# Category: groups

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

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

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

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

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

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

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

• a handle for 2 Euros,
• a mirror for 1 Euro,
• a cross-cap for 1 Euro,
• an order $n$ cone point for $(n-1)/n$ Euro,
• an order $n$ corner reflector for $(n-1)/2n$ Euro, if you have the required mirrors to install this piece. Here’s a standard brick wall, with its fundamental domain and corresponding orbifold made from a mirror piece (1 Euro), two order $2$ corner reflectors (each worth $.25$ Euro), and one order $2$ cone point (worth $.5$ Euro). That is, this orbifold will cost you exactly $2$ Euros.

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

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

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

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

Clearly I have no desire to introduce cohomology, but it may still be possible to get the underlying idea across. Let’s take the Penrose triangle (all pictures below taken from Penrose’s paper) The idea is to break up such a picture in several parts, each of which we do know to construct in 3D (that is, we take a particular cover of our figure). We can slice up the Penrose triangle in three parts, and if you ever played with Lego you’ll know how to construct each one of them. Next, position the constructed pieces in space as in the picture and decide which of the two ends is closer to you. In $Q_1$ it is clear that point $A_{12}$ is closer to you than $A_{13}$, so we write $A_{12} < A_{13}$.

Similarly, looking at $Q_2$ and $Q_3$ we see that $A_{23} < A_{21}$ and that $A_{31} < A_{32}$.

Next, if we try to reassemble our figure we must glue $A_{12}$ to $A_{21}$, that is $A_{12}=A_{21}$, and similarly $A_{23}=A_{32}$ and $A_{31}=A_{13}$. But, then we get
$A_{13}=A_{31} < A_{32}=A_{23} < A_{21}=A_{12} < A_{13}$ which is clearly absurd.

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

In this series I’ll mention some books I found entertaining, stimulating or comforting during these Corona times. Read them at your own risk. This must have been the third time I’ve read The genius in by basement – The biography of a happy man by Alexander masters.

I first read it when it came out in 2011.

Then, in conjunction with Genius at play – The Curious Mind of John Horton Conway Conway’s biography by Siobhan Roberts, in july 2017, which is probably the best way to read this book.

And, then again last week, as Simon Norton‘s work pops up wherever I look, as in the previous post.

It takes some time to get used to the rather chaotic style (probably used because that’s how Masters perceives Norton), and all attempts at explaining Simon’s mathematics can better be skipped.

The book tries to find an answer as to why a child prodigy and genius like Simon Norton failed to secure a safe place in academics.

Page 328:

Simon’s second explanation of his loss of mathematical direction is heartbreaking. Now that Conway has fled to America, there is no one in the mathematical world who will work with him.

They say he is too peculiar, too shabby, too old.

His interests are fixed in mathematics that has had its day. His brilliance is frigid. His talent, perfectly suited to an extraordinary moment in algebraic history (the symmetry work at Cambridge during the early 1970s and 1980s) is out of fashion.

This may give the impression that Norton stopped doing good math after Conway left for Princeton in 1985. This is far from true.

Norton’s Wikipedia page mentions only post 1995 publications, which in itself is deplorable, as it leaves out his contributions to the ATLAS and his seminal paper with Conway on Monstrous moonshine.

Here’s Alexander Masters talking about ‘Genius in my basement’

I’ll leave you with a nice quote, comparing Monstrous Moonshine to a Sainsbury’s bag on Jupiter.

Page 334:

This much I do know: Monstrous Moonshine links the Monster to distant mathematics and the structure of space in ways that are as awe-inspiring to a man like Simon as it would be to an astronaut to step out of his space machine on Jupiter, and find a Sainsbury’s bag floating past. That’s why it’s called ‘Moonshine’, because mathematicians can even now hardly believe it.

‘I think’, said Simon, standing up from his berth and shaking crumbs and clotted blobs of oil and fish off his T-shirt onto the covers, ‘I can explain to you what Moonshine is in one sentence.’

When he really tries, Simon can be a model of clarity.

‘It is,’ he said, ‘the voice of God.’

Ps, wrt. SNORT.

An integral $n$-dimensional lattice $L$ is the set of all integral linear combinations
$L = \mathbb{Z} \lambda_1 \oplus \dots \oplus \mathbb{Z} \lambda_n$
of base vectors $\{ \lambda_1,\dots,\lambda_n \}$ of $\mathbb{R}^n$, equipped with the usual (positive definite) inner product, satisfying
$(\lambda, \mu ) \in \mathbb{Z} \quad \text{for all \lambda,\mu \in \mathbb{Z}.}$
But then, $L$ is contained in its dual lattice $L^* = Hom_{\mathbb{Z}}(L,\mathbb{Z})$, and if $L = L^*$ we say that $L$ is unimodular.

If all $(\lambda,\lambda) \in 2 \mathbb{Z}$, we say that $L$ is an even lattice. Even unimodular lattices (such as the $E_8$-lattice or the $24$ Niemeier lattices) are wonderful objects, but they can only live in dimensions $n$ which are multiples of $8$.

Just like the Conway group $Co_0 = .0$ is the group of rotations of the Leech lattice $\Lambda$, one might ask whether there is a very special lattice on which the Monster group $\mathbb{M}$ acts faithfully by rotations. If such a lattice exists, it must live in dimension at least $196883$. Simon Norton (1952-2019) – Photo Credit

A first hint of such a lattice is in Conway’s original paper A simple construction for the Fischer-Griess monster group (but not in the corresponding chapter 29 of SPLAG).

Conway writes that Simon Norton showed ‘by a very simple computations that does not even require knowledge of the conjugacy classes, that any $198883$-dimensional representation of the Monster must support an invariant algebra’, which, after adding an identity element $1$, we now know as the $196884$-dimensional Griess algebra.

Further, on page 529, Conway writes:

Norton has shown that the lattice $L$ spanned by vectors of the form $1,t,t \ast t’$, where $t$ and $t’$ are transposition vectors, is closed under the algebra multiplication and integral with respect to the doubled inner product $2(u,v)$. The dual quotient $L^*/L$ is cyclic of order some power of $4$, and we believe that in fact $L$ is unimodular.

Here, transposition vectors correspond to transpositions in $\mathbb{M}$, that is, elements of conjugacy class $2A$.

In his post, Adam considers the $196883$-dimensional lattice $L’ = L \cap 1^{\perp}$ (which has $\mathbb{M}$ as its rotation symmetry group), and asks for the minimal norm (squared) of a lattice point, which he believes is $448$, and for the number of minimal vectors in the lattice, which might be
$2639459181687194563957260000000 = 9723946114200918600 \times 27143910000$
the number of oriented arcs in the Monster graph.

Here, the Monster graph has as its vertices the elements of $\mathbb{M}$ in conjugacy class $2A$ (which has $9723946114200918600$ elements) and with an edge between two vertices if their product in $\mathbb{M}$ again belongs to class $2A$, so the valency of the graph must be $27143910000$, as explained in that old post the monster graph and McKay’s observation.

When I asked Adam whether he had more information about his lattice, he kindly informed me that Borcherds told him that the Norton lattice $L$ didn’t turn out to be unimodular after all, but that a unimodular lattice with monstrous symmetry had been constructed by Scott Carnahan in the paper A Self-Dual Integral Form of the Moonshine Module. Scott Carnahan – Photo Credit

The major steps (or better, the little bit of it I could grasp in this short time) in the construction of this unimodular $196884$-dimensional monstrous lattice might put a smile on your face if you are an affine scheme aficionado.

Already in his paper Vertex algebras, Kac-Moody algebras, and the Monster, Richard Borcherds described an integral form of any lattice vertex algebra. We’ll be interested in the lattice vertex algebra $V_{\Lambda}$ constructed from the Leech lattice $\Lambda$ and call its integral form $(V_{\Lambda})_{\mathbb{Z}}$.

One constructs the Moonshine module $V^{\sharp}$ from $V_{\Lambda}$ by a process called ‘cyclic orbifolding’, a generalisation of the original construction by Frenkel, Lepowsky and Meurman. In fact, there are now no less than 51 constructions of the moonshine module.

One starts with a fixed point free rotation $r_p$ of $\Lambda$ in $Co_0$ of prime order $p \in \{ 2,3,5,7,13 \}$, which one can lift to an automorphism $g_p$ of the vertex algebra $V_{\Lambda}$ of order $p$ giving an isomorphism $V_{\Lambda}/g_p \simeq V^{\sharp}$ of vertex operator algebras over $\mathbb{C}$.

For two distinct primes $p,p’ \in \{ 2,3,5,7,13 \}$ if $Co_0$ has an element of order $p.p’$ one can find one such $r_{pp’}$ such that $r_{pp’}^p=r_{p’}$ and $r_{pp’}^{p’}=r_p$, and one can lift $r_{pp’}$ to an automorphism $g_{pp’}$ of $V_{\Lambda}$ such that $V_{\Lambda}/g_{pp’} \simeq V_{\Lambda}$ as vertex operator algebras over $\mathbb{C}$.

Problem is that these lifts of automorphisms and the isomorphisms are not compatible with the integral form $(V_{\Lambda})_{\mathbb{Z}}$ of $V_{\Lambda}$, but ‘essentially’, they can be performed on
$(V_{\Lambda})_{\mathbb{Z}} \otimes_{\mathbb{Z}} \mathbb{Z}[\frac{1}{pp’},\zeta_{2pp’}]$
where $\zeta_{2pp’}$ is a primitive $2pp’$-th root of unity. These then give a $\mathbb{Z}[\tfrac{1}{pp’},\zeta_{2pp’}]$-form on $V^{\sharp}$.

Next, one uses a lot of subgroup information about $\mathbb{M}$ to prove that these $\mathbb{Z}[\tfrac{1}{pp’},\zeta_{2pp’}]$-forms of $V^{\sharp}$ have $\mathbb{M}$ as their automorphism group.

Then, using all his for different triples in $\{ 2,3,5,7,13 \}$ one can glue and use faithfully flat descent to get an integral form $V^{\sharp}_{\mathbb{Z}}$ of the moonshine module with monstrous symmetry and such that the inner product on $V^{\sharp}_{\mathbb{Z}}$ is positive definite.

Finally, one looks at the weight $2$ subspace of $V^{\sharp}_{\mathbb{Z}}$ which gives us our Carnahan’s $196884$-dimensional unimodular lattice with monstrous symmetry!

Beautiful as this is, I guess it will be a heck of a project to deduce even the simplest of facts about this wonderful lattice from running through this construction.

For example, what is the minimal length of vectors? What is the number of minimal length vectors? And so on. All info you might have is very welcome.

Two lattices $L$ and $L’$ in the same vector space are called neighbours if their intersection $L \cap L’$ is of index two in both $L$ and $L’$.

In 1957, Martin Kneser gave a method to find all unimodular lattices (of the same dimension and signature) starting from one such unimodular lattice, finding all its neighbours, and repeating this with the new lattices obtained.

In other words, Kneser’s neighbourhood graph, with vertices the unimodular lattices (of fixed dimension and signature) and edges between them whenever the lattices are neighbours, is connected. Martin Kneser (1928-2004) – Photo Credit

Last time, we’ve constructed the Niemeier lattice $(A_1^{24})^+$ from the binary Golay code $\mathcal{C}_{24}$
$L = (A_1^{24})^+ = \mathcal{C}_{24} \underset{\mathbb{F}_2}{\times} (A_1^{24})^* = \{ \tfrac{1}{\sqrt{2}} \vec{v} ~|~\vec{v} \in \mathbb{Z}^{\oplus 24},~v=\vec{v}~mod~2 \in \mathcal{C}_{24} \}$
With hindsight, we know that $(A_1^{24})^+$ is the unique neighbour of the Leech lattice in the Kneser neighbourhood graph of the positive definite, even unimodular $24$-dimensional lattices, aka the Niemeier lattices.

Let’s try to construct the Leech lattice $\Lambda$ from $L=(A_1^{24})^+$ by Kneser’s neighbour-finding trick. Sublattices of $L$ of index two are in one-to-one correspondence with non-zero elements in $L/2L$. Take $l \in L – 2L$ and $m \in L$ such that the inner product $l.m$ is odd, then
$L_l = \{ x \in L~|~l.x~\text{is even} \}$
is an index two sublattice because $L = L_l \sqcup (L_l+m)$. By definition $l.x$ is even for all $x \in L_l$ and therefore $\frac{l}{2} \in L_l^*$. We have this situation
$L_l \subsetneq L = L^* \subsetneq L_l^*$
and $L_l^*/L_l \simeq \mathbb{F}_2 \oplus \mathbb{F}_2$, with the non-zero elements represented by $\{ \frac{l}{2}, m, \frac{l}{2}+m \}$. That is,
$L_l^* = L_l \sqcup (L_l+m) \sqcup (L_l+\frac{l}{2}) \sqcup (L_l+(\frac{l}{2}+m))$
This gives us three lattices
$\begin{cases} M_1 &= L_l \sqcup (L_l+m) = L \\ M_2 &= L_l \sqcup (L_l+\frac{l}{2}) \\ M_3 &= L_l \sqcup (L_l+(\frac{l}{2}+m)) \end{cases}$
and all three of them are unimodular because
$L_l \subsetneq M_i \subseteq M_i^* \subsetneq L_l^*$
and $L_l$ is of index $4$ in $L_l^*$.

Now, let’s assume the norm of $l$, that is, $l.l \in 4 \mathbb{Z}$. Then, either the norm of $\frac{l}{2}$ is odd (but then the norm of $\frac{l}{2}+m$ must be even), or the norm of $\frac{l}{2}$ is even, in which case the norm of $\frac{l}{2}+m$ is odd.

That is, either $M_2$ or $M_3$ is an even unimodular lattice, the other one being an odd unimodular lattice.

Let’s take for $l$ and $m$ the vectors $\lambda = \frac{1}{\sqrt{2}} (1,1,\dots,1) \in L – 2L$ and $\mu = \sqrt{2}(1,0,\dots,0) \in L$, then
$\lambda.\lambda = \frac{1}{2}\times 24 = 12 \quad \text{and} \quad \mu.\lambda = 1$
Because $\frac{\lambda}{2}.\frac{\lambda}{2} = \frac{12}{4}=3$ is odd, we have that
$\Lambda = L_{\lambda} \sqcup (L_{\lambda} + (\frac{\lambda}{2} + \mu))$
is an even unimodular lattice, which is the Leech lattice, and
$\Lambda_{odd} = L_{\lambda} \sqcup (L_{\lambda} + \frac{\lambda}{2})$
is an odd unimodular lattice, called the odd Leech lattice. John Leech (1926-1992) – Photo Credit

Let’s check that these are indeed the Leech lattices, meaning that they do not contain roots (vectors of norm two).

The only roots in $L = (A_1^{24})^+$ are the $48$ roots of $A_1^{24}$ and they are of the form $\pm \sqrt{2} [ 1, 0^{23} ]$, but none of them lies in $L_{\lambda}$ as their inproduct with $\lambda$ is one. So, all non-zero vectors in $L_{\lambda}$ have norm $\geq 4$.

As for the other part of $\Lambda$ and $\Lambda_{odd}$
$(L_{\lambda} + \frac{\lambda}{2}) \sqcup (L_{\lambda} + \mu + \frac{\lambda}{2}) = (L_{\lambda} \sqcup (L_{\lambda}+\mu))+\frac{\lambda}{2} = L + \frac{\lambda}{2}$
From the description of $L=(A_1^{24})^+$ it follows that every coordinate of a vector in $L + \frac{\lambda}{2}$ is of the form
$\frac{1}{\sqrt{2}}(v+\frac{1}{2}) \quad \text{or} \quad \frac{1}{\sqrt{2}}(v+\frac{3}{2})$
with $v \in 2 \mathbb{Z}$, with the second case instances forming a codeword in $\mathcal{C}_{24}$. In either case, the square of each of the $24$ coordinates is $\geq \frac{1}{8}$, so the norm of such a vector must be $\geq 3$, showing that there are no roots in this region either.

If one takes for $l$ a vector of the form $\frac{1}{\sqrt{2}} v = \frac{1}{\sqrt{2}}[1^a,0^{24-a}]$ where $a=8,12$ or $16$ and $v \in \mathcal{C}_{24}$, takes $m=\mu$ as before, and repeats the construction, one gets the other Niemeier-neighbours of $(A_1^{24})^+$, that is, the lattices $(A_2^{12})^+$, $(A_3^8)^+$ and $(D_4^6)^+$.

For $a=12$ one needs a slightly different argument, see section 0.2 of Richard Borcherds’ Ph.D. thesis.

Here’s the upper part of Kneser‘s neighbourhood graph of the Niemeier lattices: The Leech lattice has a unique neighbour, that is, among the $23$ remaining Niemeier lattices there is a unique one, $(A_1^{24})^+$, sharing an index two sub-lattice with the Leech.

How would you try to construct $(A_1^{24})^+$, an even unimodular lattice having the same roots as $A_1^{24}$?

The root lattice $A_1$ is $\sqrt{2} \mathbb{Z}$. It has two roots $\pm \sqrt{2}$, determinant $2$, its dual lattice is $A_1^* = \tfrac{1}{\sqrt{2}} \mathbb{Z}$ and we have $A_1^*/A_1 \simeq C_2 \simeq \mathbb{F}_2$.

Thus, $A_1^{24}= \sqrt{2} \mathbb{Z}^{\oplus 24}$ has $48$ roots, determinant $2^{24}$, its dual lattice is $(A_1^{24})^* = \tfrac{1}{\sqrt{2}} \mathbb{Z}^{\oplus 24}$ and the quotient group $(A_1^{24})^*/A_1^{24}$ is $C_2^{24}$ isomorphic to the additive subgroup of $\mathbb{F}_2^{\oplus 24}$.

A larger lattice $A_1^{24} \subseteq L$ of index $k$ gives for the dual lattices an extension $L^* \subseteq (A_1^{24})^*$, also of index $k$. If $L$ were unimodular, then the index has to be $2^{12}$ because we have the situation
$A_1^{24} \subseteq L = L^* \subseteq (A_1^{24})^*$
So, Kneser’s glue vectors form a $12$-dimensional subspace $\mathcal{C}$ in $\mathbb{F}_2^{\oplus 24}$, that is,
$L = \mathcal{C} \underset{\mathbb{F}_2}{\times} (A_1^{24})^* = \{ \tfrac{1}{\sqrt{2}} \vec{v} ~|~\vec{v} \in \mathbb{Z}^{\oplus 24},~v=\vec{v}~mod~2 \in \mathcal{C} \}$
Because $L = L^*$, the linear code $\mathcal{C}$ must be self-dual meaning that $v.w = 0$ (in $\mathbb{F}_2$) for all $v,w \in \mathcal{C}$. Further, we want that the roots of $A_1^{24}$ and $L$ are the same, so the minimal number of non-zero coordinates in $v \in \mathcal{C}$ must be $8$.

That is, $\mathcal{C}$ must be a self-dual binary code of length $24$ with Hamming distance $8$. Marcel Golay (1902-1989) – Photo Credit

We now know that there is a unique such code, the (extended) binary Golay code, $\mathcal{C}_{24}$, which has

• one vector of weight $0$
• $759$ vectors of weight $8$ (called ‘octads’)
• $2576$ vectors of weight $12$ (called ‘dodecads’)
• $759$ vectors of weight $16$
• one vector of weight $24$

The $759$ octads form a Steiner system $S(5,8,24)$ (that is, for any $5$-subset $S$ of the $24$-coordinates there is a unique octad having its non-zero coordinates containing $S$).

Witt constructed a Steiner system $S(5,8,24)$ in his 1938 paper “Die $5$-fach transitiven Gruppen von Mathieu”, so it is not unthinkable that he checked the subspace of $\mathbb{F}_2^{\oplus 24}$ spanned by his $759$ octads to be $12$-dimensional and self-dual, thereby constructing the Niemeier-lattice $(A_1^{24})^+$ on that sunday in 1940.

John Conway classified all nine self-dual codes of length $24$ in which the weight
of every codeword is a multiple of $4$. Each one of these codes $\mathcal{C}$ gives a Niemeier lattice $\mathcal{C} \underset{\mathbb{F}_2}{\times} (A_1^{24})^*$, all but one of them having more roots than $A_1^{24}$.

Vera Pless and Neil Sloan classified all $26$ binary self-dual codes of length $24$.

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 $D_{16}^*/D_{16}$:
$\begin{cases} =(0, \dots,0) &~\text{of norm 0} \\ =(\tfrac{1}{2},\dots,\tfrac{1}{2}) &~\text{of norm 4} \\ =(0,\dots,0,1) &~\text{of norm 1} \\ =(\tfrac{1}{2},\dots,\tfrac{1}{2},-\tfrac{1}{2}) &~\text{of norm 4} \end{cases}$
and he verifies that the determinant of $D_{16}^+=D_{16}+(+D_{16})$ is indeed one (btw. adding coset $$ 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.

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);
[ f1, f2 ] -> [ (2,5)(3,4)(7,10)(8,9), (1,5,4,3,2)(6,10,9,8,7) ]
gap> IsomorphismGroups(D,s10);
[ 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);
fail
gap> IsCyclic(s10);
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.

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.