Extending McKay’s E8 graph?

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

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

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

This is the content of John McKay’s E(8)-observation : there should be a precise relation between the nodes of the extended Dynkin diagram and these 9 conjugacy classes in such a way that the order of the class corresponds to the component of the fundamental root. More precisely, one conjectures the following correspondence:

John Duncan found such a connection by considering carefully the corresponding moonshine groups and their inter-relation. For more on this, look at the old post E8 from moonshine groups. The extended Dynkin diagram with these moonshine groups as vertices is:

Duncan does this by assigning numbers to moonshine groups: the dimension is the order of the corresponding monster element and the valency is one more than the copies of $C_2$ generated by the Atkin-Lehner involutions in the moonshine group.

One might ask whether there is a graph on all 171 moonshine groups, compatible with the valencies of every vertex.

Now, even for the 9 groups in McKay’s question, the valencies do not determine the graph uniquely and Duncan proceeds with an ad hoc condition on the edges.

There is a partition on the 9 groups by the property whether or not the index of the intersection with $\Gamma_0(2)$ is at most two. Then Duncan declares that there cannot be an edge between two groups belonging to the same class.

His motivation for this property comes from classical McKay-correspondence for the binary icosahedral group (where the vertices correspond to simple representations $S$, and the edges from $S$ to factors of $S \otimes V_2$, where $V_2$ is the restriction of the standard $2$-dimensional simple for $SU(2)$).

Of the $9$ simples there are only $4$ faithful ones, $5$ come from simples of $A_5$. Because $\Gamma_0(2)$ is a subgroup of the modular group of index 2, he then views $\Gamma_0(2)$ as similar to the subgroup $A_5$ in the binary icosahedral group, and declares a moonshine group to be faithful if its index in the intersection with $\Gamma_0(2)$ is at most two.

One might ask whether there is another, more natural, definition for having an edge (or multiple ones) between arbitrary moonshine groups.

And, what is the full graph on the 171 groups?

a morning in the mountains

For mornings like this I happily drive a day to be here.

the moonshine picture – at last

The monstrous moonshine picture is the subgraph of Conway’s big picture consisting of all lattices needed to describe the 171 moonshine groups.

It consists of:

– exactly 218 vertices (that is, lattices), out of which

– 97 are number-lattices (that is of the form $M$ with $M$ a positive integer), and

– 121 are proper number-like lattices (that is of the form $M \frac{g}{h}$ with $M$ a positive integer, $h$ a divisor of $24$ and $1 \leq g \leq h$ with $(g,h)=1$).

The $97$ number lattices are closed under taking divisors, and the corresponding Hasse diagram has the following shape

Here, number-lattices have the same colour if they have the same local structure in the moonshine picture (that is, have a similar neighbourhood of proper number-like lattices).

There are 7 different types of local behaviour:

The white numbered lattices have no proper number-like neighbours in the picture.

The yellow number lattices (2,10,14,18,22,26,32,34,40,68,80,88,90,112,126,144,180,208 = 2M) have local structure

\[
\xymatrix{M \ar@{-}[r] & \color{yellow}{2M} \ar@{-}[r] & M \frac{1}{2}} \]

which involves all $2$-nd (square) roots of unity centered at the lattice.

The green number lattices (3,15,21,39,57,93,96,120 = 3M) have local structure

\[
\xymatrix{& M \ar@[red]@{-}[d] & \\ M \frac{1}{3} \ar@[red]@{-}[r] & \color{green}{3M} \ar@[red]@{-}[r] & M \frac{2}{3}} \]

which involve all $3$-rd roots of unity centered at the lattice.

The blue number lattices (4,16,20,28,36,44,52,56,72,104 = 4M) have as local structure

\[
\xymatrix{M \frac{1}{2} \ar@{-}[d] & & M \frac{1}{4} \ar@{-}[d] \\
2M \ar@{-}[r] & \color{blue}{4M} \ar@{-}[r] & 2M \frac{1}{2} \ar@{-}[d] \\
M \ar@{-}[u] & & M \frac{3}{4}} \]

and involve the $2$-nd and $4$-th root of unity centered at the lattice.

The purple number lattices (6,30,42,48,60 = 6M) have local structure

\[
\xymatrix{& M \frac{1}{3} \ar@[red]@{-}[d] & 2M \frac{1}{3} & M \frac{1}{6} \ar@[red]@{-}[d] & \\
M \ar@[red]@{-}[r] & 3M \ar@{-}[r] \ar@[red]@{-}[d] & \color{purple}{6M} \ar@{-}[r] \ar@[red]@{-}[u] \ar@[red]@{-}[d] & 3M \frac{1}{2} \ar@[red]@{-}[r] \ar@[red]@{-}[d] & M \frac{5}{6} \\
& M \frac{2}{3} & 2M \frac{2}{3} & M \frac{1}{2} & } \]

and involve all $2$-nd, $3$-rd and $6$-th roots of unity centered at the lattice.

The unique brown number lattice 8 has local structure

\[
\xymatrix{& & 1 \frac{1}{4} \ar@{-}[d] & & 1 \frac{1}{8} \ar@{-}[d] & \\
& 1 \frac{1}{2} \ar@{-}[d] & 2 \frac{1}{2} \ar@{-}[r] \ar@{-}[d] & 1 \frac{3}{4} & 2 \frac{1}{4} \ar@{-}[r] & 1 \frac{5}{8} \\
1 \ar@{-}[r] & 2 \ar@{-}[r] & 4 \ar@{-}[r] & \color{brown}{8} \ar@{-}[r] & 4 \frac{1}{2} \ar@{-}[d] \ar@{-}[u] & \\
& & & 1 \frac{7}{8} \ar@{-}[r] & 2 \frac{3}{4} \ar@{-}[r] & 1 \frac{3}{8}} \]

which involves all $2$-nd, $4$-th and $8$-th roots of unity centered at $8$.

Finally, the local structure for the central red lattices $12,24 = 12M$ is

\[
\xymatrix{
M \frac{1}{12} \ar@[red]@{-}[dr] & M \frac{5}{12} \ar@[red]@{-}[d] & M \frac{3}{4} \ar@[red]@{-}[dl] & & M \frac{1}{6} \ar@[red]@{-}[dr] & M \frac{1}{2} \ar@[red]@{-}[d] & M \frac{5}{6} \ar@[red]@{-}[dl] \\
& 3M \frac{1}{4} \ar@{-}[dr] & 2M \frac{1}{6} \ar@[red]@{-}[d] & 4M \frac{1}{3} \ar@[red]@{-}[d] & 2M \frac{1}{3} \ar@[red]@{-}[d] & 3M \frac{1}{2} \ar@{-}[dl] & \\
& 2M \frac{1}{2} \ar@[red]@{-}[r] & 6M \frac{1}{2} \ar@{-}[dl] \ar@[red]@{-}[d] \ar@{-}[r] & \color{red}{12M} \ar@[red]@{-}[d] \ar@{-}[r] & 6M \ar@[red]@{-}[d] \ar@{-}[dr] \ar@[red]@{-}[r] & 2M & \\
& 3M \frac{3}{4} \ar@[red]@{-}[dl] \ar@[red]@{-}[d] \ar@[red]@{-}[dr] & 2M \frac{5}{6} & 4M \frac{2}{3} & 2M \frac{2}{3} & 3M \ar@[red]@{-}[dl] \ar@[red]@{-}[d] \ar@[red]@{-}[dr] & \\
M \frac{1}{4} & M \frac{7}{12} & M \frac{11}{12} & & M \frac{1}{3} & M \frac{2}{3} & M}
\]

It involves all $2$-nd, $3$-rd, $4$-th, $6$-th and $12$-th roots of unity with center $12M$.

No doubt this will be relevant in connecting moonshine with non-commutative geometry and issues of replicability as in Plazas’ paper Noncommutative Geometry of Groups like $\Gamma_0(N)$.

Another of my pet follow-up projects is to determine whether or not the monster group $\mathbb{M}$ dictates the shape of the moonshine picture.

That is, can one recover the 97 number lattices and their partition in 7 families starting from the set of element orders of $\mathbb{M}$, applying some set of simple rules?

One of these rules will follow from the two equivalent notations for lattices, and the two different sets of roots of unities centered at a given lattice. This will imply that if a number lattice belongs to a given family, certain divisors and multiples of it must belong to related families.

If this works out, it may be a first step towards a possibly new understanding of moonshine.

Knights and Knaves, the Heyting way

(image credit: Joe Blitzstein via Twitter)

Smullyan’s Knights and Knaves problems are classics. On an island all inhabitants are either Knights (who only tell true things) and Knaves (who always lie). You have to determine their nature from a few statements. Here’s a very simple problem:

“Abercrombie met just two inhabitants, A and B. A made the following statement: “Both of us are Knaves.” What is A and what is B?”

Now, this one is simple enough to solve, but for more complicated problems a generic way to solve the puzzles is to use propositional calculas, as explained in Smullyan’s Logical Labyrinths”, chapter 8 “Liars, truth-tellers and propositional logic’.

If an inhabitants $A$ asserts a proposition $P$, and if $k_A$ is the assertion ‘$A$ is a Knight’, then the statement can be rephrased as

\[
k_A \Leftrightarrow P \]

for if $A$ is a Knight, $P$ must be true and if $A$ is a Knave $P$ must be false.

Usually, one can express $P$ as a propositional statement involving $k_A,k_B,k_C,\dots$.
The example above can be rephrased as

\[
k_A \Leftrightarrow (\neg k_A \wedge \neg k_B) \]

Assigning truth values to $k_A$ and $k_B$ and setting up the truth-table for this sentence, one sees that the only possibility for this to be true is that $k_A$ equals $0$ and $k_B$ equals $1$. So, $A$ is a Knave and $B$ is a Knight.

Clearly, one only requires this approach for far more difficult problems.

In almost all Smullyan puzzles, the only truth values are $0$ and $1$. There’s a short excursion to Boolean algebras (sorry, Boolean islands) in chapter 9 ‘Variable Liars’ in Logical Labyrinths. But then, the type of problems are about finding equivalent notions of Boolean algebras, rather that generalised Knights&Knaves puzzles.

Did anyone pursue the idea of Smullyanesque puzzles with truth values in a proper Heyting algebra?

I only found one blog-post on this: Non-Classical Knights and Knaves by Jason Rosenhouse.

He considers three valued logic (the Heyting algebra corresponding to the poset 0-N-1, and logical connectives as in the example on the Wiki-page on Heyting algebras.

On his island the natives cycle, repeatedly and unpredictably, between the two states. They are knights for a while, then they enter a transitional phase during which they are partly knight and partly knave, and then they emerge on the other side as knaves.

“If Joe is in the transitional phase, and you say, “Joe is a knight,” or “Joe is a knave,” what truth value should we assign to your statement? Since Joe is partly knight and partly knave, neither of the classical truth values seems appropriate. So we shall assign a third truth value, “N” to such statements. Think of N as standing for “neutral” or “neither true nor false.” On the island, vague statements are assigned the truth value N.

Just to be clear, it’s not just any statement that can be assigned the truth value N. It is only vague statements that receive that truth value, and for now our only examples of such statements are attributions of knight-hood and knave-hood to people in the transitional phase.

For the natives, entering the transitional phase implied a disconcerting loss of identity. Uncertain of how to behave, they hedged their bets by only making statements with truth value N. People in the transitional phase were referred to as neutrals. So there are now three kinds of people: Knights, who only make true statements; Knaves, who only make false statements; and Neutrals, who only make statements with the truth value N.”

He gives one example of a possible problem:

“Suppose you meet three people, named Dave, Evan and Ford. They make the following statements:

Dave: Evan is a knight.
Evan: Ford is a knave.
Ford: Dave is a neutral.

Can you determine the types of all three people?”

If you know of more of these Smullanesque problems using Heyting algebras, please leave a comment.

a SNORTgo endgame

SNORT, invented by Simon NORTon is a map-coloring game, similar to COL. Only, this time, neighbours may not be coloured differently.

SNORTgo, similar to COLgo, is SNORT played with go-stones on a go-board. That is, adjacent stones must have the same colour.

SNORT is a ‘hot’ game, meaning that each player is eager to move as most moves will improve your position. In COL players are reluctant to move, because a move limits your next moves.

For this reason, SNORT positions are much harder to evaluate, and one needs the full force of Conways’s ONAG.

Here’s a SNORTgo endgame. Who has a winning strategy?, and what is the first move in that strategy?

The method to approach such an endgame is similar to that in COLgo. First we remove all dead spots from the board.

What remains, are a 4 spots available only to Right (white) and 5 spots available only to Left (bLack). Further, there a 3 ‘live’ regions: the upper righthand corner and the two lower corners.

The value of these corners must be computed inductively.

Here’s the answer:

For example, Right’s best option in the left-most game (corresponding to the upper righthand corner of the endgame) is to put his stone on N12, resulting in a game in which neither player can move (the zero game).

On the other hand, Left can put a stone at either N11, N12 or N13 leaving a game in which she has two more moves, whereas Right han none (the $2$ game).

The other positions are computed similarly.

To get the value of the endgame we have to sum up all these values.

This can either be done using the addition rule given in ONAG, or by using programs in combinatorial game theory.

There’s Combinatorial Game Suite, developed by Aaron Siegel. But, for some reason I can no longer use it on macOS High Sierra.

Fortunately, the older program David Wolfe’s toolkit is still available, and runs on my MacBook.

The sum game evaluates to $\{ \{3|2 \}|-1 \}$, which is a ‘fuzzy’ game, that is, its value is confused with $0$.

This means that the first player to move has a winning strategy in the endgame.

Can you spot the (unique) winning move for Right (white) and one (of two) winning move for Left (bLack)?

Spoiler alert : solution in the comments.