“Convex polyhedra are among the oldest mathematical objects. Indeed the five platonic solids, which constitute the climax of Euclid’s books, were already known to the ancient people of Scotland some 4,000 years ago; see Figure 1.”

It sure would make a good story, the (ancient) Scotts outsmarting the Greek in discovering the five Platonic solids. Sadly, the truth is different.

Once again, hat tip to +David Roberts on Google+ for commenting on the AMS announcement and for linking to a post by John Baez and a couple of older posts here refuting this claim.

in which I tell the story of the original post and its aftermath. The bottom-line is this:

Summarizing : the Challifour photograph is not taken at the Ashmolean museum, but at the National Museum of Scotland in Edinburgh and consists of 5 of their artifacts (or 4 if ball 3 and 4 are identical) vaguely resembling cube, tetrahedron, dodecahedron (twice) and octahedron. The fifth Platonic solid, the icosahedron, remains elusive.

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 $n=2$ 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

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 $GL_2(\mathbb{A}_{\mathbb{Q}})/GL_2(\mathbb{Q})$.

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.

As far as I know there are no recent developments in the story of Grothendieck’s Lasserre writings.

Since may 2017 the Mormoiron part of the notes, saved by Jean Malgoire, are scanned and made available at the Archives Grothendieck.

Some of Grothendieck’s children were present at the opening ceremony, and an interview was made with Alexandre jr. :

Rather than going into Grothendieck’s mathematics, he speaks highly of his father’s role in the ecological (Survivre et vivre) and anti-nuclear movements of the early 70ties.

The full story of Survivre et Vivre, and Grothendieck’s part in it, can be read in the thesis by Celine Pessis:

Here’s her talk at the IHES: “L’engagement d’Alexandre Grothendieck durant la première moitié des années 1970”.

Returning to Montpellier’s Archives Grothendieck, Mateo Carmona G started a project to transcribe ‘La Longue Marche à travers la Théorie de Galois’ at GitHub.

From an email: “I am specially interested in Cote n° 149 that seems to contain Grothendieck separate notes on anabelian geometry.”

If you want to read up on the story of Grothendieck’s gribouillis, here are some older posts, in chronological order:

sending a sequence $a = (a_0,a_1,\dots)$ to its first entry $a_0$.

It’s a bit more difficult to see that $\Re(\mathbb{Z})$ also has a co-multiplication

\[
\Delta~:~\Re(\mathbb{Z}) \rightarrow \Re(\mathbb{Z}) \otimes_{\mathbb{Z}} \Re(\mathbb{Z}) \]
with properties dual to those of usual multiplication.

To describe this co-multiplication in general will have to await another post. For now, we will describe it on the easier ring $\Re(\mathbb{Q})$ of all rational linear recursive sequences.

For such a sequence $q = (q_0,q_1,q_2,\dots) \in \Re(\mathbb{Q})$ we consider its Hankel matrix. From the sequence $q$ we can form symmetric $k \times k$ matrices such that the opposite $i+1$-th diagonal consists of entries all equal to $q_i$
\[
H_k(q) = \begin{bmatrix} q_0 & q_1 & q_2 & \dots & q_{k-1} \\
q_1 & q_2 & & & q_k \\
q_2 & & & & q_{k+1} \\
\vdots & & & & \vdots \\
q_{k-1} & q_k & q_{k+1} & \dots & q_{2k-2} \end{bmatrix} \]
The Hankel matrix of $q$, $H(q)$ is $H_k(q)$ where $k$ is maximal such that $det~H_k(q) \not= 0$, that is, $H_k(q) \in GL_k(\mathbb{Q})$.

Let $S(q)=(s_{ij})$ be the inverse of $H(q)$, then the co-multiplication map
\[
\Delta~:~\Re(\mathbb{Q}) \rightarrow \Re(\mathbb{Q}) \otimes \Re(\mathbb{Q}) \]
sends the sequence $q = (q_0,q_1,\dots)$ to
\[
\Delta(q) = \sum_{i,j=0}^{k-1} s_{ij} (D^i q) \otimes (D^j q) \]
where $D$ is the shift operator on sequence
\[
D(a_0,a_1,a_2,\dots) = (a_1,a_2,\dots) \]

If $a \in \Re(\mathbb{Z})$ is such that $H(a) \in GL_k(\mathbb{Z})$ then the same formula gives $\Delta(a)$ in $\Re(\mathbb{Z})$.

For the Fibonacci sequences $F$ the Hankel matrix is
\[
H(F) = \begin{bmatrix} 1 & 1 \\ 1& 2 \end{bmatrix} \in GL_2(\mathbb{Z}) \quad \text{with inverse} \quad S(F) = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} \]
and therefore
\[
\Delta(F) = 2 F \otimes ~F – DF \otimes F – F \otimes DF + DF \otimes DF \]
There’s a lot of number theoretic and Galois-information encoded into the co-multiplication on $\Re(\mathbb{Q})$.

To see this we will describe the co-multiplication on $\Re(\overline{\mathbb{Q}})$ where $\overline{\mathbb{Q}}$ is the field of all algebraic numbers. One can show that

Here, $\overline{\mathbb{Q}}[ \overline{\mathbb{Q}}_{\times}^{\ast}]$ is the group-algebra of the multiplicative group of non-zero elements $x \in \overline{\mathbb{Q}}^{\ast}_{\times}$ and each $x$, which corresponds to the geometric sequence $x=(1,x,x^2,x^3,\dots)$, is a group-like element
\[
\Delta(x) = x \otimes x \quad \text{and} \quad \epsilon(x) = 1 \]

$\overline{\mathbb{Q}}[d]$ is the universal Lie algebra of the $1$-dimensional Lie algebra on the primitive element $d = (0,1,2,3,\dots)$, that is
\[
\Delta(d) = d \otimes 1 + 1 \otimes d \quad \text{and} \quad \epsilon(d) = 0 \]

Finally, the co-algebra maps on the elements $S_i$ are given by
\[
\Delta(S_i) = \sum_{j=0}^i S_j \otimes S_{i-j} \quad \text{and} \quad \epsilon(S_i) = \delta_{0i} \]

That is, the co-multiplication on $\Re(\overline{\mathbb{Q}})$ is completely known. To deduce from it the co-multiplication on $\Re(\mathbb{Q})$ we have to consider the invariants under the action of the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ as
\[
\Re(\overline{\mathbb{Q}})^{Gal(\overline{\mathbb{Q}}/\mathbb{Q})} \simeq \Re(\mathbb{Q}) \]

Unlike the Fibonacci sequence, not every integral linear recursive sequence has an Hankel matrix with determinant $\pm 1$, so to determine the co-multiplication on $\Re(\mathbb{Z})$ is even a lot harder, as we will see another time.