# Category: absolute

‘Gabriel’s topos’ (see here) is the conjectural, but still elusive topos from which the validity of the Riemann hypothesis would follow.

It is the latest attempt in Alain Connes’ 20 year long quest to tackle the RH (before, he tried the tools of noncommutative geometry and later those offered by the field with one element).

For the last 5 years he hopes that topos theory might provide the missing ingredient. Together with Katia Consani he introduced and studied the geometry of the Arithmetic site, and later the geometry of the scaling site.

If you look at the points of these toposes you get horribly complicated ‘non-commutative’ spaces, such as the finite adele classes $\mathbb{Q}^*_+ \backslash \mathbb{A}^f_{\mathbb{Q}} / \widehat{\mathbb{Z}}^{\ast}$ (in case of the arithmetic site) and the full adele classes $\mathbb{Q}^*_+ \backslash \mathbb{A}_{\mathbb{Q}} / \widehat{\mathbb{Z}}^{\ast}$ (for the scaling site).

In Vienna, Connes gave a nice introduction to the arithmetic site in two lectures. The first part of the talk below also gives an historic overview of his work on the RH

The second lecture can be watched here.

However, not everyone is as optimistic about the topos-approach as he seems to be. Here’s an insightful answer on MathOverflow by Will Sawin to the question “What is precisely still missing in Connes’ approach to RH?”.

Other interesting MathOverflow threads related to the RH-approach via the field with one element are Approaches to Riemann hypothesis using methods outside number theory and Riemann hypothesis via absolute geometry.

About a month ago, from May 10th till 14th Alain Connes gave a series of lectures at Ohio State University with title “The Riemann-Roch strategy, quantizing the Scaling Site”.

The accompanying paper has now been arXived: The Riemann-Roch strategy, Complex lift of the Scaling Site (joint with K. Consani).

Especially interesting is section 2 “The geometry behind the zeros of $\zeta$” in which they explain how looking at the zeros locus inevitably leads to the space of adele classes and why one has to study this space with the tools from noncommutative geometry.

Perhaps further developments will be disclosed in a few weeks time when Connes is one of the speakers at Toposes in Como.

The Fibonacci sequence reappears a bit later in Dan Brown’s book ‘The Da Vinci Code’ where it is used to login to the bank account of Jacques Sauniere at the fictitious Parisian branch of the Depository Bank of Zurich.

Last time we saw that the Hankel matrix of the Fibonacci series $F=(1,1,2,3,5,\dots)$ is invertible over $\mathbb{Z}$
$H(F) = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} \in SL_2(\mathbb{Z})$
and we can use the rule for the co-multiplication $\Delta$ on $\Re(\mathbb{Q})$, the algebra of rational linear recursive sequences, to determine $\Delta(F)$.

For a general integral linear recursive sequence the corresponding Hankel matrix is invertible over $\mathbb{Q}$, but rarely over $\mathbb{Z}$. So we need another approach to compute the co-multiplication on $\Re(\mathbb{Z})$.

Any integral sequence $a = (a_0,a_1,a_2,\dots)$ can be seen as defining a $\mathbb{Z}$-linear map $\lambda_a$ from the integral polynomial ring $\mathbb{Z}[x]$ to $\mathbb{Z}$ itself via the rule $\lambda_a(x^n) = a_n$.

If $a \in \Re(\mathbb{Z})$, then there is a monic polynomial with integral coefficients of a certain degree $n$

$f(x) = x^n + b_1 x^{n-1} + b_2 x^{n-2} + \dots + b_{n-1} x + b_n$

such that for every integer $m$ we have that

$a_{m+n} + b_1 a_{m+n-1} + b_2 a_{m+n-2} + \dots + b_{n-1} a_{m+1} + a_m = 0$

Alternatively, we can look at $a$ as defining a $\mathbb{Z}$-linear map $\lambda_a$ from the quotient ring $\mathbb{Z}[x]/(f(x))$ to $\mathbb{Z}$.

The multiplicative structure on $\mathbb{Z}[x]/(f(x))$ dualizes to a co-multiplication $\Delta_f$ on the set of all such linear maps $(\mathbb{Z}[x]/(f(x)))^{\ast}$ and we can compute $\Delta_f(a)$.

We see that the set of all integral linear recursive sequences can be identified with the direct limit
$\Re(\mathbb{Z}) = \underset{\underset{f|g}{\rightarrow}}{lim}~(\frac{\mathbb{Z}[x]}{(f(x))})^{\ast}$
(where the directed system is ordered via division of monic integral polynomials) and so is equipped with a co-multiplication $\Delta = \underset{\rightarrow}{lim}~\Delta_f$.

Btw. the ring structure on $\Re(\mathbb{Z}) \subset (\mathbb{Z}[x])^{\ast}$ comes from restricting to $\Re(\mathbb{Z})$ the dual structures of the co-ring structure on $\mathbb{Z}[x]$ given by
$\Delta(x) = x \otimes x \quad \text{and} \quad \epsilon(x) = 1$

From this description it is clear that you need to know a hell of a lot number theory to describe this co-multiplication explicitly.

As most of us prefer to work with rings rather than co-rings it is a good idea to begin to study this co-multiplication $\Delta$ by looking at the dual ring structure of
$\Re(\mathbb{Z})^{\ast} = \underset{\underset{ f | g}{\leftarrow}}{lim}~\frac{\mathbb{Z}[x]}{(f(x))}$
This is the completion of $\mathbb{Z}[x]$ at the multiplicative set of all monic integral polynomials.

This is a horrible ring and very little is known about it. Some general remarks were proved by Kazuo Habiro in his paper Cyclotomic completions of polynomial rings.

In fact, Habiro got interested is a certain subring of $\Re(\mathbb{Z})^{\ast}$ which we now know as the Habiro ring and which seems to be a red herring is all stuff about the field with one element, $\mathbb{F}_1$ (more on this another time). Habiro’s ring is

$\widehat{\mathbb{Z}[q]} = \underset{\underset{n|m}{\leftarrow}}{lim}~\frac{\mathbb{Z}[q]}{(q^n-1)}$

and its elements are all formal power series of the form
$a_0 + a_1 (q-1) + a_2 (q^2-1)(q-1) + \dots + a_n (q^n-1)(q^{n-1}-1) \dots (q-1) + \dots$
with all coefficients $a_n \in \mathbb{Z}$.

Here’s a funny property of such series. If you evaluate them at $q \in \mathbb{C}$ these series are likely to diverge almost everywhere, but they do converge in all roots of unity!

Some people say that these functions are ‘leaking out of the roots of unity’.

If the ring $\Re(\mathbb{Z})^{\ast}$ is controlled by the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, then Habiro’s ring is controlled by the abelianzation $Gal(\overline{\mathbb{Q}}/\mathbb{Q})^{ab} \simeq \hat{\mathbb{Z}}^{\ast}$.

In The Da Vinci Code, Dan Brown feels he need to bring in a French cryptologist, Sophie Neveu, to explain the mystery behind this series of numbers:

13 – 3 – 2 – 21 – 1 – 1 – 8 – 5

The Fibonacci sequence, 1-1-2-3-5-8-13-21-34-55-89-144-… is such that any number in it is the sum of the two previous numbers.

It is the most famous of all integral linear recursive sequences, that is, a sequence of integers

$a = (a_0,a_1,a_2,a_3,\dots)$

such that there is a monic polynomial with integral coefficients of a certain degree $n$

$f(x) = x^n + b_1 x^{n-1} + b_2 x^{n-2} + \dots + b_{n-1} x + b_n$

such that for every integer $m$ we have that

$a_{m+n} + b_1 a_{m+n-1} + b_2 a_{m+n-2} + \dots + b_{n-1} a_{m+1} + a_m = 0$

For the Fibonacci series $F=(F_0,F_1,F_2,\dots)$, this polynomial can be taken to be $x^2-x-1$ because
$F_{m+2} = F_{m+1}+F_m$

The set of all integral linear recursive sequences, let’s call it $\Re(\mathbb{Z})$, is a beautiful object of great complexity.

For starters, it is a ring. That is, we can add and multiply such sequences. If

$a=(a_0,a_1,a_2,\dots),~\quad \text{and}~\quad a’=(a’_0,a’_1,a’_2,\dots)~\quad \in \Re(\mathbb{Z})$

then the sequences

$a+a’ = (a_0+a’_0,a_1+a’_1,a_2+a’_2,\dots) \quad \text{and} \quad a \times a’ = (a_0.a’_0,a_1.a’_1,a_2.a’_2,\dots)$

are again linear recursive. The zero and unit in this ring are the constant sequences $0=(0,0,\dots)$ and $1=(1,1,\dots)$.

So far, nothing terribly difficult or exciting.

It follows that $\Re(\mathbb{Z})$ has a co-unit, that is, a ring morphism

$\epsilon~:~\Re(\mathbb{Z}) \rightarrow \mathbb{Z}$

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

$\Re(\overline{\mathbb{Q}}) \simeq (\overline{\mathbb{Q}}[ \overline{\mathbb{Q}}_{\times}^{\ast}] \otimes \overline{\mathbb{Q}}[d]) \oplus \sum_{i=0}^{\infty} \overline{\mathbb{Q}} S_i$

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.

Reference: Richard G. Larson, Earl J. Taft, ‘The algebraic structure of linearly recursive sequences under Hadamard product’

Once in every six months there’s a flurry of online excitement about Mochizuki’s alleged proof of the abc-conjecture.

It seems to be that time of the year again.

The twitter-account of the ever optimistic @math_jin is probably the best source for (positive) news about IUT/ABC. He now announces the latest version of Yamashita’s ‘summary’ of Mochizuki’s proof:

Another informed source is Ed Frenkel. He sometimes uses his twitter-account @edfrenkel to broadcast Ivan Fesenko‘s enthusiasm.

Googling further, I stumbled upon an older (newspaper) article on the subject: das grosse ABC by Marlene Weiss, for which she got silver at the 2017 science journalism awards.

In case you prefer an English translation: The big ABC.

Here’s her opening paragraph:

“In a children’s story written by the Swiss author Peter Bichsel, a lonely man decides to invent his own language. He calls the table “carpet”, the chair “alarm clock”, the bed “picture”. At first he is enthusiastic about his idea and always thinks of new words, his sentences sound original and funny. But after a while, he begins to forget the old words.”

The article is less optimistic than other recent popular accounts of Mochizuki’s story, including:

Monumental proof to torment mathematicians for years to come in Nature by Davide Castelvecchi.

Hope Rekindled for Perplexing Proof in Quanta-magazine by Kevin Hartnett.

Baffling ABC maths proof now has impenetrable 300-page ‘summary’ in the New Scientist by Timothy Revell.

Marlene Weiss fears a sad ending:

“Table is called “carpet”, chair is called “alarm clock”, bed is called “picture”. In the story by Peter Bichsel, the lonely man ends up having so much trouble communicating with other people that he speaks only to himself. It is a very sad story.”

Perhaps things will turn out for the better, and we’ll hear about it sometime.

In six months, I’d say…

In several of his talks on #IUTeich, Mochizuki argues that usual scheme theory over $\mathbb{Z}$ is not suited to tackle problems such as the ABC-conjecture.

The idea appears to be that ABC involves both the additive and multiplicative nature of integers, making rings into ‘2-dimensional objects’ (and clearly we use both ‘dimensions’ in the theory of schemes).

So, perhaps we should try to ‘dismantle’ scheme theory, and replace it with something like geometry over the field with one element $\mathbb{F}_1$.

The usual $\mathbb{F}_1$ mantra being: ‘forget all about the additive structure and only retain the multiplicative monoid’.

So perhaps there is yet another geometry out there, forgetting about the multiplicative structure, and retaining just the addition…

In the forgetting can’t be that hard, can it?-post we have seen that the forgetful functor

$F_{+,\times}~:~\mathbf{rings} \rightarrow \mathbf{sets}$

(that is, forgetting both multiplicative and additive information of the ring) is representable by the polynomial ring $\mathbb{Z}[x]$.

So, what about our ‘dismantling functors’ in which we selectively forget just one of these structures:

$F_+~:~\mathbf{rings} \rightarrow \mathbf{monoids} \quad \text{and} \quad F_{\times}~:~\mathbf{rings} \rightarrow \mathbf{abelian~groups}$

Are these functors representable too?

Clearly, ring maps from $\mathbb{Z}[x]$ to our ring $R$ give us again the elements of $R$. But now, we want to encode the way two of these elements add (or multiply).

This can be done by adding extra structure to the ring $\mathbb{Z}[x]$, namely a comultiplication $\Delta$ and a counit $\epsilon$

$\Delta~:~\mathbb{Z}[x] \rightarrow \mathbb{Z}[x] \otimes \mathbb{Z}[x] \quad \text{and} \quad \epsilon~:~\mathbb{Z}[x] \rightarrow \mathbb{Z}$

The idea of the comultiplication being that if we have two elements $r,s \in R$ with corresponding ring maps $f_r~:~\mathbb{Z}[x] \rightarrow R \quad x \mapsto r$ and $f_s~:~\mathbb{Z}[x] \rightarrow R \quad x \mapsto s$, composing their tensorproduct with the comultiplication

$f_v~:~\mathbb{Z}[x] \rightarrow^{\Delta} \mathbb{Z}[x] \otimes \mathbb{Z}[x] \rightarrow^{f_r \otimes f_s} R$

determines another element $v \in R$ which we can take either the product $v=r.s$ or sum $v=r+s$, depending on the comultiplication map $\Delta$.

The role of the counit is merely sending $x$ to the identity element of the operation.

Thus, if we want to represent the functor forgetting the addition, and retaining the multiplication we have to put on $\mathbb{Z}[x]$ the structure of a biring

$\Delta(x) = x \otimes x \quad \text{and} \quad \epsilon(x) = 1$

(making $x$ into a ‘group-like’ element for Hopf-ists).

The functor $F_{\times}$ forgetting the multiplication but retaining the addition is represented by the Hopf-ring $\mathbb{Z}[x]$, this time with

$\Delta(x) = x \otimes 1 + 1 \otimes x \quad \text{and} \quad \epsilon(x) = 0$

(that is, this time $x$ becomes a ‘primitive’ element).

Perhaps this adds another feather of weight to the proposal in which one defines algebras over the field with one element $\mathbb{F}_1$ to be birings over $\mathbb{Z}$, with the co-ring structure playing the role of descent data from $\mathbb{Z}$ to $\mathbb{F}_1$.

As, for example, in my note The coordinate biring of $\mathbf{Spec}(\mathbb{Z})/\mathbb{F}_1$.

Some weeks ago, Robert Kucharczyk and Peter Scholze found a topological realisation of the ‘hopeless’ part of the absolute Galois group $\mathbf{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. That is, they constructed a compact connected space $M_{cyc}$ such that etale covers of it correspond to Galois extensions of the cyclotomic field $\mathbb{Q}_{cyc}$. This gives, at least in theory, a handle on the hopeless part of the Galois group $\mathbf{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}_{cyc})$, see the previous post in this series.

Here, we will get halfway into constructing $M_{cyc}$. We will try to understand the topology of the prime ideal spectrum $\mathbf{Spec}(\mathbb{C}[\overline{\mathbb{Q}}^{\times}])$ of the complex group algebra of the multiplicative group $\overline{\mathbb{Q}}^{\times}$ of all non-zero algebraic numbers.

[section_title text=”Pontryagin duals”]

Take an Abelian locally compact group $A$ (for example, an Abelian group equipped with the discrete topology), then its Pontryagin dual $A^{\vee}$ is the space of all continuous group morphisms $A \rightarrow \mathbb{S}^1$ to the unit circle $\mathbb{S}^1$ endowed with the compact open topology.

There are these topological properties of the locally compact group $A^{\vee}$:

– $A^{\vee}$ is compact if and only if $A$ has the discrete topology,

– $A^{\vee}$ is connected if and only if $A$ is a torsion free group,

– $A^{\vee}$ is totally disconnected if and only if $A$ is a torsion group.

If we take the additive group of rational numbers with the discrete topology, the dual space $\mathbb{Q}^{\vee}$ is the one-dimensional solenoid

It is a compact and connected group, but is not path connected. In fact, it path connected components can be identified with the finite adele classes $\mathbb{A}_f/\mathbb{Q} = \widehat{\mathbb{Z}}/\mathbb{Z}$ where $\widehat{\mathbb{Z}}$ is the ring of profinite integers.

Keith Conrad has an excellent readable paper on this fascinating object: The character group of $\mathbb{Q}$. Or you might have a look at this post.

[section_title text=”The multiplicative group of algebraic numbers”]

A torsion element $x$ in the multiplicative group $\overline{\mathbb{Q}}^{\times}$ of all algebraic numbers must satisfy $x^N=1$ for some $N$ so is a root of unity, so we have the exact sequence of Abelian groups

$0 \rightarrow \pmb{\mu}_{\infty} \rightarrow \overline{\mathbb{Q}}^{\times} \rightarrow \overline{\mathbb{Q}}^{\times}_{tf} \rightarrow 0$

where the last term is the maximal torsion-free quotient of $\overline{\mathbb{Q}}^{\times}$. By Pontryagin duality this gives us an exact sequence of compact topological groups

$0 \rightarrow (\overline{\mathbb{Q}}^{\times}_{tf})^{\vee} \rightarrow (\overline{\mathbb{Q}}^{\times})^{\vee} \rightarrow \pmb{\mu}^{\vee}_{\infty} \rightarrow 0$

Here, the left-most space is connected and $\pmb{\mu}^{\vee}_{\infty}$ is totally disconnected. That is, the connected components of $(\overline{\mathbb{Q}}^{\times})^{\vee}$ are precisely the translates of the connected subgroup $(\overline{\mathbb{Q}}^{\times}_{tf})^{\vee}$.

[section_title text=”Prime ideal spectra”]

The short exact sequence of Abelian groups gives a short exact sequence of the corresponding group schemes

$0 \rightarrow \mathbf{Spec}(\mathbb{C}[\overline{\mathbb{Q}}^{\times}_{tf}]) \rightarrow \mathbf{Spec}(\mathbb{C}[\overline{\mathbb{Q}}^{\times}] \rightarrow \mathbf{Spec}(\mathbb{C}[\pmb{\mu}_{\infty}]) \rightarrow 0$

The torsion free abelian group $\overline{\mathbb{Q}}^{\times}_{tf}$ is the direct limit $\underset{\rightarrow}{lim}~M_i$ of finitely generated abelian groups $M_i$ and as the corresponding group algebra $\mathbb{C}[M_i] = \mathbb{C}[x_1,x_1^{-1},\cdots, x_k,x_k^{-1}]$, we have that $\mathbf{Spec}(\mathbb{C}[M_i])$ is connected. But then this also holds for

$\mathbf{Spec}(\mathbb{C}[\overline{\mathbb{Q}}^{\times}_{tf}]) = \underset{\leftarrow}{lim}~\mathbf{Spec}(\mathbb{C}[M_i])$

The underlying group of $\mathbb{C}$-points of $\mathbf{Spec}(\mathbb{C}[\pmb{\mu}_{\infty}])$ is $\pmb{\mu}_{\infty}^{\vee}$ and is therefore totally disconnected. But then we have

$\pi_0(\mathbf{Spec}(\mathbb{C}[\overline{\mathbb{Q}}^{\times}]) \simeq \pi_0(\mathbf{Spec}(\mathbb{C}[\pmb{\mu}_{\infty}]) \simeq \pmb{\mu}_{\infty}^{\vee}$

and, more importantly, for the etale fundamental group

$\pi_1^{et}(\mathbf{Spec}(\mathbb{C}[\overline{\mathbb{Q}}^{\times}],x) \simeq \pi_1^{et}(\mathbf{Spec}(\mathbb{C}[\overline{\mathbb{Q}}^{\times}_{tf}],y)$

So, we have to compute the latter one. Again, write the torsion-free quotient as a direct limit of finitely generated torsion-free Abelian groups and recall that connected etale covers of $\mathbf{Spec}(\mathbb{C}[M_i])=\mathbf{Spec}(\mathbb{C}[x_1,x_1^{-1},\cdots,x_k,x_k^{-1}])$ are all of the form $\mathbf{Spec}(\mathbb{C}[N])$, where $N$ is a subgroup of $M_i \otimes \mathbb{Q}$ that contains $M_i$ with finite index (that is, adjoining roots of the $x_i$).

Again, this goes through the limit and so a connected etale cover of $\mathbf{Spec}(\mathbb{C}[\overline{\mathbb{Q}}^{\times}_{tf}])$ would be determined by a subgroup of the $\mathbb{Q}$-vectorspace $\overline{\mathbb{Q}}^{\times}_{tf} \otimes \mathbb{Q}$ containing $\overline{\mathbb{Q}}^{\times}_{tf}$ with finite index.

But, $\overline{\mathbb{Q}}^{\times}_{tf}$ is already a $\mathbb{Q}$-vectorspace as we can take arbitrary roots in it (remember we’re using the multiplicative structure). That is, $\mathbf{Spec}(\mathbb{C}[\overline{\mathbb{Q}}^{\times}])$ is simply connected!

[section_title text=”Bringing in the Galois group”]

Now, we’re closing in on the mysterious space $M_{cyc}$. Clearly, it cannot be the complex points of $\mathbf{Spec}(\mathbb{C}[\overline{\mathbb{Q}}^{\times}])$ as this has no proper etale covers, but we still have to bring the Galois group $\mathbf{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}_{cyc})$ into the game.

The group algebra $\mathbb{C}[\overline{\mathbb{Q}}^{\times}]$ is a commutative and cocommutative Hopf algebra, and all the elements of the Galois group act on it as Hopf-automorphisms, so it is natural to consider the fixed Hopf algebra

$H_{cyc}=\mathbb{C}[\overline{\mathbb{Q}}^{\times}]^{\mathbf{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}_{cyc})}$

This Hopf algebra has an interesting alternative description as a subalgebra of the Witt ring $W(\mathbb{Q}_{cyc})$, bringing it into the realm of $\mathbb{F}_1$-geometry.

This ring of Witt vectors has as its underlying set of elements $1 + \mathbb{Q}_{cyc}[[t]]$ of formal power series in $\mathbb{Q}_{cyc}[[t]]$. Addition on this set is defined by multiplication of power series. The surprising fact is that we can then put a ring structure on it by demanding that the product $\odot$ should obey the rule that for all $a,b \in \mathbb{Q}_{cyc}$ we have

$(1-at) \odot (1-bt) = 1 – ab t$

In this mind-boggling ring the Hopf algebra $H_{cyc}$ is the subring consisting of all power series having a rational expression of the form

$\frac{1+a_1t+a_2t^2+ \cdots + a_n t^n}{1+b_1 t + b_2 t^2 + \cdots + b_m t^m}$

with all $a_i,b_j \in \mathbb{Q}_{cyc}$.

We can embed $\pmb{\mu}_{\infty}$ by sending a root of unity $\zeta$ to $1 – \zeta t$, and then the desired space $M_{cyc}$ will be close to

$\mathbf{Spec}(H_{cyc} \otimes_{\mathbb{Z}[\pmb{\mu}_{\infty}]} \mathbb{C})$

but I’ll spare the details for another time.

In case you want to know more about the title-picture, quoting from John Baez’ post The Beauty of Roots:

“Sam Derbyshire decided to to make a high resolution plot of some roots of polynomials. After some experimentation, he decided that his favorite were polynomials whose coefficients were all 1 or -1 (not 0). He made a high-resolution plot by computing all the roots of all polynomials of this sort having degree ≤ 24. That’s $2^{24}$ polynomials, and about $24 \times 2^{24}$ roots — or about 400 million roots! It took Mathematica 4 days to generate the coordinates of the roots, producing about 5 gigabytes of data.”

We know embarrassingly little about the symmetries of the roots of all polynomials with rational coefficients, or if you prefer, the absolute Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$.

In the title picture the roots of polynomials of degree $\leq 4$ with small coefficients are plotted and coloured by degree: blue=4, cyan=3, red=2, green=1. Sums and products of roots are again roots and by a symmetry we mean a map on all roots, sending sums to sums and products to products and leaving all the green dots (the rational numbers) fixed.

John Baez has an excellent post on the beauty of roots, including a picture of all polynomials of degree $\leq 5$ with integer coefficients between $-4$ and $4$ and, this time, colour-coded by: grey=2, cyan=3, red=4 and black=5.

In both pictures there’s a hint of the unit circle, black in the title picture and spanning the ‘white gaps’ in the picture above.

If we’d only consider the sub-picture of all (sums and products of) roots including the rational numbers on the horizontal axis and the roots of unity on the unit circle we’d get the cyclotomic field $\mathbb{Q}_{cyc} = \mathbb{Q}(\mu_{\infty})$. Here we know all symmetries: they are generated by taking powers of the roots of unity. That is, we know all about the Galois group $Gal(\mathbb{Q}_{cyc}/\mathbb{Q})$.

The ‘missing’ symmetries, that is the Galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q}_{cyc})$ remained a deep mystery, until last week…

[section_title text=”The oracle speaks”]

On september 15th, Robert Kucharczyk and Peter Scholze (aka the “oracle of arithmetic” according to Quanta-magazine) arXived their paper Topological realisations of absolute Galois groups.

They discovered a concrete compact connected Hausdorff space $M_{cyc}$ such that Galois extensions of $\mathbb{Q}_{cyc}$ correspond to connected etale covers of $M_{cyc}$.

Let’s look at a finite field $\mathbb{F}_p$. Here, Galois extensions of $\mathbb{F}_p$ (and there is just one such extension of degree $n$, upto isomorphism) correspond to connected etale covers of the circle $S^1$.

An etale map $X \rightarrow S^1$ is such that every circle point has exactly $n$ pre-images. Here again, up to homeomorphism, there is a unique such $n$-fold cover of $S^1$ (the picture on the left gives the cover for $n=2$).

.

If we replace $\mathbb{F}_p$ by the cyclotomic field $\mathbb{Q}_{cyc}$ then the compact space $M_{cyc}$ replaces the circle $S^1$. So, if we take a splitting polynomial of degree $n$ with coefficients in $\mathbb{Q}_{cyc}$, then there is a corresponding etale $n$-fold cover $X \rightarrow M_{cyc}$ such that for a specific point $p$ in $M_{cyc}$ its pre-images correspond to the roots of the polynomial. Nice!

Sadly, there’s a catch. Even though we have a concrete description of $M_{cyc}$ it turns out to be a horrible infinite dimensional space, it is connected but not path-connected, and so on.

Even Peter Scholze says it’s unclear whether new results can be proved from this result (see around 39.15 in his Next Generation Outreach Lecture).

Btw. if your German is ok, this talk is a rather good introduction to classical Galois theory and etale fundamental groups, including the primes=knots analogy.

[section_title text=”the imaginary field with one element”]

Of course there’s no mention of it in the Kucharczyk-Scholze paper, but this result is excellent news for those trying to develop a geometry over the imaginary field with one element $\mathbb{F}_1$ and hope to apply this theory to problems in number theory.

As a side remark, some of these people have just published a book with the EMS Publishing House: Absolute arithmetic and $\mathbb{F}_1$-geometry

The basic idea is that the collection of all prime numbers, $\mathbf{Spec}(\mathbb{Z})$ is far too large an object to be a terminal object (as it is in schemes). One should therefore extend the setting of schemes to so called $\mathbb{F}_1$-schemes, in which $\mathbf{Spec}(\mathbb{Z})$ is some higher dimensional object.

Initially, one hoped that $\mathbf{Spec}(\mathbb{Z})/\mathbb{F}_1$ might look like a curve, so that one could try to mimick Weil’s proof of the Riemann hypothesis for curves to prove the genuine Riemann hypothesis.

But, over the last decade it became clear that $\mathbf{Spec}(\mathbb{Z})/\mathbb{F}_1$ looks like an infinite dimensional space, a bit like the space $M_{cyc}$ above.

I’ll spare this to a couple of follow-up posts, but for now I’ll leave you with the punchline:

The compact connected Hausdorff space $M_{cyc}$ of Kucharczyk and Scholze is nothing but the space of complex points of $\mathbf{Spec}(\mathbb{Q}_{cyc})/\mathbb{F}_1$!

“Sometimes ideas, like men, jump up and say ‘hello’. They introduce themselves, these ideas, with words. Are they words? These ideas speak so strangely.”

“All that we see in this world is based on someone’s ideas. Some ideas are destructive, some are constructive. Some ideas can arrive in the form of a dream. I can say it again: some ideas arrive in the form of a dream.”

Here’s such an idea.

It all started when Norma wanted to compactify her twisted-prime-fruit pies. Norma’s pies are legendary in Twin Peaks, but if you never ate them at Double R Diner, here’s the concept.

Start with a long rectangular strip of pastry and decorate it vertically with ribbons of fruit, one fruit per prime, say cherry for 2, huckleberry for 3, and so on.

For elegance, I argued, the $p$-th ribbon should have width $log(p)$.

“That may very well look natural to you,” she said, “but our Geometer disagrees”. It seems that geometers don’t like logs.

Whatever. I won.

That’s Norma’s basic pie, or the $1$-pie as we call it. Next, she performs $n$ strange twists in one direction and $m$ magical operations in another, to get one of her twisted-pies. In this case we would order it as her $\frac{m}{n}$-pie.

Marketing-wise, these pies are problematic. They are infinite in length, so Norma can serve only a finite portion, limiting the number of fruits you can taste.

That’s why Norma wants to compactify her pies, so that you can hold the entire pastry in your hands, and taste the infinite richness of our local fruits.

“Wait!”, our Geometer warned, “You can never close them up with ordinary scheme-dough, the laws of scheme-pastry prohibit this!” He suggested to use a ribbon of marzipan, instead.

“Fine, then… Margaret, before you start complaining again, how much marzipan should I use?”, Norma asked.

“Well,” I replied, “ideally you’d want it to have zero width, but that wouldn’t close anything. So, I’d go for the next best thing, the log being zero. Take a marzipan-ribbon of width $1$.”

The Geometer took a $1$-pie, closed it with marzipan of width $1$, looked at the pastry from every possible angle, and nodded slowly.

“Yes, that’s a perfectly reasonable trivial bundle, or structure sheaf if you want. I’d sell it as $\mathcal{O}_{\overline{\mathbf{Spec}(\mathbb{Z})}}$ if I were you.”

“In your dreams!  I’ll simply call this  a $1$-pastry, and an $\frac{m}{n}$-pie closed with a $1$-ribbon of marzipan can be ordered from now on as an $\frac{m}{n}$-pastry.”

“I’m afraid this will not suffice,” our Geometer objected, ” you will have to allow pastries having an arbitrary marzipan-width.”

“Huh? You want me to compactify an $\frac{m}{n}$-pie  with marzipan of every imaginable width $r$ and produce a whole collection of … what … $(\frac{m}{n},r)$-pastries? What on earth for??”

“Well, take an $\frac{m}{n}$-pastry and try to unravel it.”

Oh, here we go again, I feared.

Whereas Norma’s pies all looked and tasted quite different to most of us, the Geometer claimed they were all the same, or ‘isomorphic’ as he pompously declared.

“Just reverse the operations Norma performed and you’ll end up with a $1$-pie”, he argued.

So Norma took an arbitrary $\frac{m}{n}$-pastry and did perform the reverse operations, which was a lot more difficult that with pies as now the marzipan-bit produced friction. The end-result was a $1$-pie held together with a marzipan-ribbon of width strictly larger or strictly smaller than $1$, but never gave back the $1$-pastry. Strange!

“Besides”, the Geometer added, “if you take two of your pastries, which I prefer to call $\mathcal{L}$ and $\mathcal{M}$, rather than use your numerical system, then their product $\mathcal{L} \otimes \mathcal{M}$ is again a pastry, though with variable marzipan-width.

In the promotional stage it might be nice to give the product for free to anyone ordering two pastries.”

“And how should I produce such a product-pastry?”

“Well, I’m too lazy to compute such things, it must follow trivially from elementary results in Picard-pastry. Surely, our log lady will work out the details in your notation. No doubt it will involve lots of logs…”

And so I did the calculations in my dreams, and I wrote down all formulas in the Double R Diner log-book, for Norma to consult whenever a customer ordered a product, or power of pastries.

A few years ago we had a Japanese tourist visiting Twin Peaks. He set up office in the Double R Diner, consulted my formulas, observed Norma’s pastry production and had endless conversations with our Geometer.

I’m told he categorified Norma’s pastry-bizness, probably to clone the concept to the Japanese market, replacing pastries by sushi-rolls.

When he left, he thanked me for working out the most trivial of examples, that of the Frobenioid of $\mathbb{Z}$…