Category: featured

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.

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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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the Bost-Connes coset space

Published January 17, 2008 by lievenlb

By now, everyone remotely interested in Connes’ approach to the Riemann hypothesis, knows the _one line mantra_

one can use noncommutative geometry to extend Weil’s proof of the Riemann-hypothesis in the function field case to that of number fields

But, can one go beyond this sound-bite in a series of blog posts? A few days ago, I was rather optimistic, but now, after reading-up on the Connes-Consani-Marcolli project, I feel overwhelmed by the sheer volume of their work (and by my own ignorance of key tools in the approach). The most recent account takes up half of the 700+ pages of the book Noncommutative Geometry, Quantum Fields and Motives by Alain Connes and Matilde Marcolli…

So let us set a more modest goal and try to understand one of the first papers Alain Connes wrote about the RH : Noncommutative geometry and the Riemann zeta function. It is only 24 pages long and relatively readable. But even then, the reader needs to know about class field theory, the classification of AF-algebras, Hecke algebras, etc. etc. Most of these theories take a book to explain. For example, the first result he mentions is the main result of local class field theory which appears only towards the end of the 200+ pages of Jean-Pierre Serre’s Local Fields, itself a somewhat harder read than the average blogpost…

Anyway, we will see how far we can get. Here’s the plan : I’ll take the heart-bit of their approach : the Bost-Connes system, and will try to understand it from an algebraist’s viewpoint. Today we will introduce the groups involved and describe their cosets.

For any commutative ring $R $ let us consider the group of triangular $2 \times 2 $ matrices of the form

$P_R = { \begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix}~|~b \in R, a \in R^* } $

(that is, $a $ in an invertible element in the ring $R $). This is really an affine group scheme defined over the integers, that is, the coordinate ring

$\mathbb{Z}[P] = \mathbb{Z}[x,x^{-1},y] $ becomes a Hopf algebra with comultiplication encoding the group-multiplication. Because

$\begin{bmatrix} 1 & b_1 \\ 0 & a_1 \end{bmatrix} \begin{bmatrix} 1 & b_2 \\ 0 & a_2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \times b_2 + b_1 \times a_2 \\ 0 & a_1 \times a_2 \end{bmatrix} $

we have $\Delta(x) = x \otimes x $ and $\Delta(y) = 1 \otimes y + y \otimes x $, or $x $ is a group-like element whereas $y $ is a skew-primitive. If $R \subset \mathbb{R} $ is a subring of the real numbers, we denote by $P_R^+ $ the subgroup of $P_R $ consisting of all matrices with $a > 0 $. For example,

$\Gamma_0 = P_{\mathbb{Z}}^+ = { \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}~|~n \in \mathbb{Z} } $

which is a subgroup of $\Gamma = P_{\mathbb{Q}}^+ $ and our first job is to describe the cosets.

The left cosets $\Gamma / \Gamma_0 $ are the subsets $\gamma \Gamma_0 $ with $\gamma \in \Gamma $. But,

$\begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix} \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & b+n \\ 0 & a \end{bmatrix} $

so if we represent the matrix $\gamma = \begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix} $ by the point $~(a,b) $ in the right halfplane, then for a given positive rational number $a $ the different cosets are represented by all $b \in [0,1) \cap \mathbb{Q} = \mathbb{Q}/\mathbb{Z} $. Hence, the left cosets are all the rational points in the region between the red and green horizontal lines. For fixed $a $ the cosets correspond to the rational points in the green interval (such as over $\frac{2}{3} $ in the picture on the left.

Similarly, the right cosets $\Gamma_0 \backslash \Gamma $ are the subsets $\Gamma_0 \gamma $ and as

$\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix} = \begin{bmatrix} 1 & b+na \\ 0 & a \end{bmatrix} $

we see similarly that the different cosets are precisely the rational points in the region between the lower red horizontal and the blue diagonal line. So, for fixed $a $ they correspond to rational points in the blue interval (such as over $\frac{3}{2} $) $[0,a) \cap \mathbb{Q} $. But now, let us look at the double coset space $\Gamma_0 \backslash \Gamma / \Gamma_0 $. That is, we want to study the orbits of the action of $\Gamma_0 $, acting on the right, on the left-cosets $\Gamma / \Gamma_0 $, or equivalently, of the action of $\Gamma_0 $ acting on the left on the right-cosets $\Gamma_0 \backslash \Gamma $. The crucial observation to make is that these actions have finite orbits, or equivalently, that $\Gamma_0 $ is an almost normal subgroup of $\Gamma $ meaning that $\Gamma_0 \cap \gamma \Gamma_0 \gamma^{-1} $ has finite index in $\Gamma_0 $ for all $\gamma \in \Gamma $. This follows from

$\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix} \begin{bmatrix} 1 & m \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & b+m+an \\ 0 & a \end{bmatrix} $

and if $n $ varies then $an $ takes only finitely many values modulo $\mathbb{Z} $ and their number depends only on the denominator of $a $. In the picture above, the blue dots lying on the line over $\frac{2}{3} $ represent the double coset

$\Gamma_0 \begin{bmatrix} 1 & \frac{2}{3} \\ 0 & \frac{2}{3} \end{bmatrix} $ and we see that these dots split the left-cosets with fixed value $a=\frac{2}{3} $ (that is, the green line-segment) into three chunks (3 being the denominator of a) and split the right-cosets (the line-segment under the blue diagonal) into two subsegments (2 being the numerator of a). Similarly, the blue dots on the line over $\frac{3}{2} $ divide the left-cosets in two parts and the right cosets into three parts.

This shows that the $\Gamma_0 $-orbits of the right action on the left cosets $\Gamma/\Gamma_0 $ for each matrix $\gamma \in \Gamma $ with $a=\frac{2}{3} $ consist of exactly three points, and we denote this by writing $L(\gamma) = 3 $. Similarly, all $\Gamma_0 $-orbits of the left action on the right cosets $\Gamma_0 \backslash \Gamma $ with this value of a consist of two points, and we write this as $R(\gamma) = 2 $.

For example, on the above picture, the black dots on the line over $\frac{2}{3} $ give the matrices in the double coset of the matrix

$\gamma = \begin{bmatrix} 1 & \frac{1}{7} \\ 0 & \frac{2}{3} \end{bmatrix} $

and the gray dots on the line over $\frac{3}{2} $ determine the elements of the double coset of

$\gamma^{-1} = \begin{bmatrix} 1 & -\frac{3}{14} \\ 0 & \frac{3}{2} \end{bmatrix} $

and one notices (in general) that $L(\gamma) = R(\gamma^{-1}) $. But then, the double cosets with $a=\frac{2}{3} $ are represented by the rational b’s in the interval $[0,\frac{1}{3}) $ and those with $a=\frac{3}{2} $ by the rational b’s in the interval $\frac{1}{2} $. In general, the double cosets of matrices with fixed $a = \frac{r}{s} $ with $~(r,s)=1 $ are the rational points in the line-segment over $a $ with $b \in [0,\frac{1}{s}) $.

That is, the Bost-Connes double coset space $\Gamma_0 \backslash \Gamma / \Gamma_0 $ are the rational points in a horrible fractal comb. Below we have drawn only the part of the dyadic values, that is when $a = \frac{r}{2^t} $ in the unit inverval

and of course we have to super-impose on it similar pictures for rationals with other powers as their denominators. Fortunately, NCG excels in describing such fractal beasts…

UPDATE : here is a slightly beter picture of the coset space, drawing the part over all rational numbers contained in the 15-th Farey sequence. The blue segments of length one are at 1,2,3,…

Quiver-superpotentials

Published January 14, 2008 by lievenlb

It’s been a while, so let’s include a recap : a (transitive) permutation representation of the modular group $\Gamma = PSL_2(\mathbb{Z}) $ is determined by the conjugacy class of a cofinite subgroup $\Lambda \subset \Gamma $, or equivalently, to a dessin d’enfant. We have introduced a quiver (aka an oriented graph) which comes from a triangulation of the compactification of $\mathbb{H} / \Lambda $ where $\mathbb{H} $ is the hyperbolic upper half-plane. This quiver is independent of the chosen embedding of the dessin in the Dedeking tessellation. (For more on these terms and constructions, please consult the series Modular subgroups and Dessins d’enfants).

Why are quivers useful? To start, any quiver $Q $ defines a noncommutative algebra, the path algebra $\mathbb{C} Q $, which has as a $\mathbb{C} $-basis all oriented paths in the quiver and multiplication is induced by concatenation of paths (when possible, or zero otherwise). Usually, it is quite hard to make actual computations in noncommutative algebras, but in the case of path algebras you can just see what happens.

Moreover, we can also see the finite dimensional representations of this algebra $\mathbb{C} Q $. Up to isomorphism they are all of the following form : at each vertex $v_i $ of the quiver one places a finite dimensional vectorspace $\mathbb{C}^{d_i} $ and any arrow in the quiver
[tex]\xymatrix{\vtx{v_i} \ar[r]^a & \vtx{v_j}}[/tex] determines a linear map between these vertex spaces, that is, to $a $ corresponds a matrix in $M_{d_j \times d_i}(\mathbb{C}) $. These matrices determine how the paths of length one act on the representation, longer paths act via multiplcation of matrices along the oriented path.

A necklace in the quiver is a closed oriented path in the quiver up to cyclic permutation of the arrows making up the cycle. That is, we are free to choose the start (and end) point of the cycle. For example, in the one-cycle quiver

[tex]\xymatrix{\vtx{} \ar[rr]^a & & \vtx{} \ar[ld]^b \\ & \vtx{} \ar[lu]^c &}[/tex]

the basic necklace can be represented as $abc $ or $bca $ or $cab $. How does a necklace act on a representation? Well, the matrix-multiplication of the matrices corresponding to the arrows gives a square matrix in each of the vertices in the cycle. Though the dimensions of this matrix may vary from vertex to vertex, what does not change (and hence is a property of the necklace rather than of the particular choice of cycle) is the trace of this matrix. That is, necklaces give complex-valued functions on representations of $\mathbb{C} Q $ and by a result of Artin and Procesi there are enough of them to distinguish isoclasses of (semi)simple representations! That is, linear combinations a necklaces (aka super-potentials) can be viewed, after taking traces, as complex-valued functions on all representations (similar to character-functions).

In physics, one views these functions as potentials and it then interested in the points (representations) where this function is extremal (minimal) : the vacua. Clearly, this does not make much sense in the complex-case but is relevant when we look at the real-case (where we look at skew-Hermitian matrices rather than all matrices). A motivating example (the Yang-Mills potential) is given in Example 2.3.2 of Victor Ginzburg’s paper Calabi-Yau algebras.

Let $\Phi $ be a super-potential (again, a linear combination of necklaces) then our commutative intuition tells us that extrema correspond to zeroes of all partial differentials $\frac{\partial \Phi}{\partial a} $ where $a $ runs over all coordinates (in our case, the arrows of the quiver). One can make sense of differentials of necklaces (and super-potentials) as follows : the partial differential with respect to an arrow $a $ occurring in a term of $\Phi $ is defined to be the path in the quiver one obtains by removing all 1-occurrences of $a $ in the necklaces (defining $\Phi $) and rearranging terms to get a maximal broken necklace (using the cyclic property of necklaces). An example, for the cyclic quiver above let us take as super-potential $abcabc $ (2 cyclic turns), then for example

$\frac{\partial \Phi}{\partial b} = cabca+cabca = 2 cabca $

(the first term corresponds to the first occurrence of $b $, the second to the second). Okay, but then the vacua-representations will be the representations of the quotient-algebra (which I like to call the vacualgebra)

$\mathcal{U}(Q,\Phi) = \frac{\mathbb{C} Q}{(\partial \Phi/\partial a, \forall a)} $

which in ‘physical relevant settings’ (whatever that means…) turn out to be Calabi-Yau algebras.

But, let us return to the case of subgroups of the modular group and their quivers. Do we have a natural super-potential in this case? Well yes, the quiver encoded a triangulation of the compactification of $\mathbb{H}/\Lambda $ and if we choose an orientation it turns out that all ‘black’ triangles (with respect to the Dedekind tessellation) have their arrow-sides defining a necklace, whereas for the ‘white’ triangles the reverse orientation makes the arrow-sides into a necklace. Hence, it makes sense to look at the cubic superpotential $\Phi $ being the sum over all triangle-sides-necklaces with a +1-coefficient for the black triangles and a -1-coefficient for the white ones. Let’s consider an index three example from a previous post

[tex]\xymatrix{& & \rho \ar[lld]_d \ar[ld]^f \ar[rd]^e & \\
i \ar[rrd]_a & i+1 \ar[rd]^b & & \omega \ar[ld]^c \\
& & 0 \ar[uu]^h \ar@/^/[uu]^g \ar@/_/[uu]_i &}[/tex]

In this case the super-potential coming from the triangulation is

$\Phi = -aid+agd-cge+che-bhf+bif $

and therefore we have a noncommutative algebra $\mathcal{U}(Q,\Phi) $ associated to this index 3 subgroup. Contrary to what I believed at the start of this series, the algebras one obtains in this way from dessins d’enfants are far from being Calabi-Yau (in whatever definition). For example, using a GAP-program written by Raf Bocklandt Ive checked that the growth rate of the above algebra is similar to that of $\mathbb{C}[x] $, so in this case $\mathcal{U}(Q,\Phi) $ can be viewed as a noncommutative curve (with singularities).

However, this is not the case for all such algebras. For example, the vacualgebra associated to the second index three subgroup (whose fundamental domain and quiver were depicted at the end of this post) has growth rate similar to that of $\mathbb{C} \langle x,y \rangle $…

I have an outlandish conjecture about the growth-behavior of all algebras $\mathcal{U}(Q,\Phi) $ coming from dessins d’enfants : the algebra sees what the monodromy representation of the dessin sees of the modular group (or of the third braid group).
I can make this more precise, but perhaps it is wiser to calculate one or two further examples…

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the crypto lattice

Published January 12, 2008 by lievenlb

Last time we have seen that tori are dual (via their group of characters) to lattices with a Galois action. In particular, the Weil descent torus $R_n=R^1_{\mathbb{F}_{p^n}/\mathbb{F}_p} \mathbb{G}_m $ corresponds to the permutation lattices $R_n^* = \mathbb{Z}[x]/(x^n-1) $. The action of the generator $\sigma $ (the Frobenius) of the Galois group $Gal(\mathbb{F}_{p^n}/\mathbb{F}_p) $ acts on the lattice by multiplication with $x $.

An old result of Masuda (1955), using an even older lemma by Speiser (1919), asserts than whenever the character-lattice $T^* $ of a torus $T $ is a permutation-lattice, the torus is rational, that is, the function-field
of the torus $\mathbb{F}_p(T) $ is purely trancendental

$\mathbb{F}_p(y_1,\ldots,y_d) = \mathbb{F}_p(T) = (\mathbb{F}_{q^n}(T^*))^{Gal} $

(recall from last time that the field on the right-hand side is the field of fractions of the $Gal $-invariants of the group-algebra of the free Abelian group $T^* = \mathbb{Z} \oplus \ldots \oplus \mathbb{Z} $ where the rank is equal to the dimension $d $ of the torus).

The basic observation made by Rubin and Silverberg was that the known results on crypto-compression could be reformulated in the language of algebraic tori as : the tori $T_2 $ (LUC-system) and $T_6 $ (CEILIDH-system) are rational! So, what about the next cryptographic challenges? Are the tori $T_{30} $, $T_{210} $ etc. also rational varieties?

Recall that as a group, the $\mathbb{F}_p $-points of the torus $T_n $, is the subgroup of $\mathbb{F}_{p^n}^* $ corresponding to the most crypto-challenging cyclic subgroup of order $\Phi_n(p) $ where $\Phi_n(x) $ is the n-th cyclotomic polynomial. The character-lattice of this crypto-torus $T_n $ we call the crypto-lattice and it is

$T_n^* = \mathbb{Z}[x]/(\Phi_n(x)) $

(again the action of the Frobenius is given by multiplication with $x $) and hence has rank $\phi(n) $, explaining that the torus $T_n $ has dimension $\phi(n) $ and hence that we can at best expect a compression from $n $-pits to $\phi(n) $-pits. Note that the lattice $T_n^* $ is no longer a permutation lattice, so we cannot use the Masuda-Speiser result to prove rationality of $T_n $.

What have mathematicians proved on $T_n $ before it became a hot topic? Well, there is an old conjecture by V. E. Voskresenskii asserting that all $T_n $ should be rational! Unfortunately, he could prove this only when $n $ is a prime power. Further, he proved that for all $n $, the lattice $T_n $ is at least stably-rational meaning that it is rational upto adding free parameters, that is

$\mathbb{F}_p(T_n)(z_1,\ldots,z_l) = \mathbb{F}_p(y_1,\ldots,y_{d+l}) $

which, sadly, is only of cryptographic-use if $l $ is small (see below). A true rationality result on $T_n $ was proved by A.A. Klyashko : $T_n $ is rational whenever $n=p^a.q^b $ a product of two prime powers.But then, $30=2 \times 3 \times 5 $ the first unknown case…

At Crypto 2004, Marten van Dijk and David Woodruff were able to use an explicit form of Voskresenskii stable rationality result to get an asymptotic optimal crypto-compression rate of $n/\phi(n) $, but their method was of little practical use in the $T_{30} $, for what their method gave was a rational map

$T_{30} \times \mathbb{A}^{32}_{\mathbb{F}_p} \rightarrow \mathbb{A}^{40}_{\mathbb{F}_p} $

and the number of added parameters (32) is way too big to be of use.

But then, one can use century-old results on cyclotomic polynomials to get a much better bound, as was shown in the paper Practical cryptography in high dimensional tori by the collective group of all people working (openly) on tori-cryptography. The idea is that whenever q is a prime and a is an integer not divisible by q, then on the level of cyclotomic polynomials we have the identity

$\Phi_{aq}(x) \Phi_a(x) = \Phi_a(x^q) $

On the level of tori this equality implies (via the character-lattices) an ismorphism (with same assumptions)

$T_{aq}(\mathbb{F}_p) \times T_a(\mathbb{F}_p) \simeq (R^1_{\mathbb{F}_{p^q}/\mathbb{F}_p} T_a)(\mathbb{F}_p) = T_a(\mathbb{F}_{p^q}) $

whenever aq is not divisible by p. Apply this to the special case when $q=5,a=6 $ then we get

$T_{30}(\mathbb{F}_p) \times T_6(\mathbb{F}_p) \simeq R^1_{\mathbb{F}_{p^5}/\mathbb{F}_p} T_6(\mathbb{F}_p) $

and because we know that $T_6 $ is a 2-dimensional rational torus we get, using Weil descent, a rational map

$T_{30} \times \mathbb{A}^2_{\mathbb{F}_p} \rightarrow \mathbb{A}^{10}_{\mathbb{F}_p} $

which can be used to get better crypto-compression than the CEILIDH-system!

This concludes what I know of the OPEN state of affairs in tori-cryptography. I’m sure ‘people in hiding’ know a lot more at the moment and, if not, I have a couple of ideas I’d love to check out. So, when I seem to have disappeared, you know what happened…

Weil descent

Published January 9, 2008 by lievenlb

A classic Andre Weil-tale is his narrow escape from being shot as a Russian spy

The war was a disaster for Weil who was a conscientious objector and so wished to avoid military service. He fled to Finland, to visit Rolf Nevanlinna, as soon as war was declared. This was an attempt to avoid being forced into the army, but it was not a simple matter to escape from the war in Europe at this time. Weil was arrested in Finland and when letters in Russian were found in his room (they were actually from Pontryagin describing mathematical research) things looked pretty black. One day Nevanlinna was told that they were about to execute Weil as a spy, and he was able to persuade the authorities to deport Weil instead.

However, Weil’s wikipedia entry calls this a story too good to be true, and continues

In 1992, the Finnish mathematician Osmo Pekonen went to the archives to check the facts. Based on the documents, he established that Weil was not really going to be shot, even if he was under arrest, and that Nevanlinna probably didn’t do – and didn’t need to do – anything to save him. Pekonen published a paper on this with an afterword by Andre Weil himself. Nevanlinna’s motivation for concocting such a story of himself as the rescuer of a famous Jewish mathematician probably was the fact that he had been a Nazi sympathizer during the war. The story also appears in Nevanlinna’s autobiography, published in Finnish, but the dates don’t match with real events at all. It is true, however, that Nevanlinna housed Weil in the summer of 1939 at his summer residence Korkee at Lohja in Finland – and offered Hitler’s Mein Kampf as bedside reading.

This old spy-story gets a recent twist now that it turns out that Weil’s descent theory of tori has applications to cryptography. So far, I haven’t really defined what tori are, so let us start with some basics.

The simplest (and archetypical) example of an algebraic torus is the multiplicative group(scheme) $\mathbb{G}_m $ over a finite field $\mathbb{F}_q $ which is the affine variety

$\mathbb{V}(xy-1) \subset \mathbb{A}^2_{\mathbb{F}_q} $. that is, the $\mathbb{F}_q $ points of $\mathbb{G}_m $ are precisely the couples ${ (x,\frac{1}{x})~:~x \in \mathbb{F}_q^* } $ and so are in one-to-one correspondence with the non-zero elements of $\mathbb{F}_q $. The coordinate ring of this variety is the ring of Laurant polynomials $\mathbb{F}_q[x,x^{-1}] $ and the fact that multiplication induces a group-structure on the points of the variety can be rephrased by saying that this coordinate ring is a Hopf algebra which is just the Hopf structure on the group-algebra $\mathbb{F}_q[\mathbb{Z}] = \mathbb{F}_q[x,x^{-1}] $. This is the first indication of a connection between tori defined over $\mathbb{F}_q $ and lattices (that is free $\mathbb{Z} $-modules with an action of the Galois group $Gal(\overline{F}_q/F_q) $. In this correspondence, the multiplicative group scheme $\mathbb{G}_m $ corresponds to $\mathbb{Z} $ with the trivial action.

Now take a field extension $\mathbb{F}_q \subset \mathbb{F}_{q^n} $, is there an affine variety, defined over $\mathbb{F}_q $ whose $\mathbb{F}_q $-points are precisely the invertible elements $\mathbb{F}_{q^n}^* $? Sure! Just take the multiplicative group over $\mathbb{F}_{q^n} $ and write the elements x and y as $x = x_1 + x_2 a_2 + \ldots + x_n a_n $ (and a similar expression for y with ${ 1,a_2,\ldots,a_n }$ being a basis of $\mathbb{F}_{q^n}/\mathbb{F}_q $ and write the defning equation $xy-1 $ out, also with respect to this basis and this will then give you the equations of the desired variety, which is usually denoted by $R^1_{\mathbb{F}_{q^n}/\mathbb{F}_q} \mathbb{G}_m $ and called the Weil restriction of scalars torus.

A concrete example? Take $\mathbb{F}_9 = \mathbb{F}_3(\sqrt{-1}) $ and write $x=x_1+x_2 \sqrt{-1} $ and $y=y_1+y_2 \sqrt{-1} $, then the defining equation $xy-1 $ becomes

$~(x_1y_1-x_2y_2) + (x_1y_2-x_2y_1) \sqrt{-1} = 1 $

whence $R^1_{\mathbb{F}_9/\mathbb{F}_3} = \mathbb{V}(x_1y_1-x_2y_2-1,x_1y_2-x_2y_1) \subset \mathbb{A}^4_{\mathbb{F}_3} $, the intersection of two quadratic hypersurfaces in 4-dimensional space.

Why do we call $R^1 \mathbb{G}_m $ a _torus_? Well, as with any variety defined over $\mathbb{F}_q $ we can also look at its points over a field-extension, for example over the algebraic closure $\overline{\mathbb{F}}_q $ and then it is easy to see that

$R^1_{\mathbb{F}_{q^n}/\mathbb{F}_q} \mathbb{G}_m (\overline{\mathbb{F}}_q) = \overline{\mathbb{F}}_q^* \times \ldots \times \overline{\mathbb{F}}_q^* $ (n copies)

and such algebraic groups are called tori. (To understand terminology, the compact group corresponding to $\mathbb{C}^* \times \mathbb{C}^* $ is $U_1 \times U_1 = S^1 \times S^1 $, so a torus).

In fact, it is already the case that the $\mathbb{F}_{q^n} $ points of the restriction of scalar torus are $\mathbb{F}_{q^n}^* \times \ldots \times \mathbb{F}_{q^n}^* $ and therefore we call this field a splitting field of the torus.

This is the general definition of an algebraic torus : a torus T over $\mathbb{F}_q $ is an affine group scheme over $\mathbb{F}_q $ such that, if we extend scalars to the algebraic closure (and then it already holds for a finite extension) we get an isomorphism of affine group schemes

$T \times_{\mathbb{F}_q} \overline{\mathbb{F}}_q = \overline{\mathbb{F}}_q^* \times \ldots \times \overline{\mathbb{F}}_q^* = (\overline{\mathbb{F}}_q^*)^{n} $

in which case we call T a torus of dimension n. Clearly, the Galois group $Gal(\overline{\mathbb{F}}_q^*/\mathbb{F}_q) $ acts on the left hand side in such a way that we recover $T $ as the orbit space for this action.

Hence, anther way to phrase this is to say that an algebraic torus is the Weil descent of an action of the Galois group on the algebraic group $\overline{\mathbb{F}}_q^* \times \ldots \times \overline{\mathbb{F}}_q^* $.

Of course we can also rephrase this is more algebraic terms by looking at the coordinate rings. The coordinate ring of the algebraic group $~(\overline{\mathbb{F}}_q^*)^n $ is the group-algebra of the rank n lattice $\mathbb{Z}^n = \mathbb{Z} \oplus \ldots \oplus \mathbb{Z} $ (the free Abelian group of rank n), that is,
$\overline{\mathbb{F}}_q [ \mathbb{Z}^n ] $. Now the Galois group acts both on the field $\overline{\mathbb{F}}_q $ as on the lattice $\mathbb{Z}^n $ coming from the action of the Galois group on the extended torus $T \times_{\mathbb{F}_q} \overline{\mathbb{F}}_q $. In fact, it is best to denote this specific action on $\mathbb{Z}^n $ by $T^* $ and call $T^* $ the character group of $T $. Now, we recover the coordinate ring of the $\mathbb{F}_q $-torus $T $ as the ring of invariants

$\mathbb{F}_q[T] = \overline{\mathbb{F}}_q [T^*]^{Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q)} $

Hence, the restriction of scalars torus $R^1_{\mathbb{F}_{q^n}/\mathbb{F}_q} \mathbb{G}_m $ is an n-dimensional torus over $\mathbb{F}_q $ and its corresponding character group is the free Abelian group of rank n which can be written as $\mathbb{Z}[x]/(x^n-1) = \mathbb{Z}1 \oplus \mathbb{Z}x \oplus \ldots \oplus \mathbb{Z}x^{n-1} $ and where the action of the cyclic Galois group $Gal(\mathbb{F}_{q^n}/\mathbb{F}_q) = C_n = \langle \sigma \rangle $ s such that the generator $\sigma $ as as multiplication by $x $. That is, in this case the character group is a permutation lattice meaning that the $\mathbb{Z} $-module has a basis which is permuted under the action of the Galois group. Next time we will encounter more difficult tori sich as the crypto-torus $T_n $.

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