neverendingbooks

Lockdown reading : Centenal

Published March 21, 2021 by lievenlb

In this series I’ll mention some books I found entertaining, stimulating or comforting during these Corona times. Read them at your own risk.

The Centenal Cycle is a trilogy written by Malka Older.

A Centenal is the basic political unit of a future micro-democracy. It is a neighbourhood consisting of 100.000 people which can vote for any government it wants, from anywhere in the world.

“Centenal-based microdemocracy naturally requires extensive use of technology. In my book, it’s provided through a massive international bureaucracy known as Information, which offers voters data about the thousands of possible governments and helps those governments manage what may be far-flung territories once they’re elected.” (Malka Older)

In this trilogy Malka Older draws from her own life: she obtained a Ph. D. from Sciences Po exploring the dynamics of multi-level governance and disaster response, and has more than a decade of experience in humanitarian aid and development.

The Centenal Cycle consists of these three books:

Infomocracy (2016) (link containing excerpts).

It’s been twenty years and two election cycles since Information, a powerful search engine monopoly, pioneered the switch from warring nation-states to global micro-democracy. The corporate coalition party Heritage has won the last two elections. With another election on the horizon, the Supermajority is in tight contention, and everything’s on the line.

Null States (2017).

The future of democracy is about to implode.

After the last controversial global election, the global infomocracy that has ensured thirty years of world peace is fraying at the edges. As the new Supermajority government struggles to establish its legitimacy, agents of Information across the globe strive to keep the peace and maintain the flows of data that feed the new world order.

State Tectonics (2018) (link containing excerpts).

The future of democracy must evolve or die.

The last time Information held an election, a global network outage, two counts of sabotage by major world governments, and a devastating earthquake almost shook micro-democracy apart. Five years later, it’s time to vote again, and the system that has ensured global peace for 25 years is more vulnerable than ever.

Here’s a short interview with Malka Older on Sci-Fi, AI and its possible uses in the writing process.

Here’s a longer clip in which she talks about ‘Speculative Resistance’ at the Personal Democracy Forum 2018.

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Witt and his Niemeier lattices

Published March 19, 2021 by lievenlb

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:

use our knowledge of Niemeier lattices to figure out which of these Witt was most likely to stumble upon, and
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.

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the bongcloud attack

Published March 18, 2021 by lievenlb

In this neverending pandemic there’s a shortage of stories putting a lasting smile on my face. Here’s one.

If you are at all interested in chess, you’ll know by now that some days ago IGMs (that is, international grandmasters for the rest of you) Magnus Carlsen and Hikaru Nakamura opened an official game with a double bongcloud, and couldn’t stop laughing.

The bongcloud attack is the chess opening in which white continues after

1. e2-e4, e7-e5

with

2. Ke1-e2 !

thereby blocking the diagonals for the bishop and queen, losing the ability to castle, and putting its king in danger.

If you are left clueless, you should download the free e-book Winning with the bongcloud immediately.

If you are (or were) a chess player, it is the perfect parody to all those books you had to suffer through in order to build up an ‘opening repertoire’.

If you are new to chess (perhaps after watching The queen’s gambit), it gives you a nice selection of easy mate-in-one problems.

Every possible defence against the bongcloud is illustrated with a ‘game’ illustrating the massive advantage the attack gives, ending with a situation in which … black(!) has a one-move mate.

One example:

In the two knight Copacabana tango defense against the bongcloud, that is the position after

the Haight-Asbury (yeah, well) game (Linares, 1987) continued with:

4. Kg3, Nxe4?
5. Kh3, d6+
6. Qg4!, Bxg4
7. Kxg4, Qf6??
8. Ne2, h5+??
9. Kh3!, Nxf2
10. Kg3!

giving this position

which ‘Winning with the bongcloud’ evaluates as:

White continues his textbook execution of a “pendulum,” swinging back and forth between g3 and h3 to counter every Black threat. With the dynamically placed King, Black’s attack teeters on the edge of petering out. Nxh1 will trap the Black Knight and extinguish the threat.

In the actual game, play took a different turn as Black continued his h-file pressure. Nonetheless, this game is an excellent example of how 2. …Nc6 is often a wasted tempo in the Bongcloud.

Here’s Nakamura philosophising over the game and the bongcloud. Try to watch at least the first 30 seconds or so to see the commentators reaction to the actual Carlsen-Nakamura game.

Now, that put a smile on your face, didn’t it?

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de Bruijn’s pentagrids (2)

Published March 17, 2021 by lievenlb

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.

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Kasha-eating dragons

Published March 15, 2021 by lievenlb

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.

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Teapot supremacy

Published March 13, 2021 by lievenlb

No, this is not another timely post about the British Royal family.

It’s about Richard Borcherds’ “teapot test” for quantum computers.

A lot of money is being thrown at the quantum computing hype, causing people to leave academia for quantum computing firms. A recent example (hitting the press even in Belgium) being the move of Bob Coecke from Oxford University to Cambridge Quantum Computing.

Sure, quantum computing is an enticing idea, and we have fantastic quantum algorithms such as Shor’s factorisation algorithm and Grover’s search algorithm.

The (engineering) problem is building quantum computers with a large enough number of qubits, which is very difficult due to quantum decoherence. To an outsider it may appear that the number of qubits in a working quantum computer is growing at best linearly, if not logarithmic, in sharp contrast to Moore’s law for classical computers, stating that the number of transistors in an integrated circuit doubles every two years.

Quantum computing evangelists assure us that this is nonsense, and that we should replace Moore’s law by Neven’s law claiming that the computing power of quantum computers will grow not just exponentially, but doubly exponentially!

What is behind these exaggerated claims?

In 2015 the NSA released a policy statement on the need for post-quantum cryptography. In the paper “A riddle wrapped in an enigma”, Neil Koblitz and Alfred Menezes carefully examined NSA’s possible strategies behind this assertion.

Can the NSA break PQC? Can the NSA break RSA? Does the NSA believes that RSA-3072 is much more quantum-resistant than ECC-256 and even ECC-384?, and so on.

Perhaps the most plausible of all explanations is this one : the NSA is using a diversion strategy aimed at Russia and China.

Suppose that the NSA believes that, although a large-scale quantum computer might eventually be built, it will be hugely expensive. From a cost standpoint it will be less analogous to Alan Turing’s bombe than to the Manhattan Project or the Apollo program, and it will be within the capabilities of only a small number of nation-states and huge corporations.

Suppose also that, in thinking about the somewhat adversarial relationship that still exists between the U.S. and both China and Russia, especially in the area of cybersecurity, the NSA asked itself “How did we win the Cold War? The main strategy was to goad the Soviet Union into an arms race that it could not afford, essentially bankrupting it. Their GNP was so much less than ours, what was a minor set-back for our economy was a major disaster for theirs. It was a great strategy. Let’s try it again.”

This brings us to the claim of quantum supremacy, that is, demonstrating that a programmable quantum device can solve a problem that no classical computer can solve in any feasible amount of time.

In 2019, Google claimed “to have reached quantum supremacy with an array of 54 qubits out of which 53 were functional, which were used to perform a series of operations in 200 seconds that would take a supercomputer about 10,000 years to complete”. In December 2020, a group based in USTC reached quantum supremacy by implementing a type of Boson sampling on 76 photons with their photonic quantum computer. They stated that to generate the number of samples the quantum computer generates in 20 seconds, a classical supercomputer would require 600 million years of computation.

Richard Borcherds rants against the type of problems used to claim quantum ‘supremacy’. He proposes the ‘teapot problem’ which a teapot can solve instantaneously, but will be impossibly hard for classical (and even quantum) computers. That is, any teapot achieves ‘teapot supremacy’ over classical and quantum computers!

Another point of contention are the ‘real-life applications’ quantum computers are said to be used for. Probably he is referring to Volkswagen’s plan for traffic optimization with a D-Wave quantum computer in Lisbon.

“You could give these guys a time machine and all they’d use it for was going back to watch some episodes of some soap opera they missed”

Enjoy!

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de Bruijn’s pentagrids

Published March 11, 2021 by lievenlb

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.