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Author: lievenlb

Moonshine’s green anaconda

The largest snake in the moonshine picture determines the moonshine group $(24|12)$ and is associated to conjugacy class $24J$ of the monster.

It contains $70$ lattices, about one third of the total number of lattices in the moonshine picture.

The anaconda’s backbone is the $(288|1)$ thread below (edges in the $2$-tree are black, those in the $3$-tree red and coloured numbers are symmetric with respect to the $(24|12)$-spine and have the same local snake-structure.

\[
\xymatrix{9 \ar@{-}[r] \ar@[red]@{-}[d] & \color{green}{18} \ar@{-}[r] \ar@[red]@{-}[d] & \color{yellow}{36} \ar@{-}[r] \ar@[red]@{-}[d] & \color{yellow}{72} \ar@{-}[r] \ar@[red]@{-}[d] & \color{green}{144} \ar@{-}[r] \ar@[red]@{-}[d] & 288 \ar@[red]@{-}[d] \\
3 \ar@{-}[r] \ar@[red]@{-}[d] & \color{blue}{6} \ar@{-}[r] \ar@[red]@{-}[d] & \color{red}{12} \ar@{-}[r] \ar@[red]@{-}[d] & \color{red}{24} \ar@{-}[r] \ar@[red]@{-}[d] & \color{blue}{48} \ar@{-}[r] \ar@[red]@{-}[d] & 96 \ar@[red]@{-}[d] \\
1 \ar@{-}[r] & \color{green}{2} \ar@{-}[r] & \color{yellow}{4} \ar@{-}[r] & \color{yellow}{8} \ar@{-}[r] & \color{green}{16} \ar@{-}[r] & 32 } \]

These are the only number-lattices in the anaconda. The remaining lattices are number-like, that is of the form $M \frac{g}{h}$ with $M$ an integer and $1 \leq g < h$ with $(g,h)=1$.
There are

– $12$ with $h=2$ and $M$ a divisor of $72$.

– $12$ with $h=3$ and $M$ a divisor of $32$.

– $12$ with $h=4$ and $M$ a divisor of $18$.

– $8$ with $h=6$ and $M$ a divisor of $8$.

– $8$ with $h=12$ and $M=1,2$.

The non-number lattices in the snake are locally in the coloured numbers:

In $2,16,18,144=2M$

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

In $4,8,36,72=4M$

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

In $6,48=6M$

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

In $12,24=12M$ the local structure looks like

Here, we used the commutation relations to reach all lattices at distance $log(6)$ and $log(12)$ by first walking the $2$-adic tree and postpone the last step for the $3$-tree.

Perhaps this is also a good strategy to get a grip on the full moonshine picture:

First determine subsets of the moonshine thread with the same local structure, and then determine for each class this local structure.

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a COLgo endgame

COL is a map-colouring game, attibuted to Colin Vout. COLgo is COL played with Go-stones on a go-board.

The two players, bLack (left) and white (right) take turns placing a stone of their colour on the board, but two stones of the same colour may not be next to each other.

The first player unable to make a legal move looses this game.

As is common in combinatorial game theory we do not specify which player has the move. There are $4$ different outcomes, the game is called:

– positive, if there is a winning strategy for Left (bLack),
– negative, if there is a winning strategy for Right (white),
– zero, if there is a winning strategy for the second-player,
– fuzzy, if there is a winning strategy for the first player.

Here’s an endgame problem: who wins this game?

Spoiler alert: solution below.

First we can exclude all spots which are dead, that is, are excluded for both players. Example, F11 is dead because it neighbors a black as well as a white stone, but F10 is alive as it can be played by white (Right).

If we remove all dead spots, we are left with 4 regions (the four extremal corners of the board) as well as 5 spots, 3 for white and 2 for black.

That is, the game reduces to this “sum”-game, in which a player chooses one of the regions and does a legal move in that component, or takes a stone of its own colour from the second row.

Next, we have to give a value to each of the region-games.

– the right-most game has value $0$ as the second player has a winning strategy by reflecting the first player’s move with respect to the central (dead) spot.

– the left-most game is equivalent to one black stone. Black can make two moves in the game, independent of the only move that white can make. So it has value $+1$.

– the sum-game of the two middle games has value zero. The second player can win by mirroring the first player’s move in the other component. This is called the Tweedledee-Tweedledum argument.

But then, the total value of the endgame position is

zero, so the first player to move looses the game!

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a wintry chataigneraie

It took us some time to clear the array of old chestnut trees.

But it paid off. A good harvest easily gives half a ton of chestnuts, including some rare, older varieties.

In the autumn, the coloured leaves make a spectacular view. Now it looks rather desolate.

But, the wind piles up the fallen leaves in every nook, cranny, corner or entrance possible.

My day in the mountains: ‘harvesting’ fallen leaves…

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