Category: featured

Chinese remainders and adele classes

Published January 31, 2008 by lieven

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.

adeles and ideles

Published January 29, 2008 by lieven

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…

BC stands for Bi-Crystalline graded

Published January 26, 2008 by lieven

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…