[
D_{16} = \{ (x_1,\dots,x_{16}) \in \mathbb{Z}^{16}~|~x_1+ \dots + x_{16} \in 2 \mathbb{Z} \} \]
has determinant $4$ so he needs to add further lattice points (necessarily contained in the dual lattice $D_{16}^*$) to get it unimodular. He knows the coset representatives of $D_{16}^*/D_{16}$:
\[
\begin{cases}
[0]=(0, \dots,0) &~\text{of norm $0$} \\
[1]=(\tfrac{1}{2},\dots,\tfrac{1}{2}) &~\text{of norm $4$} \\
[2]=(0,\dots,0,1) &~\text{of norm $1$} \\
[3]=(\tfrac{1}{2},\dots,\tfrac{1}{2},-\tfrac{1}{2}) &~\text{of norm $4$}
\end{cases}
\]
and he verifies that the determinant of $D_{16}^+=D_{16}+([1]+D_{16})$ is indeed one (btw. adding coset $[3]$ gives an isomorphic lattice). Witt calls this procedure to arrive at the correct lattices forced (‘zwangslaufig’).

So, how do you think Witt would go about finding even unimodular lattices in dimension $24$?

To me it is clear that he would start with a direct sum of root lattices whose dimensions add up to $24$, compute the determinant of the Gram matrix and, if necessary, add coset classes to arrive at a unimodular lattice.

Today we would call this procedure ‘adding glue’, after Martin Kneser, who formalised this procedure in 1967.

On January 28th 1940, Witt writes that he found more than $10$ different classes of even unimodular lattices in dimension $24$ (without giving any details) and mentioned that the determination of the total number of such lattices will not be entirely trivial (‘scheint nicht ganz leicht zu sein’).

The complete classification of all $24$ even unimodular lattices in dimension $24$ was achieved by Hans Volker Niemeier in his 1968 Ph.D. thesis “Definite quadratische Formen der Dimension 24 und Diskriminante 1”, under the direction of Martin Kneser. Naturally, these lattices are now known as the Niemeier lattices.

Which of the Niemeier lattices were known to Witt in 1940?

There are three obvious certainties: $E_8 \oplus E_8 \oplus E_8$, $E_8 \oplus D_{16}^+$ (both already unimodular, the second by Witt’s work) and $D_{24}^+$ with a construction analogous to the one of $D_{16}^+$.

To make an educated guess about the remaining Witt-Niemeier lattices we can do two things:

  1. use our knowledge of Niemeier lattices to figure out which of these Witt was most likely to stumble upon, and
  2. imagine how he would adapt his modular form approach in dimension $16$ to dimension $24$.

Here’s Kneser’s neighbourhood graph of the Niemeier lattices. Its vertices are the $24$ Niemeiers and there’s an edge between $L$ and $M$ whenever the intersection $L \cap M$ is of index $2$ in both $L$ and $M$. In this case, $L$ and $M$ are called neighbours.



Although the theory of neighbours was not known to Witt, the graph may give an indication of how likely it is to dig up a new Niemeier lattice by poking into an already discovered one.
The three certainties are the three lattices at the bottom of the neighborhood graph, making it more likely for the lattices in the lower region to be among Witt’s list.

For the other approach, the space of modular forms of weight $12$ is two dimensional, with a basis formed by the series
\[
\begin{cases}
E_6(q) = 1 + \tfrac{65520}{691}(q+2049 q^2 + 177148 q^3+4196353q^4+\dots \\
\Delta(q) = q-24 q^2+252q^3-1472q^4+ \dots
\end{cases}
\]

If you are at all with me, Witt would start with a lattice $R$ which is a direct sum of root lattices, so he would know the number of its roots (the norm $2$ vectors in $R$), let’s call this number $r$. Now, he wants to construct an even unimodular lattice $L$ containing $R$, so the theta series of both $L$ and $R$ must start off with $1 + r q^2 + \dots$. But, then he knows
\[
\theta_L(q) = E_6(q) + (r-\frac{65520}{691})\Delta(q) \]
and comparing coefficients of $\theta_L(q)$ with those of $\theta_R(q)$ will give him an idea what extra vectors he has to throw in.

If we’re generous to Witt (and frankly, why shouldn’t we), we may believe that he used his vast knowledge of Steiner systems (a few years earlier he wrote the definite paper on the Mathieu groups, and a paper on Steiner systems) to construct in this way the lattices $(A_1^{24})^+$ and $(A_2^{12})^+$.

The ‘glue’ for $(A_1^{24})^+$ is coming from the extended binary Golay code, which itself uses the Steiner system $S(5,8,24)$. $(A_2^{12})^+$ is constructed using the extended ternary Golay code, based on the Steiner system $S(5,6,12)$.

The one thing that would never have crossed his mind that sunday in 1940 was to explore the possibility of an even unimodular 24-dimensional lattice $\Lambda$ without any roots!

One with $r=0$, and thus with a theta series starting off as
\[
\theta_{\Lambda}(q) = 1 + 196560 q^4 + 16773120 q^6 + \dots \]
No, he did not find the Leech lattice that day.

If he would have stumbled upon it, it would have simply blown his mind.

It would have been so much against all his experiences and intuitions that he would have dropped everything on the spot to write a paper about it, or at least, he would have mentioned in his ‘more than $10$ lattices’-claim that, surprisingly, one of them was an even unimodular lattice without any roots.

Leave a Comment

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
\[
\mathbb{Z} \vec{e}_1 + \dots + \mathbb{Z} \vec{e}_n \subset \mathbb{R}^n \]
is the hyperoctahedral group of all signed $n \times n$ permutation matrices
\[
B_n = C_2^n \rtimes S_n \]
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 $(\vec{a},\pi) \in B_n$ is given by
\[
T_{ij} = T(\vec{a},\pi)_{ij} = a_j \delta_{i,\pi(j)} \]
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
\[
\tau((\vec{a},\pi))(k) = \begin{cases} \pi(k)+n \delta_{-1,a_k}~\text{if $1 \leq k \leq n$} \\
\pi(k-n)+n(1-\delta_{-1,a_{k-n}})~\text{if $n+1 \leq k \leq 2n$} \end{cases} \]
the image is generated by the permutations
\[
\begin{cases}
\alpha = (1,2)(n+1,n+2), \\
\beta=(1,2,\dots,n)(n+1,n+2,\dots,2n), \\
\gamma=(n,2n)
\end{cases}
\]
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 $D_5 = \langle x,y~|~x^5=e=y^2,~xyx=y \rangle$ 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 $\tau^{-1}$.
\[
\begin{array}{c|c}
\sigma & (\vec{a},\pi) \\
\hline
(2,5)(3,4)(7,10)(8,9) & ((1,1,1,1,1),(2,5)(3,4)) \\
(1,5,4,3,2)(6,10,9,8,7) & ((1,1,1,1,1)(1,5,4,3,2)) \\
(1,6)(2,5)(3,4)(7,10)(8,9) & ((-1,1,1,1,1),(2,5)(3,4)) \\
(1,10,9,3,2)(4,8,7,6,5) & ((-1,1,1,-1,1),(1,5,4,3,2))
\end{array}
\]
giving the signed permutation matrices
\[
\begin{array}{c|cc}
& x & y \\
\hline
A & \begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 \end{bmatrix} &
\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \end{bmatrix} \\
\hline
B & \begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
-1 & 0 & 0 & 0 & 0 \end{bmatrix} &
\begin{bmatrix}
-1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \end{bmatrix}
\end{array}
\]
$D_5$ has $4$ conjugacy classes with representatives $e,y,x$ and $x^2$. the
character table of $D_5$ is
\[
\begin{array}{c|cccc}
& (1) & (2) & (2) & (5) \\
& 1_a & 5_1 & 5_2 & 2_a \\
D_5 & e & x & x^2 & y \\
\hline
T & 1 & 1 & 1 & 1 \\
V & 1 & 1 & 1 & -1 \\
W_1 & 2 & \tfrac{-1+ \sqrt{5}}{2} & \tfrac{-1 -\sqrt{5}}{2} & 0 \\
W_2 & 2 & \tfrac{-1 -\sqrt{5}}{2} & \tfrac{-1+\sqrt{5}}{2} & 0
\end{array}
\]
Using the signed permutation matrices it is easy to determine the characters of the $5$-dimensional representations $A$ and $B$
\[
\begin{array}{c|cccc}
D_5 & e & x & x^2 & y \\
\hline
A & 5 & 0 & 0 & 1 \\
B & 5 & 0 & 0 & -1
\end{array}
\]
decomosing into $D_5$-irreducibles as
\[
A \simeq T \oplus W_1 \oplus W_2 \quad \text{and} \quad B \simeq V \oplus W_1 \oplus W_2 \]
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 $\mathbb{R}^5=A \rightarrow W_1 = \mathbb{R}^2$.

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
\[
\begin{cases}
v_0 = (1,1,1,1,1) \\
v_1 = (1,\zeta,\zeta^2,\zeta^3,\zeta^4) \\
v_2 =(1,\zeta^2,\zeta^4,\zeta,\zeta^3) \\
v_3 = (1,\zeta^3,\zeta,\zeta^4,\zeta^2) \\
v_4 = (1,\zeta^4,\zeta^3,\zeta^2,\zeta)
\end{cases}
\]
The action of $y$ on these vectors is given by $y.v_i = v_{5-i}$ because
\[
x.(y.v_i) = (xy).v_i=(yx^{-1}).v_i=y.(x^{-1}.v_i) = y.(\zeta^{-i} v_i) = \zeta^{-1} (y.v_i) \]
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
\[
T = \mathbb{C} v_0, \quad W_1 = \mathbb{C} v_1 + \mathbb{C} v_4, \quad \text{and} \quad W_2 = \mathbb{C} v_2 + \mathbb{C} v_3 \]
But we want to consider $A$ as a real representation. As $\zeta^j = cos(\tfrac{2 \pi j}{5})+i~sin(\tfrac{2 \pi j}{5}) = c_j + i s_j$ hebben we can take the vectors in $\mathbb{R}^5$
\[
\begin{cases}
\frac{1}{2}(v_1+v_4) = (1,c_1,c_2,c_3,c_4)= u_1 \\
-\frac{1}{2}i(v_1-v_4) = (0,s_1,s_2,s_3,s_4) = u_2 \\
\frac{1}{2}(v_2+v_3) = (1,c_2,c_4,c1,c3)= w_1 \\
-\frac{1}{2}i(v_2-v_3) = (0,s_2,s_4,s_1,s_3)= w_2
\end{cases}
\]
and $A$ decomposes as a real $D_5$-representation with
\[
T = \mathbb{R} v_0, \quad W_1 = \mathbb{R} u_1 + \mathbb{R} u_2, \quad \text{and} \quad W_2 = \mathbb{R} w_1 + \mathbb{R} w_2 \]
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 $\pi_{W_1}~:~\mathbb{R}^5=A \rightarrow W_1=\mathbb{R}^2$ 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
\[
\begin{cases}
\sum_{i=0}^4 x_i = 0 \\
\sum_{i=0}^4 c_{2i} x_i = 0 \\
\sum_{i=0}^4 s_{2i} x_i = 0
\end{cases}
\]



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 $\vec{k}=(k_0,\dots,k_4) \in \mathbb{Z}^5$ and define the open hypercube $H_{\vec{k}}$ corresponding to $\vec{k}$ as the set of points
\[
(x_0,\dots,x_4) \in \mathbb{R}^5~:~\forall 0 \leq i \leq 4~:~k_i – 1 < x_i < k_i \] From the vector $\vec{\gamma} = (\gamma_0,\dots,\gamma_4)$ determining the Rhombic tiling we define the $2$-dimensional plane $P_{\vec{\gamma}}$ in $\mathbb{R}^5$ given by the equations \[ \begin{cases} \sum_{i=0}^4 x_i = 0 \\ \sum_{i=0}^4 c_{2i} (x_i - \gamma_i) = 0 \\ \sum_{i=0}^4 s_{2i} (x_i - \gamma_i) = 0 \end{cases} \] The point being that $P_{\vec{\gamma}}$ is the linear plane $W_1$ in $\mathbb{R}^5$ translated over the vector $\vec{\gamma}$, 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 $\vec{\gamma}=(\gamma_0,\dots,\gamma_4)$ are the points
\[
\sum_{i=0}^4 k_i (c_i,s_i) \]
with $\vec{k}=(k_0,\dots,k_4) \in \mathbb{Z}^5$ such that $H_{\vec{k}} \cap P_{\vec{\gamma}} \not= \emptyset$.

To see this, let $\vec{x} = (x_0,\dots,x_4) \in P_{\vec{\gamma}}$, then $\vec{x}-\vec{\gamma} \in W_1$, but then there is a vector $\vec{y} \in \mathbb{R}^2$ such that
\[
x_j – \gamma_j = \vec{y}.\vec{v}_j \quad \forall~0 \leq j \leq 4 \]
But then, with $k_j=\lceil \vec{y}.\vec{v}_j + \gamma_j \rceil$ 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 $\vec{\gamma}=(\gamma_0,\dots,\gamma_4)$.

Here we used regularity of the pentagrid in order to have that $k_j=\vec{y}.\vec{v}_j + \gamma_j$ 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 $\mathbb{R}^5$, preserving the hypercube-lattice $\mathbb{Z}^5$ in $\mathbb{R}^5$. If the $2$-plane $P_{\vec{\gamma}}$ is left stable under these rotations, then because rotations preserve distances, also the subset of lattice-points
\[
S_{\vec{\gamma}} = \{ (k_0,\dots,k_4)~|~H_{\vec{k}} \cap P_{\vec{\gamma}} \not= \emptyset \} \subset \mathbb{Z}^5 \]
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 $\pi : A \rightarrow W_1$, 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 $P_{\vec{\gamma}}$ left stable under all rotations of $D_5$ is when $\vec{\gamma} = \vec{0}$, 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.

Leave a Comment

Kasha-eating dragons

This semester I’m teaching a first course in representation theory. On campus, IRL! It’s a bit strange, using a big lecture room for a handful of students, everyone wearing masks, keeping distances, etc.



So far, this is their only course on campus, so it has primarily a social function. The breaks in between are infinitely more important than the lectures themselves. I’d guess breaks take up more than one third of the four hours scheduled.

At first, I hoped to make groups and their representations relevant by connecting to the crisis at hand, whence the the symmetries of Covid-19 post, and the Geometry of Viruses series of posts.

Not a great idea. I guess most of us are by now over-saturated with Corona-related news, and if students are allowed to come to campus just one afternoon per week, the last thing they want to hear about is, right, Covid.

So I need to change tactics. By now we’ve reached the computation of character tables, and googling around I found this MathOverflow-topic: Fun applications of representations of finite groups.

The highest rated answer, by Vladimir Dotsenko, suggests a problem attributed to Kirillov:

An example from Kirillov’s book on representation theory: write numbers 1,2,3,4,5,6 on the faces of a cube, and keep replacing (simultaneously) each number by the average of its neighbours. Describe (approximately) the numbers on the faces after many iterations.

A bit further down the list, the Lecture notes on representation theory by Vera Serganova are mentioned. They start off with a variation of Kirillov’s question (and an extension of it to the dodecahedron):

Hungry knights. There are n hungry knights at a round table. Each of them has a plate with certain amount of food. Instead of eating every minute each knight takes one half of his neighbors servings. They start at 10 in the evening. What can you tell about food distribution in the morning?

Breakfast at Mars. It is well known that marsians have four arms, a standard family has 6 persons and a breakfast table has a form of a cube with each person occupying a face on a cube. Do the analog of round table problem for the family of marsians.

Supper at Venus. They have five arms there, 12 persons in a family and sit on the faces of a dodecahedron (a regular polyhedron whose faces are pentagons).

Perhaps the nicest exposition of the problem (and its solution!) is in the paper Dragons eating kasha by Tanya Khovanova.

Suppose a four-armed dragon is sitting on every face of a cube. Each dragon has a bowl of kasha in front of him. Dragons are very greedy, so instead of eating their own kasha, they try to steal kasha from their neighbors. Every minute every dragon extends four arms to the four neighboring faces on the cube and tries to get the kasha from the bowls there. As four arms are fighting for every bowl of
kasha, each arm manages to steal one-fourth of what is in the bowl. Thus each
dragon steals one-fourth of the kasha of each of his neighbors, while at the same
time all of his own kasha is stolen. Given the initial amounts of kasha in every
bowl, what is the asymptotic behavior of the amounts of kasha?

I can give them quick hints to reach the solution:

  • the amounts of kasha on each face gives a vector in $\mathbb{R}^6$, which is an $S_4$-representation,
  • calculate the character of this kasha-representation,
  • use the character table of $S_4$ to decompose the representation into irreducibles,
  • identify each of the irreducible factors as instances in the kasha-representation,
  • check that the food-grabbing operation is an $S_4$-morphism,
  • remember Schur’s lemma, and compute the scaling factors on each irreducible component,
  • conclude!

But, I can never explain it better than Khovanova’s treatment of the kasha-eating dragons problem.

Leave a Comment

de Bruijn’s pentagrids

In a Rhombic tiling (aka a Penrose P3 tiling) we can identify five ribbons.

Opposite sides of a rhomb are parallel. We may form a ribbon by attaching rhombs along opposite sides. There are five directions taken by sides, so there are five families of ribbons that do not intersect, determined by the side directions.

Every ribbon determines a skeleton curve through the midpoints of opposite sides of the rhombs. If we straighten these skeleton curves we get a de Bruijn’s pentagrid.



A pentagrid is a grid of the plane consisting of five families of parallel lines, at angles multiples of $72^o = 360^o/5$, and with each family consisting of parallel lines with a spacing given by a number $\gamma_i$ for $0 \leq i \leq 4$, satisfying
\[
\gamma_0+\gamma_1+\gamma_2+\gamma_3+\gamma_4=0 \]

The points of the plane in the $j$-th family of the pentagrid are
\[
\{ \vec{x} \in \mathbb{R}^2~|~\vec{x}.\vec{v}_j + \gamma_j \in \mathbb{Z} \} \]
where $\vec{v}_j = \zeta^j = (cos(2 \pi j/5),sin(2 \pi j/5))$.

A pentagrid is regular if no point in $\mathbb{R}^2$ belongs to more than two of the five grids. For almost all choices $\gamma_0,\dots,\gamma_4$ the corresponding pentagrid is regular. The pentagrid coordinates of a point $\vec{x} \in \mathbb{R}^2$ are the five integers $(K_0(\vec{x}),\dots,K_4(\vec{x}) \in \mathbb{Z}^5$ defined by
\[
K_j(\vec{x}) = \lceil \vec{x}.\vec{v}_j + \gamma_j \rceil \]
with $\lceil r \rceil$ the smallest integer greater of equal to $r$, and clearly the function $K_j(\vec{x})$ is constant in regions between the $j$-th grid lines.



With these pentagrid-coordinates one can associate a vertex to any point $\vec{x} \in \mathbb{R}^2$
\[
V(\vec{x}) = \sum_{j=0}^4 K_j(\vec{x}) \vec{v}_j \]
which is constant in regions between grid lines.

Here’s de Bruijn‘s result:

For $\vec{x}$ running over $\mathbb{R}^2$, the vertices $V(\vec{x})$ coming from a pentagrid determine a Rhombic tiling of the plane. Even better, every Rhombic tiling comes from a pentagrid.



We’ll prove this for regular pentagrids (the singular case follows from a small deformation).

Any intersection point $\vec{x}_0$ of the pentagrid belongs to exactly two families of parallel lines, so we have two integers $r$ and $s$ with $0 \leq r < s \leq 4$ and integers $k_r,k_s \in \mathbb{Z}$ such that $\vec{x}_0$ is determined by the equations \[ \begin{cases} \vec{x}.\vec{v}_r + \gamma_r = k_r \\ \vec{x}.\vec{v}_s + \gamma_s = k_s \end{cases} \] In a small neighborhood of $\vec{x}_0$, $V(\vec{x})$ takes the values of four vertices of a rhomb. These vertices are associated to the $5$-tuples in $\mathbb{Z}^5$ given by \[ (K_0(\vec{x}_0),\dots,K_4(\vec{x}_0))+\epsilon_1 (\delta_{0r},\dots,\delta_{4r})+\epsilon_2 (\delta_{0s},\dots,\delta_{4s}) \] with $\epsilon_1,\epsilon_2 \in \{ 0,1 \}$. So, the intersection points of the regular pentagrid lines correspond to rhombs, and the regions between the grid lines (which are called meshes) correspond to vertices, whose positions are given by $V(\vec{x})$.

Observe that these four vertices give a thin rhombus if the angle between $\vec{v}_r$ and $\vec{v}_s$ is $144^o$ and determine a thick rhombus if the angle is $72^o$.

Not all pentagrid coordinates $(k_0,\dots,k_4)$ occur in the tiling though. For $\vec{x} \in \mathbb{R}^2$ we have
\[
K_j(\vec{x}) = \vec{x}.\vec{v}_j + \gamma_j + \lambda_j(\vec{x}) \]
with $0 \leq \lambda_j(\vec{x}) < 1$. In a regular pentagrid at most two of the $\lambda_j(\vec{x})$ can be zero and so we have \[ 0 < \lambda_0(\vec{x}) + \dots + \lambda_4(\vec{x}) < 5 \] and we have assumed that $\gamma_0 + \dots + \gamma_4 = 0$, giving us \[ \sum_{j=0}^4 K_j(\vec{x}) = \vec{x}.(\sum_{j=0}^4 \vec{v}_j) + \sum_{j=0}^4 \gamma_j + \sum_{j=0}^4 \lambda_j(\vec{x}) = \sum_{j=0}^4 \lambda_j(\vec{x}) \] The left hand side must be an integers and the right hand side a number strictly between $0$ and $5$. This defines the index of a vertex
\[
ind(\vec{x}) = \sum_{j=0}^4 K_j(\vec{x}) \in \{ 1,2,3,4 \} \]
Therefore, every vertex in the tiling may be represented as
\[
k_0 \vec{v}_0 + \dots + k_4 \vec{v}_4 \]
with $(k_0,\dots,k_4) \in \mathbb{Z}^5$ satisfying $\sum_{j=0}^4 k_j \in \{ 1,2,3,4 \}$. If we move a point along the edges of a rhombus, we note that the index increases by $1$ in the directions of $\vec{v}_0,\dots,\vec{v}_4$ and decreases by $1$ in the directions $-\vec{v}_0,\dots,-\vec{v}_4$.



It follows that the index-values of the vertices of a thick rhombus are either $1$ and $3$ at the $72^o$ angles and $2$ at the $108^o$ angles, or they are $2$ and $4$ at the $72^o$ angles, and $3$ at the $108^o$ angles. For a thin rhombus the index-values must be either $1$ and $3$ at the $144^o$ angles and $2$ at the $36^o$ angles, or $2$ and $4$ at the $144^o$ angles and $3$ at the $36^o$ angles.

We still have to show that this gives a legal Rhombic tiling, that is, that the gluing restrictions of Penrose’s rhombs are satisfied.



We do this by orienting and colouring the edges of the thin and thick rhombus towards the vertex defined by the semi-circles on the rhombus-pieces. Edges connecting a point of index $3$ to a point of index $2$ are coloured red, edges connecting a point of index $1$ to a point of index $2$, or connecting a point of index $3$ to one of index $4$ are coloured blue.

We orient green edges pointing from index $2$ to index $1$, or from index $3$ to index $4$. As this orientation depends only on the index-values of the vertices, two rhombs sharing a common green edge also have the same orientation on that edge.

On each individual rhomb, knowing the green edges and their orientation forced the red edges and their orientation, but we still have to show that if two rhombs share a common red edge, this edge has the same orientation on both (note in the picture above that a red edge can both flow from $2$ to $3$ as well as from $3$ to $2$).

From the gluing conditions of Penrose’s rhombs we see that if $PQ$ be a red edge, in common to two rhombs, and these rhombs have angle $\alpha$ resp. $\beta$ in $P$, then $\alpha$ and $\beta$ must be both less that $90^o$ or both bigger than $90^o$. We translate this in terms of the pentragrid.



The two rhombs correspond to two intersection points $A$ and $B$, and we may assume that they both lie on a line $l$ of family $0$ of parallel lines (other cases are treated similarly by cyclic permutation), and that $A$ is an intersection with a family $p$-line and $B$ with a family $q$-line, with $p,q \in \{ 1,2,3,4 \}$. The interval $AB$ crosses the unique edge common to the two rhombs determined by $A$ and $B$, and we call the interval $AB$ red if $\sum_j K_j(\vec{x})$ is $2$ on one side of $AB$ and $3$ on the other side. The claim about the angles $\alpha$ and $\beta$ above now translates to: if $AB$ is red, then $p+q$ is odd (in the picture above $p=1$ and $q=2$).

By a transformation we may assume that $\gamma_0=0$ and that $l$ is the $Y$-axis. For every $y \in \mathbb{R}$ we then get
\[
\begin{cases}
K_1(0,y) = \lceil y .sin(2 \pi/5) + \gamma_1 \rceil \\
K_2(0,y) = \lceil y .sin(4 \pi/5) + \gamma_2 \rceil \\
K_3(0,y) = \lceil -y .sin(4 \pi/5) + \gamma_3 \rceil \\
K_4(0,y) = \lceil -y .sin(2 \pi/5) + \gamma_4 \rceil
\end{cases}
\]
and $\gamma_1+\gamma_4$ and $\gamma_2+\gamma_3$ are not integers (otherwise the pentagrid is not regular). If $y$ runs from $-\infty$ to $+\infty$ we find that
\[
K_1(0,y)+K_4(0,y) – \lceil \gamma_1 + \gamma_4 \rceil = \begin{cases} 0 \\ 1 \end{cases} \]
with jumps from $0$ to $1$ at places where $(\lceil \gamma_1 + \gamma_4 \rceil – \gamma_1)/sin(2 \pi/5) \in \mathbb{Z}$ and from $1$ to $0$ when $\gamma_4/sin(2 \pi/5) \in \mathbb{Z}$. (A similar result holds replacing $K_1,K_4,\gamma_1,\gamma_4,sin(2 \pi/5)$ by $K_2,K_3,\gamma_2,\gamma_3,sin(4 \pi/5)$.

Because the points on $l$ intersecting with $1$-family and $4$-family gridlines alternate (and similarly for $2$- and $3$-family gridlines) we know already that $p \not= q$. If we assume that $p+q$ is even, we have two possibilities, either $\{p,g \}=\{ 1,3 \}$ or $\{ 2,4 \}$. As $\gamma_0=0$ we have $\gamma_1+\gamma_2+\gamma_3+\gamma_4=0$ and therefore
\[
\lceil \gamma_1 + \gamma_4 \rceil + \lceil \gamma_2+\gamma_3 \rceil = 1 \]
It then follows that $K_1(0,y)+K_2(0,y)+K_3(0,y)+K_4(0,y) = 1$ or $3$ between the points $A$ and $B$. Therefore $K_0(y,0) + \dots + K_4(0,y)$ is either $1$ on the left side and $2$ on the right side, or is $3$ on the left side and $4$ on the right side, so $AB$ must be green, a contradiction. Therefore, $p+q$ is odd, and the orientation of the red edge in both rhombs is the same. Done!

An alternative way to see this correspondence between regular pentagrids and Rhombic tilings as as follows. To every intersection of two gridlines we assign a rhombus, a thin one if they meet at an acute angle of $36^o$ and a thich one if this angle is $72^o$, with the long diagonal dissecting the obtuse angle.



Do this for all intersections, surrounding a given mesh.



Attach the rhombs by translating them towards the mesh, and finally draw the colours of the edges.



Another time, we will connect this to the cut-and-project method, using the representation theory of $D_5$.

Leave a Comment

Borcherds’ favourite numbers

Whenever I visit someone’s YouTube or Twitter profile page, I hope to see an interesting banner image. Here’s the one from Richard Borcherds’ YouTube Channel.

Not too surprisingly for Borcherds, almost all of these numbers are related to the monster group or its moonshine.

Let’s try to decode them, in no particular order.

196884

John McKay’s observation $196884 = 1 + 196883$ was the start of the whole ‘monstrous moonshine’ industry. Here, $1$ and $196883$ are the dimensions of the two smallest irreducible representations of the monster simple group, and $196884$ is the first non-trivial coefficient in Klein’s j-function in number theory.

$196884$ is also the dimension of the space in which Robert Griess constructed the Monster, following Simon Norton’s lead that there should be an algebra structure on the monster-representation of that dimension. This algebra is now known as the Griess algebra.

Here’s a recent talk by Griess “My life and times with the sporadic simple groups” in which he tells about his construction of the monster (relevant part starting at 1:15:53 into the movie).

1729

1729 is the second (and most famous) taxicab number. A long time ago I did write a post about the classic Ramanujan-Hardy story the taxicab curve (note to self: try to tidy up the layout of some old posts!).

Recently, connections between Ramanujan’s observation and K3-surfaces were discovered. Emory University has an enticing press release about this: Mathematicians find ‘magic key’ to drive Ramanujan’s taxi-cab number. The paper itself is here.

“We’ve found that Ramanujan actually discovered a K3 surface more than 30 years before others started studying K3 surfaces and they were even named. It turns out that Ramanujan’s work anticipated deep structures that have become fundamental objects in arithmetic geometry, number theory and physics.”

Ken Ono

24

There’s no other number like $24$ responsible for the existence of sporadic simple groups.

24 is the length of the binary Golay code, with isomorphism group the sporadic Mathieu group $M_24$ and hence all of the other Mathieu-groups as subgroups.

24 is the dimension of the Leech lattice, with isomorphism group the Conway group $Co_0 = .0$ (dotto), giving us modulo its center the sporadic group $Co_1=.1$ and the other Conway groups $Co_2=.2, Co_3=.3$, and all other sporadics of the second generation in the happy family as subquotients (McL,HS,Suz and $HJ=J_2$)



24 is the central charge of the Monster vertex algebra constructed by Frenkel, Lepowski and Meurman. Most experts believe that the Monster’s reason of existence is that it is the symmetry group of this vertex algebra. John Conway was one among few others hoping for a nicer explanation, as he said in this interview with Alex Ryba.

24 is also an important number in monstrous moonshine, see for example the post the defining property of 24. There’s a lot more to say on this, but I’ll save it for another day.

60

60 is, of course, the order of the smallest non-Abelian simple group, $A_5$, the rotation symmetry group of the icosahedron. $A_5$ is the symmetry group of choice for most viruses but not the Corona-virus.

3264

3264 is the correct solution to Steiner’s conic problem asking for the number of conics in $\mathbb{P}^2_{\mathbb{C}}$ tangent to five given conics in general position.



Steiner himself claimed that there were $7776=6^5$ such conics, but realised later that he was wrong. The correct number was first given by Ernest de Jonquières in 1859, but a rigorous proof had to await the advent of modern intersection theory.

Eisenbud and Harris wrote a book on intersection theory in algebraic geometry, freely available online: 3264 and all that.

248

248 is the dimension of the exceptional simple Lie group $E_8$. $E_8$ is also connected to the monster group.

If you take two Fischer involutions in the monster (elements of conjugacy class 2A) and multiply them, the resulting element surprisingly belongs to one of just 9 conjugacy classes:

1A,2A,2B,3A,3C,4A,4B,5A or 6A

The orders of these elements are exactly the dimensions of the fundamental root for the extended $E_8$ Dynkin diagram.

This is yet another moonshine observation by John McKay and I wrote a couple of posts about it and about Duncan’s solution: the monster graph and McKay’s observation, and $E_8$ from moonshine groups.

163

163 is a remarkable number because of the ‘modular miracle’
\[
e^{\pi \sqrt{163}} = 262537412640768743.99999999999925… \]
This is somewhat related to moonshine, or at least to Klein’s j-function, which by a result of Kronecker’s detects the classnumber of imaginary quadratic fields $\mathbb{Q}(\sqrt{-D})$ and produces integers if the classnumber is one (as is the case for $\mathbb{Q}(\sqrt{-163})$).

The details are in the post the miracle of 163, or in the paper by John Stillwell, Modular Miracles, The American Mathematical Monthly, 108 (2001) 70-76.

Richard Borcherds, the math-vlogger, has an entertaining video about this story: MegaFavNumbers 262537412680768000

His description of the $j$-function (at 4:13 in the movie) is simply hilarious!

Borcherds connects $163$ to the monster moonshine via the $j$-function, but there’s another one.

The monster group has $194$ conjugacy classes and monstrous moonshine assigns a ‘moonshine function’ to each conjugacy class (the $j$-function is assigned to the identity element). However, these $194$ functions are not linearly independent and the space spanned by them has dimension exactly $163$.

One Comment

Conway’s musical sequences

Before we’ll come to applications of quasicrystals to viruses it is perhaps useful to illustrate essential topics such as deflation, inflation, aperiodicity, local isomorphism and the cut-and project method in the simplest of cases, that of $1$-dimensional tilings.

We want to tile the line $\mathbb{R}^1$ with two kinds of tiles, short ($S$) and ($L$) long intervals, differing by a golden ratio factor $\tau=\tfrac{1}{2}(1+\sqrt{5}) \approx 1.618$.



Clearly, no two tiles may overlap and we impose a gluing restriction: there can be no two consecutive $S$-intervals in the tiling, and no three consecutive $L$-intervals.

The code of a tiling is a doubly infinite word in $S$ and $L$ such that there are no two consecutive $S$’s nor three consecutive $L$’s. For example
\[
\sigma = \dots LSLLSLS\underline{L}LSLLSL \dots \]
We underline one tile to distinguish the sequence from shifts of it.

Conway’s musical sequences will be special codes (or tilings), allowing for the inverse operations of inflation and deflation, terms coined by John Conway in relation to Penrose tilings. The musical sequences are important to understand Conway’s worms (sometimes called “wormholes”) which are strings of Long and Short bow ties in a Penrose tiling, and to measure the distances between Amman bars. In fact, many of the properties of Penrose tilings and $3$-dimensional quasicrystals (for example, local isomorphism) have their counterparts for tilings having a musical sequence as code.



Conway’s investigations of Penrose tiles held up work on the ATLAS-project and caused some problems at home:

“In pursuing his investigations, he unsurped some of his wife Eileen’s territory, covering the dining table with an infinite nuisance of tiles. He cut them out himself, causing his right hand to hurt with cramps for days. To Eileen’s dismay, he studied the dining table mosaic for a year, relegating family meals to the kitchen and prohibiting dinner parties.”

From “Genius at Play – The curious mind of John Horton Conway” by Siobhan Roberts

Let’s investigate inflation and deflation of these tilings.

The point of the golden factor $\tau$ is to allow for deflation. That is, we can replace a tiling by another one with tiles $S$ and $L$ both a factor $\tfrac{1}{\tau}=\tau-1 \approx 0.618$ smaller than the original tiles. If the original $S$-tile has length $a$ (and the $L$-tile length $\tau a$), then the new tile $S$ wil have length $\tfrac{1}{\tau}a$ and the new $L$-tile length $a$.
We do this by replacing each old $S$-tile by a new $L$-tile, and to break up any old $L$-tile in a new $L$ and new $S$-tile, as $\tau a = a + \tfrac{1}{\tau}a$ (note that $\tau^2=\tau+1$)



To get the code of the deflated tiling we replace each letter $S$ by a letter $L$ and each $L$ by $LS$. The underlined letter will be the first letter of the deflated underlined letter in the original sequence. The deflated sequence of the one above is
\[
def(\sigma) = \dots LSLLSLSLLSL\underline{L}SLSLLSLSLLS \dots \]
and it is easy to see that the deflated tiling satisfies again the gluing condition.

Certain of these tilings (not all!) allow for an inverse to deflation, called inflation, increasing the size of the tiles by a factor $\tau$.

Starting from a tiling we divide each $L$-tile in half and these mid-points will be end-points of the tiles in the new tiling, erasing all endpoints of the original one. The inflated tiling will have two sorts of tiles, a new short one $S$ of length $\tau a$ obtained from the end-half of an original $L$-tile, followed b the start-half of an original $L$-tile, and a new long tile $L$ of length $\tau^2 a = (\tau+1) a$, made of the end-half of an original $L$, followed by an original $S$, followed by the start-half of an original $L$.



We get the code of the inflated tiling by replacing first each $L$ by $ll$ and subsequently replace each word $lSl$ by a letter $L$ and each $ll$ by $S$. An example,
\[
\sigma = \dots LSLLSLSLLSL\underline{L}SLSLLSLSLLS \dots \\
\dots llSllllSllSllllSlll\underline{l}SllSllllSllSllllS \dots \\
inf(\sigma) = \dots (l)LSLLSLS\underline{L}LSLLS(lS) \dots \]

But, the inflated tiling may no longer satisfy the gluing condition. An example
\[
\dots LSLSLSL \dots \mapsto \dots llSllSllSll \dots \mapsto \dots (l)LLL(l) \dots \]

A Conway musical sequence is the code of a tiling $\sigma$ such that all its consecutive inflations $inf^n(\sigma)$ satisfy the gluing condition. For the corresponding Conway tiling $\sigma$ we have that
\[
def(inf(\sigma))=\sigma=inf(def(\sigma)) \]

Let’s construct at least two such Conway tilings (later we’ll see that there are uncountably many). Take $C_n=def^n(LS\underline{L})$ and write it in a special form to highlight symmetries.
\begin{eqnarray*}
C_0 =& (L.S)\underline{L} \\
C_1 =& L(S.L)\underline{L}S \\
C_2 =& LSL(L.S)\underline{L}SL \\
C_3 =& LSLLSL(S.L)\underline{L}SLLS \\
C_4 =& LSLLSLSLLSL(L.S)\underline{L}SLLSLSL
\end{eqnarray*}

The even terms have middle-part $(L.S)$ and the odd ones $(S.L)$. The remaing left and right parts are each others reflexion (or part of it). This is easily seen by induction as are the inclusions
\[
C_0 \subset C_2 \subset C_4 \subset \dots \subset C_{even} \quad \text{and} \quad C_1 \subset C_3 \subset C_5 \subset \dots \subset C_{odd} \]

$C_{even}$ and $C_{odd}$ are special Conway musical sequences, called the middle $C$-sequences, and are each others inflation and deflation. If you are familiar with Penrose tilings, these are the $1$-dimensional counterparts of the cartwheel Penrose tiling (here with the $10$ Conway worms emanating from the center, and with the borders of the first few cartwheels drawn).



A direct consequence of inflation on Conway’s musical sequences is that the corresponding tiling is aperiodic, that is, it has no translation symmetry.

For, inflation only depends on the local configuration of tiles, so if translation by $R$ is a symmetry of a musical sequence $\sigma$ then it is also a symmetry of $inf(\sigma)$, and so also of $inf^n(\sigma)$. But for large $n$ we will have that $R < \tau^n a$ (with $a$ the size of the tiles in $\sigma$). But then a tile in $inf^n(\sigma)$ and its translation by $R$ must overlap which is impossible if $+R$ is a translation symmetry of $inf^n(\sigma)$. Done!

Returning to the middle C-sequences, what was the point of starting with $C_0 = LSL$? Well, it follows directly from the gluing restrictions that any letter in a musical sequence is part of a subword $LSL$ of $\sigma$. But then, every finite subword $W$ of $\sigma$ is also a subword of $C_{2n}$ for some large $n$.

For, let $d$ be the length of the interval corresponding to $W$ and choose $n$ such that $d > \tau^{2n} a$ then the interval of the line corresponding to $W$ is contained in a single tile in $inf^{2n}(\sigma)$ and this tile belongs to a subword $LSL$ of $inf^{2n}(\sigma)$. But then $W$ will be a subword of the $2n$-th deflation of that interval $LSL \subset inf^{2n}(\sigma)$, which is $C_{2n}$.

Or, as Conway would phrase it with respect to Penrose tilings (quote again from Siobhan Roberts’ book)

“Every points is in the cartwheel somewhere. If you jab your finger anywhere, on any point anywhere on teh pattern, you are part of a cartwheel. The whole ting is overlapping cartwheels.”

An immediate consequence is the local isomorphism theorem: Every subword of a musical sequence $\sigma$ appears infinitely many times as subword of any other musical sequence. That is, one cannot distinguish two tilings of the line with musical sequence codes from each other by looking at finite intervals!

The argument is similar to the one above. The finite interval corresponding to the subword lies in a unique tile of $inf^n(\sigma)$ for $n$ large enough. Now, take another musical sequence $\mu$ and consider any of the infinitely many tiles of the same type in $inf^n(\mu)$, then $def^n$ of such a tile will contain the subword in $\phi$.

Another time, we’ll see that musical sequences can be produced by the ‘cut-and-project’-method (what I called the ‘windows’-method before).
This time we will project parts of the standard $2$-dimensional lattice $\mathbb{Z}^2$ onto the line, which is a lot easier to visualise than de Bruijn’s projection from $\mathbb{R}^5$ to produce Penrose tilings or the projection from six dimensional space to harvest quasicrystals.

Leave a Comment

GoV 2 : Viruses and quasi-crystals

If you look around for mathematical theories of the structure of viruses, you quickly end up with the work of Raidun Twarock and her group at the University of York.



We’ve seen her proposal to extend the Caspar-Klug classification of viruses. Her novel idea to distribute proteins on the viral capsid along Penrose-like tilings shouldn’t be taken too literally. The inherent aperiodic nature of Penrose tiles doesn’t go together well with perfect tilings of the sphere.

Instead, the observation that these capsid tilings resemble somewhat Penrose tilings is a side-effect of another great idea of the York group. Recently, they borrowed techniques from the theory of quasicrystals to gain insight in the inner structure of viruses, in particular on the interaction of the capsid with the genome.

By the crystallographic restriction theorem no $3$-dimensional lattice can have icosahedral symmetry. But, we can construct aperiodic structures (quasicrystals) which have local icosahedral structure, much like Penrose tilings have local $D_5$-symmetry

This is best explained by de Bruijn‘s theory of pentagrids (more on that another time). Here I’ll just mention the representation-theoretic idea.

The isometry group of the standard $5$-dimensional lattice $\mathbb{Z}^5$ is the group of all signed permutation $5 \times 5$ matrices $B_5$ (Young’s hyperoctahedral group). There are two distinct conjugacy classes of subgroups in $B_5$ isomorphic to $D_5$, one such subgroup generated by the permutation matrices
\[
x= \begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 \end{bmatrix} \qquad \text{and} \qquad
y = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \end{bmatrix} \]
The traces of $x,x^2$ and $y$, together with the character table of $D_5$ tell us that this $5$-dimensional $D_5$-representation splits as the direct sum of the trivial representation and of the two irreducible $2$-dimensional representations.
\[
\mathbb{R}^5 = A \simeq T \oplus W_1 \oplus W_2 \]
with $T = \mathbb{R} d$, $W_1 = \mathbb{R} u_1 + \mathbb{R} u_2$ and $W_2 = \mathbb{R} w_1 + \mathbb{R} w_2$ where
\[
\begin{cases}
(1,1,1,1,1)=d \\
(1,c_1,c_2,c_3,c_4)= u_1 \\
(0,s_1,s_2,s_3,s_4) = u_2 \\
(1,c_2,c_4,c1,c3)= w_1 \\
(0,s_2,s_4,s_1,s_3)= w_2
\end{cases}
\]
and $c_j=cos(2\pi j/5)$ and $s_j=sin(2 \pi/5)$. We have a $D_5$-projection
\[
\pi : A \rightarrow W_1 \quad (y_0,\dots,y_4) \mapsto \sum_{i=0}^4 y_i(c_i u_1+s_i u_2) \]
The projection maps the vertices of the $5$-dimensional hypercube to a planar configuration with $D_5$-symmetry.



de Bruijn’s results say that if we take suitable ‘windows’ of lattice-points in $\mathbb{Z}^5$ and project them via the $D_5$-equivariant map $\pi$ onto the plane, then the images of these lattice points become the vertices of a rhombic Penrose tiling (and we get all such tilings by choosing our window carefully).



This explains why Penrose tilings have a local $D_5$-symmetry. I’ll try to come back to de Bruijn’s papers in future posts.

But, let’s go back to viruses and the work of Twarock’s group using methods from quasicrystals. Such aperiodic structures with a local icosahedral symmetry can be constructed along similar lines. This time one starts with the standard $6$-dimensional lattice $\mathbb{Z^6}$ with isometry group $B_6$ (signed $6 \times 6$ permutation matrices).

This group has three conjugacy classes of subgroups isomorphic to $A_5$, but for only one of them this $6$-dimensional representation decomposes as the direct sum of the two irreducible $3$-dimensional representations of $A_5$ (the decompositions in the two other cases contain an irreducible of dimension $4$ or $5$ together with trivial factor(s)). A representant of the crystallographic relevant case is given by the signed permutation matrices
\[
x= \begin{bmatrix}
0 & 1 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & -1
\end{bmatrix} \qquad \text{and} \qquad y=
\begin{bmatrix}
0 & 0 & 1 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & -1 & 0 & 0
\end{bmatrix} \]

Again, using suitable windows of $\mathbb{Z}^6$-lattice points and using the $A_5$-equivariant projection to one of the two $3$-dimensional components, one obtains quasicrystals with local $A_5$-symmetry.

In this $3$-dimensional case the replacements of the thick and thin rhombi are these four parallellepipeda, known as the Amman blocks



which must be stacked together obeying the gluing condition that dots of the same colour must be adjacent.

Has anyone looked at a possible connection between the four Amman blocks (which come in pairs) and the four (paired) nucleotides in DNA? Just an idle thought…

These blocks grow into quasicrystals with local icosahedral symmetry.



The faces on the boundary of such a sphere-like quasicrystal then look a lot like a Penrose tiling.

How can we connect these group and representation-theoretic ideas to the structure of viruses? Here’s another thought-provoking proposal coming from the York group.

Take the $A_5$ subgroup of the hyperoctahedral group in six dimensiona $B_6$ generated by the above two matrices (giving a good $A_5$-equivariant projection $\pi$ to three dimensional space) and consider an intermediate group
\[
A_5 \subsetneq G \subseteq B_6 \]
Take a point in $\mathbb{R}^6$ and look at its orbit under the isometries of $G$, then all these points have the same distance from the origin in $\mathbb{R}^6$. Now, project this orbit under $\pi$ to get a collection of points in $\mathbb{R}^3$.

As $\pi$ is only $A_5$-equivariant (and not $G$-equivariant) the image points may lie in different shells from the origin. We can try to relate these shells of points to observational data on the inner structures of viruses.

Here’s a pretty convincing instance of such a correlation, taken from the thesis by Emilio Zappa “New group theoretical methods for applications in virology and quasicrystals”.



This is the inner structure of the Hepatitis B virus, showing the envelope (purple), capsid protein (cream) and genome (light blue). The coloured dots are the image points in the different shells around the origin.

Do viruses invade us from the sixth dimension??

Leave a Comment

GoV 1 : Geometry of viruses

As you may have guessed from the symmetries of Covid-19 post, I did spend some time lately catching up with the literature on the geometric structure and symmetries of viruses. It may be fun to run a little series on this.

A virus is a parasite, so it cannot reproduce on its own and needs to invade a host cell to replicate. All information needed for this replication process is stored in a fragile DNA or RNA string, the viral genome.

This genome needs to be protected by a coating made of proteins, the viral capsid. Most viruses have an additional fatty protection layer, the envelope, decorated by virus (glyco)proteins (such as the ‘spikes’ needed to infiltrate the host cell).



Most viruses are extremely small (between 20 and 200nm), our friend the corona-virus measures between 80 and 120nm. So, its genome is also pretty small (the corona genome has around 30.000 base pairs). To maximise its information, the volume of the protective capsid must be as large as possible, and must be formed by just a few different proteins (to free as much space in the code of the genome for other operations) and clusters of them are distributed over the polyhedral capsid, as symmetric as possible.

This insight led Watson and Crick, the discoverers of the structure of DNA, to the ‘genetic economy’-proposal that most sphere-like viruses will have an icosahedral capsid because the icosahedron is the Platonic solid with the largest volume and rotational symmetry group. They argued that the capsid is most likely constructed from a single subunit (capsomere), which is repeated many times to form the protein shell.



Little is known about capsid formation, that is the process in which the capsid proteins self-assemble into an icosahedral shape, nor about the precise interplay between the genome and the capsid proteins. If we would understand these two things better it might open new possibilities for anti-viral drugs, by either blocking the self-assembly process or by breaking the genome-capsid interaction.

A first proposal for the capsid structure was put forward by Caspar and Klug. Their quasi-equivalence principle asserts that each of the 20 triangular faces of the icosahedron is subdivided in 3 subunits, each consisting of at least one protein.

Most viruses have much more than 60 proteins in their capsid, so Caspar and Klug introduced their $T$-number giving the number of proteins per subunit. One superimposes the triangulation of the icosahedron with the hexagonal plane lattice, then $T$ is the number of sub-triangles of these hexagons contained in each subunit. For $T = 7$ we have the following situation



Folding back the triangulation to form the icosahedron one then obtains a tiling consisting of hexagons (the green regions) and pentagons (the blue regions)



It turned out that many viruses with icosahedral symmetry consist of subunits having a different number of proteins, such as dimers (2 proteins), trimers (3 proteins), or pentamers (5 proteins) and these self-organise around a 2, 3, or 5-fold rotational axis of the icosahedron.

This led Reidun Twarock around 2000 to propose her virus tiling theory. This is a generalisation of the Caspar-Klug theory in which one superimposese the triangulation of the icosahedron with other tilings of the plane, consisting of two or more non-congruent tiles. Here an example which looks a bit like the aperiodic Penrose tilings of the plane.



Here’s a recent Quanta-Magazine article on Twarock’s work and potential consequences: The illuminating geometry of viruses.

And here’s an LMS Popular Lecture, from 2008, by Raidun Twarock herself: “Know your enemy – viruses under the mathematical microscope”.

Leave a Comment

Scholze’s condensed sets and Mazzola’s path to creativity

Some months ago, Peter Scholze wrote a guest post on the Xena-blog: Liquid tensor experiment, proposing a challenge to formalise the proof of one of his results with Dustin Clausen on condensed mathematics.

Scholze and Clausen ran a masterclass in Copenhagen on condensed mathematics, which you can binge watch on YouTube starting here

Scholze also gave two courses on the material in Bonn of which the notes are available here and here.

Condensed mathematics claims that topological spaces are the wrong definition, and that one should replace them with the slightly different notion of condensed sets.

So, let’s find out what a condensed set is.

Definition: Condensed sets are sheaves (of sets) on the pro-étale site of a point.

(there’s no danger we’ll have to rewrite our undergraduate topology courses just yet…)

In his blogpost, Scholze motivates this paradigm shift by observing that the category of topological Abelian groups is not Abelian (if you put a finer topology on the same group then the identity map is not an isomorphism but doesn’t have a kernel nor cokernel) whereas the category of condensed Abelian groups is.

It was another Clausen-Scholze result in the blogpost that caught my eye.

But first, for something completely different.

In “Musical creativity”, Guerino Mazzola and co-authors introduce a seven steps path to creativity.



Here they are:

  1. Exhibiting the open question
  2. Identifying the semiotic context
  3. Finding the question’s critical sign
  4. Identifying the concept’s walls
  5. Opening the walls
  6. Displaying extended wall perspectives
  7. Evaluating the extended walls

Looks like a recipe from distant flower-power pot-infused times, no?

In Towards a Categorical Theory of Creativity for Music, Discourse, and Cognition, Mazzola, Andrée Ehresmann and co-authors relate these seven steps to the Yoneda lemma.

  1. Exhibiting the open question = to understand the object $A$
  2. Identifying the semiotic context = to describe the category $\mathbf{C}$ of which $A$ is an object
  3. Finding the question’s critical sign = $A$ (?!)
  4. Identifying the concept’s walls = the uncontrolled behaviour of the Yoneda functor
    \[
    @A~:~\mathbf{C} \rightarrow \mathbf{Sets} \qquad C \mapsto Hom_{\mathbf{C}}(C,A) \]
  5. Opening the walls = finding an objectively creative subcategory $\mathbf{A}$ of $\mathbf{C}$
  6. Displaying extended wall perspectives = calculate the colimit $C$ of a creative diagram
  7. Evaluating the extended walls = try to understand $A$ via the isomorphism $C \simeq A$.

(Actually, I first read about these seven categorical steps in another paper which might put a smile on your face: The Yoneda path to the Buddhist monk blend.)

Remains to know what a ‘creative’ subcategory is.

The creative moment comes in here: could we not find a subcategory
$\mathbf{A}$ of $\mathbf{C}$ such that the functor
\[
Yon|_{\mathbf{A}}~:~\mathbf{C} \rightarrow \mathbf{PSh}(\mathbf{A}) \qquad A \mapsto @A|_{\mathbf{A}} \]
is still fully faithful? We call such a subcategory creative, and it is a major task in category theory to find creative categories which are as small as possible.

All the ingredients are here, but I had to read Peter Scholze’s blogpost before the penny dropped.

Let’s try to view condensed sets as the result of a creative process.

  1. Exhibiting the open question: you are a topologist and want to understand a particular compact Hausdorff space $X$.
  2. Identifying the semiotic context: you are familiar with working in the category $\mathbf{Tops}$ of all topological spaces with continuous maps as morphisms.
  3. Finding the question’s critical sign: you want to know what differentiates your space $X$ from all other topological spaces.
  4. Identifying the concept’s walls: you can probe your space $X$ with continuous maps from other topological spaces. That is, you can consider the contravariant functor (or presheaf on $\mathbf{Tops}$)
    \[
    @X~:~\mathbf{Tops} \rightarrow \mathbf{Sets} \qquad Y \mapsto Cont(Y,X) \]
    and Yoneda tells you that this functor, up to equivalence, determines the space $X$ upto homeomorphism.
  5. Opening the walls: Tychonoff tells you that among all compact Hausdorff spaces there’s a class of pretty weird examples: inverse limits of finite sets (or a bit pompous: the pro-etale site of a point). These limits form a subcategory $\mathbf{ProF}$ of $\mathbf{Tops}$.
  6. Displaying extended wall perspectives: for every inverse limit $F \in \mathbf{ProF}$ (for ‘pro-finite sets’) you can look at the set $\mathbf{X}(F)=Cont(F,X)$ of all continuous maps from $F$ to $X$ (that is, all probes of $X$ by $F$) and this functor
    \[
    \mathbf{X}=@X|_{\mathbf{ProF}}~:~\mathbf{ProF} \rightarrow \mathbf{Sets} \qquad F \mapsto \mathbf{X}(F) \]
    is a sheaf on the pre-etale site of a point, that is, $\mathbf{X}$ is the condensed set associated to $X$.
  7. Evaluating the extended walls: Clausen and Scholze observe that the assignment $X \mapsto \mathbf{X}$ embeds compact Hausdorff spaces fully faithful into condensed sets, so we can recover $X$ up to homeomorphism as a colimit from the condenset set $\mathbf{X}$. Or, in Mazzola’s terminology: $\mathbf{ProF}$ is a creative subcategory of $\mathbf{(cH)Tops}$ (all compact Hausdorff spaces).

It would be nice if someone would come up with a new notion for me to understand Mazzola’s other opus “The topos of music” (now reprinted as a four volume series).



Leave a Comment

The symmetries of Covid-19

A natural question these days might be: “what are the rotational symmetries of the Covid-19 virus?”



Most illustrations show a highly symmetric object, suggesting it might have icosahedral symmetry. In fact, many viruses do have icosahedral symmetry as a result of the ‘genetic economy principle’ proposed by Watson and Crick in 1956, resulting in the Caspar-Klug classification of viral capsids.

But then, perhaps this icosahedral illusion is a result of design decisions illustrators made turning scientific data into pictures. Veronica Falconieri Hays wrote a beautiful article describing the effort going into this: How I built a 3d-model of the coronavirus for Scientific American. Here’s her final picture



And yes, icosahedral symmetry was one of her design decisions:

The M proteins form pairs, and it is estimated that there are 16–25 M proteins per spike on the surface of the virus. I ended up modeling 10 M protein pairs (so 20 M proteins) per spike in my model. Some researchers hypothesize that the M proteins form a lattice within the envelope (interacting with an underlying lattice of N proteins; see below). I decided to use an icosahedral sphere to create a regular distribution of the M protein dimers to hint at this hypothesis.

The spikes (or S-proteins) are the tentacles in these pictures, and one of the few hard figures on Corona is that ‘on average’ there are 74 of them.

This fact is enough to rule out icosahedral symmetry.

If the icosahedral rotation group (of order $60$, isomorphic to $A_5$) acts on the $74$ spikes, then each orbit consists of $60$ spikes unless that spike lies on a twofold, threefold or fivefold rotation axis, in which cases the number of spikes in its orbit are respectively $30$, $20$ or $12$. So, we can’t get a total number of $74$ spikes!

However, just looking at the number of spikes we cannot rule out octahedral symmetry!

The octahedral rotation group (of order $24$, isomorphic to $S_4$) will have orbits of size $24$ unless the spike lies on a twofold, threefold or fourfold rotation axis, giving orbits of size $12$, $8$ and $6$ respectively (the midpoints of edges, the vertices and the midpoints of faces of the octahedron), and

\[
74 = 24+24+12+8+6 \]

The most symmetric arrangement of spikes would be to subdivide each of the $8$ triangular faces of the octahedron into $6$ triangles with vertices the midpoint of the face, a vertex and a midpoint of an edge, and then to position the spikes on the axis through the vertices and midpoints of these smaller triangles.

Googling around I found very few references to symmetries of Covid-19, probably because it has an helical RNA-coil, which seems not to go well with Caspar-Klug type polyhedral viral capsids.

Here’s an exception: A structural model for the Coronavirus nucleocapsid by Federico Coscio, Alejandro D. Nadra, and Diego U. Ferreiro.



They propose a truncated octahedron as capsid (in transparent brown) with interior a continuous coil packing of blue and cyan helices. The virus membrane with the spikes and M proteins is drawn in blue.

If you have better info or references on the (conjectural) symmetries of Covid-19, please leave a comment.

One Comment