MSci Presentation this Week

Another hiatus in blog posts: I've been attending Prof Shahn Majid's course in Noncommutative Geometry at the London Taught Course Center Additionally, this whole month I've been busy writing my MSci project and preparing for my presentation. I still have so much to do, and little more than a week left! Exam revision comes afterwards, so I don't expect to have any more exciting maths post during that period either. Though I am hoping to write a post or two about 1) Ergodic theory 2) Cornelissen and Marcolli's paper on Bost-Connes systems and isomorphisms of number fields, 3) Bora Yaklinoglu's proof of a full arithemtic subalgebra for BC-systems (assuming I get a chance to see it!) and/or 4) Things I've learned from the LTCC course.

In the meantime, I thought I'd post a link to draft of my slides for my MSci presentation.

Happy Pi day!

Fun and Games with X and f: A talk on Ergodic Theory

While I'm not running the UCL Undergrad colloquium anymore, I promised the new management that I could be their back-up speaker in case someone else drops out. Someone did, and I had an opportunity give an undergrad-level talk on motivations for studying Ergodic theory. I tried to present it as a game: trying to determine orbits of increasingly complex systems. You can download my notes as a pdf or just read them here:



1. First Game: Sets and Functions

Let $X$ be an arbitrary set (e.g. points in space, animals in a zoo.) and let $f: X \rightarrow X$ be some function on $X$. For some $x \in X$, we're gonna look at what happens to the set $\{ x, f(x), f(f(x)), \ldots \}$. We'll call this set the orbit of $x$, and denote it $\text{orb}(x)$. We'll write $f^0(x)$ for $x$, $f^2(x)$ for $f(f(x))$, $f^3(x)$ for $f(f(f(x)))$, et cetera. In this way we have an action of the natural numbers $\mathbb{N}$ on $X$. To be explicit, each $n \in \mathbb{N}$ acts on each $x \in X$ by

$\displaystyle n \mapsto f^n(x)$

It might be worth pointing out that:

$\displaystyle n+m \mapsto f^{n+m}(x) = f^n ( f^m(x))$

If $f$ is invertible, then we have an action of the group $\mathbb{Z}$ on $X$. In fact, we can play our game with any group $G$ that acts on $X$, but for now we're only considering $\mathbb{N}$.
We're gonna play three games with the pair $(X, f)$. The first has to do with sets and functions, the second with topological spaces and continuous functions, and the third with a measurable space $X$. Each game will get progressively harder, but progressively more rewarding.
So here's the first game: given a set $X$, we have to find an $x \in X$ and a $f: X \rightarrow X$ so that as we cycle through $x$, $f(x)$, $f^2(x)$, \ldots we end up with all of X. Id est, $\text{orb}(x) = X$.
At first glance it may seem impossible to tell without more information about X. But we can already exclude an entire category of sets $X$. Can you see it?
That's right, uncountable sets won't play. $\text{orb}(x) = \{ f^n(x) \; : \; n \in \mathbb{N}\}$, thus its always countable. Let's try it for a really easy set: the natural numbers $\mathbb{N}$ themselves. Can you find a number $n$ and a function $f: \mathbb{N} \rightarrow \mathbb{N}$ such that $\text{orb}(n) = \mathbb{N}$ ?
I hope no one has trouble seeing that $n=0$ and $f(n) = n + 1$ does the trick. What about for the integers $X = \mathbb{Z}$? Can one find a function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ and an integer $n$ such that $\text{orb}(n) = \mathbb{Z}$ ? How about for $\mathbb{Q}$?
Now lets try to finish the game: Can we do this for an arbitrary countably infinite set $X$?
Since $X$ is countably infinite, we know we have a bijection $\phi: X \rightarrow \mathbb{N}$. Let $x \in X$, and choose $\phi$ so that $\phi (x) = 0$. Now remember our function for the natural numbers? We'll rename it $p(n) = n+1$ here. We have something like this:

$\displaystyle X \xrightarrow{\phi} \mathbb{N} \xrightarrow{\phi^{-1}} X$

So we pass into the natural numbers, use the orbit there, and then pass back into $X$. Id est, let $f : X \rightarrow X$ by $f(x)= \phi^{-1} \circ p \circ \phi (x)$. Then $\text{orb}(x) = X$. As it turns out, the condition that there exists an $f: X \rightarrow X$ and an $x \in X$ such that $\text{orb}(x) = X$ is exactly the same condition that X is countably infinite, as $\text{orb}(x)$ puts $X$ in a bijection with the natural numbers. This game was rigged so that we could win!
But while we're at it, let's try a slightly harder game: given a countably infinite set $X$, can we find a function $f: X \rightarrow X$ such that $\text{orb}(x) = X$ for all $x \in X$ ?
No! Can you see why not? Let $x, y \in X$. since $\text{orb}(x) = \text{orb}(y) = X$, there exist numbers $n, m$ such that $f^n(x) = y$ and $f^m(y) = x$. Hence $f^m(f^n(x)) = x$. In other words, $f^{n+m}(x) = x$, meaning the action of the natural numbers on $X$ is periodic, and $|\text{orb}(x)| \leq n+m$. The orbits of an element tell us something special about that element, and unless we're in a finite set, only a few elements can visit the entire set.

2. Second Game: Orbits for Topological Spaces

So as we stated, the last game (finding a function and element such that the orbit is the entire set) was rigged so that we always win. But it's helpful to see just how badly its rigged. In the category of sets, the only ``structure'' we have to preserve is the cardinality of the set. Bijective functions obviously preserve cardinality. Since countably infinite sets are all bijective with the natural numbers by definition they are, in a sense, all the same. The category of countably infinite sets has only one element, $\mathbb{N}$. Anything we prove about $\mathbb{N}$ as a countably infinite set is automatically true for all other countably infinite sets.
So to make the game a bit more interesting, we're going to have to play in a more exciting category. Instead of looking at countably infinite sets, we're going to look at topological spaces. Loosely speaking, a topological space is our most general notion of a space. It gives us a mathematical language for describing when points are ``near each other'', or ``in a neighborhood''. If we say that $X$ is a topological space, we mean that there exists a collection of open subsets $U \subset X$, and these subsets must behave in a particular way. (For more information, see a decent book on topology, like those by Munkres or Armstrong, or talk to students in our General Topology study group.)
Instead of dealing with arbitrary functions $f: X \rightarrow X$, we'll also want a new condition on $f$. Since $X$ is a topological space, we want functions that live on topological spaces, in the same way a linear map lives on vector spaces, or group homomorphism live on groups. These functions are the ``continuous functions''. We require $f: X \rightarrow X$ to be continuous, that is, if $U \subset X$ is open, then $f^{-1} (U) = \{ x \in X : f(x) \in U \}$ must be open, too.
Now that we have $X$ a topological space and $f$ a continuous function (we'll call the pair $(X, f)$ a topological dynamical system), we can get back to our game. When is $\text{orb}(x) = X$ ?
Almost never. Topological spaces (like $\mathbb{R}$ or $\mathbb{C}$) are almost always uncountable, and as we already discussed, this is impossible. So we have to modify our game a bit. If $\text{orb}(x) \neq X$, what is the next best thing? (If you've not taken measure theory, functional analysis, or general topology, you may be forgiven for not knowing.)
In this game, we want $\text{orb}(x)$ to be dense in $X$. If you've never seen dense sets before, think of the rationals $\mathbb{Q}$ sitting in $\mathbb{R}$. In $\mathbb{R}$, the open sets are exactly the open intervals. And any open interval in $\mathbb{R}$ intersects $\mathbb{Q}$. If you've had more analysis, another way to say this is that a subset of $X$ is dense in $X$ if its closure is all of $X$.
So this is the game: given a pair $(X,f)$, $X$ a topological space and $f$ a continuous function, can you find an element $x \in X$ such that $\overline{\text{orb}(x)} = X$ (id est, the closure of $\text{orb}(x)$ is $X$).
This game is a lot harder, and it'll require some new tools. Let me give you a definition and a few lemmas.

Def1 1 Let $(X, f)$ be a topological dynamical system. We call the system minimal if given a closed set $V \subset X$ such that $f(V) = V$, then $V = X$ or $V = \emptyset$. (That is, if the only invariant closed sets are the whole space or the null set.)

Minimality allows us to win the game. Actually, its equivalent to winning; consider the following proposition:
Let $(X, f)$ be a topological dynamical system. The system is minimal if and only if $\overline{\text{orb}(x)} = X$ for all $x \in X$.
Proof: Assume that $(X, f)$ is minimal. $\overline{\text{orb}(x)}$ is clearly closed, invariant under $f$, and non-empty. Thus it must be X. Conversely, assume X is not minimal, and let $V$ be a closed, non-empty $f$-invariant subset of X. Let $x \in V$. Since V is $f$-invariant, $\text{orb}(x) \subset V$, and hence its closure can't be all of X. $\Box$

This game is called Topological Dynamics. Another lemma, not stated here, says that any topological space $X$ has an minimal subsystem. The proof is a simple application of Zorn's lemma. We can actually state (and prove!) a key theorem in the subject:

Thrm 2 (Birkhoff Recurrence Theorem) Let $(X,f)$ be a minimal topological dynamical system. Then there is some $x \in X$ and a sequence $1 \leq n_1 < n_2 < \ldots $ of natural numbers such that $f^{n_k}(x) \rightarrow x$ as $k \rightarrow \infty$.

Proof: Let $x \in X$. Since the system is minimal, we know that $\text{orb}(x)$ is dense in $X$. If there is some $n$ such that $f^n(x)=x$, then we're done. Otherwise, $\text{orb}(x)$ dense means that its closure is the whole space, which means that we can find some sequence $y_k \in \text{orb}(x)$ such that $y_n \rightarrow x$. Naturally, each $y_k = f^{n_k}(x)$. So we're done. $\Box$

Who else plays this game? It's actually been used successfully to prove theorems in number theory. More specifically, it's been used to prove a Ramsey-type problem about colouring the integers. I'll state it.

Thrm 3 (Van der Waerden's Theorem) Suppose that the integers are coloured in r colours. Then for every $k \geq 2$ there is a monochromatic arithmetic progression of length k.

The proof uses a generalization of the above Recurrence Theorem. It also uses compact metric spaces instead of general topological spaces. But, of course, compact metric spaces is a sub-category of topological spaces, and they're much easier to work with (general topological spaces can be quite pathological.)
So we've seen how to make the game more interesting my looking at more exciting categories. We've stated an equivalent condition to winning the game (minimality), and though I haven't shown you any specific examples of the game, I've given you some of the rewards (e.g. Van der Waerden's Theorem) of winning. Let's look at the third game:

3. Third game: Measurable Systems

Now we're going to look at another category of sets and functions. This time $X$ must be a compact metric space. But we want a bit more structure. $X$ must be measurable, that is, for certain ``well behave'' subsets $U \subset X$, we have a function, called a measure, that assigns a ``size'' to U, $\mu(U) \in [0,1]$, and we require the size of the total space to be 1, id est, $\mu(X) = 1$. The subsets for with $\mu$ is defined are called the measurable sets. If you've taken measure theory or probability, then you should recognize that we want $X$ to be a probability space. You can check those courses for more details on this category.
The function $f: X \rightarrow X$ must meet new requirements, too. It must ``live'' on measurable spaces, id est, $f$ must be a measurable map. The definition is similar to that for continuous functions: $f$ is measurable if for all measurable subsets $U \subset X$, $f^{-1}(U)$ is also a measurable set. The measuring function $\mu$ must also be $f$-invariant. That is, $\mu(f^{1}(U) = \mu (U)$.
We call the triple $(X, \mu, f)$ a measure-preserving system. Like in the last two games, we want to know what happens to $\text{orb}(x)$. But things get very subtle here. Like in the last game, we had a condition (minimal systems) that told us when we were onto something. We have a similar condition here:

Def1 4 We call the measure preserving system $(X, \mu, f)$ ergodic if the only $f$-invariant sets have measure 0 or measure 1.

This game is called Ergodic Theory. Who do you think plays it and why? Can you prove a result similar to our proposition on minimality for ergodicity?
At the core, this game is concerned with the statistical study paths of motion of points in some space, whether it be the phase space of a Hamiltonian in physics, or the state spaces of Markov chains (random processes) in statistics.
The idea is to think of the action $n \in \mathbb{N}$ as a particular time, and then $\text{orb}(x) = \{ f^n(x) : n\mathbb{N}\}$ is the time-path of state $x$. This allows us to get a ``time average'' and a ``space average'' of functions on the measure preserving system. Let $\phi: X \rightarrow \mathbb{C}$ be an integrable function.

Def1 5 Let $(X, \mu, f)$ be a measure preserving system. The time average of $\phi$ starting from $x \in X$ is denoted

$\displaystyle <\phi>_x = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=0}^{n} \phi(f^i(x))$

Def1 6 The space average of $\phi$ is denoted

$\displaystyle \overline{\phi} = \int \phi(x) d\mu$

(This is the integral with respect to the measure $\mu$. If you've not seen this before, talk to someone who has taken measure theory.)

To see think about these definitions, consider a subset $U \in X$, and the indicator function on $A$ (that's the function $I: X \rightarrow \mathbb{C}$, $I(x) = 1$ is $x \in UA$, otherwise $I(x)=0$). The time average of $<I>_x$ is time that the orbit of $x$ will spend in A, while the space average is the probability that a random state $x$ is in $U$.
One of the important results about measure theory is the following theorem:

Thrm 7 For an ergodic measure preserving system $(X, \mu, f)$ and $\phi: X \rightarrow \mathbb{C}$ a measurable function, the limit $<\phi>_x$ exists and is equal to $\overline{\phi}$ for almost all $x \in X$.

Since the two are almost always equal, almost all paths cover the state space in the same way. In other words, this theorem tells us that for a sufficiently large amount of time (id est, a sufficiently large sample) we can learn information about the entire system (or entire population.) Think ``law of large numbers''.
While we can't play too many ergodic games, I do want to give an example.
Let $X = \mathbb{R}^2/\mathbb{Z}^2$, that is, let $X$ be a torus, and let $\alpha \in \mathbb{R}^2$. Let $f: X \rightarrow X$ be the function $f(x) = x + \alpha$. With a bit of work from measure theory, you can show that this is an ergodic measure preserving system.
When $\alpha = (\sqrt{2},\sqrt{3})$ and $x = (0,0)$, then $\text{orb}(x)$ is dense in $X$.
If $\alpha = (\frac{1}{3},\frac{2}{5})$ and $x = (0,0)$, then $\overline{\text{orb}(x)}$ is a finite set of fifteen points.
If $\alpha = (\sqrt{2},\sqrt{2})$ and $x = (0,0)$, then $\text{orb}(x)$ is dense in the subspace $\{ (x,x) : x \in \mathbb{R}/\mathbb{Z} \}$.
These weird facts pop out of a much deeper theorem, Ratner's theorem, that I cannot even state, but it roughly says that orbit closures are ``algebraic sets''. Ratner actually has several related theorems, and these theorems are related to the proof of the Oppenheim conjecture. The Oppenheim conjecture is an important statement about analytic number theory and real quadratic forms in several variables. It was proven using Ergodic theorems on Lie Groups. The proof of this theorem could actually make a third year project.

4. Sources

I lifted material for this talk from several sources, most prominently:

1. Ben Green's notes for Ergodic theory at Cambridge.
2. Terrence Tao's Blog
3. Wikipedia
4. Cosma Shalizi's notebook
   

The Global Artin Map for $\mathbb{Q}$

I'm discussing the global Artin homomorphism $\theta : \mathcal{C}_\mathbb{Q} \rightarrow \text{Gal}(\mathbb{Q}^\text{ab}/\mathbb{Q})$ used in our key statement about the Bost-Connes system, as well as the Adeles, Ideles, and other components needed to consider it.

In my last significant post I stated the global Artin map for the rationals intertwined with the Galois action on the values of Bost-Connes KMS states. I wanted to talk about the Artin map in a bit more detail. The general case for any number field $\mathbb{K}$ is a bit too complicated for me to discuss right now, but the case for $\mathbb{Q}$ isn't bad at all. Let's start with $\mathcal{C}_\mathbb{Q} = \mathbb{A}^{\ast}_\mathbb{Q}/\mathbb{Q}^{\ast}$, the Idele class group of $\mathbb{Q}$.
First recall the definition of the Adeles, $\mathbb{A}_\mathbb{Q}$.

$ \displaystyle \mathbb{A}_\mathbb{Q} = \mathbb{R} \times \prod_{p \; \text{prime}}^{\prime} \mathbb{Q}_p $

Where $\mathbb{Q}_p$ is the completed field of $p$-adic numbers. The idea here is that we're looking at all possible completions of the number field $\mathbb{Q}$, including the standard absolute value and all $p$-adic valuations. Each of these completed fields is a ``local'' field, containing information about $\mathbb{Q}$ at a particular prime, and the ring of Adeles combines all that information into one giant ring. The $\prime$ on the product $\prod^{\prime}$ indicates that this product is restricted over $\mathbb{Z}_p$, id est, it has the condition that

$ \displaystyle \mathbb{R} \times \prod_{p \; \text{prime}}^{\prime} \mathbb{Q}_p = \{ a=(a_\infty,a_2,a_3,\ldots) \in \mathbb{A}_\mathbb{Q} \; \text{iff} \; a_p \in \mathbb{Z}_p \; \text{for all but finitely many p}\$

$\mathbb{A}^{\ast}_\mathbb{Q}$ is the set of all invertible elements in the ring, we can write it $\mathbb{R}^\ast \times \prod_p^\prime \mathbb{Q}_p^\ast$. As $\mathbb{Q}_p$ is the field of fractions for $\mathbb{Z}_p$, we have an isomorphism $\mathbb{Q}_p \cong \mathbb{Z}_p[\frac{1}{p}] \cong p^\mathbb{Z} \times \mathbb{Z}_p$. If we identify $p^\mathbb{Z}$ with $\mathbb{Z}$ we can write $\mathbb{Q}_p^\ast \cong \mathbb{Z} \times \mathbb{Z}^\ast_p$ and

$ \displaystyle \mathbb{A}_\mathbb{Q}^\ast = \mathbb{R}^\ast \times \prod_{p \; \text{prime}}^{\prime} \mathbb{Q}_p^\ast \cong \{\pm1\} \times \mathbb{R}_{>0} \times \prod \mathbb{Z}_p^\ast \times \bigoplus_p \mathbb{Z}$

Now for an $r \in \mathbb{Q}^\ast$, we can write $r = \text{sgn}(r) \prod_p p^n(p)$, so we can take $\mathbb{Q}^\ast$ to $\{\pm1\} \times \bigoplus_p \mathbb{Z}$ by $r \mapsto (\text{sgn}(r),n(p)_p)$. By noted that $\mathbb{R}_{>0} \cong \mathbb{R}$ by logarithm, we can write:

$ \displaystyle \mathbb{A}_\mathbb{Q}^\ast \cong \mathbb{Q}^\ast \times \mathbb{R} \times \prod_p \mathbb{Z}_p^\ast$

Now recall our favourite profinite group $\hat{\mathbb{Z}} = \varprojlim_k \mathbb{Z}/k\mathbb{Z} \cong \text{End}(\mathbb{Q}/\mathbb{Z})$. Since we can write each $k$ as $p_1^{n(p_1)} \ldots p_l^{n(p_l)}$, we also have

$ \displaystyle \mathbb{Z}/k\mathbb{Z} = \prod_{i=1}^{l} \mathbb{Z}/p_i^{n(p_i)} \mathbb{Z}$

After passing to the inverse limit we have:

$ \displaystyle \hat{\mathbb{Z}} = \varprojlim_k \mathbb{Z}/k\mathbb{Z} = \prod_p \varprojlim_n \mathbb{Z}/p^n\mathbb{Z} = \prod_p \mathbb{Z}_p$

Thus $ \mathbb{A}_\mathbb{Q}^\ast \cong \mathbb{Q}^\ast \times \mathbb{R} \times \hat{\mathbb{Z}}^\ast $ and finally:

$ \displaystyle \mathcal{C}_\mathbb{Q} \cong \mathbb{R} \times \hat{\mathbb{Z}}^\ast$

Now I cannot prove it, but class field theory tells us that the map $\theta:\mathcal{C}_\mathbb{Q} \rightarrow \text{Gal}(\mathbb{Q}^\text{ab}/\mathbb{Q})$ is surjective, and the kernel is the connected component of the identity on $\mathcal{C}_\mathbb{Q}$, which we denote $\mathcal{D}_\mathbb{Q} = \mathbb{R}$. Hence $\mathcal{C}_\mathbb{Q} / \mathcal{D}_\mathbb{Q} \cong \hat{\mathbb{Z}}^\ast$ and we have an isomorphism:

$ \displaystyle \theta : \hat{\mathbb{Z}}^\ast \rightarrow \text{Gal}(\mathbb{Q}^\text{ab}/\mathbb{Q})$

We can see that these two groups are isomorphic assuming only the Kronecker-Weber theorem (which says that $\mathbb{Q}^\text{ab} = \mathbb{Q}^\text{cycl}$) and some facts about inverse limits and Galois theory. $\mathbb{Q}^\text{cycl} = \bigcup_n \mathbb{Q}(\zeta_n)$, where $\zeta_n$ is the n-th root of unity. Recall that $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^\ast$. Hence we have

$ \displaystyle \text{Gal}(\mathbb{Q}^\text{cycl}/\mathbb{Q}) = \varprojlim_n \text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong \varprojlim_n (\mathbb{Z}/n\mathbb{Z})^\ast = \hat{\mathbb{Z}}^\ast$

Thanks to the following sources for help on this post:

• Conversation with Prof Minhyong Kim
• Lenstra, Profinite groups and The Idele Class Group
• Fields and Galois Theory, JS Milne
• p-adic Numbers, Fernando Gouvea
  

Thoughts on a QSM proof the Kronecker-Weber Theorem

As I discussed in my last posts, our whole interest in Bost-Connes systems is due to its interaction with the class field theory of certain number fields. In the case of $\mathbb{Q}$, which I'm covering in my MSci project, we can recover the full class field theory and get a description of the maximal abelian/cyclotomic extension of $\mathbb{Q}$ without reference to field extensions. To be precise, we have

Thrm 1 (Prop 3.33 Connes & Marcolli) The quotient $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]/(\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \cap J)$ is a field isomorphic to $\mathbb{Q}^\text{cycl}$ and the ideal $(\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \cap J)$ in $C^\ast(\mathbb{Q}/\mathbb{Z})$ is equal to ideal generated by the $\pi_m$ in $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$, where $J$ and $\pi_m$ are defined in my last post.

Despite this result, Connes & Marcolli state that its still an open problem to find a proof the of the Kronecker-Weber theorem using methods from Quantum Statistical Mechanics. After some googling today, I couldn't find any more information on this, and [perhaps due to my ignorance] I can't figure out why it would be so difficult. I must not understand something. But it would seem to be that given a finite abelian extention of the rationals $\mathbb{K}$ our task would be to find an ideal $K$ such that $(\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \cap J) \subset K$ and so that $\mathbb{K} \sim \mathbb{Q}[\mathbb{Q}/\mathbb{Z}]/K \subset \mathbb{Q}[\mathbb{Q}/\mathbb{Z}]/(\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \cap J) \sim \mathbb{Q}^\text{cycl}$. I obviously don't know much about the details of implementing it, but it might be something to think about as I study the proof of the that main class field theory result of the Bost Connes system. Any thoughts?

If you ask me where I’m going, I’ll tell you where I’ve been.

This post is a brief recap of my work on my MSci project at UCL since July 2010. This post was hastily written. Please excuse any grammatical errors or typos.

Since I've now stated the significance of the Bost-Connes system, its time to consider the next step. But it's now mid-January, and due-dates are fast approaching for my MSci project. Thus, I thought I would take some time to write a short summary of everything I've covered since I started in July. I've actually not done much, and I'm ashamed of how little work I managed to do in such a long time. In any case, lets go over where I've been:

• Quantum Statistical Mechanics
  Basic definitions (no examples), KMS states, symmetry groups. There's a lot of details I missed, particularly about the relationship between GNS representations and extremal KMS states. I have books by Bratteli and Robinson on Operator Algebras and Quantum Statistical Mechanics and Haag's Local Quantum Physics to help me out there.
  
  
• $\mathbb{Q}$-lattices and Commensurability
  Basic definitions, and conversations with Professor Kim to help build some intuition, inversion and division.
  
  
• $C^\ast$-algebras
  Basic definitions and few examples, though I'm missing quite a few details that I should like, particularly GNS representations. (Gelfand-Naimark Theorem would be nice, but not really needed here.)
  
  
• Profinite Completion of $\mathbb{Z}$
  I spent quite a few hours trying to figure out Pontryagin duality and how $\text{Hom}(\mathbb{Q}/\mathbb{Z},\mathbb{Q}/\mathbb{Z})$. I would like to go over this again, using more sophisticated methods (category theory) and better understand how $C(\hat{\mathbb{Z}}) = C^\ast(\mathbb{Q}/\mathbb{Z})$
  
  
• Category Theory, Groupoids, and \'{E}tale Groupids
  Categories, Functors, Groupoids, a few examples, \'{E}tale Groupids (a topology imposed on a groupoid), which I don't think I fully understand and would like to review.
  
  
• Convolution Algebras and $C^\ast$ groupoid algebras for an \'{E}tale Groupid
  Basic definition on convolution, general discussion of the $C^\ast$ completion of a group/oid algebra via a Haar measure (without rigorously defining the Haar measure), discussion of why an Etale Groupid does not need this, and the completion of a $C^\ast$ groupoid algebra this way via a maximal norm of the image of a representation onto a Hilbert space.
  
  
• Generators and Relations for Bost Connes System via semigroup crossed products
  Proof that the $C^\ast$ groupoid algebra for the Bost Connes System is a semigroup crossed product (I don't yet understand the proof fully, despite conversations with Dr Javier Lopez), which is vitally important, because later research shows that this is the ``right'' way to describe Bost-Connes type systems. Proof of the generators and relations via the semigroup crossed product.
  
  
• (Briefly) Hecke Algebras and the Bost Connes System
  Brief definition and description of the Bost Connes system via Hecke Algebras, as they were used in the original paper.
  
  
• (Briefly) The Arithmetic Subalgebra and its analysis inspired from Eisenstein Series
  I didn't spend too much time on this, because I \underline{think} it's more important for generalizations of the Bost-Connes system then for the 1-dim case, and I don't plan to discuss the generalizations in detail. In any case, I went over the results of trigonometric analogue of the Eisenstein series to study the arithmetic subalgebra of the Bost-Connes system. Its easiest to see this algebra exist from the Hecke algebra perspective (I think). Weil's boook on Elliptic Functions contains a lot of the ideas needed to do this in detail.
  
  
• Class Field Theory and the action of the KMS states of the Bost Connes System
  Very brief discussion of adeles, ideles, and the global Artin map, and statement about how the Bost Connes System can recover a presentation of the maximal abelian extension of $\mathbb{Q}$ without reference to field extensions. Plan to generalize this result. See my last post, as well my notes from my October talk at the Undergrad Maths Colloquium
  
  

So what's next? Besides filling in the details for the above (I'm embarrassed that despite taking so long to cover that, I glossed over a lot of details), I essentially have two items left I want to cover: 1) the proof of the theorem in my last post and 2) a summary of the on-going work on generalizations of the BC system. The latter may require learning about Shimura varieties. Both will certainly require a better understanding of class field theory: the adeles, ideles and Artin maps. So that is where I'm going!

Class Field Theory & the Bost-Connes System

In this post I describe the classification of the KMS states of the Bost-Connes system and their relationship to the class field theory of $\mathbb{Q}$, additionally, we show how the rational subalgebra gives us a presentation of $\mathbb{Q}^\text{cycl}$ without any reference to field extensions. Finally, we describe the ``plan'' for generalizing these methods to an abitrary number field.

In all this talk about groupoids and algebras, one may have forgotten that the Bost Connes system $\mathcal{A}_1 = C^{\ast}(\mathcal{G}_1/\mathbb{R}_{+})$ (with $\mathcal{G}_1$ being the groupoid of the commensurability classes of 1dQLs) is a Quantum Statistical Mechanical system with time evolution $\sigma_t$. The ``proper'' way to study the QSM system $(\mathcal{A}_1,\sigma_t)$ is to look at its KMS states. As oft stated in my blog, the KMS states intertwine with the Galois group of the maximal Abelian extension of $\mathbb{Q}$. We can finally state this result properly, but first I need to pull a few quick ideas from Class Field Theory (ideas that I don't understand properly myself). First we want to define the ring of adeles $\mathbb{A}_\mathbb{Q}$ of $\mathbb{Q}$ (I'll define this for a general number field in another blog).
Consider the $p$-adic integers, $\mathbb{Z}_p$, which is the inverse limit of $\mathbb{Z}/p^n\mathbb{Z}$. Each $\mathbb{Z}/p^n\mathbb{Z}$ has the discrete topology, so the Tychonov's theorem says that $\mathbb{Z}_p$ has a compact topology, it also has field of fractions $\mathbb{Q}_p$. The adeles $\mathbb{A}_\mathbb{Q}$ are the restricted product of $\mathbb{Q}_p$ over $\mathbb{Z}_p$ for all primes $p$, where ``restricted'' means that elements of $\mathbb{A}_\mathbb{Q}$ are the infinite sequences $a = (a_2, a_3, \ldots)$ such that all by finitely many $a_p \in \mathbb{Z}_p$.
The group of Id\'{e}les is the invertible elements $\mathbb{A}^{\ast}_\mathbb{Q}$. But this isn't enough for us. We need the id\'{e}le class group $\mathcal{C}_\mathbb{Q} = \mathbb{A}^{\ast}_\mathbb{Q} / \mathbb{Q}$. Class Field/Algebraic Number Theory says that we have a group homomorphism, the global Artin homomorphism:

$\displaystyle \theta : \mathcal{C}_\mathbb{Q} \rightarrow \text{Gal}(\mathbb{Q}^\text{cycl}/\mathbb{Q})$

This homomorphism is surjective, and its kernel is a connected component of the identity in $\mathcal{C}_\mathbb{Q}$, which we'll denote $\mathcal{D}_\mathbb{Q}$. This means we have an isomorphism. For $\mathbb{Q}$, it turns out that $\mathcal{C}_\mathbb{Q} / \mathcal{D}_\mathbb{Q} = \hat{\mathbb{Z}}^{\ast}$, so we have the following isomorphism:

$\displaystyle \theta: \hat{\mathbb{Z}}^{\ast} \rightarrow \text{Gal}(\mathbb{Q}^\text{cycl}/\mathbb{Q})$

One should note that I don't understand any of the above class field theory, and I've only looked up enough details to be able to state the isomorphism. We're now ready to state the classification of the KMS states and the main result of the Bost-Connes system (not to mention my project!) This is Theorem 3.32 in the monograph by Connes and Marcolli, first stated in Hecke algebras, type III factors and phase transitions with spontaneous symmetry breaking in number theory by Bost and Connes. Recall that that $\xi_\beta$ denotes the set of extremal KMS states at inverse temperature $\beta$.

Thrm 1 The KMS states for the Quantum Stastical Mechanical System $(\mathcal{A}_1,\sigma_t)$ behave as follows:

• $\xi_\beta$ is a singleton for all $0 < \beta \leq 1$. The KMS state, when restricted to the subalgebra $\mathcal{B}_\mathbb{Q}$ has values:

  $\displaystyle \phi_\beta(e(a/b)) = b^{-\beta} \prod_{p \; \text{prime},\; p|b} \left( \frac{1-p^{\beta-1}}{1-p^{-1}} \right)$

  
  
• For $1 < \beta \leq \infty$, the elements of $\xi_\beta$ are indexed by invertible 1dQLs modulo scaling $\rho \in \hat{\mathbb{Q}}$. On $\mathcal{B}_\mathbb{Q}$ they take the values ($\zeta_{a/b}$ being roots of unity):

  $\displaystyle \phi_{\beta,\rho}(e(a\b)) = \frac{1}{\zeta(\beta)} \sum_{n=1}^{\infty} n^{-\beta} \rho ( \zeta_{a/b}^n)$

  Hence the partition function of the system is the Riemann zeta function. (I do not understand the argument or its significance.)
  
• The extreme zero-temperature KMS states $\xi_\infty$ have the property that:

  $\displaystyle \phi(\mathcal{A}_{1,\mathbb{Q}}) \subset \mathbb{Q}^{\text{cycl}} \; \text{for all}\; \phi \in \xi_\infty$

  Moreover, the global Artin isomorphim interacts with the Galois action on the values of the KMS state as follows:

  $\displaystyle \gamma(\phi(f)) = \phi ( \theta^{-1}(\gamma) f)$

  
  

A proof of that theorem is not given in the monograph (but it is found elsewhere, at the least in terms of Hecke algebras in the aforementioned paper!) I do intend to cover the proof in my project. Additionally we can actual use this idea to give a presentation of $\mathbb{Q}^\text{cycl}$. Consider $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \subset C^\ast(\mathbb{Q}/\mathbb{Z}) \sim C(\hat{\mathbb{Z}})$.
If you let $u(r)$ be the canonical basis of $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$, and let

$\displaystyle \pi_m = \frac{1}{m} \sum_{r \in \mathbb{Q}/\mathbb{Z},\; mr=0} u(r)$

For $m > 1$ be idempotents. Finally, let $J \subset C^\ast(\mathbb{Q}/\mathbb{Z})$ by the ideal generated by the $\pi_m$. I won't prove it here, but it turns out that $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \cap J$ is equal to the ideal generated by the $\pi_m$ in $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$. Moreover, $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] / (\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \cap J)$ is a field isomorphic to $\mathbb{Q}^\text{cycl}$
Connes and Marcolli give a ``plan of action'' for extending this result to study the class field theory of a general number field $\mathbb{K}$. I give it below:

1. For $\mathbb{K}$, construct a quantum statistical mechanical system $(\mathcal{A}_\mathbb{K},\sigma_t)$ that has the Dedekind zeta function $\zeta_\mathbb{K}(\beta)$ as a partition function, and whose symmetries induce an action of $\mathcal{C}_\mathbb{K}/\mathcal{D}_\mathbb{K}$ on the KMS states.
   
2. Find an arithmetic subalgebra $\mathcal{A}^\text{arithm}_\mathbb{K}$ that has the aforementioned interaction with the class id\'{e}le group and global Artin isomorphism: $\phi \circ \alpha(f) = \sigma(\alpha)\phi(f)$
   
3. Compute the presentation of the algebra $\mathcal{A}^\text{arithm}_\mathbb{K}$ and the extremal zero-temperature KMS states in terms of their values on $\mathcal{A}^\text{arithm}_\mathbb{K}$.
   

In addition to proving the above theorem, I also hope to discuss (in a high level) the recent work on implementing that plan.

Describing the key Sub-algebra of the Bost-Connes System - Part III

Third in a three part series of me trying to describe the Bost-Connes algebra. Part I: $\mathbb{Q}$-Lattices and the Presentation. Part II: Equivalent description in terms of a Hecke Algebra. Part III: Describing the key Sub-algebra.

In my last post I briefly described how the Bost-Connes system was originally described as a Hecke algebra $\mathcal{H}_{\mathbb{C}}(P_{\mathbb{Q}}^{+},P_{\mathbb{Z}}^{+})$, where $P_R$ is the functor that takes an abelian ring $R$ to a group of 2 by 2 matrices over $R$. If you'll recall, we arrived at that Hecke algebra by taking the complexification of $\mathcal{H}_{\mathbb{Q}}(P_{\mathbb{Q}}^{+},P_{\mathbb{Z}}^{+})$. It turns out that this algebra is where all the KMS & Galois group intertwining-magic happens (which I hope to describe in my next post). The purpose in this post (and in section 4.4 of the monograph by Connes & Marcolli, which this post is following) is to describe the algebra in terms of functions on $\mathbb{Q}$-lattices, as this approach is the one used in generalizations. We'll denote this subalgebra by $\mathcal{A}_{1,\mathbb{Q}}$.
The description here is actually quite similar to modular forms: a function $f$ on the space of 1dQL is homogeneous of weight $k$ if it satisfies $f(\lambda(\Lambda, \phi)) = \lambda^{-k} f(\Lambda,\phi)$. E.g., the space of functions on 1dQLs of weight zero is $C(\hat{\mathbb{Z}})$. For $a \in \mathbb{Q}/\mathbb{Z}$, we can define a class of functions on 1dQLs of weight zero by:

$\displaystyle e_{1,a}(\Lambda,\phi) = c(\Lambda) \sum_{y \in \Lambda + \phi(a)} y^{-1$

When $\phi(a) \neq 0$. $e_{1,a}(\Lambda,\phi) = 0$ for $\phi(a) = 0$ or $\phi(a) \in \Lambda$. Also, $c(\Lambda) = \frac{1}{2\pi i} | \Lambda |$, $|\Lambda|$ the covolume of the lattice. These $e_{1,a}$'s are the key functions.

Thrm 1 (Connes & Marcolli 3.30) The functions $e_{1,a}$ have the following properties:

1. The $e_{1,a}$ for $a \in \mathbb{Q}/\mathbb{Z}$ generate the group ring $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$.
2. $\mathcal{A}_{1,\mathbb{Q}}$ is generated by the $e_{1,a}$ and the $\mu_n$ and $\mu_n^{\ast}$ from the previous post.
3. The complexification of $\mathcal{A}_{1,\mathbb{Q}}$ is the $C^{*}$-algebra $\mathcal{A}_1$, generated by the above elements.
   

The proof relies on relations between the $e_{1,a}$ and their generalizations (I do not fully understand the details), and is related to similar work on Eisenstien series (Weil, Elliptic Functions according to Eisenstein and Kronecker contains similar work). While I cannot prove much of anything here, I hope to describe those relations and give a sketch of their proofs.
We can generalize these functions as follows: for $a \in \mathbb{Q}/\mathbb{Z}$, define:

$\displaystyle \epsilon_{k,a}(\Lambda, \phi) = \sum_{y \in \Lambda + \phi(a)} y^{-k$

When $\phi(a) \neq 0$, otherwise $\epsilon_{k,a}(\Lambda, \phi) = \lambda_k c(\Lambda)^{-k}$, with $\lambda_k = (2^k -1)\gamma_k$, and $\gamma_k$ defined by the generating series:

$\displaystyle \frac{1}{e^x - 1} = 1 - \frac{x}{2} - \sum_{j=1}^{\infty} \gamma_{2j}x^{2j$

The $\epsilon_{k,a}$ are of weight $k$. We define the weight $0$ functions by $e_{k,a} = c^k \epsilon_{k,a}$.
The functions $e_{k,a}$ can actually be expressed as polynomials of $e_{1,a}$. For instance, if you let $p_1(u)=u$ and $p_{k+1}(u) = \frac{1}{k}(u^2 - \frac{1}{4}) p^{\prime}_k (u)$, then $e_{k,a} = p_k(e_{1,a})$. The proof of this statement apparently follows from basic formulas for the trigonometric analog of the Eisenstein series. Connes & Marcolli cite the book by Weil, ch2, but I've yet to digest that material (or the proof that they give).
Our $e_{k,a}$ can also be expressed in terms of our generators $e(r)$ of $\mathcal{A}_1$. First, consider them as functions on 1dQLs by $e(r)(\Lambda,\phi) = \text{exp}(2 \pi i \rho(r))$ where $(\Lambda,\phi) = \lambda(\mathbb{Z},\rho)$. For $a \in \mathbb{Q}/\mathbb{Z}$ and $n > 0$ such that $na = 0$, we have the following relation:

$\displaystyle e_{1,a} = \sum_{k=1}^{n-1}\left( \frac{k}{n} - \frac{1}{2} \right) e(ka) $

I have one more relation I'd like to state, but before I can I need to define a new concept. Let $\pi_n$ by the characteristic function of the set of all 1dQL's $(\Lambda,\phi)$ such that the restriction of the map $\phi$ to the $n$-torsion points of $\mathbb{Q}/\mathbb{Z}$, $\phi_n$, is 0. (If this is the case, we say that the $\mathbb{Q}$-lattice $(\Lambda,\phi)$ is divisible by $n$.) Now for a given $N > 0$, the functions $e_{k,a}$ satisfy:

$\displaystyle \sum_{Na = 0} e_{k,a} = \gamma_k \sum_{d | N} C_{k,d} \pi_d $

Where

$\displaystyle C_{k,d} = (2^k - 2)f_1(d) + N^k f_{1-k}(d)$

$f_1$ the Euler totient function, and

$\displaystyle f_k (n) = n^k \, \prod_{p \, \text{prime}, \, p|n} (1-p^{-k})$

The proof of the main theorem relies on these 3 relations. If we let $\mathcal{B}_{\mathbb{Q}}$ be the algebra generated over $\mathbb{Q}$ by the $e_{1,a}$, then our polynomial relation says that $e_{k,a} \in \mathcal{B}_{\mathbb{Q}}$. Our relation with the $e(r)$ says that $\mathcal{B}_{\mathbb{Q}}$ is a subalgebra of the group ring $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$. We use that last relation to show that our characteristic functions $\pi_n$ belong to $\mathcal{B}_{\mathbb{Q}}$, and then to show that all $e(r)$ are in $\mathcal{B}_{\mathbb{Q}}$ as well, so that $\mathcal{B}_{\mathbb{Q}} = \mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$. The second and third parts of the theorem follow easily from that.
This section has a lot of details with Eisenstein series, et cetera, that I did not work out. As this section is more to do with how to generalize the Bost-Connes system, I didn't want to spend too much time on it; it may be tangential to my goals for my project. Though, I may come back to it later on. In the meantime, my next post will be describing the interaction with class field theory.

Equivalent description in terms of a Hecke Algebra - Part II

Second in a three part series of me trying to describe the Bost-Connes algebra. Part I: $\mathbb{Q}$-Lattices and the Presentation. Part II: Equivalent description in terms of a Hecke Algebra. Part III: Describing the key Sub-algebra.

In this post I hope to very briefly describe the Hecke Algebra formulation of the Bost-Connes System. I'm embarrassed to say it, but I was initially afraid of the Hecke Algebra description (the wikipedia page didn't include much information I could understand). Now that I've read through it, I've realized that 1) it's not that difficult and 2) it's not really used in the more interesting generalizations to Complex and Real Multiplication (Shimura varieties and ${\mathbb{Q}}$-lattices seem much more important.) Nevertheless, I thought I'd talk about because it was the way the system was formulated in the original 1995 Paper ``Hecke Algebras, Type III Factors and Phase Transitions with Spontaneous Symmetry Breaking in Number Theory''. So for any ring $R$, we define:

$\displaystyle P_R := \left\{ \begin{pmatrix} 1 & b \\ 0 & a \end{pmatrix} : a, b \in R, a\; \text{invertible} \right\}$

And let $\Gamma_0 = P_\mathbb{Z}^+ = \left\{ \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} : n \in \mathbb{N} \right\}$ and $\Gamma = P_\mathbb{Q}^+ = \left\{ \begin{pmatrix} 1 & a \\ 0 & k \end{pmatrix} : a, k \in \mathbb{Q}^+ \right\}$. We'll be looking at the coset space $\Gamma/\Gamma_0$ (actually the double coset $\Gamma_0 \char`\\ \Gamma/\Gamma_0$, but we aren't so worried about that). You have to be careful to watch the left and right cosets here. First note that the left action of $\Gamma_0$ on $\Gamma/\Gamma_0$ has finite orbits. To see this, let $\gamma = \begin{pmatrix} 1 & a \\ 0 & k \\ \end{pmatrix} \in \Gamma$. Then $\begin{pmatrix} 1 & n \\ 0 & 1 \\ \end{pmatrix} \gamma \begin{pmatrix} 1 & m \\ 0 & 1 \\ \end{pmatrix} = \begin{pmatrix} 1 & n + a + mk \\ 0 & k \end{pmatrix}}$ for $n,m \in \mathbb{N}$. As $m$ varies, $mk$ only takes finitely many values modulo $\mathbb{Z}$, the number depending on the $b$, where $k = \frac{a}{b}$. Thus we do get a finite orbit for $\gamma \Gamma_0$. In fact, the same holds for the right action. (``Hecke Algebras, Type III Factors and Phase Transitions with Spontaneous Symmetry Breaking in Number Theory'' pg 17) We define the Hecke algebra $\mathcal{H}_\mathbb{Q} (\Gamma,\Gamma_0)$ as the convolution algebra of finitely supported functions $f: \Gamma_0 \char`\\ \Gamma \rightarrow \mathbb{Q}$ such that

$\displaystyle f(\gamma \gamma_0) = f(\gamma), \;\;\; \text{for all}\;\gamma\in\Gamma, \; \gamma_0\in\Gamma_0$

. The convolution is given by:

$\displaystyle f_1 f_2 (\gamma) = \sum_{g \in \Gamma_0 \char`\\ \Gamma} f_1(\gamma g^{-1})f_2(g)$

And the involution by:

$\displaystyle f^*(\gamma) = \overline{f(\gamma^{-1})}$

We can complexify this algebra using a tensor product with $\mathbb{C}$ to get:

$\displaystyle \mathcal{H}_\mathbb{C}(\Gamma,\Gamma_0) = \mathcal{H}_\mathbb{Q}(\Gamma,\Gamma_0) \otimes_\mathbb{Q} \mathbb{C}$

Similiar to before, we have a representation on the Hilbert space $\mathcal{H} = \ell^2 (\Gamma_0/\Gamma)$ defined by

$\displaystyle (\pi(f)\psi)(\gamma) = \sum_{g \in \Gamma_0/\Gamma} f(\gamma g^{-1}) \psi(g)$

Where $\psi : \Gamma_0/\Gamma \rightarrow \mathbb{C} \in \ell^2(\Gamma_0/\Gamma)$. This allows the same $C^*$-algebra completition as before, and the time evolution is given by

$\displaystyle \sigma_t(f)(\gamma) = \left(\frac{L(\gamma)}{R(\gamma)} \right)^{-it} f(\gamma)$

. Where $L(\gamma)$ is the cardinality of the left $\Gamma_0$ orbit of $\gamma \in \Gamma/\Gamma_0$ and $R(\gamma) = L(\gamma^{-1})$. The aforementioned paper actually derives the same set of generators and relations that I described in my last post, hence this Hecke algebra is the same as the algebra we built from commensurability on 1dQL's modulo scaling.
Next post will describe the key subalgebra of the BC system.

Presentations of the Bost-Connes Algebra - Part I

First in a three part series of me trying to describe the Bost-Connes algebra. Part I: $\mathbb{Q}$-Lattices and the Presentation. Part II: Equivalent description in terms of a Hecke Algebra. Part III: Describing the key Sub-algebra.

Our goal here is to present the algebra of the Bost-Connes system in terms of generators and relations (Prop 3.23, Noncommutative Geometry, Quantum Fields, and Motives).
Let $\mathcal{G}=\{(r,\rho) : r \in \mathbb{Q}^{*}_{+}, \rho \in \hat{\mathbb{Z}}\; \text{such that} \; r \rho \in \hat{\mathbb{Z}}\}$ be our groupoid of commensurable 1dQL's modulo scaling, and let $C^{\ast}[\mathcal{G}]$ be the $C^{\ast}$-completion of the convolution algebra of continuous complex-valued functions on $\mathcal{G}$ with compact support. It's not too difficult to see that $C^{\ast}[\mathcal{G}]$ contains $C(\hat{\mathbb{Z}})$ (continuous complex-valued function on $\hat{\mathbb{Z}}$). Pontryagin duality (which I do not fully understand, but I'm not worried about that at this point) gives us an isomorphism between $C(\hat{\mathbb{Z}})$ and $C^{\ast}$ group algebra $C^{\ast}[\mathbb{Q}/\mathbb{Z}]$. So let $e_{\gamma}, \gamma \in \mathbb{Q}/\mathbb{Z}$ be the canonical additive bases for $C^{\ast}[\mathbb{Q}/\mathbb{Z}]$. E.g., for an $f \in C^{\ast}[\mathbb{Q}/\mathbb{Z}]$, we can write

$\displaystyle f = \sum_{\gamma \in \mathbb{Q}/\mathbb{Z}} \lambda_{\gamma} e_{\gamma}\;\;\; \lambda_{\gamma} \in \mathbb{C}$

So that $f(\gamma) := \lambda_{\gamma}$. Our $e_{\gamma}$ can define a function $e_{\gamma} : \mathcal{G} \rightarrow \mathbb{C}$ by

$\displaystyle e_{\gamma}(r,\rho) = \begin{cases} \text{exp}(2\pi i \rho(\gamma)) & r=1 \\ 0 & \text{otherwise} \end{cases}$

Hence $e_{\gamma} \in C^{\ast}[\mathcal{G}]$. Moreover, they behave under convolution in $C^{\ast}[\mathcal{G}]$ just as they would as the basis for the C* group algebra, namely, we have:

1. $e_{\gamma_1} e_{\gamma_2} = e_{\gamma_1 + \gamma_2}$
2. $e_{0} = \mu_1$, id est, $e_{0}$ maps to unity in $C^{\ast}[\mathcal{G}]$. (See edit in previous post)
3. For an $f \in C^{\ast}[\mathcal{G}]$, the involution of $f$ is defined as $f^{\ast}(r,\rho) := \overline{f(r^{-1},r \rho)}$. This gives us $e_{\gamma}^{\ast} = e_{-\gamma}$.
   

The proofs of the above are straightforward, just following from the definitions (keep in mind that the multiplication is convolution in $C^{\ast}[\mathcal{G}]$).
Now, recall that the space of 1dQL's modulo scaling looks like $\hat{\mathbb{Z}}$ and that our groupoid $\mathcal{G}$ describes the commensurability relation of 1dQL's modulo scaling, so that $C(\hat{\mathbb{Z}}) \simeq C^{\ast}[\mathbb{Q}/\mathbb{Z}]$ captures quite a bit of the information in $C^{\ast}[\mathcal{G}]$. The commensurability condition is captured by a semigroup cross product with $\mathbb{N}$. I have not been able to figure out how, nor have I found a description in any of the literature I've read. But I can give a description of the action. For an $f \in C(\hat{\mathbb{Z}})$, let

$\displaystyle \alpha_n(f) (r, \rho) = \begin{cases} f(n^{-1} \rho) & \rho \in n \hat{\mathbb{Z}} \\ 0 & \text{otherwise} \end{cases}$

Recall the $\mu_n$'s from the last post, id est,

$\displaystyle \mu_n(r,\rho) = \begin{cases} 1 & r=n \\ 0 & \text{otherwise} \end{cases}$

By treating our $f \in C(\hat{\mathbb{Z}})$ as $f(r,\rho) := f(\rho)$, we can conjugate $f$ with $\mu_n$ and its involution and get:

$\displaystyle \mu_n f \mu_n^{\ast} = \alpha_n(f)$

Again, the multiplication here is convolution in $C^{\ast}[\mathcal{G}]$, and the proof is straightforward. As discussed in my last post, the $\mu_n$'s also behave nicely under the following relations (proofs are still straightforward):

3. $\mu_n \mu_m = \mu_{nm}$
4. $\mu_n^{\ast} \mu_n = \mu_1$, where $\mu_1$ is the unity/multiplicative identity.
   

So we see then that out $e_{\gamma}$'s describe $C(\hat{\mathbb{Z}})$, and that our $\mu_n$'s capture the semigroup action to implement commensurability, hence together the two are sufficient to describe $C^{\ast}[\mathcal{G}]$. We're just missing one last relation: while we can conjugate $f \in C(\hat{\mathbb{Z}})$, we haven't shown what happens when we conjugate a basis element $e_{\gamma}$ by $\mu_n$. The proof of this relation is slightly more involved than the previous ones, so I'll describe $\mu_n e_{\gamma} \mu_n^{\ast}$ step by step. Lets do the far right convolution first:

$\displaystyle e_{\gamma}\mu_n^{\ast}(r,\rho) = \sum_{r_1 r_2 = r} e_{\gamma}(r_1,r_2 \rho) \mu_n^{\ast}(r_2,\rho) = \sum_{r_1 r_2 = r} e_{\gamma}(r_1,r_2 \rho) \mu_n(r_2^{-1},r_2 \rho)$

We see that $r_2 = n^{-1}$, $r_1 = 1$ and $r=n^{-1}$ for the sum to be nonzero. So we have:

$\displaystyle e_{\gamma}\mu_n^{\ast}(r,\rho) = \begin{cases} \text{exp}(2\pi i (\frac{1}{n} \rho)(\gamma) ) & r = n^{-1} \\ 0 & \text{otherwise} \end{cases}$

For the full conjugation, we have:

$\displaystyle \mu_n e_{\gamma}\mu_n^{\ast}(r,\rho) = \sum_{r_1 r_2 = r} \mu_n(r_1,r_2 \rho) \; e_{\gamma}\mu_n^{\ast}(r_2,\rho)$

We see again that $r_2 = n^{-1}$, $r_1 = n$ and thus $r=1$ for the sum to be nonzero. So we have

$\displaystyle \mu_n e_{\gamma}\mu_n^{\ast}(r,\rho) = \begin{cases} \text{exp}(2\pi i (\frac{1}{n} \rho)(\gamma) ) & r = 1 \\ 0 & \text{otherwise} \end{cases}$

We can express this purely in terms of our $e_{\gamma}$'s. Note that for $\gamma = \frac{a}{b} + \mathbb{Z}$ (I'll henceforth omit the $+ \mathbb{Z}$), we have $n$ $\delta$'s such that $n\delta = \gamma$, namely $\delta_k = \frac{a + kb}{nb}$ for $k=0, \ldots, n-1$. For such a $\delta$, $e_{\delta}(1,\rho) = \text{exp}(2\pi i \rho (\frac{1}{n} \gamma) ) = \text{exp}(2\pi i (\frac{1}{n} \rho) (\gamma) )$. Hence we have

$\displaystyle \mu_n e_{\gamma}\mu_n^{\ast}(r,\rho) = \frac{1}{n} \sum_{n \delta = \gamma} e_{\delta}(r,\rho) = \begin{cases} \text{exp}(2\pi i (\frac{1}{n} \rho)(\gamma) ) & r = 1 \\ 0 & \text{otherwise} \end{cases} $

That, along with relations 1-5, describe the Bost-Connes algebra. The time evolution in terms of this description is given by:

$\sigma_t(\mu_n) = n^{it}\mu_n, \;\;\; \sigma_t(e_{\gamma}) = e_{\gamma}$

Additionally, we have that $C^{\ast}[\mathbb{G}] \simeq C(\hat{\mathbb{Z}}) \rtimes \mathbb{N}$. It is also stated in Noncommutative Geometry, Quantum Fields, and Motives that $C^{\ast}[\mathbb{G}]$ is Morita equivalent to $C_0(\mathbb{A}_f) \rtimes \mathbb{Q}^{\ast}_{+}$, as shown in Laca's ``From Endomorphisms to Automorphisms And Back: Dilations and Full Corners'', but I'm not at all worried about that.
I am worried about how the cross product with $\mathbb{N}$ that I've described above implements the commensurability condition on 1dQL's modulo scaling. In fact, I don't even know what it means for the crossed product action to implement commensurability (my initial guess was that $\alpha_n(f)$ would be constant on commensurable 1dQLs, but I've yet to make sense of this condition). The main monograph states the action in terms of a function on 1dQL's:

$\displaystyle \alpha_n(f)(\Lambda,\phi) = f(n\Lambda,\phi)$

For $(\Lambda, \phi)$ divisible by $n$ in a specific sense, $0$ otherwise.. This implementation is mentioned in ``From Physics to Number theory via Noncommutative Geometry'', ``$\mathbb{Q}$-Lattices: Quantum Statistical Mechanics and Galois Theory'', and ``Lectures on Arithmetic Noncommutative Geometry''. I've read a few papers describing the transitions from Hecke algebras to the semigroup crossed product $C(\hat{\mathbb{Z}}) \rtimes \mathbb{N}$, including Laca's ``Semigroups of $\ast$-Endomorphisms, Dirichlet Series, and Phase Transitions'' and ``A Semigroup Crossed Product Arising In Number Theory'' but nothing describing the groupoid of the commensurability relation 1dQL's modulo scaling. I would appreciate any help anyone can provide. (As well as any comments on the rest of the maths in this post - I was fumbling about quite a bit.)
Next post will be a brief account of an equivalent formulation via Hecke Algebras.

Confusion in the Presentation of the Bost-Connes System

I had planned to write a long glorious post about the generators and relations of the Bost Connes system, the same formulated via Hecke algebras, and the key arithmetic subalgebra. Alas, I got stuck on a rather silly point: verifying the relations. Recall that we're considering the groupoid $\mathcal{G} = \{(r,\rho) : r \in \mathbb{Q}^{*}_{+}, \rho \in \hat{\mathbb{Z}}\; \text{such that} \; r \rho \in \hat{\mathbb{Z}}\}$ and the $C^{\ast}$-completion $C^{\ast}[\mathcal{G}]$ of its convolution algebra:

$\displaystyle \{f:\mathcal{G} \rightarrow \mathbb{C} \;|\; f\; \text{is continuous and with compact support.}\}$

Let me start by stating the generating function I'm having trouble with: for all $n \in \mathbb{N}$, let $\mu_n : \mathcal{G} \rightarrow {\mathbb C}$ be the functions:

$\displaystyle \mu_n(r,\rho) = \begin{cases} 1 & r=n \\ 0 & \text{otherwise} \end{cases} $

These functions, under some relations (and plus another set of functions), are suppose to generate $C^{\ast}[\mathcal{G}]$. Nevermind how that is suppose to happen, I don't even understand the relations! Specifically, it's stated that

$\displaystyle \mu_n^{\ast} \mu_n = 1 \;\text{for all}\; n \in \mathbb{N}$

(There are several other relations, only this one is causing trouble.)

The multiplication here is convolution, not pointwise multiplication; it's clearly 0 under pointwise multiplication. So lets try this out. First recall that for $f \in C^{\ast}[\mathcal{G}]}$, we have $f^{\ast}(g) = \overline{f(g^{-1})}$, and convolution means that $f_1 f_2 (g) = \sum_{g_1 g_2 = g} f_1(g_1) f_2(g_2)$. Next, recall what inversion means in $\mathcal{G}$. A $g \in \mathcal{G}$ means that $g = (r,\rho)$, and if you write down the multiplication $(r_1,\rho_1) \circ (r_2,\rho_2) = (r_1 r_2, \rho_2)$ when $r_2 \rho_2 = \rho_1$, it's not heard to see that $(r,\rho)^{-1} = (r^{-1},r\rho)$. Assuming I didn't make in error in that, we can write down $\mu_n^{\ast} \mu_n$:

$\displaystyle \mu_n^{\ast} \mu_n (r,\rho) = \sum_{(r_1,\rho_1) \circ (r_2,\rho_2) = (r, \rho)} \mu_n^{\ast}(r_1,\rho_1)\mu_n (r_2,\rho_2)$

Now, the condition on the sum, $(r_1,\rho_1) \circ (r_2,\rho_2) = (r, \rho)$ clearly means that $\rho_2 = \rho$, and the condition to perform the multiplication then forces $\rho_1 = r_2 \rho$. So we have $(r_1,r_2 \rho) \circ (r_2,\rho) = (r, \rho)$ and we can restate this as:

$\displaystyle \mu_n^{\ast} \mu_n (r,\rho) = \sum_{r_1 r_2 = r} \mu_n^{\ast}(r_1)\mu_n (r_2) = \sum_{r_1 r_2 = r} \mu_n(r_1^{-1})\mu_n (r_2)$

Where $r_1 r_2 \rho, r_2 \rho \in \hat{\mathbb{Z}}$. I've dropped the $\rho$'s as the $\mu_n$'s don't care about them. The term $\mu_n(r_1^{-1})\mu_n (r_2)$ is 0 except when $n=r_1^{-1}=r_2$, but since $r = r_1 r_2$, we require $r=1$ in order to get a nonzero sum. We end up with:

$\displaystyle \mu_n^{\ast} \mu_n (r,\rho) = \mu_1 (r,\rho) \; \text{instead of} \; \mu_n^{\ast} \mu_n (r,\rho) = 1 $

Where am I going wrong?

Edit: I'm not going wrong. The problem is that I don't understand unity in a convolution algebra. Here, $1$ is the function such that for all $f \in C^{\ast}[\mathcal{G}]$, $1f = f1 = f$ under convolution. Writing down the convolution of $\mu_1$ shows that it works. Thanks to Professor Kim for questioning me on that. Right, now back to the rest of the writeup!

Note: definitions for the functions came from ``Fun with $\mathbb{F}_1$'' page 11 and ``Noncommutative Geometry, Quantum Fields, and Motives'' page 417-418, 424. I didn't use Hecke algebras here, but that formulation is in Proposition 18 of ``Hecke Algebras, Type III Factors and Phase Transitions with Spontaneous Symmetry Breaking in Number Theory'' by Bost and Connes.

Groupoid Convolution Algebras and Their Completions to C∗-algebras

This post is also online as a PDF. You can download it here.

So we were working with the groupoid $\mathcal{U}_1 = \{(r,\rho) : r \in \mathbb{Q}^{*}_{+}, \rho \in \hat{\mathbb{Z}}\: \text{such that} \: r \rho \in \hat{\mathbb{Z}}\}$, where composition is defined by $(r_1,\rho_1) \circ (r_2,\rho_2) = (r_1 r_2, \rho_2)$ when $r_2 \rho_2 = \rho_1$. Much in the same way we can define a group algebra, we can also define a groupoid convolution algebra. To get some intuition, let us first talk about the group algebra. For some field $\mathbb{F}$ and [finite] group $G$, the group algebra $\mathbb{F}[G]$ is simply the $\mathbb{F}$-vector space with the group elements as it's basis. Since $G$ is the basis, for each $x \in \mathbb{F}[G]$, we can write

$\displaystyle x = \sum_{g \in G} x_g g \; \text{where} \; x_g \in \mathbb{F} $

The product is simply the group operation (acting on the basis, and thus all elements of the algebra) and is obviously only commutative if the underlying group is Abelian. To make it explicit, let $x$ be as above, $y = \sum_{g \in G} y_g g$ with $y_g \in \mathbb{F}$. Then

$\displaystyle x y = \sum_{g, h \in G} (x_g y_h) gh$

Now, any $x \in \mathbb{F}[G]$ defines a function $f_x : G \rightarrow \mathbb{F}$ by $ g \mapsto x_g$. Conversely, any function $f : G \rightarrow \mathbb{F}$ defines an element in the group algebra $f \mapsto \sum_{g \in G} f(g) g$. So that we can also describe the group algebra $\mathbb{F}[G]$ as the set $ \{ f: G \rightarrow \mathbb{F} \} $. The product on this algebra of functions then looks like:

$\displaystyle f_x \ast f_y (g) = \sum_{hk =g} x_h y_k = \sum_{hk=g} f_x(h) f_y(k)$

It's not hard to generalize this to a topological group, we simply need the sum $\sum_{hk=g} f_x(h) f_y(k)$ to converge in the field $\mathbb{F}$. Hence we're going to need to talk about the topology of $\mathbb{F}$. So for simplicity, let's just deal with the complex numbers $\mathbb{C}$ with the standard topology. Then $\mathbb{C}[G]$ almost makes sense, we just require that the functions $f \in \mathbb{C}[G]$ have finite support, id est, that they are nonzero only on a finite number of points.

Problem 1 Is the product on $\mathbb{C}[G]$ still finite for a topological group $G$ if we extend it to functions $f$ which have compact support?

Now, for an Etale groupoid $\mathcal{G}$, we can use the above to define its convolution algebra:

Definition 2 (Convolution Algebra of an Etale Groupoid) Let $\mathcal{G}$ be an Etale groupoid. The groupoid (or convolution) algebra $\mathbb{C}[\mathcal{G}]$ is the set

$\displaystyle \{ f: \mathcal{G} \rightarrow \mathbb{C} \suchthat \text{f is continuous with compact support} \} $

With pointwise addition, scalar multiplication, and the product

$\displaystyle f_1 \ast f_2 (g) = \sum_{g_1 \circ g_2=g} f_1(g_1)f_2(g_2)$

Problem 3 Why is continuity needed here? And what is it about Etale groupoids that allow finite sums when the functions only have compact, not finite, support?

Note that since $f \in \mathbb{C}[\mathcal{G}]$ is simply a function $\mathcal{G} \rightarrow \mathbb{C}$, we can define a $\ast$-operation by complex conjugation, id est, $f^{\ast}(g) = \overline{f(g)}$.

0.1. Completion into a $C^{\ast}$-algebra

According to Connes and Marcolli, we can complete $\mathbb{C}[\mathcal{G}]$ into a $C^{\ast}$-algebra via the following procedure, but first we need to define a few extra terms. Recall from my writeup on groupoids that $\mathcal{G}^0$ denotes set of objects of the groupoid, id est, $\mathcal{G}^0 = \{ \gamma \circ \gamma^{-1} \suchthat \gamma \in \mathcal{G}\}$. For all $y \in \mathcal{G}^0$, let $\mathcal{G}_y$ denote the set of all morphisms whose target and source is $y$. Then all elements in $\mathcal{G}_y$ can be composed with each other, and the category-theoretic definition of a groupoid gives us an identity and inverse, so $\mathcal{G}_y$ is a group.

Definition 4 (Isotropy Group) $\mathcal{G}_y$ is called the Isotropy Group at $y \in \mathcal{G}^0$.

Now for each isotropy group $\mathcal{G}_y$ (or any group, really), we can associate to it a Hilbert space

$\displaystyle \ell^2(\mathcal{G}_y) = \{ \alpha:\mathcal{G}_y \rightarrow \mathbb{C} \suchthat \sum_{g \in \mathcal{G}_y} |\alpha(g)|^2 < \infty \}$

With the inner product:

$\displaystyle <\alpha_1,\alpha_2> = \sum_{g \in \mathcal{G}_y} \alpha_1(g)\overline{\alpha_2(g)}.$

Let $f \in \mathbb{C}[\mathcal{G}]$. For all $\alpha \in \ell^2(\mathcal{G}_y)$ we can define a map:

$\displaystyle \beta_{\alpha} : \mathcal{G}_y \rightarrow \mathbb{C} \; \text{by} \; \beta(g) = \sum_{\substack{g_1 g_2 = g \\ g_1,g_2 \in \mathcal{G}_y}} f(g_1)\alpha(g_2)$

Problem 5 Show that $\beta_{\alpha} (g)$ converges, and show that $\beta_{\alpha} \in \ell^2(\mathcal{G}_y)$.

Problem 6 Hence we have a map $B_f:\ell^2(\mathcal{G}_y) \rightarrow \ell^2(\mathcal{G}_y)$ by $B(\alpha) = \beta_{\alpha}$. Show that $B_f \in \mathcal{B}(\ell^2(\mathcal{G}_y))$, id est, show that $B_f$ is a bounded [linear] operator.

Problem 7 Now we have a map $\pi_y : \mathbb{C}[\mathcal{G}] \rightarrow \mathcal{B}(\ell^2(\mathcal{G}_y))$ by $\pi_y(f) = B_f$ for all $y \in \mathcal{G}^0$. Show that this map is a representation of the groupoid convolution algebra, id est, that it's a homomorphism of algebras.

This representation allows us to define a norm on the groupoid convolution algebra. Specifically, we have:

$\displaystyle \|f\| = \sup_{y\in\mathcal{G}^0} \| \pi_y(f) \|_{\mathcal{B}(\ell^2(\mathcal{G}_y))$

This has all been a bit confusing. Let us recap briefly: we noted that at each object in the groupoid we can find an isotropy group. We built a Hilbert space on that isotropy group and then constructed a representation on $\mathbb{C}[\mathcal{G}]$ to that Hilbert space. We defined the norm of an $f \in \mathbb{C}[\mathcal{G}]$ in our groupoid convolution algebra as the supremum of the operator norms of all such representations of $f$.
Connes and Marcolli state that if $\mathcal{G}^0$ is compact then $\mathbb{C}[\mathcal{G}]$ has a unit, while Khalkahali holds that $\mathbb{C}[\mathcal{G}]$ has a unit if and only if $\mathcal{G}^0$ is finite. Does compactness of the objects in an Etale groupoid imply that the objects are finite? In any case, since the algebra has a unit, we can complete it. Call its completion $C^{\ast}[\mathcal{G}]$.

Problem 8 Show that $\mathbb{C}[\mathcal{G}]$ is incomplete.

Problem 9 Prove that $C^{\ast}[\mathcal{G}]$ is a $C^{\ast}$-algebra, namely, show that the $C^{\ast}$-identity, $\| f f^{\ast} \| = \| f \|^2$, holds. We've established that it's a complete $\ast$-algebra.

Problem 10 Show that our groupoid $\mathcal{U}_1$ is unital.

In a previous post I described a time evolution on $C^{\ast}[\mathcal{U}_1]$ using the ratio of the covolumes of the underlying pair of commensurable $\mathbb{Q}$-lattices. Thus, except for the problems mentioned here, I've fully constructed the $C^{\ast}$-algebra Bost-Connes system.

0.2. What's next?

Considering my embarrassingly slow rate of progress, I think it's unreasonable for me to expect that I would progress beyond a detailed, in-depth description of the Bost-Connes system and it's interaction with the Class Field Theory of $\mathbb{Q}$. (I would very much appreciate advice from both of my supervisors on this point!) Which is unfortunate, as I would eventually like to discuss the two other systems mentioned in the Connes and Marcolli monograph (including the system that interacts with the Class Field Theory of an imaginary number field). Moreover, I'd especially like to visit the work of Ha and Paugam in their paper "Bost-Connes-Marcolli systems for Shimura varieties" which generalizes the construction of these systems in the way described in Connes' monograph. But again, I think it's unlikely that I would have time for that after finishing my study, and filling in the details, of the original Bost-Connes system. To that end, I still need to:

1. State and prove the explicit presentation of the Bost-Connes System. I've been wholly unable to make sense of the proof given in the Connes and Marcolli monograph, and have spent some time staring at the Hecke algebra description (I still don't know what a "Hecke algebra" is.) given in the 1994 paper by Bost and Connes. Additionally, Dr Javier Lopez, who has graciously offered to assist me, suggested that I have a look through Fun with $\mathbb{F}_1$ to gather more intuition.
2. Describe the ``arithmetic sub-algebra'' of the Bost-Connes System.
3. Class Field Theory. I figure I need to understand at least enough CFT to be able to prove the Kronecker-Weber theorem and to state properly its interactions with the CFT of $\mathbb{Q}$, which I mentioned in in my talk at the Undergrad Maths Colloquium.
4. Prove the statements about such interactions.
   

My next post (on 1.) should come soon.

Convolution algebras are commutative.

Edit: Nope, they aren't.

I didn't realize this until I was writing my talk.  I wanted to post last night, but I didn't have time.  Indeed, I barely have time now!

In any case, let $\mathcal{U}$ be an $\'{E}tale$ groupoid.  We define its convolution algebra as the set of all functions $f: \mathcal{U} \to \mathbb{C}$ with the convolution product

$f_1 \ast f_2 (g) = \sum_{g_1 g_2=g} f_1(g_1) f_2(g_2)$

The problem isn't so strikingly obvious if we let $\mathcal{U}$ be the graph $X \cross X$ of an equivalence relation, e.g., the algebra of the Bost-Connes system.  In this case, the convolution is:

$f_1 \ast f_2 (x,y) = \sum_{x \sim r \sim y} f_1(x,r) f_2 (r,y)$

The pairs $(x,r)$ and $(r,y)$ are always going to be swapped at some point, so this product is commutative too.

Does the completion into a $C^{\ast}$-algebra somehow not preserve commutativity?  That would just be weird.

In any case, I don't have much more time to keep thinking about it, as I have to run to UCL to give my talk on the Bost Connes system!   Can anyone help me out here?  It wouldn't damage the analysis of the states on the Bost Connes system if it were commutative, but it would certainly weaken the motivation for using KMS states, et cetera.  (Not to mention make my talk seem a bit silly.  I suppose if Connes can present in a commutative algebra while giving motivation for a noncommutative one, then so can I!)

Talk at the UCL Undergrad Maths Colloquium on the Bost Connes system

Apologies again for the lack of posts.  I'm currently working on a rather long writeup, detailing C* algebras, the completion of the convolution algebras, the presentation of and the arithmetic subalgebra of the bost connes system, and it's interaction with the class field theory of Q.  Perhaps its a bit too ambitious for a single blog post. I'm trying to pin down exactly what I'm confused on.  I've not been able to prove the presentation of the system, or even understand what I read on it.

In any case, I'm giving a talk at the UCL Undergrad Maths Colloquium on my project, and thought I'd post my notes here.  You can download them from the colloquium from this link.

Another Hiatus

My blog has been a bit sleepy these past few days.  It will sleepy for a bit longer, I'm not doing any lecture or project work in preparation for this weeks mathematics GRE subject test on October 9th.  I'll be taking it in Leeds.

I figure I need a particularly high score, given my weak exam results.  I took my first practice test while ago and I was only in the 60th percentile.  I hope with some more work this week I can bring it up to at least the 85th!

Terry Tao on Measure Theory

any UCL folks who have taken or are about to take Measure Theory could probably benefit from following Tao's notes on the subject. good stuff!

http://terrytao.wordpress.com/2010/08/31/course-announcement-245a-real-analysis/

http://terrytao.wordpress.com/2010/09/04/245a-prologue-the-problem-of-measure/

http://terrytao.wordpress.com/2010/09/09/245a-notes-1-lebesgue-measure/

http://terrytao.wordpress.com/2010/09/19/245a-notes-2-the-lebesgue-integral/

http://terrytao.wordpress.com/2010/09/25/245a-notes-3-integration-on-abstract-measure-spaces-and-the-convergence-theorems/

And the category link:

http://terrytao.wordpress.com/category/teaching/245a-real-analysis/

Groupoids and Equivelance Relations

So in working out the details and the structure of the groupoids of commensurability relations on 1dQLs, I realized I could generalize the method a bit more.

Given a set S with an equivalence relation ~, we can form a groupoid consisting of all ordered pairs of equivalent elements, i.e., all pairs $(s_1,s_2)$ where $s_1 \sim s_2$. The composition is then $(s_1,s_2) \circ (s_3,s_4) = (s_1,s_4)$ defined when $s_2 = s_3$.

It's such an obvious example of a groupoid that I'm embarrassed that I missed it. I'm sure it's in a textbook somewhere. I would have saved a lot of time if I had found and read that textbook a week ago.

Edit
Oh! Look! There it is! It's on wikipedia. fml.

The Étale Groupoids of the Bost-Connes System

PDF version of the past 3 posts.

Recall that a 1-dimensional $ {\mathbb{Q}}$-lattice (henceforth called a 1dQL) can be denoted by $ {(\frac{1}{\lambda} \mathbb{Z}, \frac{1}{\lambda} \rho)}$ with $ {\lambda}$ a positive real and $ {\rho \in End(\mathbb{Q}/\mathbb{Z})=\hat{\mathbb{Z}}}$. Let $ {\mathcal{Q}_1}$ denote the set of commensurability relations of such lattices. Then $ {\mathcal{Q}_1}$ consists of ordered pairs $ {(\Lambda_1,\Lambda_2)}$ of commensurable 1dQLs. There's a natural composition of elements, namely:

$ {(\Lambda_1,\Lambda_2) \circ (\Lambda_3, \Lambda_4) = (\Lambda_1,\Lambda_4)}$ defined for $ {\Lambda_2=\Lambda_3}$.

It's not too hard to see that each element has an inverse. Indeed $ {(\Lambda_1,\Lambda_2)^{-1} = (\Lambda_2,\Lambda_1)}$, so that $ {(\Lambda_2,\Lambda_1) \circ (\Lambda_1,\Lambda_4) = (\Lambda_3, \Lambda_4)}$. Hence we have a groupoid. The units $ {\mathcal{Q}_1^0}$ consists of the pairs $ {(\Lambda, \Lambda)}$, id est, just the set of 1dQLs.
We can actually give a much more clear description of this groupoid. Let $ {\mathcal{G}_1 = \{(r,\rho,\lambda) : r \in \mathbb{Q}^{*}_{+}, \, \rho \in \hat{\mathbb{Z}}, \, \lambda \in \mathbb{R}^{*}_{+} \; such \; that \; r \rho \in \hat{\mathbb{Z}} \}}$ be the groupoid with the composition

$ {(r_1,\rho_1,\lambda_1) \circ (r_2,\rho_2,\lambda_2) = (r_1 r_2, \rho_2, \lambda_2)}$ defined for $ {r_2 \rho_2 = \rho_1}$, $ {r_2 \lambda_2 = \lambda_1}$.

and inverse elements

$ \displaystyle (r,\rho,\lambda)^{-1} = \left(\frac{1}{r},r \rho, r \lambda \right) $

Proposition. The map $ {\phi : \mathcal{G}_1 \rightarrow \mathcal{Q}_1}$ by

$ \displaystyle (r,\rho,\lambda) \mapsto \left( \left( \frac{1}{r \lambda} \mathbb{Z}, \frac{1}{\lambda} \rho \right), \left( \frac{1}{\lambda} \mathbb{Z}, \frac{1}{\lambda} \rho \right) \right)$

is a groupoid isomorphism. (Connes, Marcolli; Noncommutative Geometry, Quantum Fields, and Motives. Lemma 3.21-2.)
Proof: First note that the two 1dQLs, $ {\left( \frac{1}{r \lambda} \mathbb{Z}, \frac{1}{\lambda} \rho \right)}$ and $ {\left( \frac{1}{\lambda} \mathbb{Z}, \frac{1}{\lambda} \rho \right)}$ are commensurable. Indeed, $ {\mathbb{Q} \frac{1}{r \lambda} \mathbb{Z} = \frac{1}{\lambda} \mathbb{Q} \mathbb{Z}}$.

Second, note that every pair of commensurable 1dQLs is of this form. To see this, let $ {(\frac{1}{\lambda_1} \mathbb{Z}, \frac{1}{\lambda_1} \rho_1)}$ and $ {(\frac{1}{\lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2)}$ be another pair of commensurable 1dQLs. $ {\frac{1}{\lambda_2}}$ then must be a rational multiple of $ {\frac{1}{\lambda_1}}$, id est, $ {\lambda_1 = r \lambda_2}$ for a strictly positive rational $ {r=\frac{a}{b}}$ (see previous posts on $ {\mathbb{Q}}$-lattices). Additionally, we must have $ {\frac{1}{\lambda_1} \rho_1 = \frac{1}{\lambda_2} \rho_2 \mod (\frac{1}{\lambda_1} \mathbb{Z} + \frac{1}{\lambda_2} \mathbb{Z})}$ Factoring out by $ {\lambda_2}$, this yields $ {\frac{1}{r} \rho_1 = \rho_2 \mod \frac{1}{a} \mathbb{Z}}$, and then $ {a\rho_2 - b \rho_1 = 0}$ and finally $ {\rho_1 = r \rho_2}$, so that $ {\frac{1}{\lambda_1}\rho_1=\frac{1}{r \lambda_2} r \rho_2 = \frac{1}{\lambda_2}\rho_2}$. Hence the map is surjective.

Finally, note that the map preserves the groupoid composition and inversion. The composition

$ \displaystyle \left( (\frac{1}{r_1 \lambda_1} \mathbb{Z}, \frac{1}{\lambda_1} \rho_1),(\frac{1}{\lambda_1} \mathbb{Z}, \frac{1}{\lambda_1} \rho_1) \right) \circ \left( (\frac{1}{r_2 \lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2),(\frac{1}{\lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2) \right)$

is only defined in $ {\mathcal{Q}_1}$ if $ {\lambda_1=r_2 \lambda_2}$ and $ {\rho_1 = r_2 \rho_2}$. In that case, it equals

$ \displaystyle \left( (\frac{1}{r_1 \lambda_1} \mathbb{Z}, \frac{1}{\lambda_1} \rho_1),(\frac{1}{\lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2) \right)=\left( (\frac{1}{r_1 r_2 \lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2),(\frac{1}{\lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2) \right) $

Which is exactly $ {\phi(r_1 r_2, \rho_2,\lambda_2)}$. Additionally,

$ \displaystyle \phi(\frac{1}{r},r \rho, r \lambda ) = \left( (\frac{1}{\lambda} \mathbb{Z}, \frac{1}{\lambda} \rho), (\frac{1}{r \lambda} \mathbb{Z}, \frac{1}{\lambda} \rho) \right)$

Which is the inverse of $ {\left( (\frac{1}{r \lambda} \mathbb{Z}, \frac{1}{\lambda} \rho), (\frac{1}{\lambda} \mathbb{Z}, \frac{1}{\lambda} \rho) \right)}$ $ \Box$


Proposition. $ {\mathcal{G}_1}$ is an Étale groupoid. (I do not yet know the significance of this statement.)

Proof: Since $ {\mathcal{G}_1}$ has the product topology, $ {\mathbb{Q}_{+}^{*}}$ has the subspace topology, multiplication in all spaces is continuous, and division in $ {\mathbb{Q}_{+}^{*}}$ is continuous, the groupoid operations are homeomorphisms. $ \Box$


But we're not done yet. All the action in the Bost-Connes system comes not from $ {\mathcal{G}_1}$, but from the commensurability relations of 1dQLs modulo scaling, id est, from $ {\mathcal{Q}_1/\mathbb{R}_{+}^{*}}$. The unit set $ {(\mathcal{Q}_1/\mathbb{R}_{+}^{*})^0}$ is simply $ {\hat{\mathbb{Z}}}$ in this case. As you might expect, we can model this groupoid with an Étale groupoid similar to $ {\mathcal{G}_1}$: Let

$ \displaystyle \mathcal{U}_1 = \{ (r,\rho) : r \in \mathbb{Q}_{+}^{*})^0, \rho \in \hat{\mathbb{Z}} \; such \; that \; r \rho \in \hat{\mathbb{Z}}\}$

With composition

$ { (r_1,\rho_1) \circ (r_2,\rho_2) = (r_1 r_2,\rho_2)}$ defined when $ {\rho_1 = r_2 \rho_2}$.

.
Proposition. The map $ {\gamma : \mathcal{U}_1 \rightarrow \mathcal{Q}_1/\mathbb{R}_{+}^{*}}$ by

$ \displaystyle (r,\rho) \mapsto \left( (\frac{1}{r}\mathbb{Z},\rho), (\mathbb{Z},\rho) \right)$

is a groupoid isomorphism. (Connes, Marcolli; Noncommutative Geometry, Quantum Fields, and Motives. Prop 3.22.)
Proof: The proof of this is similar to the last one, so I won't repeat myself. Oddly enough, Connes & Marcolli give a short proof of the first isomorphism, and a long proof of this one. $ \Box$

Since the topology of the groupoids largely comes from $ {\mathbb{R}}$, they are also locally compact (at least, I think that's the reason.)

1. Why This Matters

It's important to keep the larger goal of the project in mind when working out these smaller details. My goal is to describe the construction of the Bost-Connes system, as I did here. Connes & Marcolli give an explicit description in terms of generators and relations of the Bost-Connes algebra in Noncommutative Geometry, Quantum Fields, Motives [1]. Additionally, Marcolli gives the same description in Lectures on Arithmetic Noncommutative Geometry [2], Connes & Marcolli give it again in $ {\mathbb{Q}}$-Lattices: Quantum Statistical Mechanics and Galois Theory [3], and Bost & Connes give it in Hecke Algebras, Type III Factors and Phase Transitions with Spontaneous Symmetry Breaking in Number Theory [4]. However, only [1] and [4] provide a proof for the description, and the latter only in terms of Hecke Algebras. [1], however, starts from convolution algebras of these Étale groupoids, and shows the relation with the Hecke algebra description in the paper by Bost & Connes [4]. Hence, it's important that I understand and am able to describe these groupoids.

2. What's next

The next step is a careful study of the convolution and C* groupoid algebras of the above structures, leading to a proof of the explicit description of the Bost-Connes system (Prop. 3.23 in Noncommutative Geometry, Quantum Fields, Motives). Following that, I'll need to understand the relation with the Hecke Algebras given by Bost & Connes and the symmetries of the system. Then I can begin a study of the class field theory of $ {\mathbb{Q}}$ and learn how to apply the Bost-Connes system.

Eventually, I'll study the two other Quantum Statistical Mechanical systems, but first I need to finish the Bost-Connes system and its applications, and learn its corresponding sub algebras and Shimura varieties, which will be vital to generalizing the techniques used to apply it.

3. Sources

Noncommutative Geometry, Quantum Fields, Motives. Connes, Marcolli
Lectures on Arithmetic Noncommutative Geometry. Marcolli (the usual two)
A Homology Theory for Étale groupoids. Marius Crainic and Ieke Moerdijk
Abstract Algebra. Dummit, Foote (appendix on category theory.)
Category Theory. Steve Awodey
Wikipedia on Groupoids

A Minor Problem with the Étale groupoids of Bost-Connes

EDIT: I was confused and delusional when writing this blog; the problem discussed below does not actually exist. I didn't know how to take the inverse in the groupoids mentioned. It's actually quite simple and will be discussed in a later post. This entry can be safely ignored.

I had hoped to move beyond my study of the groupoids mentioned in the construction of the Bost Connes system on Monday. Yet it wasn't til Tuesday that I had fully understood what an Étale groupoid was. Though I had thought I had understood the aforementioned groupoids and the proofs of their isomorphisms, I realized in the middle of my writeup last night that I was still confused about things. I worked out some details on paper, but there's one thing left that I'm hung up on. It has to do with the inversion of the elements in the groupoid. Here's the detail isolated all by itself:

Given a $\rho \in \hat{\mathbb{Z}}$ and a positive rational $r$ such that $r \rho \in \hat{\mathbb{Z}}$. I need that $\frac{1}{r} \rho \in \hat{\mathbb{Z}}$ as well. But I don't see how this could possible be true, at least not when I'm thinking of $\hat{\mathbb{Z}}$ as an inverse limit. However....

As I proved earlier, $\hat{\mathbb{Z}}$ is isomorphic to $End(\mathbb{Q}/\mathbb{Z})$, so we can think of $\rho$ as being a homomorphism $\rho : \mathbb{Q}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$. Now if $ r \rho$ is still such a homomorphism ($(r \rho )(x) =r \rho(x)$) then I can't see any reason why $\frac{1}{r} \rho$ wouldn't be. It would preserve addition and the identity on $\mathbb{Q}/\mathbb{Z}$ just fine (I think).

So that brings me to another question...is there any positive rational $r$ and homomorphism $\rho \in End(\mathbb{Q}/\mathbb{Z})$ such that $r \rho$ is not a homomorphism? If not, then why does Connes state that requirement?

Some thoughts immediately after writing this post
A homomorphism $\rho \in End(\mathbb{Q}/\mathbb{Z})$ must preserve torsion, id est, there's a smallest $n$ such that an $x \in \mathbb{Q}/\mathbb{Z}$ will have $n x = 0$. $\rho$ must preserve $n \rho(x) = 0$ if it's going to be a homomorphism. Multiplication by a natural number $k$ is an a homomorphism on $\mathbb{Q}/\mathbb{Z}$, but multiplication by $\frac{1}{k}$ is not. Hence I have my $r=k$ where $r \rho \in End(\mathbb{Q}/\mathbb{Z})$ but $\frac{1}{r} \rho$ is not. And thus I still have a problem inverting elements in the groupoids mentioned. I'll have to spell out this problem more explicitly in another post.

Groupoids

I'm continuing a series of posts working out some details regarding some details in the groupoids used to construct the Bost-Connes system.

Groupoids

A groupoid is essentially a group, but one where the binary operation is not defined for all pairs of elements. That is, a groupoid is set G such that for all $g \in G$, there exist an inverse $g^{-1}$, together with a partial function $ \ast : G \times G \to G$ ($\ast$ is not necessarily a binary operation; it needn't be defined for all ordered pairs.) and with the following properties:

• Associativity: If $a * b$ and $b * c$ are defined, then $(a*b)*c=a*(b*c)$ is also defined
• Inverse: $a*a^{-1}=a^{-1}*a$ is always defined.
• Units: If $a*b$ is defined, then $a*b*b^{-1}=a$ and $a^{-1}*a*b=b$

While the above definition of a groupoid is familiar to one who has any experience with groups, it's sometimes more useful to think of it as a category: Let $G$ be a groupoid, and define $G^0$ to be the set of all elements of the form $g * g^{-1}$. Now let $x, y \in G^0$. If $f \in G$ is an element such that $y * f * x$ is defined, then we say $f \in G^1$ and $f: x \to y$.

In this case, it's not to hard to see that composition works: if we have $f:x \to y$ and $g:y \to z$, then $gf$ is defined and so is $z*gf*x$. Additionally, note that for every $x \in G^0$, the identity arrow $1_x$ is simply $x$. One oddity we have is that the inverse map $x * f^{-1} * y$ is defined (remember that since $x, y \in G^0$, we have $x=a*a^{-1}$ for some $a$, so that $x^{-1}=x$.) so that we have an inverse arrow $f^{-1} : y \to x$. Thus any groupoid can be considered as a category where all arrows have inverses.

Moreover, any category where all arrows have inverses is also a groupoid, so the two characterizations are actually equivalent definitions. Here's how: Given a groupoid $C$ in the category theoretic sense (id est, we have objects $C^0$ and arrows $C^{-1}$), we can denote all arrows $f: x \to y$ by $G(x,y)$. Let G be the disjoint union of all such $G(x,y)$. Then inverses are defined for all elements, and arrow composition because the partially defined groupoid operation.

Now lets take a groupoid $G$ (with objects, now called units $G^0$ and arrows/morphisms $G^{1}$). Recall the notation $G(x,y) = \{ f \in G^1 : f:x \to y. x,y \in G^0\}$. Let $x \in G^0$. Then $G(x,x)$ is a group (recall the example of a category of a group from earlier!) We call it the isotropy group at $x$ and denote it by $I_x$.

Before we continue, lets give an example. Let $GL(\mathbb{R})$ denote all the real invertible matrices under matrix multiplication. then $GL(\mathbb{R})^0$ is in bijective correspondence with the natural numbers, as for each natural number $n$ we have the corresponding $n \times n$ identity matrix. $G(n,m)$ is empty when $n \neq m$, but $G(n,n)$ is the general linear group $GL_n(\mathbb{R})$ of all $n \times n$ invertible matrices.

Finally, we say a groupoid $G$ is a topological groupoid when $G^0$ and $G^1$ are topological spaces, and all the corresponding maps on them (e.g. for $f \in G^1$ we have the domain $d(f)$, $d : G^1 \to G^0$, arrow composition $ \circ : G^1 \times G^1 \to G^1$, et cetera) must be continuous. Or equivalently, inversion and the groupoid operation must be continuous maps. A topological groupoid $G$ is said to be Étale if the domain map $d : G^1 \to G^0$ is a local homeomorphism (and thus all the other maps too), or equivalently, if inversion and the groupoid operation are local homeomorphism. Examples of these will come in the next post.

Sources
Noncommutative Geometry, Quantum Fields, Motives. Connes
Lectures on Arithmetic Noncommutative Geometry. Marcolli (the usual two)
A Homology Theory for Étale groupoids. Marius Crainic and Ieke Moerdijk
Abstract Algebra Dummit, Foote (appendix on category theory.)
Category Theory Steve Awodey

Categories and Functors

In my last post, when describing the setup for the Bost-Connes system, I displayed my lack of understanding of groupoids. In the next few posts I want to clear that up, describe Étale groupoids, and provide some more details on the proofs of the groupoid isomorphisms in Connes' monograph that I mentioned in the last construction. First let me talk about categories.


Categories and Functors
To describe a groupoid, it's good to have the language of category theory in your toolkit. Category theory abstracts the notion of functions, homomorphisms, et cetera and allow you to make statements about those relations. More formally, by a category (actually, a small category) $\mathcal{C}$ I mean a set objects $\mathcal{C}^0$, together with a set of "morphisms" or "arrows" between these objects $\mathcal{C}^1$. For each arrow $f \in \mathcal{C}^1$ we have the domain $d(f) \in \mathcal{C}^0$ and the range $r(f) \in \mathcal{C}^0$. We write $f: A \to B$ to denote that $A=d(f)$ and $B=r(f)$. The arrows are functions on the objects, and we can compose them. Formally, if $f : A\to B$ and $g: B \to C$, then there is an arrow $g \circ f : A \to C$. Additionally, for every $A \in \mathcal{C}^0$ there is an identity array $1_A : A \to A$ such that $f \circ 1_A = f = 1_B \circ f$. Composition is, of course, associative.

As an example, lets pretend we have a well-defined notion of the set of all groups, call it $\mathcal{G}^0$. Between any two groups there exist some homomorphism $\phi$. $\phi$ may be trivial, sending everything to to the identity, but it exists nevertheless. Collect all such $\phi$'s into the set $\mathcal{G}^1$. Then $\mathcal{G}$ is a category.

We can also look at a single group, $G$, and consider it as a category. The object set, $G^0$ contains only the identity element $e$. All the fun happens in the set $G^1$, which contains all the other elements of of $G$. Then "arrow" composition is not only associative, but also has an inverse and the identity is $1_e$. This example will make more sense after I discuss groupoids, but keep it in mind!

One can think of categories as being purely algebraic structures. And like any algebraic structure, we'll want some structure-preserving map between them. This brings us to functors. A functor $F: \mathcal{C} \to \mathcal{D}$ between categories $\mathcal{C}$ and $\mathcal{D}$ is a mapping of objects to objects and arrows to arrows such that it preserves all the expected notions. To be more explicit, a functor meets:

• $F(f:A \to B) = F(f) : F(A) \to F(B)$
• $F(1_A) = 1_{F(A)}$
and finally
• $F(g \circ f) = F(g) \circ F(f)$

So in the case of the category of groups and homomorphisms between them, a functor preserves the groups themselves, but also all homomorphisms of those groups. In the case of the category of a group, a functor is actually just a group isomorphism.

Sources
Noncommutative Geometry, Quantum Fields, Motives. Connes
Lectures on Arithmetic Noncommutative Geometry. Marcolli (the usual two)
A Homology Theory for Étale groupoids. Marius Crainic and Ieke Moerdijk
Abstract Algebra Dummit, Foote (appendix on category theory.)
Category Theory Steve Awodey