Noncommutative de Rham cohomology of finite fields

As my lack of blogs might suggest, I've not made much progress on the Ph.D. front for the past several weeks. I've been reading papers by Drinfeld, et al, trying to crack the Grothendieck-Tiechm\"{u}ller group. I've not understood much. While I'm not giving up on those ideas, I am taking my supervisor's suggestion to work more closely with him on a smaller project. I'm computing the Noncommutative de Rham cohomology of extensions of finite fields.

Finite Fields and Differential Calculi

As a specific example of this, let's look at the finite field ${\mathbb{F}_2}$ and a field extension of degree 2. Readers familiar with basic field theory should know that, in order to extend this field, we need a degree 2 irreducible polynomial in ${\mathbb{F}_2}$, say ${\mu^2 + \mu + 1}$. This polynomial will generate a prime ideal in the polynomial ring ${\mathbb{F}_2[\mu]}$, and the quotient ring ${\mathbb{F}_2[\mu] / (\mu^2 + \mu + 1)}$ will be our new field.So what does this field look like? Essentially, elements of this field are polynomials with the condition that ${\mu^2 = \mu + 1}$. The reader should see that all the elements of the new field are ${0, 1, \mu, \; \text{and}\; \mu+1}$. What we've really done here is adjoined the root of the polynomial ${\mu^2 + \mu + 1}$ to our field. Call this new field ${\mathbb{F}_4}$. It can also be thought of as a vector space over ${\mathbb{F}_2}$ with basis ${\{1, \mu \}}$.Our ``space'' is the finite line, which we're modeling with ${A = \mathbb{F}_2[x]}$. Now how do we get to noncommutative de Rham cohomology? First, we need the differential calculus of 1-forms. Recall from my early post on NCG black holes that we can define a differential calculus over an algebra ${A}$ as the pair ${(d, \Omega^1)}$, with ${\Omega^1}$ a bimodule over ${A}$ and ${d: A \rightarrow \Omega^1}$ a linear map that obeys the product rule ${d(ab) = d(a)\cdot b + a\cdot d(b)}$. Additionally, we require that the set ${\{a \cdot d(b) : a, b \in A\}}$ spans our calculus ${\Omega^1}$.Now, ${\mathbb{F}_2[x]}$ has an obvious action on the polynomial ring ${\mathbb{F}_4[x]}$, so we can think of ${\Omega^1 = \mathbb{F}_4[x]}$ as our calculus of one forms. For an ${f \in A = \mathbb{F}_2[x]}$, the derivative looks like:

$\displaystyle df = \left( f(x + \mu) - f(x) \right) \mu^{-1}$

We also have an NCG notion of an exterior algebra, which gives us spaces of n-forms extended from ${\Omega^1}$. This is how we get a de Rham cohomology. In our case we have a complex:

$\displaystyle \mathbb{F}_2[x] = \Omega^0 \xrightarrow{d^0} \Omega^1 = \mathbb{F}_4[x] \xrightarrow{d^1} \Omega^2 \xrightarrow{d^2} \ldots $

Where ${d^n = d |_{\Omega^n}}$ is the derivative restricted to the calculus of n-forms (exempli gratia, ${d^0}$ is the ${d}$ we defined above.) The n-th de Rham cohomology is then ${H^n(A) = \text{ker}(d^n) / \text{Im}(d^{n-1})}$.But what are ${\Omega^2, d^1}$, et cetera? We'll save that topic for another post. The zeroth cohomology group is just ${H^0 = \text{ker}(d^0)}$ and for now we'll just worry about calculating that.

Finding ${H^0}$

So ${H^0}$ will consist of exactly the polynomials ${f \in F_2[x]}$ such that:

$\displaystyle f(x+u) - f(x) = 0$

Obviously constants fit this. Because we're working in a field of characteristic ${p=2}$, we know that ${(x + a)^p = x^p + a^p}$ is the Frobenius automorphism. This causes me to suspect that only polynomials of the form:

$\displaystyle f_n(x) = x^{2^n} + a_{n-1} x^{2^{n-1}} + \ldots + a_0 x$

can lie in the kernel. We'll prove this later, but first let's see what exactly happens with such a polynomial:

$\displaystyle f_n(x + \mu) = (x+\mu)^{2^n} + a_{n-1} (x+\mu)^{2^{n-1}} + \ldots + a_0(x+\mu) $
$
= x^{2^n} + \mu^{2^n} + a_{n-1}( x^{2^{n-1}} + \mu^{2^{n-1}}) + \ldots + a_0 x + a_0 \mu $

Hence we have

$\displaystyle f_n(x + \mu) - f_n(x) = \mu^{2^n} + a_{n-1} \mu^{2^{n-1}}+ \ldots + a_0 \mu$

From earlier, we have that ${\mu^2 = \mu + 1 = \mu^{-1}}$. So ${\mu^3 = 1}$, more explicitly,

$\displaystyle \mu^r = \begin{cases} 1 & r \equiv 0 \mod 3 \\ \mu & r \equiv 1 \mod 3 \\ \mu^{-1} & r \equiv 2 \mod 3 \end{cases} $

Moreover, we have that ${2^r \equiv 1 \mod 3}$ when ${r}$ is odd, and ${2 \mod 3}$ when ${r}$ is even. Combing all this, we have that for even ${n}$:

$\displaystyle f_n(x+\mu) - f_n(x) = \mu (a_{n-1} + a_{n_3} + \ldots + a_1) + \mu^{-1} (1 + a_{n-2} + \ldots + a_0)$

Hence ${df_{2k} = 0}$ when ${\sum_{i=1}^{k} a_{2(k-i) + 1} = 0}$ and ${\sum_{i=1}^{k} a_{2(k-i)} = 1}$. Similarly, when ${n= 2k + 1}$, we have that ${df_{2k+1} = 0}$ when ${\sum_{i=1}^{k} a_{2(k-i) + 1} = 1}$ and ${\sum_{i=0}^{k} a_{2(k-i)} = 0}$.We can use these formulas to write down a basis for some things in kernel by concentrating on the shortest polynomials with the highest powers. The above formulas say that the shortest possible polynomial in the kernel will have two elements of degree 2 to the power of the same parity. Hence our basis is ${\{ x^{2^n} + x^{2^{n-2}} : n \geq 2\}}$.Now we need to prove that such things are the only thing in the kernel. To see this, let's calculate the action of...

Update 18 Mar 2012

The proof I originally posted here was wrong. I made two mistakes: first, my calculation of $f(x + \mu) - f(x)$ neglected constant terms. Second, I never bothered to check that the coefficient of $x^k$ is zero when $f$ isn't spanned by our powers-of-two basis (I only checked to see if there is a nonzero term, but since we're working in a field of characteristic 2, the sum of nonzero terms can still be zero.) I spent several days trying to correct this proof, but I can't. The statement itself is wrong. ${\{ x^{2^n} + x^{2^{n-2}} : n \geq 2\}}$ is not the basis. I have several counter examples, exempli gratia $f(x) = x^{12} + x^9 + x^6 + x^3$ and $f(x) = x^{20} + x^{17} + x^5 + x^2$ and I believe I can find polynomials in the kernel that have 8, 16, et cetera terms that aren't the sum of smaller things in the kernel. So...I still have some work to do before I can find $H^0$.

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Constructing the Grothendieck-Teichmuller Group

So for the past six or seven months I've been trying to get a copy of the paper ``On quasitriangular Quasi-Hopf algebras and a group closely connected with ${\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})}$'' by V.G. Drinfel'd. My online search uncovered only a Russian copy (which, unfortunately, I don't read). My library searches were equally unfruitful: the British library's copy is ``unavailable'' and I was unable to find a copy at another UK institution. My supervisor was sure he had a copy, but when he discovered that it mysteriously disappeared he encouraged me to request an inter-library loan with my university. I did, and my university said they found a copy in Leeds, but the last ten pages are missing. So - if any readers have a copy of the aforementioned paper: I'd greatly appreciate your lending me it!

In the meanwhile, I've been reading ``On Associators and the Grothendieck-Teichmuller Group'' by Dror Bar-Natan. Among other things, that paper described the construction of said group via a ``Parenthesized Braid Category''. I'll discuss that construction in this post.

1. Braids and Categories

I really don't want to give a precise definition of a braid (I've not yet seen one that doesn't make my head hurt), but hopefully I won't be too wrong if I propose the following:

Def1 1 A braid on ${n}$ strands is a set of ${n}$ oriented curves with distinct starting and ending points where the starting points lie on the same line, each endpoint lies on a separate line, and no curve contains a loop.

For example, the following is a braid:

Clearly we can enumerate the starting and ending points, and treat a braid as a permutation on them. In this way, braids form a category where the braid itself is the morphism and the objects are simply sets to be permuted.

We're going to change our class of objects slightly - we're going to consider parenthesizations of sets. Id est, pretend that the sets have some underlying multiplicative structure. We want to order the sets and denote the order of multiplication. Exempli gratia, the ordering ${(1 \, 2 \, 3)}$ has two parenthesizations: ${((1 \, 2) \, 3)}$ and ${(1 \, (2 \, 3))}$. Our braid morphisms are allowed to act on any parenthesizations.

We aren't done complicated things yet: we're going to modify the morphisms themselves so that they become formal sums:

$\displaystyle \sum_{i=1}^{k} \beta_i B_i$

Where each ${B_i}$ is a braid on ${n}$ strands, all ${B_i}$'s have the same effective permutation, and ${\beta_i}$ belongs to some ${\mathbb{Q}}$-algebra, usually ${\mathbb{Q}}$ or ${\mathbb{C}}$. We're forming the ``algebriod'' over ${\mathbb{Q}}$ or ${\mathbb{C}}$ over the set of braids on ${n}$ strands, with composition defined to be a bilinear map in the same way you would do for a group algebra.

This category is our Parenthesized Braid Category ${\textbf{PaB}}$.

2. Fibred Linear Categories

Our category is fibred. To see this, consider the category ${\textbf{PaP}}$ of parenthesized objects as before, but with simple permutations as the morphisms. Then consider the functor ${\textbf{S} : \textbf{PaB} \rightarrow \textbf{PaP}}$ that maps each object to itself, but takes each formal sum of braids to that one effective permutation. Clearly the set of morphisms in ${\text{PaB}}$ from ${O_1}$ to ${O_2}$ is:

$\displaystyle \text{Hom}_\textbf{PaB}(O_1, O_2) = \textbf{S}^{-1} \left( \text{Hom}_\textbf{PaP}(O_1, O_2) \right)$

Where P is the effective permutation.

One might also note that for a fixed permutation ${P}$, the fibre of morphisms ${\textbf{S}^{-1}(P)}$ is a vector space (actually, its an algebra, because we can compose morphisms as our product). This allows us to define some very ``linear'' notions, for instance:

We can define subcategories: a category is a subcategory of ${\textbf{PaB}}$ if it shares the same collection of objects and each ${\text{Hom}(O_1, O_2)}$ is a subspace of the corresponding ${\text{Hom}_\textbf{PaB}(O_1, O_2)}$.

A subcategory is an ideal if whenever one of two composable ${B_1}$ or ${B_2}$ belongs to it then the composition belongs to it, too. Exempli gratia, let ${\textbf{I}}$ be the ideal where all morphisms ${\sum \beta_i B_i}$ have ${\sum \beta_i = 0}$. Then we can define the quotient ${\textbf{PaB}/\textbf{I}}$ in an obvious way: each set of morphisms ${\text{Hom}(O_1, O_2)}$ is the corresponding quotient in the bigger set of morphisms.

Additionally, we can define idealic powers: ${\textbf{I}^n}$ has morphisms that can be written as compositions of ${n}$ morphisms in ${\textbf{I}}$. We can continue this and take an inverse systems with an inverse limits, in fact, we can even do the ${\textbf{I}}$-${adic}$ completion:

$\displaystyle \widehat{\textbf{PaB}} = \varprojlim_n \textbf{PaB}/\textbf{I}^n$

Finally, we can do a tensor product of ${\textbf{PaB}}$ with itself (in the same way we would do tensor powers of an algebra). For ${\textbf{PaB}^{2\otimes}}$ we define the morphism as the disjoint union:

$\displaystyle \text{Hom}_{\textbf{PaB}^{2\otimes}}(O_1,O_2) = \coprod_{P \in \text{Hom}_{\textbf{PaB}}(O_1, O_2)} \textbf{S}^{-1}(P) \otimes \textbf{S}^{-1}(P)$

Now we can even do a coproduct ${\Delta: \textbf{PaB} \rightarrow \textbf{PaB}^{2\otimes}}$ by ${B \mapsto B \otimes B}$. This coproduct is a functor. (I suspect that there is a bialgebra/hopf algebra structure hidden somewhere in here but I've not thought about it enough yet. Not for this post.)

3. Some Functors

In addition to ${\textbf{S} : \textbf{PaB} \rightarrow \textbf{PaP}}$, we also have a few functors ${\textbf{PaB} \rightarrow \textbf{PaB}}$. We can describe these by their action on a single braid and one can expand them as a linear action on the formal sums we use. First, we have the extension functors ${d_0}$ and ${d_{n+1}}$ that add a single straight strand to the left or right of the braid, respectively.

Second, we have cabling functors ${d_i}$ for ${1 \leq i \leq n}$ on a braid of n strands. This functor simply doubles the ${i}$-th strand.

Finally, we have a strand-removal functor ${s_i}$, which, as you might guess, removes the ${i}$-th strand.

We can now define the Grothendieck-Teichmuller group ${\widehat{\text{GT}}}$. This is the group of all invertible functors ${a: \textbf{PaB} \rightarrow \textbf{PaB}}$ such that ${\textbf{S} \circ a = S}$, ${d_i \circ a = a \circ d_i}$, ${s_i \circ a = a \circ s_i}$, ${\Delta \circ a = a \otimes a \circ \Delta}$. We also require that $a$ leaves the braid constant:

Right! There we are. Sooner or later I actually do some maths, but for now I'm stuck understanding trying to understand definitions, so hence this post.

4. Source

This is all just a reiteration of the material in the aforementioned paper by Bar-Natan.

Climbing Out of Black Holes

Last month I published a two questions that had me stuck inside Majid's ``almost commutative black hole''. We'll start today be answering those questions while trying to avoid insulting myself and my readers, and then move on to calculate the Grassmann connection on our Black Hole Algebra.

1. NCG Vector Bundles For the Schwarzschild Solution

Our ``space'' is the algebra $ {A = k[x_1, x_2, x_3, r, r^{-1}]}$, our ``vector bundle'' projective module $ {\mathcal{E}}$ is the image of the operator given by the idempotent matrix $ {P \in \text{M}_3(A)}$ with $ {(P)_{ij} = \delta_{ij} - \frac{x_i x_j}{r^2}}$, and our calculus of 1-forms $ {\Omega_1}$ is the bimodule generated by symbols $ {\text{d}x_i}$ for $ {i = 1, 2, 3}$. The question was how can $ {\mathcal{E} \subset A^{\oplus 3}}$ by spanned by $ {\omega_i = \text{d}x_i - \frac{x_i \text{d}r}{r}}$ for $ {i = 1, 2, 3}$.
The key here is to realize that the $ {\Omega_1}$ is isomorphic to $ {A^{\oplus 3}}$. Consider $ {A^{\oplus 3}}$ to be the space of row 3-vectors $ {v = (a_1 \; a_2 \; a_3)}$, and give it the ``standard basis'' $ {e_i = (\delta_{1 i} \; \delta_{2 i} \; \delta_{3 i} )}$. In this case, the idempotent acts on the right, and its action on the standard basis is $ {e_i \cdot P = P_{ik}e_k}$. Under the isomorphism given by $ {e_i \mapsto \text{d}x_i}$, we see that $ {\mathcal{E}}$ is spaned by $ {e_i \cdot P \mapsto P_{ik}\text{d}x_k = \text{d}x_i - \frac{x_i \text{d}r}{r} = \omega_i}$ as desired.

2. Grassmann Connections in NCG

Last time we defined a connection on an NCG vector bundle $ {\mathcal{E}}$ as a linear map $ {\nabla:\mathcal{E} \rightarrow \Omega_1 \otimes_{A} \mathcal{E}}$ that obeys certain conditions. We also defined the ``Grassmann Connection'' a map $ {\nabla_P = p \circ \text{d}}$, where $ {P}$ is the idempotent of the projective module. More explicitly, the Grassmann connection is, for a $ {v \in A^{\oplus n}}$

$ \displaystyle \nabla_P (v \cdot P) = \text{d} ( v \cdot P ) \cdot P = v \cdot \text{d} (P) \cdot P + \text{d}v \cdot P$

Where $ {\text{d}}$ acts on $ {v}$ and $ {P}$ component-wise. We had asked how this definition makes sense, as its image appears to lie in $ {\Omega_1^{\oplus n}}$.
My chief problem here was that I don't understand tensor products over a ring (or over an algebra, in this case). The key out of this problem is to realize two facts about tensor product spaces: Given module (over $ {A}$) $ {M}$, $ {N}$, and $ {R}$,

1. $ {(M \oplus N) \otimes R}$ is isomorphic to $ {M \otimes R \oplus N \otimes R}$, by $ {(m, n) \otimes r \mapsto (m \otimes r, n \otimes r)}$, and
2. $ {M \otimes A}$ is isomorphic to $ {M}$ by $ {m \otimes a \mapsto ma}$.
   

Given these two facts, it's not hard to see that $ {\Omega_1^{\oplus n} \cong (\Omega_1 \otimes A)^{\oplus n} \cong \Omega_1 \otimes A^{\oplus n}}$. Then it's easy to see that the map $ {P \circ \text{d}}$ maps to the right space.

3. The Grassmann Connection for our Black Hole Model

We'll calculate the action of the Grassmann Connection on our spanning set $ {\omega_i}$. In this section, we'll identify (by the previously defined isomorphims) the ``standard basis'' $ {e_i}$ with $ {\text{d}x_i}$, so that $ {\omega_i = e_i \cdot P = P_{ik}\text{d}x_k = \text{d}x_i - \frac{x \text{d}r}{r}}$. From the above definition, we have:

$ \displaystyle \nabla_P (\omega_i) = \text{d}e_i \cdot P + e_i \text{d}P \cdot P $

$ {\text{d}e_i = 0}$, as $ {e_i}$ is constant (or equivalently, because $ {\text{d}^2=0}$). So the bulk of the problem lies in calculating the matrix product $ {\text{d}P \cdot P}$. Let's write it out:

$ \displaystyle (\text{d}P \cdot P)_{ij} = (\text{d}P)_{ik} P_{kj} = -\text{d}\left( \frac{x_i x_k}{r^2} \right) \left( \delta_{kj} - \frac{x_k x_j}{r^2} \right) = -\frac{x_i}{r^2} P_{kj}\text{d}x_k$

So that $ {e_i \text{d}P \cdot P = -\frac{x_i}{r^2} P_{kl} P_{lj} dx_k \otimes dx_j = -\frac{x_i}{r^2} \omega_l \otimes \omega_l}$. Here I used the fact that P is idempotent and the generators of our algebra commute. This gives us our connection:

$ \displaystyle \nabla_P (\omega_i) = -\frac{x_i}{r^2} \omega_l \otimes \omega_l$

4. Next Steps

We won't actually be using this connection much. Rather, we're going to define a new one based on the idea that $ {\nabla (\text{d}x_i)}$ should be $ {0}$. From here, we'll begin discussing NCG metrics, and write down one for our black hole.

5. Sources

1. Section 4 of Almost commutative Riemannian geometry: wave operators
2. Email conversations with Majid
3. Atiyah and MacDonald, Introduction to Commutative Algebra
   

Confusion in Black Holes

I've been trying to make my way through Majid's paper, ``Almost commutative Riemannian geometry: wave operators'', particularly the section where he constructs the model for a Schwarzschild black hole. I've not had much success. I'm meant to reconstruct the model over a $ {\mathbb{F}_p}$, but I'm stuck on basic definitions. I'll discuss two of those things in this blog, first the vector bundle (aka projective module) and then the Grassmann connection.

1. NCG Vector Bundle for The Schwarzschild Solution

To construct the model, we start by reconsidering our notion of 3-dimensional space. Rather than thinking of coordinates $ {(x_1, x_2, x_3)}$, we're going to recast ``space'' as a ``coordinate algebra'', in particular, an algebra of polynomials $ {k[x_1, x_2, x_3]}$ over a field $ {k}$ (we'll let $ {k = \mathbb{R}}$ for now, but my task is to redo this section of the paper with $ {k=\mathbb{F}_p}$. Moreover, we're working in a sphere, so we also request that our algebra contain functions rational in $ {r}$, where $ {r^2 = x_1^2 + x_2^2 + x_3^2}$. Hence our ``space'' is the algebra $ {A = k[x_1, x_2, x_3, r, r^{-1}]}$ modded out by the aforementioned relation.

For such an NCG space (nevermind that $ {A}$ is actually commutative here), we define a vector bundle as a projective module. An easy way to get a projective module is to take a few copies of $ {A}$ under the image of an idempotent $ {E \in M_n(A)}$, e.g, let $ {E}$ be such an idempotent, then $ {\mathcal{E} = Im(E)}$ is our vector bundle. In this case we're taking the 3 by 3 matrix:

$ \displaystyle E = \begin{pmatrix} 1 - \frac{x_1^2}{r^2} & - \frac{x_1 x_2}{r^2} & - \frac{x_1 x_3}{r^2} \\ - \frac{x_2 x_1}{r^2} & 1 - \frac{x_2^2}{r^2} & - \frac{x_2 x_3}{r^2} \\ - \frac{x_3 x_1}{r^2} & - \frac{x_3 x_2}{r^2} & 1 - \frac{x_3^2}{r^2} \end{pmatrix}$

Thus our vector bundle is the subspace $ {\mathcal{E} = Im(E) \subset A^3}$. We expect elements of this vector bundle to be ``3-vectors'' with entries from $ {A}$. Yet in the paper, Majid states that $ {\omega_i = \text{d}x_i - \frac{x_i \text{d}r}{r}}$ for $ {i = 1, \, 2, \, 3}$ spans the 2-dimensional bundle $ {\mathcal{E}}$. But (judging by the $ {\text{d}x_i}$ and $ {\text{d}r}$ terms) each $ {\omega_i}$ is in our bimodule of 1-forms $ {\Omega_1}$. Where am I going wrong?

2. Grassmann Connections in NCG

Let's assume I'm not hopelessly confused about that vector bundle thing. Recall that a connection on an NCG vector bundle is a linear map $ {\nabla_{\mathcal{E}}: \mathcal{E} \rightarrow \Omega_1 \otimes \mathcal{E}}$ that obeys the following rule:

$ \displaystyle \nabla_{\mathcal{E}} (a s) = \text{d} a \otimes s + a\nabla_{\mathcal{E}}(s) \; \forall a \in A \; s \in \mathcal{E} $

According to a proposition I've read in Majid's lecture notes, if $ {\mathcal{E} = Im(E)}$ for a projector $ {E \in M_n(A)}$, then we have a ``Grassmann connection''

$ \displaystyle \nabla_{\mathcal{E}} (E v) = E \text{d}(Ev) = E\left(\text{d}(E)v + E(\text{d}v)\right) = E\text{d}(E)v + E(\text{d}v)$

Where $ {v \in A^n}$ and $ {\text{d}}$ acts on $ {v}$ and $ {E}$ component-wise. But the image of the connection is suppose to live in $ {\Omega_1 \otimes \mathcal{E}}$. $ {\text{d}v}$ lives in $ {\Omega_1^n}$, and $ {\text{d}E}$ lives in $ {M_n(\Omega_1)}$. How do we get to our tensor product space $ {\Omega_1 \otimes \mathcal{E}}$ ?

3. Sources

1. Section 4 of Almost commutative Riemannian geometry: wave operators
   

Starting my PhD - Noncommutative Geometry and Black Holes

I've had a great summer holiday. I've done a few programming projects. I've restarted my position with Universal Pictures International. It's time to start writing about my mathematics consistently. My PhD supervisor has given me a small project based on this paper: he has this noncommutative geometric model of a black hole, and I'm suppose to see what happens if I re-do each step over a finite field. It's been two weeks and, as usual, I've made very little progress on it. I don't actually understand what's going on. So let's spend a moment or two trying to clear the fog and see what a ``noncommutative geometric model'' means.

1. Noncommutative Geometry Re-visited

Regular readers (ha! as if there are any...) will know that my previous experience with NCG was with the Bost-Connes system, a topic I hope to take up again soon. My supervisor's work; however, is considerably different. We keep the general philosophy of starting from an algebra and trying to construct a geometry from it, but instead of working over Operator Algebras we're sticking to less-analytic algebraic structures and imposing additional objects on them that are meant to emmulate geometric notions. Let's go over this in detail and discuss some of these objects - keeping in mind that I don't know any geometry, canot provide any motivation for these concepts, and generally don't have a clue what I'm doing.

1.1. Differential Forms

Most readers will know that one can impose a Differentiable Structure on a topological manifold and use it to start doing some geometry. We'll dispose of the topological manifold and replace it with an algebra $ {A}$ and impose on it the notion of a 1-form. NCG 1-forms live in ``differential calculus`` $ {\Omega_1}$, which we define to be a bimodule (id est, a module on both sides) over $ {A}$. The ``differentiable structure'' comes in the form of a linear map $ {d: A \rightarrow \Omega_1}$ that obeys the product rule:

$ \displaystyle d(ab) = d(a) \cdot b + a \cdot d(b) \; \forall a, b \in A$

Additionally, we require that $ { (a, b) \mapsto a \cdot d(b) }$ spans the bimodule $ {\Omega_1}$.
Some readers may wonder what sort of algebras $ {A}$ can have a differential calculus. Actually, each algebra $ {A}$ necessarily has one, but we won't discuss that in this post.

1.2. Vector Bundles & Connections

My knowledge of modern geometry ends at this point and I am solely trusting the NCG literature. Initially, when seeing these terms, I think of the long definitions needed in Differential Geometry and start chasing down the terms in various textbooks. We don't need that here, and can describe these objects with simple algebraic ideas. That said, the next tool in our discussion is the NCG notion of a Vector Bundle. The definition here is a chain of algebraic structures: a NCG vector bundle is a finitely generated projective module over $ {A}$. A projective module is the image of a free module under a projection/idempotent. A free module is a module with basis vectors. E.g., $ {A^n}$ is a free module and if $ {E \in M_n(A)}$ is idempotent, then $ {E A^n}$ is a vector bundle.
Now if $ {\mathcal{E}}$ is an NCG vector bundle, and $ {\Omega_1}$ a differential calculus, both over $ {A}$, we can also define the NCG notion of a connection: A connection is a linear map

$ \displaystyle \nabla : \mathcal{E} \rightarrow \Omega_1 \otimes \mathcal{E}$

with the following condition:

$ \displaystyle \nabla (as) = d(a) \otimes s + a \nabla(s) \; \forall a\in A, \; s \in \mathcal{E}$

Next time we'll (Lord willing) discuss metrics, curvature, and how they all fit together to get a model for a black hole.

2. Sources

1. LTCC Lecture notes in NCG by S Majid
2. Section 4 of Almost commutative Riemannian geometry: wave operators
3. Section 2 of Noncommutative Riemannian geometry on graphs
   

You don't understand something until you think it's obvious.

I learned a valuable life lesson in the course of completing my MSci degree. It's not so much about mathematics as it is about understanding mathematics or any other complicated subject.

Often in studying maths, I would spend hours or days stuck on a problem, proof, or section of a textbook. After I finally found a solution or an epiphany in what I read I would feel terrible. "I'm so stupid!" I'd exclaim. I would throw out or scratch out pages of notes, trying to blot out any evidence of the embarrassingly long time it took me to grasp the concept.

I had the same feeling just a few hours ago - but not with mathematics, with programming. I had spent the past three days trying to figure out the macro system in the Racket dialect of Lisp. I managed to write a simple message-passing object system, based on the one used in Structure and Interpretation of Computer Programs, just to see if I can. I patched together pieces of code from various samples, cargo-cult programming, until I had something working. I spent the next two days trying to understand what I wrote - noting my progress in another blog. Bit by bit, things started to come together - lot's of little epiphanies. "This is all so simple. It shouldn't have taken me three days to do this."

I remembered all the other times I had this feeling. Most of them had to do with mathematics or coursework. The most severe episode occurred as I wrote the final draft of my thesis. There was no sudden big epiphany. Rather, as I combed through my work, I recalled hundreds of little ones. "This is all so simple. It shouldn't have taken me a year. A *real mathematician* could have done it more quickly". I still have a tinge of embarrassment whenever I send it off to someone.

It's a common malady of myself and my friends who study mathematics to think that we're stupid because it's taken so long to understand something so simple. As I recalled all this occurrences I had a new epiphany: It suddenly seems simple because we suddenly understand it.

I stopped that last thought and restarted: "This is all so simple. I'm glad I finally get these macros." It's probably a lesson I could have learned without paying all that money for tuition. It seems so simple, after all.

Mathematical posts will resume shortly.

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.