Tag: Connes

Writing & Blogging

Published February 27, 2008 by lievenlb

Terry Tao is reworking some of his better blogposts into a book, to be published by the AMS (here’s a preliminary version of the book “What’s New?”)

After some thought, I decided not to transcribe all of my posts from last year (there are 93 of them!), but instead to restrict attention to those articles which (a) have significant mathematical content, (b) are not announcements of material that will be published elsewhere, and (c) are not primarily based on a talk given by someone else. As it turns out, this still leaves about 33 articles from 2007, leading to a decent-sized book of a couple hundred pages in length.

If you have a blog and want to turn it into a LaTeX-book, there’s no need to transcribe or copy every single post, thanks to the WPTeX tool. Note that this is NOT a WP-plugin, but a (simple at that) php-program which turns all posts into a bookcontent.tex file. This file can then be edited further into a proper book.

Unfortunately, the present version chokes on LaTeXrender-code (which is easy enough to solve doing a global ‘find-and-replace’ of the tex-tags by dollar-signs) but worse, on Markdown-code… But then, someone fluent in php-regex will have no problems extending the libs/functions.php file (I hope…).

At the moment I’m considering turning the Mathieu-games-posts into a booklet. A possible title might be Mathieumatical Games. Rereading them (and other posts) I regret to be such an impatient blogger. Often I’m interested in something and start writing posts about it without knowing where or when I’ll land. This makes my posts a lot harder to get through than they might have been, if I would blog only after having digested the material myself… Typical recent examples are the tori-crypto-posts and the Bost-Connes algebra posts.

So, I still have a lot to learn from other bloggers I admire, such as Jennifer Ouellette who maintains the Coctail Party Physics blog. At the moment, Jennifer is resident blogger-journalist at the Kavli Institute where she is running a “Journal Club” workshop giving ideas on how to write better about science.

But the KITP is also committed to fostering scientific communication. That’s where I come in. Each Friday through April 26th, I’ll be presiding over a “Journal Club” meeting focusing on some aspect of communicating science.

Her most recent talk was entitled To Blog or Not to Blog? That is the Question and you can find the slides as well as a QuickTime movie of her talk. They even plan to set up a blog for the participants of the workshop. I will surely follow the rest of her course with keen interest!

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KMS, Gibbs & zeta function

Published February 21, 2008 by lievenlb

Time to wrap up this series on the Bost-Connes algebra. Here’s what we have learned so far : the convolution product on double cosets

$\begin{bmatrix} 1 & \mathbb{Z} \\ 0 & 1 \end{bmatrix} \backslash \begin{bmatrix} 1 & \mathbb{Q} \\ 0 & \mathbb{Q}_{> 0} \end{bmatrix} / \begin{bmatrix} 1 & \mathbb{Z} \\ 0 & 1 \end{bmatrix} $

is a noncommutative algebra, the Bost-Connes Hecke algebra $\mathcal{H} $, which is a bi-chrystalline graded algebra (somewhat weaker than ‘strongly graded’) with part of degree one the group-algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $. Further, $\mathcal{H} $ has a natural one-parameter family of algebra automorphisms $\sigma_t $ defined by $\sigma_t(X_n) = n^{it}X_n $ and $\sigma_t(Y_{\lambda})=Y_{\lambda} $.

For any algebra $A $ together with a one-parameter family of automorphisms $\sigma_t $ one is interested in KMS-states or Kubo-Martin-Schwinger states with parameter $\beta $, $KMS_{\beta} $ (this parameter is often called the ‘invers temperature’ of the system) as these are suitable equilibria states. Recall that a state is a special linear functional $\phi $ on $A $ (in particular it must have norm one) and it belongs to $KMS_{\beta} $ if the following commutation relation holds for all elements $a,b \in A $

$\phi(a \sigma_{i\beta}(b)) = \phi(b a) $

Let us work out the special case when $A $ is the matrix-algebra $M_n(\mathbb{C}) $. To begin, all algebra-automorphisms are inner in this case, so any one-parameter family of automorphisms is of the form

$\sigma_t(a) = e^{itH} a e^{-itH} $

where $e^{itH} $ is the matrix-exponential of the nxn matrix $H $. For any parameter $\beta $ we claim that the linear functional

$\phi(a) = \frac{1}{tr(e^{-\beta H})} tr(a e^{-\beta H}) $

is a KMS-state.Indeed, we have for all matrices $a,b \in M_n(\mathbb{C}) $ that

$\phi(a \sigma_{i \beta}(b)) = \frac{1}{tr(e^{-\beta H})} tr(a e^{- \beta H} b e^{\beta H} e^{- \beta H}) $

$= \frac{1}{tr(e^{-\beta H})} tr(a e^{-\beta H} b) = \frac{1}{tr(e^{-\beta H})} tr(ba e^{-\beta H}) = \phi(ba) $

(the next to last equality follows from cyclic-invariance of the trace map).
These states are usually called Gibbs states and the normalization factor $\frac{1}{tr(e^{-\beta H})} $ (needed because a state must have norm one) is called the partition function of the system. Gibbs states have arisen from the study of ideal gases and the place to read up on all of this are the first two chapters of the second volume of “Operator algebras and quantum statistical mechanics” by Ola Bratelli and Derek Robinson.

This gives us a method to construct KMS-states for an arbitrary algebra $A $ with one-parameter automorphisms $\sigma_t $ : take a simple n-dimensional representation $\pi~:~A \mapsto M_n(\mathbb{C}) $, find the matrix $H $ determining the image of the automorphisms $\pi(\sigma_t) $ and take the Gibbs states as defined before.

Let us return now to the Bost-Connes algebra $\mathcal{H} $. We don’t know any finite dimensional simple representations of $\mathcal{H} $ but, sure enough, have plenty of graded simple representations. By the usual strongly-graded-yoga they should correspond to simple finite dimensional representations of the part of degree one $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $ (all of them being one-dimensional and corresponding to characters of $\mathbb{Q}/\mathbb{Z} $).

Hence, for any $u \in \mathcal{G} = \prod_p \hat{\mathbb{Z}}_p^{\ast} $ (details) we have a graded simple $\mathcal{H} $-representation $S_u = \oplus_{n \in \mathbb{N}_+} \mathbb{C} e_n $ with action defined by

$\begin{cases} \pi_u(X_n)(e_m) = e_{nm} \\ \pi_u(Y_{\lambda})(e_m) = e^{2\pi i n u . \lambda} e_m \end{cases} $

Here, $u.\lambda $ is computed using the ‘chinese-remainder-identification’ $\mathcal{A}/\mathcal{R} = \mathbb{Q}/\mathbb{Z} $ (details).

Even when the representations $S_u $ are not finite dimensional, we can mimic the above strategy : we should find a linear operator $H $ determining the images of the automorphisms $\pi_u(\sigma_t) $. We claim that the operator is defined by $H(e_n) = log(n) e_n $ for all $n \in \mathbb{N}_+ $. That is, we claim that for elements $a \in \mathcal{H} $ we have

$\pi_u(\sigma_t(a)) = e^{itH} \pi_u(a) e^{-itH} $

So let us compute the action of both sides on $e_m $ when $a=X_n $. The left hand side gives $\pi_u(n^{it}X_n)(e_m) = n^{it} e_{mn} $ whereas the right-hand side becomes

$e^{itH}\pi_u(X_n) e^{-itH}(e_m) = e^{itH} \pi_u(X_n) m^{-it} e_m = $

$e^{itH} m^{-it} e_{mn} = (mn)^{it} m^{-it} e_{mn} = n^{it} e_{mn} $

proving the claim. For any parameter $\beta $ this then gives us a KMS-state for the Bost-Connes algebra by

$\phi_u(a) = \frac{1}{Tr(e^{-\beta H})} Tr(\pi_u(a) e^{-\beta H}) $

Finally, let us calculate the normalization factor (or partition function) $\frac{1}{Tr(e^{-\beta H})} $. Because $e^{-\beta H}(e_n) = n^{-\beta} e_n $ we have for that the trace

$Tr(e^{-\beta H}) = \sum_{n \in \mathbb{N}_+} \frac{1}{n^{\beta}} = \zeta(\beta) $

is equal to the Riemann zeta-value $\zeta(\beta) $ (at least when $\beta > 1 $).

Summarizing, we started with the definition of the Bost-Connes algebra $\mathcal{H} $, found a canonical one-parameter subgroup of algebra automorphisms $\sigma_t $ and computed that the natural equilibria states with respect to this ‘time evolution’ have as their partition function the Riemann zeta-function. Voila!

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“God given time”

Published February 20, 2008 by lievenlb

If you ever sat through a lecture by Alain Connes you will know about his insistence on the ‘canonical dynamic nature of noncommutative manifolds’. If you haven’t, he did write a blog post Heart bit 1 about it.

I’ll try to explain here that there is a definite “supplément d’âme” obtained in the transition from classical (commutative) spaces to the noncommutative ones. The main new feature is that “noncommutative spaces generate their own time” and moreover can undergo thermodynamical operations such as cooling, distillation etc…

Here a section from his paper A view of mathematics :

Indeed even at the coarsest level of understanding of a space provided by measure
theory, which in essence only cares about the “quantity of points” in a space, one
ﬁnds unexpected completely new features in the noncommutative case. While it
had been long known by operator algebraists that the theory of von-Neumann
algebras represents a far reaching extension of measure theory, the main surprise
which occurred at the beginning of the seventies is that such an algebra M
inherits from its noncommutativity a god-given time evolution:

$\delta~:~\mathbb{R} \rightarrow Out(M) $

where $Out M = Aut M/Int M $ is the quotient of the group of automorphisms of M
by the normal subgroup of inner automorphisms. This led in my thesis to the
reduction from type III to type II and their automorphisms and eventually to the
classiﬁcation of injective factors.

Even a commutative manifold has a kind of dynamics associated to it. Take a suitable vectorfield, consider the flow determined by it and there’s your ‘dynamics’, or a one-parameter group of automorphisms on the functions. Further, other classes of noncommutative algebras have similar features. For example, Cuntz and Quillen showed that also formally smooth algebras (the noncommutative manifolds in the algebraic world) have natural Yang-Mills flows associated to them, giving a one-parameter subgroup of automorphisms.

Let us try to keep far from mysticism and let us agree that by ‘time’ (let alone ‘god given time’) we mean a one-parameter subgroup of algebra automorphisms of the noncommutative algebra. In nice cases, such as some von-Neumann algebras this canonical subgroup is canonical in the sense that it is unique upto inner automorphisms.

In the special case of the Bost-Connes algebra these automorphisms $\sigma_t $ are given by $\sigma_t(X_n) = n^{it} X_n $ and $\sigma_t(Y_{\lambda}) = Y_{\lambda} $.

This one-parameter subgroup is crucial in the definition of the so called KMS-states (for Kubo-Martin and Schwinger) which is our next goal.

2 Comments

i’ll take rerun requests

Published February 8, 2008 by lievenlb

If you write a comment-provoking post (such as that one), you’d better deal with the reactions.

As often, the bluntest comment came from the Granada-Antwerp commuter (aka “mewt” for ‘memories of a weird traveler’…)

Javier :

Concerning the participation on the math-related posts, it is true that what you write has become more readable along the years, but yet, being able to read one of your math posts and catch the idea of what is going on (which I think is a great thing to do) is one thing. Actually understanding the details is a completely different one. And possibly most people thinks that commenting around when you only got the general idea (if any) of some math topic would be rather bold.

Personally, with your 2 last posts concerning Connes-Bost systems I am interested on understanding the story in full detail, so I printed your first post, took it home, read it carefully, made all the computations on my own (not that I dont trust yours, but you never know!) and before I had finished getting a sound impression of what was going on, the second part was already online, so had to go through the same process (in top of usual duties) just to keep your rythm. If things go as usual, by the time I am ready to make any sensible comments, you’ll be already bored of the topic and have switched to something else, so it won’t make much sense commenting at all!If its comments what you’re lasting for, write short, one-idea posts, rather than long, technically detailed ones.

The good news is that my posts become slightly more understandable. But all things can be improved… so, here’s a request :

If there is this one post you’d love to understand if only you knew already the material I subconsciously assumed, tell me or leave a comment!

and I’ll try to improve on it…

abc on adelic Bost-Connes

Published February 2, 2008 by lievenlb

The adelic interpretation of the Bost-Connes Hecke algebra $\mathcal{H} $ is based on three facts we’ve learned so far :

The diagonal embedding of the rational numbers $\delta~:~\mathbb{Q} \rightarrow \prod_p \mathbb{Q}_p $ has its image in the adele ring $\mathcal{A} $. ( details )
There is an exact sequence of semigroups $1 \rightarrow \mathcal{G} \rightarrow \mathcal{I} \cap \mathcal{R} \rightarrow \mathbb{N}^+_{\times} \rightarrow 1 $ where $\mathcal{I} $ is the idele group, that is the units of $\mathcal{A} $, where $\mathcal{R} = \prod_p \mathbb{Z}_p $ and where $\mathcal{G} $ is the group (!) $\prod_p \mathbb{Z}_p^* $. ( details )
There is an isomorphism of additive groups $\mathbb{Q}/\mathbb{Z} \simeq \mathcal{A}/\mathcal{R} $. ( details )

Because $\mathcal{R} $ is a ring we have that $a\mathcal{R} \subset \mathcal{R} $ for any $a=(a_p)_p \in \mathcal{I} \cap \mathcal{R} $. Therefore, we have an induced ‘multiplication by $a $’ morphism on the additive group $\mathcal{A}/\mathcal{R} \rightarrow^{a.} \mathcal{A}/\mathcal{R} $ which is an epimorphism for all $a \in \mathcal{I} \cap \mathcal{R} $.

In fact, it is easy to see that the equation $a.x = y $ for $y \in \mathcal{A}/\mathcal{R} $ has precisely $n_a = \prod_p p^{d(a)} $ solutions. In particular, for any $a \in \mathcal{G} = \prod_p \mathbb{Z}_p^* $, multiplication by $a $ is an isomorphism on $\mathcal{A}/\mathcal{R} = \mathbb{Q}/\mathbb{Z} $.

But then, we can form the crystalline semigroup graded skew-group algebra $\mathbb{Q}(\mathbb{Q}/\mathbb{Z}) \bowtie (\mathcal{I} \cap \mathcal{R}) $. It is the graded vectorspace $\oplus_{a \in \mathcal{I} \cap \mathcal{R}} X_a \mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $ with commutation relation
$Y_{\lambda}X_a = X_a Y_{a \lambda} $ for the base-vectors $Y_{\lambda} $ with $\lambda \in \mathbb{Q}/\mathbb{Z} $. Recall from last time we need to use approximation (or the Chinese remainder theorem) to determine the class of $a \lambda $ in $\mathbb{Q}/\mathbb{Z} $.

We can also extend it to a bi-crystalline graded algebra because multiplication by $a \in \mathcal{I} \cap \mathcal{R} $ has a left-inverse which determines the commutation relations $Y_{\lambda} X_a^* = X_a^* (\frac{1}{n_a})(\sum_{a.\mu = \lambda} Y_{\mu}) $. Let us call this bi-crystalline graded algebra $\mathcal{H}_{big} $, then we have the following facts

For every $a \in \mathcal{G} $, the element $X_a $ is a unit in $\mathcal{H}_{big} $ and $X_a^{-1}=X_a^* $. Conjugation by $X_a $ induces on the subalgebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $ the map $Y_{\lambda} \rightarrow Y_{a \lambda} $.
Using the diagonal embedding $\delta $ restricted to $\mathbb{N}^+_{\times} $ we get an embedding of algebras $\mathcal{H} \subset \mathcal{H}_{big} $ and conjugation by $X_a $ for any $a \in \mathcal{G} $ sends $\mathcal{H} $ to itself. However, as the $X_a \notin \mathcal{H} $, the induced automorphisms are now outer!

Summarizing : the Bost-Connes Hecke algebra $\mathcal{H} $ encodes a lot of number-theoretic information :

the additive structure is encoded in the sub-algebra which is the group-algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $
the multiplicative structure in encoded in the epimorphisms given by multiplication with a positive natural number (the commutation relation with the $X_m $
the automorphism group of $\mathbb{Q}/\mathbb{Z} $ extends to outer automorphisms of $\mathcal{H} $

That is, the Bost-Connes algebra can be seen as a giant mashup of number-theory of $\mathbb{Q} $. So, if one can prove something specific about this algebra, it is bound to have interesting number-theoretic consequences.

But how will we study $\mathcal{H} $? Well, the bi-crystalline structure of it tells us that $\mathcal{H} $ is a ‘good’-graded algebra with part of degree one the group-algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $. This group-algebra is a formally smooth algebra and we study such algebras by studying their finite dimensional representations.

Hence, we should study ‘good’-graded formally smooth algebras (such as $\mathcal{H} $) by looking at their graded representations. This will then lead us to Connes’ “fabulous states”…

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Chinese remainders and adele classes

Published January 31, 2008 by lievenlb

Oystein Ore mentions the following puzzle from Brahma-Sphuta-Siddhanta (Brahma’s Correct System) by Brahmagupta :

An old woman goes to market and a horse steps on her basket and crashes the eggs. The rider offers to pay for the damages and asks her how many eggs she had brought. She does not remember the exact number, but when she had taken them out two at a time, there was one egg left. The same happened when she picked them out three, four, five, and six at a time, but when she took them seven at a time they came out even. What is the smallest number of eggs she could have had?

Here’s a similar problem from “Advanced Number Theory” by Harvey Cohn (( always, i wonder how one might ‘discreetly request’ these remainders… )) :

Exercise 5 : In a game for guessing a person’s age x, one discreetly requests three remainders : r1 when x is divided by 3, r2 when x is divided by 4, and r3 when x is divided by 5. Then x=40 r1 + 45 r2 + 36 r3 modulo 60.

Clearly, these problems are all examples of the Chinese Remainder Theorem.

Chinese because one of the first such problems was posed by Sunzi [Sun Tsu] (4th century AD)
in the book Sunzi Suanjing. (( according to ChinaPage the answer is contained in the song on the left hand side. ))

There are certain things whose number is unknown.
Repeatedly divided by 3, the remainder is 2;
by 5 the remainder is 3;
and by 7 the remainder is 2.
What will be the number?

The Chinese Remainder Theorem asserts that when $N=n_1n_2 \ldots n_k $ with the $n_i $ pairwise coprime, then there is an isomorphism of abelian groups $\mathbb{Z}/N \mathbb{Z} \simeq \mathbb{Z}/n_1 \mathbb{Z} \times \mathbb{Z}/n_2 \mathbb{Z} \times \ldots \times \mathbb{Z}/n_k \mathbb{Z} $. Equivalently, given coprime numbers $n_i $ one cal always solve the system of congruence identities

$\begin{cases} x \equiv a_1~(\text{mod}~n_1) \\ x \equiv a_2~(\text{mod}~n_2) \\ \vdots \\ x \equiv a_k~(\text{mod}~n_k) \end{cases} $

and all integer solutions are congruent to each other modulo $N=n_1 n_2 \ldots n_k $.

We will need this classical result to prove that
$\mathbb{Q}/\mathbb{Z} \simeq \mathcal{A}/\mathcal{R} $
where (as last time) $\mathcal{A} $ is the additive group of all adeles and where $\mathcal{R} $ is the subgroup $\prod_p \mathbb{Z}_p $ (i’ll drop all ‘hats’ from now on, so the p-adic numbers are $\mathbb{Q}_p = \hat{\mathbb{Q}}_p $ and the p-adic integers are denoted $\mathbb{Z}_p = \hat{\mathbb{Z}}_p $).

As we will have to do calculations with p-adic numbers, it is best to have them in a canonical form using digits. A system of digits $\mathbf{D} $ of $\mathbb{Q}_p $ consists of zero and a system of representatives of units of $\mathbb{Z}_p^* $ modulo $p \mathbb{Z}_p $. The most obvious choice of digits is $\mathbf{D} = { 0,1,2,\ldots,p-1 } $ which we will use today. (( later we will use another system of digits, the Teichmuller digits using $p-1 $-th root of unities in $\mathbb{Q}_p $. )) Fixing a set of digits $\mathbf{D} $, any p-adic number $a_p \in \mathbb{Q}_p $ can be expressed uniquely in the form

$a_p = \sum_{n=deg(a_p)}^{\infty} a_p(n) p^n $ with all ‘coefficients’ $a_p(n) \in \mathbf{D} $ and $deg(a_p) $ being the lowest p-power occurring in the description of $a_p $.

Recall that an adele is an element $a = (a_2,a_3,a_5,\ldots ) \in \prod_p \mathbb{Q}_p $ such that for almost all prime numbers p $a_p \in \mathbb{Z}_p $ (that is $deg(a_p) \geq 0 $). Denote the finite set of primes p such that $deg(a_p) < 0 $ with $\mathbf{P} = { p_1,\ldots,p_k } $ and let $d_i = -deg(a_{p_i}) $. Then, with $N=p_1^{d_1}p_2^{d_2} \ldots p_k^{d_k} $ we have that $N a_{p_i} \in \mathbb{Z}_{p_i} $. Observe that for all other prime numbers $q \notin \mathbf{P} $ we have $~(N,q)=1 $ and therefore $N $ is invertible in $\mathbb{Z}_q $.

Also $N = p_i^{d_i} K_i $ with $K_i \in \mathbb{Z}_{p_i}^* $. With respect to the system of digits $\mathbf{D} = { 0,1,\ldots,p-1 } $ we have

$N a_{p_i} = \underbrace{K_i \sum_{j=0}^{d_i-1} a_{p_i}(-d_i+j) p_i^j}_{= \alpha_i} + K_i \sum_{j \geq d_i} a_{p_i}(-d_i+j)p_i^j \in \mathbb{Z}_{p_i} $

Note that $\alpha_i \in \mathbb{Z} $ and the Chinese Remainder Theorem asserts the existence of an integral solution $M \in \mathbb{Z} $ to the system of congruences

$\begin{cases} M \equiv \alpha_1~\text{modulo}~p_1^{d_1} \\
M \equiv \alpha_2~\text{modulo}~p_2^{d_2} \\
\vdots \\ M \equiv \alpha_k~\text{modulo}~p_k^{d_k} \end{cases} $

But then, for all $1 \leq i \leq k $ we have $N a_{p_i} – M = p_i^{d_i} \sum_{j=0}^{\infty} b_i(j) p^j $ (with the $b_i(j) \in \mathbf{D} $) and therefore

$a_{p_i} – \frac{M}{N} = \frac{1}{K_i} \sum_{j=0}^{\infty} b_i(j) p^j \in \mathbb{Z}_{p_i} $

But for all other primes $q \notin \mathbf{P} $ we have that $\alpha_q \in \mathbb{Z}_q $ and that $N \in \mathbb{Z}_q^* $ whence for those primes we also have that $\alpha_q – \frac{M}{N} \in \mathbb{Z}_q $.

Finally, observe that the diagonal embedding of $\mathbb{Q} $ in $\prod_p \mathbb{Q}_p $ lies entirely in the adele ring $\mathcal{A} $ as a rational number has only finitely many primes appearing in its denominator. Hence, identifying $\mathbb{Q} \subset \mathcal{A} $ via the diagonal embedding we can rephrase the above as

$a – \frac{M}{N} \in \mathcal{R} = \prod_p \mathbb{Z}_p $

That is, any adele class $\mathcal{A}/\mathcal{R} $ has as a representant a rational number. But then, $\mathcal{A}/\mathcal{R} \simeq \mathbb{Q}/\mathbb{Z} $ which will allow us to give an adelic version of the Bost-Connes algebra!

Btw. there were 301 eggs.

One Comment

adeles and ideles

Published January 29, 2008 by lievenlb

Before we can even attempt to describe the adelic description of the Bost-Connes Hecke algebra and its symmetries, we’d probably better recall the construction and properties of adeles and ideles. Let’s start with the p-adic numbers $\hat{\mathbb{Z}}_p $ and its field of fractions $\hat{\mathbb{Q}}_p $. For p a prime number we can look at the finite rings $\mathbb{Z}/p^n \mathbb{Z} $ of all integer classes modulo $p^n $. If two numbers define the same element in $\mathbb{Z}/p^n\mathbb{Z} $ (meaning that their difference is a multiple of $p^n $), then they certainly define the same class in any $\mathbb{Z}/p^k \mathbb{Z} $ when $k \leq n $, so we have a sequence of ringmorphisms between finite rings

$ \ldots \rightarrow^{\phi_{n+1}} \mathbb{Z}/p^n \mathbb{Z} \rightarrow^{\phi_n} \mathbb{Z}/p^{n-1}\mathbb{Z} \rightarrow^{\phi_{n-1}} \ldots \rightarrow^{\phi_3} \mathbb{Z}/p^2\mathbb{Z} \rightarrow^{\phi_2} \mathbb{Z}/p\mathbb{Z} $

The ring of p-adic integers $\hat{\mathbb{Z}}_p $ can now be defined as the collection of all (infinite) sequences of elements $~(\ldots,x_n,x_{n-1},\ldots,x_2,x_1) $ with $x_i \in \mathbb{Z}/p^i\mathbb{Z} $ such that
$\phi_i(x_i) = x_{i-1} $ for all natural numbers $i $. Addition and multiplication are defined componentswise and as all the maps $\phi_i $ are ringmorphisms, this produces no compatibility problems.

One can put a topology on $\hat{\mathbb{Z}}_p $ making it into a compact ring. Here’s the trick : all components $\mathbb{Z}/p^n \mathbb{Z} $ are finite so they are compact if we equip these sets with the discrete topology (all subsets are opens). But then, Tychonov’s product theorem asserts that the product-space $\prod_n \mathbb{Z}/n \mathbb{Z} $ with the product topology is again a compact topological space. As $\hat{\mathbb{Z}}_p $ is a closed subset, it is compact too.

By construction, the ring $\hat{\mathbb{Z}}_p $ is a domain and hence has a field of fraction which we will denote by $\hat{\mathbb{Q}}_p $. These rings give the p-local information of the rational numbers $\mathbb{Q} $. We will now ‘glue together’ these local data over all possible prime numbers $p $ into adeles. So, forget the above infinite product used to define the p-adics, below we will work with another infinite product, one factor for each prime number.

The adeles $\mathcal{A} $ are the restricted product of the $\hat{\mathbb{Q}}_p $ over $\hat{\mathbb{Z}}_p $ for all prime numbers p. By ‘restricted’ we mean that elements of $\mathcal{A} $ are exactly those infinite vectors $a=(a_2,a_3,a_5,a_7,a_{11},\ldots ) = (a_p)_p \in \prod_p \hat{\mathbb{Q}}_p $ such that all but finitely of the components $a_p \in \hat{\mathbb{Z}}_p $. Addition and multiplication are defined component-wise and the restriction condition is compatible with both adition and multiplication. So, $\mathcal{A} $ is the adele ring. Note that most people call this $\mathcal{A} $ the finite Adeles as we didn’t consider infinite places, i will distinguish between the two notions by writing adeles resp. Adeles for the finite resp. the full blown version. The adele ring $\mathcal{A} $ has as a subring the infinite product $\mathcal{R} = \prod_p \hat{\mathbb{Z}}_p $. If you think of $\mathcal{A} $ as a version of $\mathbb{Q} $ then $\mathcal{R} $ corresponds to $\mathbb{Z} $ (and next time we will see that there is a lot more to this analogy).

The ideles are the group of invertible elements of the ring $\mathcal{A} $, that is, $\mathcal{I} = \mathcal{A}^{\ast} $. That s, an element is an infinite vector $i = (i_2,i_3,i_5,\ldots) = (i_p)_p $ with all $i_p \in \hat{\mathbb{Q}}_p^* $ and for all but finitely many primes we have that $i_p \in \hat{\mathbb{Z}}_p^* $.

As we will have to do explicit calculations with ideles and adeles we need to recall some facts about the structure of the unit groups $\hat{\mathbb{Z}}_p^* $ and $\hat{\mathbb{Q}}_p^* $. If we denote $U = \hat{\mathbb{Z}}_p^* $, then projecting it to the unit group of each of its components we get for each natural number n an exact sequence of groups

$1 \rightarrow U_n \rightarrow U \rightarrow (\mathbb{Z}/p^n \mathbb{Z})^* \rightarrow 1 $. In particular, we have that $U/U_1 \simeq (\mathbb{Z}/p\mathbb{Z})^* \simeq \mathbb{Z}/(p-1)\mathbb{Z} $ as the group of units of the finite field $\mathbb{F}_p $ is cyclic of order p-1. But then, the induced exact sequence of finite abalian groups below splits

$1 \rightarrow U_1/U_n \rightarrow U/U_n \rightarrow \mathbb{F}_p^* \rightarrow 1 $ and as the unit group $U = \underset{\leftarrow}{lim} U/U_n $ we deduce that $U = U_1 \times V $ where $\mathbb{F}_p^* \simeq V = { x \in U | x^{p-1}=1 } $ is the specified unique subgroup of $U $ of order p-1. All that remains is to determine the structure of $U_1 $. If $p \not= 2 $, take $\alpha = 1 + p \in U_1 – U_2 $ and let $\alpha_n \in U_1/U_n $ denote the image of $\alpha $, then one verifies that $\alpha_n $ is a cyclic generator of order $p^{n-1} $ of $U_1/U_n $.

But then, if we denote the isomorphism $\theta_n~:~\mathbb{Z}/p^{n-1} \mathbb{Z} \rightarrow U_1/U_n $ between the ADDITIVE group $\mathbb{Z}/p^{n-1} \mathbb{Z} $ and the MULTIPLICATIVE group $U_1/U_n $ by the map $z \mapsto \alpha_n^z $, then we have a compatible commutative diagram

[tex]\xymatrix{\mathbb{Z}/p^n \mathbb{Z} \ar[r]^{\theta_{n+1}} \ar[d] & U_1/U_{n+1} \ar[d] \\
\mathbb{Z}/p^{n-1} \mathbb{Z} \ar[r]^{\theta_n} & U_1/U_n}[/tex]

and as $U_1 = \underset{\leftarrow}{lim}~U_1/U_n $ this gives an isomorphism between the multiplicative group $U_1 $ and the additive group of $\hat{\mathbb{Z}}_p $. In case $p=2 $ we have to start with an element $\alpha \in U_2 – U_3 $ and repeat the above trick. Summarizing we have the following structural information about the unit group of p-adic integers

$\hat{\mathbb{Z}}_p^* \simeq \begin{cases} \hat{\mathbb{Z}}_{p,+} \times \mathbb{Z}/(p-1)\mathbb{Z}~(p \not= 2) \\ \hat{\mathbb{Z}}_{2,+} \times \mathbb{Z}/2 \mathbb{Z}~(p=2) \end{cases}$

Because every unit in $\hat{\mathbb{Q}}_p^* $ can be written as $p^n u $ with $u \in \hat{\mathbb{Z}}_p^* $ we deduce from this also the structure of the unit group of the p-adic field

$\hat{\mathbb{Q}}_p^* \simeq \begin{cases} \mathbb{Z} \times \hat{\mathbb{Z}}_{p,+} \times \mathbb{Z}/(p-1)\mathbb{Z}~(p \not= 2) \\ \mathbb{Z} \times \hat{\mathbb{Z}}_{2,+} \times \mathbb{Z}/2 \mathbb{Z}~(p=2) \end{cases} $

Right, now let us start to make the connection with the apparently abstract ringtheoretical post from last time where we introduced semigroup crystalline graded rings without explaining why we wanted that level of generality.

Consider the semigroup $\mathcal{I} \cap \mathcal{R} $, that is all ideles $i = (i_p)_p $ with all $i_p = p^{n_p} u_p $ with $u_p \in \hat{\mathbb{Z}}_p^* $ and $n_p \in \mathbb{N} $ with $n_p=0 $ for all but finitely many primes p. Then, we have an exact sequence of semigroups

$1 \rightarrow \mathcal{G} \rightarrow \mathcal{I} \cap \mathcal{R} \rightarrow^{\pi} \mathbb{N}^+_{\times} \rightarrow 1 $ where the map is defined (with above notation) $\pi(i) = \prod_p p^{n_p} $ and exactness follows from the above structural results when we take $\mathcal{G} = \prod_p \hat{\mathbb{Z}}_p^* $.

This gives a glimpse of where we are heading. Last time we identified the Bost-Connes Hecke algebra $\mathcal{H} $ as a bi-crystalline group graded algebra determined by a $\mathbb{N}^+_{\times} $-semigroup crystalline graded algebra over the group algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $. Next, we will entend this construction starting from a $\mathcal{I} \cap \mathcal{R} $-semigroup crystalline graded algebra over the same group algebra. The upshot is that we will have a natural action by automorphisms of the group $\mathcal{G} $ on the Bost-Connes algebra. And… the group $\mathcal{G} = \prod_p \hat{\mathbb{Z}}_p^* $ is the Galois group of the cyclotomic field extension $\mathbb{Q}^{cyc} $!

But, in order to begin to understand this, we will need to brush up our rusty knowledge of algebraic number theory…

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BC stands for Bi-Crystalline graded

Published January 26, 2008 by lievenlb

Towards the end of the Bost-Connes for ringtheorists post I freaked-out because I realized that the commutation morphisms with the $X_n^* $ were given by non-unital algebra maps. I failed to notice the obvious, that algebras such as $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $ have plenty of idempotents and that this mysterious ‘non-unital’ morphism was nothing else but multiplication with an idempotent…

Here a sketch of a ringtheoretic framework in which the Bost-Connes Hecke algebra $\mathcal{H} $ is a motivating example (the details should be worked out by an eager 20-something). Start with a suitable semi-group $S $, by which I mean that one must be able to invert the elements of $S $ and obtain a group $G $ of which all elements have a canonical form $g=s_1s_2^{-1} $. Probably semi-groupies have a name for these things, so if you know please drop a comment.

The next ingredient is a suitable ring $R $. Here, suitable means that we have a semi-group morphism
$\phi~:~S \rightarrow End(R) $ where $End(R) $ is the semi-group of all ring-endomorphisms of $R $ satisfying the following two (usually strong) conditions :

Every $\phi(s) $ has a right-inverse, meaning that there is an ring-endomorphism $\psi(s) $ such that $\phi(s) \circ \psi(s) = id_R $ (this implies that all $\phi(s) $ are in fact epi-morphisms (surjective)), and
The composition $\psi(s) \circ \phi(s) $ usually is NOT the identity morphism $id_R $ (because it is zero on the kernel of the epimorphism $\phi(s) $) but we require that there is an idempotent $E_s \in R $ (that is, $E_s^2 = E_s $) such that $\psi(s) \circ \phi(s) = id_R E_s $

The point of the first condition is that the $S $-semi-group graded ring $A = \oplus_{s \in S} X_s R $ is crystalline graded (crystalline group graded rings were introduced by Fred Van Oystaeyen and Erna Nauwelaarts) meaning that for every $s \in S $ we have in the ring $A $ the equality $X_s R = R X_s $ where this is a free right $R $-module of rank one. One verifies that this is equivalent to the existence of an epimorphism $\phi(s) $ such that for all $r \in R $ we have $r X_s = X_s \phi(s)(r) $.

The point of the second condition is that this semi-graded ring $A$ can be naturally embedded in a $G $-graded ring $B = \oplus_{g=s_1s_2^{-1} \in G} X_{s_1} R X_{s_2}^* $ which is bi-crystalline graded meaning that for all $r \in R $ we have that $r X_s^*= X_s^* \psi(s)(r) E_s $.

It is clear from the construction that under the given conditions (and probably some minor extra ones making everything stand) the group graded ring $B $ is determined fully by the semi-group graded ring $A $.

what does this general ringtheoretic mumbo-jumbo have to do with the BC- (or Bost-Connes) algebra $\mathcal{H} $?

In this particular case, the semi-group $S $ is the multiplicative semi-group of positive integers $\mathbb{N}^+_{\times} $ and the corresponding group $G $ is the multiplicative group $\mathbb{Q}^+_{\times} $ of all positive rational numbers.

The ring $R $ is the rational group-ring $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $ of the torsion-group $\mathbb{Q}/\mathbb{Z} $. Recall that the elements of $\mathbb{Q}/\mathbb{Z} $ are the rational numbers $0 \leq \lambda < 1 $ and the group-law is ordinary addition and forgetting the integral part (so merely focussing on the ‘after the comma’ part). The group-ring is then

$\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] = \oplus_{0 \leq \lambda < 1} \mathbb{Q} Y_{\lambda} $ with multiplication linearly induced by the multiplication on the base-elements $Y_{\lambda}.Y_{\mu} = Y_{\lambda+\mu} $.

The epimorphism determined by the semi-group map $\phi~:~\mathbb{N}^+_{\times} \rightarrow End(\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]) $ are given by the algebra maps defined by linearly extending the map on the base elements $\phi(n)(Y_{\lambda}) = Y_{n \lambda} $ (observe that this is indeed an epimorphism as every base element $Y_{\lambda} = \phi(n)(Y_{\frac{\lambda}{n}}) $.

The right-inverses $\psi(n) $ are the ring morphisms defined by linearly extending the map on the base elements $\psi(n)(Y_{\lambda}) = \frac{1}{n}(Y_{\frac{\lambda}{n}} + Y_{\frac{\lambda+1}{n}} + \ldots + Y_{\frac{\lambda+n-1}{n}}) $ (check that these are indeed ring maps, that is that $\psi(n)(Y_{\lambda}).\psi(n)(Y_{\mu}) = \psi(n)(Y_{\lambda+\mu}) $.

These are indeed right-inverses satisfying the idempotent condition for clearly $\phi(n) \circ \psi(n) (Y_{\lambda}) = \frac{1}{n}(Y_{\lambda}+\ldots+Y_{\lambda})=Y_{\lambda} $ and

$\begin{eqnarray} \psi(n) \circ \phi(n) (Y_{\lambda}) =& \psi(n)(Y_{n \lambda}) = \frac{1}{n}(Y_{\lambda} + Y_{\lambda+\frac{1}{n}} + \ldots + Y_{\lambda+\frac{n-1}{n}}) \\ =& Y_{\lambda}.(\frac{1}{n}(Y_0 + Y_{\frac{1}{n}} + \ldots + Y_{\frac{n-1}{n}})) = Y_{\lambda} E_n \end{eqnarray} $

and one verifies that $E_n = \frac{1}{n}(Y_0 + Y_{\frac{1}{n}} + \ldots + Y_{\frac{n-1}{n}}) $ is indeed an idempotent in $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $. In the previous posts in this series we have already seen that with these definitions we have indeed that the BC-algebra is the bi-crystalline graded ring

$B = \mathcal{H} = \oplus_{\frac{m}{n} \in \mathbb{Q}^+_{\times}} X_m \mathbb{Q}[\mathbb{Q}/\mathbb{Z}] X_n^* $

and hence is naturally constructed from the skew semi-group graded algebra $A = \oplus_{m \in \mathbb{N}^+_{\times}} X_m \mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $.

This (probably) explains why the BC-algebra $\mathcal{H} $ is itself usually called and denoted in $C^* $-algebra papers the skew semigroup-algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \bowtie \mathbb{N}^+_{\times} $ as this subalgebra (our crystalline semi-group graded algebra $A $) determines the Hecke algebra completely.

Finally, the bi-crystalline idempotents-condition works well in the settings of von Neumann regular algebras (such as all limits of finite dimensional semi-simples, for example $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $) because such algebras excel at idempotents galore…

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Bost-Connes for ringtheorists

Published January 23, 2008 by lievenlb

Over the last days I’ve been staring at the Bost-Connes algebra to find a ringtheoretic way into it. Ive had some chats about it with the resident graded-guru but all we came up with so far is that it seems to be an extension of Fred’s definition of a ‘crystalline’ graded algebra. Knowing that several excellent ringtheorists keep an eye on my stumblings here, let me launch an appeal for help :

What is the most elegant ringtheoretic framework in which the Bost-Connes Hecke algebra is a motivating example?

Let us review what we know so far and extend upon it with a couple of observations that may (or may not) be helpful to you. The algebra $\mathcal{H} $ is the algebra of $\mathbb{Q} $-valued functions (under the convolution product) on the double coset-space $\Gamma_0 \backslash \Gamma / \Gamma_0 $ where

$\Gamma = { \begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix}~:~a,b \in \mathbb{Q}, a > 0 } $ and $\Gamma_0 = { \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}~:~n \in \mathbb{N}_+ } $

We have seen that a $\mathbb{Q} $-basis is given by the characteristic functions $X_{\gamma} $ (that is, such that $X_{\gamma}(\gamma’) = \delta_{\gamma,\gamma’} $) with $\gamma $ a rational point represented by the couple $~(a,b) $ (the entries in the matrix definition of a representant of $\gamma $ in $\Gamma $) lying in the fractal comb

defined by the rule that $b < \frac{1}{n} $ if $a = \frac{m}{n} $ with $m,n \in \mathbb{N}, (m,n)=1 $. Last time we have seen that the algebra $\mathcal{H} $ is generated as a $\mathbb{Q} $-algebra by the following elements (changing notation)

$\begin{cases}X_m=X_{\alpha_m} & \text{with } \alpha_m = \begin{bmatrix} 1 & 0 \\ 0 & m \end{bmatrix}~\forall m \in \mathbb{N}_+ \\
X_n^*=X_{\beta_n} & \text{with } \beta_n = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{n} \end{bmatrix}~\forall n \in \mathbb{N}_+ \\
Y_{\gamma} = X_{\gamma} & \text{with } \gamma = \begin{bmatrix} 1 & \gamma \\ 0 & 1 \end{bmatrix}~\forall \lambda \in \mathbb{Q}/\mathbb{Z} \end{cases} $

Using the tricks of last time (that is, figuring out what functions convolution products represent, knowing all double-cosets) it is not too difficult to prove the defining relations among these generators to be the following (( if someone wants the details, tell me and I’ll include a ‘technical post’ or consult the Bost-Connes original paper but note that this scanned version needs 26.8Mb ))

(1) : $X_n^* X_n = 1, \forall n \in \mathbb{N}_+$

(2) : $X_n X_m = X_{nm}, \forall m,n \in \mathbb{N}_+$

(3) : $X_n X_m^* = X_m^* X_n, \text{whenever } (m,n)=1$

(4) : $Y_{\gamma} Y_{\mu} = Y_{\gamma+\mu}, \forall \gamma,mu \in \mathbb{Q}/\mathbb{Z}$

(5) : $Y_{\gamma}X_n = X_n Y_{n \gamma},~\forall n \in \mathbb{N}_+, \gamma \in \mathbb{Q}/\mathbb{Z}$

(6) : $X_n Y_{\lambda} X_n^* = \frac{1}{n} \sum_{n \delta = \gamma} Y_{\delta},~\forall n \in \mathbb{N}_+, \gamma \in \mathbb{Q}/\mathbb{Z}$

Simple as these equations may seem, they bring us into rather uncharted ringtheoretic territories. Here a few fairly obvious ringtheoretic ingredients of the Bost-Connes Hecke algebra $\mathcal{H} $

the group-algebra of $\mathbb{Q}/\mathbb{Z} $

The equations (4) can be rephrased by saying that the subalgebra generated by the $Y_{\gamma} $ is the rational groupalgebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $ of the (additive) group $\mathbb{Q}/\mathbb{Z} $. Note however that $\mathbb{Q}/\mathbb{Z} $ is a torsion group (that is, for all $\gamma = \frac{m}{n} $ we have that $n.\gamma = (\gamma+\gamma+ \ldots + \gamma) = 0 $). Hence, the groupalgebra has LOTS of zero-divisors. In fact, this group-algebra doesn’t have any good ringtheoretic properties except for the fact that it can be realized as a limit of finite groupalgebras (semi-simple algebras)

$\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] = \underset{\rightarrow}{lim}~\mathbb{Q}[\mathbb{Z}/n \mathbb{Z}] $

and hence is a quasi-free (or formally smooth) algebra, BUT far from being finitely generated…

the grading group $\mathbb{Q}^+_{\times} $

The multiplicative group of all positive rational numbers $\mathbb{Q}^+_{\times} $ is a torsion-free Abelian ordered group and it follows from the above defining relations that $\mathcal{H} $ is graded by this group if we give

$deg(Y_{\gamma})=1,~deg(X_m)=m,~deg(X_n^*) = \frac{1}{n} $

Now, graded algebras have been studied extensively in case the grading group is torsion-free abelian ordered AND finitely generated, HOWEVER $\mathbb{Q}^+_{\times} $ is infinitely generated and not much is known about such graded algebras. Still, the ordering should allow us to use some tricks such as taking leading coefficients etc.

the endomorphisms of $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $

We would like to view the equations (5) and (6) (the latter after multiplying both sides on the left with $X_n^* $ and using (1)) as saying that $X_n $ and $X_n^* $ are normalizing elements. Unfortunately, the algebra morphisms they induce on the group algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $ are NOT isomorphisms, BUT endomorphisms. One source of algebra morphisms on the group-algebra comes from group-morphisms from $\mathbb{Q}/\mathbb{Z} $ to itself. Now, it is known that

$Hom_{grp}(\mathbb{Q}/\mathbb{Z},\mathbb{Q}/\mathbb{Z}) \simeq \hat{\mathbb{Z}} $, the profinite completion of $\mathbb{Z} $. A class of group-morphisms of interest to us are the maps given by multiplication by n on $\mathbb{Q}/\mathbb{Z} $. Observe that these maps are epimorphisms with a cyclic order n kernel. On the group-algebra level they give us the epimorphisms

$\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \longrightarrow^{\phi_n} \mathbb{Q}[\mathbb{Q}/\mathbb{Z}] $ such that $\phi_n(Y_{\lambda}) = Y_{n \lambda} $ whence equation (5) can be rewritten as $Y_{\lambda} X_n = X_n \phi_n(Y_{\lambda}) $, which looks good until you think that $\phi_n $ is not an automorphism…

There are even other (non-unital) algebra endomorphisms such as the map $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \rightarrow^{\psi_n} R_n $ defined by $\psi_n(Y_{\lambda}) = \frac{1}{n}(Y_{\frac{\lambda}{n}} + Y_{\frac{\lambda + 1}{n}} + \ldots + Y_{\frac{\lambda + n-1}{n}}) $ and then, we can rewrite equation (6) as $Y_{\lambda} X_n^* = X_n^* \psi_n(Y_{\lambda}) $, but again, note that $\psi_n $ is NOT an automorphism.

almost strongly graded, but not quite…

Recall from last time that the characteristic function $X_a $ for any double-coset-class $a \in \Gamma_0 \backslash \Gamma / \Gamma_0 $ represented by the matrix $a=\begin{bmatrix} 1 & \lambda \\ 0 & \frac{m}{n} \end{bmatrix} $ could be written in the Hecke algebra as $X_a = n X_m Y_{n \lambda} X_n^* = n Y_{\lambda} X_m X_n^* $. That is, we can write the Bost-Connes Hecke algebra as

$\mathcal{H} = \oplus_{\frac{m}{n} \in \mathbb{Q}^+_{\times}}~\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] X_mX_n^* $

Hence, if only the morphisms $\phi_n $ and $\psi_m $ would be automorphisms, this would say that $\mathcal{H} $ is a strongly $\mathbb{Q}^+_{\times} $-algebra with part of degree one the groupalgebra of $\mathbb{Q}/\mathbb{Z} $.

However, they are not. But there is an extension of the notion of strongly graded algebras which Fred has dubbed crystalline graded algebras in which it is sufficient that the algebra maps are all epimorphisms. (maybe I’ll post about these algebras, another time). However, this is not the case for the $\psi_m $…

So, what is the most elegant ringtheoretic framework in which the algebra $\mathcal{H} $ fits??? Surely, you can do better than generalized crystalline graded algebra…

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the Bost-Connes Hecke algebra

Published January 19, 2008 by lievenlb

As before, $\Gamma $ is the subgroup of the rational linear group $GL_2(\mathbb{Q}) $ consisting of the matrices

$\begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix} $ with $a \in \mathbb{Q}_+ $ and $\Gamma_0 $ the subgroup of all matrices $\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} $ with $n \in \mathbb{N} $. Last time, we have seen that the double coset space $\Gamma_0 \backslash \Gamma / \Gamma_0 $ can be identified with the set of all rational points in the fractal comb consisting of all couples $~(a,b) $ with $a=\frac{m}{n} \in \mathbb{Q}_+ $ and $b \in [0,\frac{1}{n}) \cap \mathbb{Q} $

The blue spikes are at the positive natural numbers $a={ 1,2,3,\ldots } $. Over $a=1 $ they correspond to the matrices $\begin{bmatrix} 1 & \gamma \\ 0 & 1 \end{bmatrix} $ with $\gamma \in [0,1) \cap \mathbb{Q} $ and as matrix-multiplication of such matrices corresponds to addition of the $\gamma $ we see that these cosets can be identified with the additive group $\mathbb{Q}/\mathbb{Z} $ (which will reappear at a later stage as the multiplicative group of all roots of unity).

The Bost-Connes Hecke algebra $\mathcal{H} = \mathcal{H}(\Gamma,\Gamma_0) $ is the convolution algebra of all comlex valued functions with finite support on the double coset space $\Gamma_0 \backslash \Gamma / \Gamma_0 $. That is, as a vector space the algebra has as basis the functions $e_X $ with $X \in \Gamma_0 \backslash \Gamma / \Gamma_0 $ (that is, $X $ is a point of the fractal comb) and such that $e_X(X)=1 $ and $e_X(Y)=0 $ for all other double cosets $Y \not= X $. The algebra product on $\mathcal{H} $ is the convolution-product meaning that if $f,f’ $ are complex functions with finite support on the Bost-Connes space, then they can also be interpreted as $\Gamma_0 $-bi-invariant functions on the group $\Gamma $ (for this just means that the function is constant on double cosets) and then $f \ast f’ $ is the function defined for all $\gamma \in \Gamma $ by

$f \ast f'(\gamma) = \sum_{\mu \in \Gamma/ \Gamma_0} f(\mu) f'(\mu^{-1} \gamma) $

Last time we have seen that the coset-space $\Gamma / \Gamma_0 $ can be represented by all rational points $~(a,b) $ with $b<1 $. At first sight, the sum above seems to be infinite, but, f and f’ are non-zero only at finitely many double cosets and we have see last time that $\Gamma_0 $ acts on one-sided cosets with finite orbits. Therefore, $f \ast f $ is a well-defined $\Gamma_0 $-bi-invariant function with finite support on the fractal comb $\Gamma_0 \backslash \Gamma / \Gamma_0 $. Further, observe that the unit element of $\mathcal{H} $ is the function corresponding to the identity matrix in $\Gamma $.

Looking at fractal-comb picture it is obvious that the Bost-Connes Hecke algebra $\mathcal{H} $ is a huge object. Today, we will prove the surprising result that it can be generated by the functions corresponding to the tiny portion of the comb, shown below.

That is, we will show that $\mathcal{H} $ is generated by the functions $e(\gamma) $ corresponding to the double-coset $X_{\gamma} = \begin{bmatrix} 1 & \gamma \\ 0 & 1 \end{bmatrix} $ (the rational points of the blue line-segment over 1, or equivalently, the elements of the group $\mathbb{Q}/\mathbb{Z} $), together with the functions $\phi_n $ corresponding to the double-coset $X_n = \begin{bmatrix} 1 & 0 \\ 0 & n \end{bmatrix} $ for all $ n \in \mathbb{N}_+ $ (the blue dots to the right in the picture) and the functions $\phi_n^* $ corresponding to the double cosets $X_{1/n} = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{n} \end{bmatrix} $ (the red dots to the left).

Take a point in the fractal comb $X = \begin{bmatrix} 1 & \gamma \\ 0 & \frac{m}{n} \end{bmatrix} $ with $~(m,n)=1 $ and $\gamma \in [0,\frac{1}{n}) \cap \mathbb{Q} \subset [0,1) \cap \mathbb{Q} $. Note that as $\gamma < \frac{1}{n} $ we have that $n \gamma < 1 $ and hence $e(n \gamma) $ is one of the (supposedly) generating functions described above.

Because $X = \begin{bmatrix} 1 & \gamma \\ 0 & \frac{m}{n} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & m \end{bmatrix} \begin{bmatrix} 1 & n \gamma \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{n} \end{bmatrix} = X_m X_{n \gamma} X_{1/n} $ we are aiming for a relation in the Hecke algebra $\phi_m \ast e(n \gamma) \ast \phi^*_n = e_X $. This is ‘almost’ true, except from a coefficient.

Let us prove first the equality of functions $e_X \ast \phi_n = n \phi_m \ast e(n \gamma) $. To do this we have to show that they have the same value for all points $Y \in \Gamma_0 \backslash \Gamma / \Gamma_0 $ in the fractal comb. Let us first study the function on the right hand side.

$\phi_m \ast e(n \gamma) = \sum_{g \in \Gamma/\Gamma_0} \phi_m(g) e(n \gamma)(g^{-1}Y) $. Because $X_m \Gamma_0 $ is already a double coset (over $m $ we have a comb-spike of length one, so all rational points on it determine at the same time a one-sided and a double coset. Therefore, $\phi_m(g) $ is zero unless $g = X_m $ and then the value is one.

Next, let us consider the function on the left-hand side. $e_X \ast \phi_n(Y) = \sum_{g \in \Gamma / \Gamma_0} e_X(g) \phi_m( g^{-1} Y) $. We have to be a bit careful here as the double cosets over $a=\frac{m}{n} $ are different from the left cosets. Recall from last time that the left-cosets over a are given by all rational points of the form $~(a,b) $ with $ b < 1 $ whereas the double-cosets over a are represented by the rational points of the form $~(a,b) $ with $b < \frac{1}{n} $ and hence the $\Gamma_0 $-orbits over a all consist of precisely n elements g.
That is, $e_X(g) $ is zero for all $ g \in \Gamma/\Gamma_0 $ except when g is one of the following matrices

$ g \in { \begin{bmatrix} 1 & \gamma \\ 0 & \frac{m}{n} \end{bmatrix}, \begin{bmatrix} 1 & \gamma+\frac{1}{n} \\ 0 & \frac{m}{n} \end{bmatrix}, \begin{bmatrix} 1 & \gamma + \frac{2}{n} \\ 0 & \frac{m}{n} \end{bmatrix}, \ldots, \begin{bmatrix} 1 & \gamma + \frac{n-1}{n} \\ 0 & \frac{m}{n} \end{bmatrix} } $

Further, $\phi_n(g^{-1}Y) $ is zero unless $g^{-1}Y \in \Gamma_0 \begin{bmatrix} 1 & 0 \\ 0 & n \end{bmatrix} \Gamma_0 $, or equivalently, that $Y \in \Gamma_0 g \Gamma_0 \begin{bmatrix} 1 & 0 \\ 0 & n \end{bmatrix} \Gamma_0 = \Gamma_0 g \begin{bmatrix} 1 & 0 \\ 0 & n \end{bmatrix} \Gamma_0 $ and for each of the choices for g we have that

$ \begin{bmatrix} 1 & \gamma + \frac{k}{n} \\ 0 & \frac{m}{n} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & n \end{bmatrix} = \begin{bmatrix} 1 & n \gamma + k \\ 0 & m \end{bmatrix} \sim \begin{bmatrix} 1 & n\gamma \\ 0 & m \end{bmatrix} $

Therefore, the function $e_X \ast \phi_n $ is zero at every point of the fractal comb unless at $\begin{bmatrix} 1 & n \gamma \\ 0 & m \end{bmatrix} $ where it is equal to $n $. This proves the claimed identity of functions and as one verifies easily that $\phi_n^* \ast \phi_n = 1 $, it follows that all base vectors $e_X $ of $\mathcal{H} $ can be expressed in the claimed generators

$ e_X = n \phi_m \ast e(n \gamma) \ast \phi_n^* $

Bost and Connes use slightly different generators, namely with $\mu_n = \frac{1}{\sqrt{n}} \phi_n $ and $\mu_n^* = \sqrt{n} \phi_n^* $ in order to have all relations among the generators being defined over $\mathbb{Q} $ (as we will see another time). This will be important later on to have an action of the cyclotomic Galois group $Gal(\mathbb{Q}^{cycl}/\mathbb{Q}) $ on certain representations of $\mathcal{H} $.

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