Rotations of Klein’s quartic

The usual argument to show that the group of all orientation-preserving symmetries of the Klein quartic is the simple group $L_2(7)$ of order $168$ goes like this:

There are two families of $7$ truncated cubes on the Klein quartic. The triangles of one of the seven truncated cubes in the first family have as center the dots, all having the same colour. The triangles of one of the truncated cubes in the second family correspond to the squares all having the same colour.



If you compare the two colour schemes, you’ll see that every truncated cube in the first family is disjoint from precisely $3$ truncated cubes in the second family.

That is, we can identify the truncated cubes of the first family with the points in the Fano plane $\mathbb{P}^2(\mathbb{F}_2)$, and those of the second family with the lines in that plane.

The Klein quartic consists of $24$ regular heptagons, so its rotation symmetry group consists of $24 \times 7 = 168$ rotations,each preserving the two families of truncated cubes. This is exactly the same number as there are isomorphisms of the Fano plane, $PGL_3(\mathbb{F}_2) = L_2(7)$. Done!

For more details, check John Baez’ excellent page on the Klein quartic, or the Buckyball curve post.

Here’s another ‘look-and-see’ proof, starting from Klein’s own description of his quartic.



Look at the rotation $g$, counter-clockwise with angle $2\pi / 7$ fixing the center of the central blue heptagon, and a similar rotation $h$ fixing the center of one of the neighbouring red heptagons.

The two vertices of the edge shared by the blue and red heptagon are fixed by $g.h$ and $h.g$, respectively, so these rotations must have order three (there are $3$ heptagons meeting in the vertex).

That is, the rotation symmetry group $G$ of the Klein quartic has order $168$, and contains two elements $g$ and $h$ of order $7$, such that the subgroup generated by them contains elements of order $3$.

This is enough to prove that the $G$ must be simple and therefore isomorphic to $L_2(7)$!

The following elegant proof is often attributed to Igor Dolgachev.

If $G$ isn’t simple there is a maximal normal subgroup $N$ with $G/N$ simple .

The only non-cyclic simple group having less elements that $168$ is $A_5$ but this cannot be $G/N$ as $60$ does not divide $168$.

So, $G/N$ must be cyclic of order $2,3$ or $7$ (the only prime divisors of $168=2^3.3.7$).

Order $2$ is not possible as any group $N$ of order $84=2^2.3.7$ can just have one Sylow $7$-subgroup. Remember that the number of $7$-Sylows of $N$ must divide $2^2.3=12$ and must be equal to $1$ modulo $7$. And $G$ (and therefore $N$) has at least two different cyclic subgroups of order $7$.

Order $3$ is impossible as this would imply that the normal subgroup $N$ of order $2^3.7=56$ must contain all $7$-Sylows of $G$, and thus also an element of order $3$. But, $3$ does not divide $56$.

Order $7$ is a bit more difficult to exclude. This would mean that there is a normal subgroup $N$ of order $2^3.3=24$.

$N$ (being normal) must contain all Sylow $2$-subgroups of $G$ of which there are either $1$ or $3$ (the order of $N$ is $2^3.3=24$).

If there is just one $S$ it should be a normal subgroup with $G/S$ (of order $21$) having a (normal) unique Sylow $7$-subgroup, but then $G$ would have a normal subgroup of index $3$, which we have excluded.

The three $2$-Sylows are conjugated and so the conjugation morphism
\[
G \rightarrow S_3 \]
is non-trivial and must have image strictly larger than $C_3$ (otherwise, $G$ would have a normal subgroup of index $3$), so must be surjective.

But, $S_3$ has a normal subgroup of index $2$ and pulling this back, $G$ must also have a normal subgroup of index two, which we have excluded. Done!

Rarer books: Singmaster’s notes

David Singmaster‘s “Notes on Rubik’s magic cube” are a collectors item, but it is still possible to buy a copy. I own a fifth edition (august 1980).

These notes capture the Rubik craze of those years really well.

Here’s a Conway story, from Siobhan Roberts’ excellent biography Genius at Play.

The ICM in Helsinki in 1978 was Conway’s last shot to get the Fields medal, but this was the last thing on his mind. He just wanted a Rubik cube (then, iron-curtain times, only sold in Hungary), so he kept chasing Hungarians at the meeting, hoping to obtain one. Siobhan writes (p. 239):

“The Fields Medals went to Pierre Deligne, Charles Fefferman, Grigory Margulis, and Daniel Quillen. The Rubik’s cube went to Conway.”

After his Notes, David Singmaster produced a follow-up newsletter “The Cubic Circular”. Only 5 magazines were published, of which 3 were double issues, between the Autumn of 1981 and the summer of 1985.

These magazines were reproduced on this page.

the monster dictates her picture

The monstrous moonshine picture is a sub-graph of Conway’s Big Picture on 218 vertices. These vertices are the classes of lattices needed in the construction of the 171 moonshine groups. That is, moonshine gives us the shape of the picture.

(image credit Friendly Monsters)

But we can ask to reverse this process. Is the shape of the picture dictated by group-theoretic properties of the monster?

That is, can we reconstruct the 218 lattices and their edges starting from say the conjugacy classes of the monster and some simple rules?

Look at the the power maps for the monster. That is, the operation on conjugacy classes sending the class of $g$ to that of $g^k$ for all divisors $k$ of the order of $g$. Or, if you prefer, the $\lambda$-ring structure on the representation ring.

Rejoice die-hard believers in $\mathbb{F}_1$-theory, rejoice!

Here’s the game to play.

Let $g$ be a monster element of order $n$ and take $d=gcd(n,24)$.

(1) : If $d=8$ and a power map of $g$ gives class $8C$ add $(n|4)$ to your list.

(2) : Otherwise, look at the smallest power of $g$ such that the class is one of $12J,8F,6F,4D, 3C,2B$ or $1A$ and add $(n|e)$ where $e$ is the order of that class, or, if $n > 24$ and $e$ is even add $(n | \frac{e}{2})$.

A few examples:

For class 20E, $d=4$ and the power maps give classes 4D and 2B, so we add $(20|2)$.

For class 32B, $d=8$ but the power map gives 8E so we resort to rule (2). Here the power maps give 8E, 4C and 2B. So, the best class is 4C but as $32 > 24$ we add $(32|2)$.

For class 93A, $d=3$ and the power map gives 3C and even though $93 > 24$ we add $(93|3)$.

This gives us a list of instances $(n|e)$ with $n$ the order of a monster element. For $N=n \times e$ look at all divisors $h$ of $24$ such that $h^2$ divides $N$ and add to your list of lattices those of the form $M \frac{g}{h}$ with $g$ strictly smaller than $h$ and $(g,h)=1$ and $M$ a divisor of $\frac{N}{h^2}$.

This gives us a list of lattices $M \frac{g}{h}$, which is an $h$-th root of unity centered as $L=M \times h$ (see this post). If we do this for all lattices in the list we can partition the $L$’s in families according to which roots of unity are centered at $L$.

This gives us the moonshine picture. (modulo mistakes I made)

The operations we have to do after we have our list of instances $(n|e)$ is pretty straightforward from the rules we used to determine the lattices needed to describe a moonshine group.

Perhaps the oddest part in the construction are the rules (1) and (2) and the prescribed conjugacy classes used in them.

One way to look at this is that the classes $8C$ and $12J$ (or $24J$) are special. The other classes are just the power-maps of $12J$.

Another ‘rationale’ behind these classes may come from the notion of harmonics (see the original Monstrous moonshine paper page 312) of the identity element and the two classes of involutions, 2A (the Fischer involutions) and 2B (the Conway involutions).

For 1A these are : 1A,3C

For 2A these are : 2A,4B,8C

For 2B these are : 2B,4D,6F,8F,12J,24J

These are exactly the classes that we used in (1) and (2), if we add the power-classes of 8C.

Perhaps I should take some time to write all this down more formally.

The Langlands program and non-commutative geometry

The Bulletin of the AMS just made this paper by Julia Mueller available online: “On the genesis of Robert P. Langlands’ conjectures and his letter to Andre Weil” (hat tip +ChandanDalawat and +DavidRoberts on Google+).

It recounts the story of the early years of Langlands and the first years of his mathematical career (1960-1966)leading up to his letter to Andre Weil in which he outlines his conjectures, which would become known as the Langlands program.

Langlands letter to Weil is available from the IAS.

The Langlands program is a vast net of conjectures. For example, it conjectures that there is a correspondence between

– $n$-dimensional representations of the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, and

– specific data coming from an adelic quotient-space $GL_n(\mathbb{A}_{\mathbb{Q}})/GL_n(\mathbb{Q})$.

For $n=1$ this is essentially class field theory with the correspondence given by Artin’s reciprocity law.

Here we have on the one hand the characters of the abelianised absolute Galois group

\[
Gal(\overline{\mathbb{Q}}/\mathbb{Q})^{ab} \simeq Gal(\mathbb{Q}(\pmb{\mu}_{\infty})/\mathbb{Q}) \simeq \widehat{\mathbb{Z}}^{\ast} \]

and on the other hand the connected components of the idele class space

\[
GL_1(\mathbb{A}_{\mathbb{Q}})/GL_1(\mathbb{Q}) = \mathbb{A}_{\mathbb{Q}}^{\ast} / \mathbb{Q}^{\ast} = \mathbb{R}_+^{\ast} \times \widehat{\mathbb{Z}}^{\ast} \]

For $n=2$ it involves the study of Galois representations coming from elliptic curves. A gentle introduction to the general case is Mark Kisin’s paper What is … a Galois representation?.

One way to look at some of the quantum statistical systems studied via non-commutative geometry is that they try to understand the “bad” boundary of the Langlands space $GL_n(\mathbb{A}_{\mathbb{Q}})/GL_n(\mathbb{Q})$.

Here, the Bost-Connes system corresponds to the $n=1$ case, the Connes-Marcolli system to the $n=2$ case.

If $\mathbb{A}’_{\mathbb{Q}}$ is the subset of all adeles having almost all of its terms in $\widehat{\mathbb{Z}}_p^{\ast}$, then there is a well-defined map

\[
\pi~:~\mathbb{A}’_{\mathbb{Q}}/\mathbb{Q}^{\ast} \rightarrow \mathbb{R}_+ \qquad (x_{\infty},x_2,x_2,\dots) \mapsto | x_{\infty} | \prod_p | x_p |_p \]

The inverse image of $\pi$ over $\mathbb{R}_+^{\ast}$ are exactly the idele classes $\mathbb{A}_{\mathbb{Q}}^{\ast}/\mathbb{Q}^{\ast}$, so we can view them as the nice locus of the horrible complicated quotient of adele-classes $\mathbb{A}_{\mathbb{Q}}/\mathbb{Q}^*$. And we can view the adele-classes as a ‘closure’ of the idele classes.

But, the fiber $\pi^{-1}(0)$ has horrible topological properties because $\mathbb{Q}^*$ acts ergodically on it due to the fact that $log(p)/log(q)$ is irrational for distinct primes $p$ and $q$.

This is why it is better to view the adele-classes not as an ordinary space (one with bad topological properties), but rather as a ‘non-commutative’ space because it is controlled by a non-commutative algebra, the Bost-Connes algebra.

For $n=2$ there’s a similar story with a ‘bad’ quotient $M_2(\mathbb{A}_{\mathbb{Q}})/GL_2(\mathbb{Q})$, being the closure of an ‘open’ nice piece which is the Langlands quotient space $GL_2(\mathbb{A}_{\mathbb{Q}})/GL_2(\mathbb{Q})$.

A forgotten type and roots of unity (again)

The monstrous moonshine picture is the finite piece of Conway’s Big Picture needed to understand the 171 moonshine groups associated to conjugacy classes of the monster.

Last time I claimed that there were exactly 7 types of local behaviour, but I missed one. The forgotten type is centered at the number lattice $84$.

Locally around it the moonshine picture looks like this
\[
\xymatrix{42 \ar@{-}[dr] & 28 \frac{1}{3} \ar@[red]@{-}[d] & 41 \frac{1}{2} \ar@{-}[ld] \\ 28 \ar@[red]@{-}[r] & \color{grey}{84} \ar@[red]@{-}[r] \ar@[red]@{-}[d] \ar@{-}[rd] & 28 \frac{2}{3} \\ & 252 & 168} \]

and it involves all square roots of unity ($42$, $42 \frac{1}{2}$ and $168$) and $3$-rd roots of unity ($28$, $28 \frac{1}{3}$, $28 \frac{2}{3}$ and $252$) centered at $84$.

No, I’m not hallucinating, there are indeed $3$ square roots of unity and $4$ third roots of unity as they come in two families, depending on which of the two canonical forms to express a lattice is chosen.

In the ‘normal’ expression $M \frac{g}{h}$ the two square roots are $42$ and $42 \frac{1}{2}$ and the three third roots are $28, 28 \frac{1}{3}$ and $28 \frac{2}{3}$. But in the ‘other’ expression
\[
M \frac{g}{h} = (\frac{g’}{h},\frac{1}{h^2M}) \]
(with $g.g’ \equiv 1~mod~h$) the families of $2$-nd and $3$-rd roots of unity are
\[
\{ 42 \frac{1}{2} = (\frac{1}{2},\frac{1}{168}), 168 = (0,\frac{1}{168}) \} \]
and
\[
\{ 28 \frac{1}{3} = (\frac{1}{3},\frac{1}{252}), 28 \frac{2}{3} = (\frac{2}{3},\frac{1}{252}), 252 = (0 , \frac{1}{252}) \} \]
As in the tetrahedral snake post, it is best to view the four $3$-rd roots of unity centered at $84$ as the vertices of a tetrahedron with center of gravity at $84$. Power maps in the first family correspond to rotations along the axis through $252$ and power maps in the second family are rotations along the axis through $28$.

In the ‘normal’ expression of lattices there’s then a total of 8 different local types, but two of them consist of just one number lattice: in $8$ the local picture contains all square, $4$-th and $8$-th roots of unity centered at $8$, and in $84$ the square and $3$-rd roots.

Perhaps surprisingly, if we redo everything in the ‘other’ expression (and use the other families of roots of unity), then the moonshine picture has only 7 types of local behaviour. The forgotten type $84$ appears to split into two occurrences of other types (one with only square roots of unity, and one with only $3$-rd roots).

I wonder what all this has to do with the action of the Bost-Connes algebra on the big picture or with Plazas’ approach to moonshine via non-commutative geometry.