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	<title>Tits &#8211; neverendingbooks</title>
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		<title>a non-commutative Jack Daniels problem</title>
		<link>https://lievenlebruyn.github.io/neverendingbooks/a-non-commutative-jack-daniels-problem/</link>
		
		<dc:creator><![CDATA[lieven]]></dc:creator>
		<pubDate>Sat, 20 Jan 2018 13:56:06 +0000</pubDate>
				<category><![CDATA[groups]]></category>
		<category><![CDATA[math]]></category>
		<category><![CDATA[noncommutative]]></category>
		<category><![CDATA[Borcherds]]></category>
		<category><![CDATA[Conway]]></category>
		<category><![CDATA[Duncan]]></category>
		<category><![CDATA[Jack Daniels]]></category>
		<category><![CDATA[Norton]]></category>
		<category><![CDATA[Ogg]]></category>
		<category><![CDATA[Ono]]></category>
		<category><![CDATA[Pizer]]></category>
		<category><![CDATA[Plazas]]></category>
		<category><![CDATA[Ricard]]></category>
		<category><![CDATA[Tits]]></category>
		<guid isPermaLink="false">http://www.neverendingbooks.org/?p=7776</guid>

					<description><![CDATA[At a seminar at the College de France in 1975, Tits wrote down the order of the monster group \[ \# \mathbb{M} = 2^{46}.3^{20}.5^9.7^6.11^2.13^3.17·19·23·29·31·41·47·59·71 \]&#8230;]]></description>
										<content:encoded><![CDATA[<p><img decoding="async" src="https://lievenlebruyn.github.io/neverendingbooks/DATA3/jackdanielsMoonshine.jpg"></p>
<p>At a seminar at the College de France in 1975, Tits wrote down the order of the monster group</p>
<p>\[<br />
\# \mathbb{M} = 2^{46}.3^{20}.5^9.7^6.11^2.13^3.17·19·23·29·31·41·47·59·71 \]</p>
<p><a href="https://en.wikipedia.org/wiki/Andrew_Ogg">Andrew Ogg</a>, who attended the talk, noticed that the prime divisors are precisely the primes $p$ for which the characteristic $p$ super-singular $j$-invariants are all defined over $\mathbb{F}_p$.</p>
<p>Here&#8217;s Ogg&#8217;s paper on this: <a href="http://www.numdam.org/article/SDPP_1974-1975__16_1_A4_0.pdf">Automorphismes de courbes modulaires</a>,  Séminaire Delange-Pisot-Poitou. Théorie des nombres, tome 16, no 1 (1974-1975).</p>
<p>Ogg offered a bottle of Jack Daniels for an explanation of this coincidence.</p>
<p>Even <a href="https://en.wikipedia.org/wiki/Richard_Borcherds">Richard Borcherds</a> didn&#8217;t claim the bottle of Jack Daniels, though his proof of the monstrous moonshine conjecture is believed to be the best explanation, at present.</p>
<p>A few years ago, John Duncan and Ken Ono posted a paper <a href="https://arxiv.org/abs/1411.5354">&#8220;The Jack Daniels Problem&#8221;</a>, in which they prove that monstrous moonshine implies that if $p$ is not one of Ogg&#8217;s primes it cannot be a divisor of $\# \mathbb{M}$. However, the other implication remains mysterious.</p>
<p>Duncan and Ono say:</p>
<p>&#8220;This discussion does not prove that every $p ∈ \text{Ogg}$ divides $\# \mathbb{M}$. It merely explains how the first principles of moonshine suggest this implication. Monstrous moonshine is the proof. Does this then provide a completely satisfactory solution to Ogg’s problem? Maybe or maybe not. Perhaps someone will one day furnish a map from the characteristic $p$ supersingular $j$-invariants to elements of order $p$ where the group structure of $\mathbb{M}$ is apparent.&#8221;</p>
<p>I don&#8217;t know whether they claimed the bottle, anyway.</p>
<p>But then, what is the <strong>non-commutative Jack Daniels Problem</strong>?</p>
<p>A footnote on the first page of Conway and Norton&#8217;s &#8216;Monstrous Moonshine&#8217; paper says:</p>
<p>&#8220;Very recently, A. Pizer has shown these primes are the only ones that satisfy a certain conjecture of Hecke from 1936 relating modular forms of weight $2$ to quaternion algebra theta-series.&#8221;</p>
<p>Pizer&#8217;s paper is <a href="https://msp.org/pjm/1978/79-2/pjm-v79-n2-p16-s.pdf">&#8220;A note on a conjecture of Hecke&#8221;</a>.</p>
<p>Maybe there&#8217;s a connection between monstrous moonshine and the arithmetic of integral quaternion algebras. Some hints:</p>
<p>The <a href="https://lievenlebruyn.github.io/neverendingbooks/the-big-picture-is-non-commutative">commutation relations in the Big Picture</a> are reminiscent of the <a href="http://www.m-hikari.com/imf/imf-2012/41-44-2012/perngIMF41-44-2012.pdf">meta-commutation relations for Hurwitz quaternions</a>, originally due to Conway in his booklet on Quaternions and Octonions.</p>
<p>The fact that the $p$-tree in the Big Picture has valency $p+1$ comes from the fact that the Brauer-Severi of $M_2(\mathbb{F}_p)$ is $\mathbb{P}^1_{\mathbb{F}_p}$. In fact, the Big Picture should be related to the Brauer-Severi scheme of $M_2(\mathbb{Z})$.</p>
<p>Then, there&#8217;s Jorge Plazas claiming that Connes-Marcolli&#8217;s $GL_2$-system <a href="https://lievenlebruyn.github.io/neverendingbooks/nc-geometry-and-moonshine">might be related to moonshine</a>.</p>
<p>One of the first things I&#8217;ll do when I return is to run to the library and get our copy of Shimura&#8217;s &#8216;Introduction to the arithmetic theory of automorphic functions&#8217;.</p>
<p>Btw. the bottle in the title image is not a Jack Daniels but the remains of a bottle of Ricard, because I&#8217;m still in the French mountains.</p>
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			</item>
		<item>
		<title>The Big Picture is non-commutative</title>
		<link>https://lievenlebruyn.github.io/neverendingbooks/the-big-picture-is-non-commutative/</link>
		
		<dc:creator><![CDATA[lieven]]></dc:creator>
		<pubDate>Wed, 03 Jan 2018 12:22:09 +0000</pubDate>
				<category><![CDATA[groups]]></category>
		<category><![CDATA[math]]></category>
		<category><![CDATA[noncommutative]]></category>
		<category><![CDATA[Bruhat]]></category>
		<category><![CDATA[Conway]]></category>
		<category><![CDATA[p-adic tree]]></category>
		<category><![CDATA[Tits]]></category>
		<guid isPermaLink="false">http://www.neverendingbooks.org/?p=7469</guid>

					<description><![CDATA[Conway&#8217;s Big Picture consists of all pairs of rational numbers $M,\frac{g}{h}$ with $M > 0$ and $0 \leq \frac{g}{h} < 1$ with $(g,h)=1$. Recall from&#8230;]]></description>
										<content:encoded><![CDATA[<p><strong>Conway&#8217;s Big Picture</strong> consists of all pairs of rational numbers $M,\frac{g}{h}$ with $M > 0$ and $0 \leq \frac{g}{h} < 1$ with $(g,h)=1$. Recall from <a href="https://lievenlebruyn.github.io/neverendingbooks/the-defining-property-of-24">last time</a> that $M,\frac{g}{h}$ stands for the lattice<br />
\[<br />
\mathbb{Z} (M \vec{e}_1 + \frac{g}{h} \vec{e}_2) \oplus \mathbb{Z} \vec{e}_2 \subset \mathbb{Q}^2 \]<br />
and we associate to it the rational $2 \times 2$ matrix<br />
\[<br />
\alpha_{M,\frac{g}{h}} = \begin{bmatrix} M &#038; \frac{g}{h} \\ 0 &#038; 1 \end{bmatrix} \]</p>
<p>If $M$ is a natural number we write $M \frac{g}{h}$ and call the corresponding lattice <em>number-like</em>, if $g=0$ we drop the zero and write $M$.</p>
<p>The Big Picture carries a wealth of structures. Today, we will see that it can be factored as the product of <a href="https://en.wikipedia.org/wiki/Building_(mathematics)">Bruhat-Tits buildings</a> for $GL_2(\mathbb{Q}_p)$, over all prime numbers $p$.</p>
<p>Here&#8217;s the factor-building for $p=2$, which is a $3$-valent tree:</p>
<p><img decoding="async" src="https://lievenlebruyn.github.io/neverendingbooks/DATA3/BuhatTits.png" width=100% ></p>
<p>To see this, define the <em>distance</em> between lattices to be<br />
\[<br />
d(M,\frac{g}{h}~|~N,\frac{i}{j}) = log~Det(q(\alpha_{M,\frac{g}{h}}.\alpha_{N,\frac{i}{j}}^{-1})) \]<br />
where $q$ is the smallest strictly positive rational number such that $q(\alpha_{M,\frac{g}{h}}.\alpha_{N,\frac{i}{j}}^{-1}) \in GL_2(\mathbb{Z})$.</p>
<p>We turn the Big Picture into a (coloured) graph by drawing an edge (of colour $p$, for $p$ a prime number) between any two lattices distanced by $log(p)$.</p>
<p>\[<br />
\xymatrix{M,\frac{g}{h} \ar@[red]@{-}[rr]|p &#038; &#038; N,\frac{i}{j}} \qquad~\text{iff}~\qquad d(M,\frac{g}{h}~|~N,\frac{i}{j})=log(p) \]</p>
<p>The $p$-coloured subgraph is $p+1$-valent.</p>
<p>The $p$-neighbours of the lattice $1 = \mathbb{Z} \vec{e}_1 \oplus \mathbb{Z} \vec{e}_2$ are precisely these $p+1$ lattices:</p>
<p>\[<br />
p \qquad \text{and} \qquad \frac{1}{p},\frac{k}{p} \qquad \text{for} \qquad 0 \leq k < p \]

And, multiplying the corresponding matrices with $\alpha_{M,\frac{g}{h}}$ tells us that the $p$-neighbours of $M,\frac{g}{h}$ are then these $p+1$ lattices:

\[
pM,\frac{pg}{h}~mod~1 \qquad \text{and} \qquad \frac{M}{p},\frac{1}{p}(\frac{g}{h}+k)~mod~1 \qquad \text{for} \qquad 0 \leq k < p \]


Here's part of the $2$-coloured neighbourhood of $1$


<img decoding="async" src="https://lievenlebruyn.github.io/neverendingbooks/DATA3/2tree.png" width=100% ></p>
<p>To check that the $p$-coloured subgraph is indeed the Bruhat-Tits building of $GL_2(\mathbb{Q}_p)$ it remains to see that it is a tree.</p>
<p>For this it is best to introduce $p+1$ operators on lattices</p>
<p>\[<br />
p \ast  \qquad \text{and} \qquad \frac{k}{p} \ast \qquad \text{for} \qquad 0 \leq k < p \]

defined by left-multiplying $\alpha_{M,\frac{g}{h}}$ by the matrices
\[
\begin{bmatrix} p &#038; 0 \\ 0 &#038; 1 \end{bmatrix} \qquad \text{and} \qquad \begin{bmatrix} \frac{1}{p} &#038; \frac{k}{p} \\ 0 &#038; 1 \end{bmatrix} \qquad \text{for} \qquad 0 \leq k < p \]

The lattice $p \ast M,\frac{g}{h}$ lies closer to $1$ than $M,\frac{g}{h}$ (unless $M,\frac{g}{h}=M$ is a number) whereas the lattices $\frac{k}{p} \ast M,\frac{g}{h}$ lie further, so it suffices to show that the $p$ operators

\[
\frac{0}{p} \ast,~\frac{1}{p} \ast,~\dots~,\frac{p-1}{p} \ast \]

form a free non-commutative monoid.

<br />This follows from the fact that the operator<br />
\[<br />
(\frac{k_n}{p} \ast) \circ \dots \circ (\frac{k_2}{p} \ast) \circ (\frac{k_1}{p} \ast) \]<br />
is given by left-multiplication with the matrix<br />
\[<br />
\begin{bmatrix} \frac{1}{p^n} &#038; \frac{k_1}{p^n}+\frac{k_2}{p^{n-1}}+\dots+\frac{k_n}{p} \\ 0 &#038; 1 \end{bmatrix} \]<br />
which determines the order in which the $k_i$ occur.</p>
<p>A lattice at distance $n log(p)$ from $1$ can be uniquely written as<br />
\[<br />
(\frac{k_{n-l}}{p} \ast) \circ \dots \circ (\frac{k_{l+1}}{p} \ast) \circ (p^l \ast) 1 \]<br />
which gives us the unique path to it from $1$.</p>
<p>The Big Picture itself is then the product of these Bruhat-Tits trees over all prime numbers $p$. Decomposing the distance from $M,\frac{g}{h}$ to $1$ as<br />
\[<br />
d(M,\frac{g}{h}~|~1) = n_1 log(p_1) + \dots + n_k log(p_k) \]<br />
will then allow us to find minimal paths from $1$ to $M,\frac{g}{h}$.</p>
<p>But we should be careful in drawing $2$-dimensional cells (or higher dimensional ones) in this &#8216;product&#8217; of trees as the operators<br />
\[<br />
\frac{k}{p} \ast \qquad \text{and} \qquad \frac{l}{q} \ast \]<br />
for different primes $p$ and $q$ do not commute, in general. The composition<br />
\[<br />
(\frac{k}{p} \ast) \circ (\frac{l}{q} \ast) \qquad \text{with matrix} \qquad \begin{bmatrix} \frac{1}{pq} &#038; \frac{kq+l}{pq} \\ 0 &#038; 1 \end{bmatrix} \]<br />
has as numerator in the upper-right corner $0 \leq kq + l < pq$ and this number can be uniquely(!) written as
\[
kq+l = up+v \qquad \text{with} \qquad 0 \leq u < q,~0 \leq v < p \]
That is, there are unique operators $\frac{u}{q} \ast$ and $\frac{v}{p} \ast$ such that
\[
(\frac{k}{p} \ast) \circ (\frac{l}{q} \ast) = (\frac{u}{q} \ast) \circ (\frac{v}{p} \ast) \]
which determine the $2$-cells
\[
\xymatrix{ \bullet \ar@[blue]@{-}[rr]^{\frac{u}{q} \ast} \ar@[red]@{-}[dd]_{\frac{v}{p} \ast} &#038; &#038; \bullet \ar@[red]@{-}[dd]^{\frac{k}{p} \ast} \\ &#038; &#038; \\ \bullet \ar@[blue]@{-}[rr]_{\frac{l}{q} \ast} &#038; &#038; \bullet} \]
These give us the commutation relations between the free monoids of operators corresponding to different primes.

<br />For the primes $2$ and $3$, relevant in the description of the Moonshine Picture, the commutation relations are</p>
<p>\[<br />
(\frac{0}{2} \ast) \circ (\frac{0}{3} \ast) = (\frac{0}{3} \ast) \circ (\frac{0}{2} \ast), \quad<br />
(\frac{0}{2} \ast) \circ (\frac{1}{3} \ast) = (\frac{0}{3} \ast) \circ (\frac{1}{2} \ast),<br />
\quad<br />
(\frac{0}{2} \ast) \circ (\frac{2}{3} \ast) = (\frac{1}{3} \ast) \circ (\frac{0}{2} \ast) \]</p>
<p>\[<br />
(\frac{1}{2} \ast) \circ (\frac{0}{3} \ast) = (\frac{1}{3} \ast) \circ (\frac{1}{2} \ast), \quad<br />
(\frac{1}{2} \ast) \circ (\frac{1}{3} \ast) = (\frac{2}{3} \ast) \circ (\frac{0}{2} \ast),<br />
\quad<br />
(\frac{1}{2} \ast) \circ (\frac{2}{3} \ast) = (\frac{2}{3} \ast) \circ (\frac{1}{2} \ast) \]</p>
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