## Tuesday, December 14, 2010

### 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.

## Sunday, December 12, 2010

### 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):
1. $\mu_n \mu_m = \mu_{nm}$
2. $\mu_n^{\ast} \mu_n = \mu_1$, 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.

## Sunday, December 5, 2010

### 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.

## Saturday, October 30, 2010

### 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.

## Tuesday, October 19, 2010

### 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!)

## Monday, October 18, 2010

### 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.

## Monday, October 4, 2010

### 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!

## Thursday, September 16, 2010

### 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

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

## Wednesday, September 15, 2010

### 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.

## Tuesday, September 14, 2010

### 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

## Thursday, September 9, 2010

### Constructing the Bost-Connes system. Attempt 1.

At the end of my last blog on lattices I described how the space of 1-dimensional $\mathbb{Q}$-lattices can be identified with $\hat{\mathbb{Z}} \times \mathbb{R}_{+}^{*}$. It's not too hard to see that if we mod out scaling by real factors, then we just end up with $\hat{\mathbb{Z}}$. Essentially, in one dimension, if we treat $\mathbb{Q}$-lattices that are multiples of each other as the same, all we have left is the homomorphism.

Now I want to describe a few more descriptions of 1 dimensional $\mathbb{Q}$-lattices (I'll henceforth use the symbol "1dQL" to mean 1 dimensional $\mathbb{Q}$-lattices), culminating in my describing a natural time evolution, and thus constructing the Quantum Statistical Mechanical system of Bost and Connes. First off, I want to describe the structure of the commensurability relation of 1dQL. But to do that, I first need to talk about groupoids.

Without delving too deep into category theory (which I am just now getting used to) let me suffice to say that a groupoid $\mathcal{G}$ is a collection of objects which will we call $\mathcal{G}^0$, and a collection of morphisms/mappings/functions (or just "arrows") $\mathcal{G}^1$ on $\mathcal{G}^0$ . We require every morphism in $\mathcal{G}^1$ to be invertible. Without category theory, you can think of a groupoid as being a lot like a normal group, except the group operation is not a binary operation. Id est, while every element has an inverse, not every pair of elements can be composed.

Now when I talk about the structure of the commensurability relation of 1dQL, its important to note that I don't mean equivalence classes. I'm actually talking about pairs $(\Lambda_1, \phi_1)$ and $(\Lambda_2, \phi_2)$ of 1dQLs that are commensurable. In Lemma 3.21 of Connes' monograph, he posits that this is isomorphic to the étale groupoid

$\mathcal{G}_1 = \{ (r, \rho, \lambda) : r \in \mathbb{Q}_{+}^{*}, \rho \in \hat{\mathbb{Z}}, \lambda \in \mathbb{R}_{+}^{*}$ such that $r \rho \in \hat{\mathbb{Z}}\}$

with the composition $(r_1,\rho_1,\lambda_1) \circ (r_2,\rho_2,\lambda_2) = (r_1 r_2,\rho_2,\lambda_2)$ if $r_2 \rho_2 = \rho_1$ and $r_2 \lambda_2 = \lambda_1$.

Étale groupoids are just groupoids with a specific topological structure (I'm still working out the details on that.) You can see this isomorphism from the map $l: (\rho,\lambda) \mapsto (\Lambda,\phi)$. In particular:

$(r,\rho,\lambda) \in \mathcal{G}_1 \mapsto (l(r \rho, r \lambda), l(\rho, \lambda))$

Finally, via Proposition 3.22, we can go even further. The following groupoid is isomorphic to the quotient $\mathcal{G}_1/\mathbb{R}_{+}^{*}$, hence we can also describe the commensurability classes of 1dQL up to scaling.

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

The isomorphism is via the map $\gamma : \mathcal{U}_1 \to \mathcal{G}_1/\mathbb{R}_{+}^{*}$ by $(r,\rho) \mapsto ((r^{-1}\mathbb{Z},\rho),(\mathbb{Z},\rho)) \in\mathcal{G}_1/\mathbb{R}_{+}^{*}$. The proof that this is an isomorphism isn't too hard (at least it doesn't look to hard) but it is yet still one of the details I've not finished.

Now to a locally compact groupoid (and I'm trusting Connes here when he says its locally compact) $\mathcal{G}$ we can construct the convolution algebra $\mathcal{A}_{c}(\mathcal{G})$. This convolution algebra is a space of functionals (into $\mathbb{C}$) on the groupoid. In the case of $\mathcal{U}_1$, the convolution product is

$f_1 * f_2 (L_1, L_2) = \sum_{L_1 \sim K \sim L_2} f_1(L_1, K) f_2(L, L_2)$

The involution is $f*(L_1, L_2) = \overline{f(L_1, L_2)}$. The symbols $L_1, L_2, K$ are 1dQLs and $\sim$ denotes commensurability.

(Note to readers: an algebra is essentially a vector space with a multiplication. The convolution product is our multiplication. Convolution is just a fancy word for an operation on two functions that produces a third. Involution is what makes C* algebras into C* algebras; its an abstraction of complex conjugation. Indeed, $\mathbb{C}$ is not only an algebra, it's a C* algebra.)
By completing the appropriate norm (which I do not fully understand, so I won't mention it here.) we can upgrade the convolution algebra $\mathcal{A}_{c}(\mathcal{U}_1)$ to the C* groupoid algebra $C^{*}(\mathcal{U}_1)$.

Connes has another description of the C* algebra in terms of generators and relations. However, I do not yet understand the proposition or the proof. It involves semigroup cross products, Morita equivalences, adeles, and notation I've not deciphered. I'll return to it soon.

The last thing left to do before we have a full Quantum Statistical Mechanical system is to define the time evolution (Lemma 3.24 in Connes' monograph). We can actually do this using data from an element in $\mathcal{G}_1/\mathbb{R}_{+}^{*}$. Recall that an element there is actually a pair of 1dQLs $L_1 = (\Lambda_1,\phi_1)$ and $L_2 = (\Lambda_2,\phi_2)$. Since they are commensurable, the lattices $\Lambda_1$ and $\Lambda_2$ differ by a scaling factor $r$. This $r$ happens to be the ratio of covolumes of two lattices.

Recall that the covolume of a lattice is the "area" of its fundamental parallelogram (or equivalent hyper-dimensional shape). And that the we can find this area by taking the determinate of the basis vectors of the lattice. In one dimension, we only have a single basis vector (some member of $\mathbb{R}$). Now when we take the ratio of the covolumes of lattices of commensurable 1 dimensional $\mathbb{Q}$-lattices, we simply get the scaling factor between the two, id est, $r$.

Now members of $C^{*}(\mathcal{G}_1/\mathbb{R}_{+}^{*})$ are simply functions on pairs of commensurable 1dQLs, e.g., $f(\Lambda_1, \Lambda_2)$. We define the time evolution as

$\sigma_{t}(f)(\Lambda_1, \Lambda_2) = r^{i t} f(\Lambda_1, \Lambda_2)$ where $r$ is the scaling factor.

Or in terms of the equivalent $C^{*}(\mathcal{U}_1)$ (functions $f(r,\rho)$ for $(r,\rho) \in \mathcal{U}_1$) we have

$\sigma_{t}(f)(r,\rho) = r^{i t} f(r,\rho)$.

And there you have it, folks, a bona fide Quantum Statistical Mechanical system. Bit embarrassing that it took me til September!

Details I've missed and Things I don't understand

Étale groupoids
How the groupoid inversion and operation works on $\mathcal{G}_1$ and $\mathcal{U}_1$. Plus how it's locally compact.
The proof of the isomorphism.
Proof of Proposition 3.22
All of Proposition 3.23 in Connes' monograph
Construction of C* group algebras and C* groupoid algebras.

Whats next

All of the above, plus the Hecke algebra description and its equivalence to the above. Then on to symetries of the system and the class field theory of $\mathbb{Q}$.

## Tuesday, September 7, 2010

### End(Q/Z) is isomorphic to \hat{Z}

So I've, more or less, finished my work set out in my last post: caught up on some algebra and category theory and wrapped up my study of the construction of the Bost Connes system (though I've not touched the Hecke algebra description yet.) I've been spending the past few days trying to write something coherent on my reading and notes, and today I got stuck on a rather small detail for an embarrassingly long time. To make matters worse, it was something I had discussed with my project supervisor just a month ago.

In any case, I think I've cleared it up, and I thought I'd post my work here for review.

Proposition $End(\mathbb{Q}/\mathbb{Z})$ is isomorphic to $\hat{\mathbb{Z}}$

First some background. $End(\mathbb{Q}/\mathbb{Z})$ is the set of all endomorphism of Q/Z. Id est, all homomorphisms from $\mathbb{Q}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$. $\mathbb{Q}/\mathbb{Z}$ is an Abelian [torsion] group, and hence a $\mathbb{Z}$-module, which makes $End(\mathbb{Q}/\mathbb{Z})$ a $\mathbb{Z}$-module as well (its group operation is addition of the mappings, e.g., if $\phi_1 \in End(\mathbb{Q}/\mathbb{Z})$ and $\phi_2 \in End(\mathbb{Q}/\mathbb{Z})$, then $(\phi_1 + \phi_2)(r) = \phi_1(r) + \phi_2(r)$) If you recall my earlier post on $\mathbb{Q}$-lattices, you'll know that End(Q/Z) is vital to describing the structure of the commensurability equivalence classes of 1 dimensional $\mathbb{Q}$-lattices.

$\hat{\mathbb{Z}}$ is a profinite ring (in case you want to google it.) It's the inverse limit, $\varprojlim \mathbb{Z}/n\mathbb{Z}$. The inverse system of groups and homomorphisms is so easy to work with that I can give an explicit description of the resulting profinite ring:

$\hat{\mathbb{Z}} = \{a=(a_1, a_2, \dots ) : \forall m|n, a_n \equiv a_m (mod m)\}$

These two rings are apparently isomorphic, so we can describe our 1 dimensional $\mathbb{Q}$-lattices with $\hat{\mathbb{Z}}$. The minor detail that had me stuck all day was the proof of that fact. So let me propose a proof now.

Given an $a \in \hat{\mathbb{Z}}$, I'll define a homomorphism $\phi_a : \mathbb{Q} / \mathbb{Z} \to \mathbb{Q} / \mathbb{Z}$ by $\phi_a (\frac{m}{n}) = a_n \frac{m}{n}$. The map $a \mapsto \phi_a$ is a homomorphism between $\hat{\mathbb{Z}}$ and $End(\mathbb{Q}/\mathbb{Z})$ (it's pretty clear that it respects the group operation, which is just component-wise modular arithmetic in the former ring, and addition of functions in the latter.) I still need to check that $\phi_a$ is an endomorphism of $\mathbb{Q}/\mathbb{Z}$, however. Let's do that:

$\phi_a ( \frac{m_1}{n_1} + \frac{m_2}{n_2}) = \phi_a (\frac{n_2 m_1 + n_1 m_2}{n_1 n_2}) = a_{n_1 n_2} \frac{n_2 m_1 + n_1 m_2}{n_1 n_2}$.

And

$\phi_a( \frac{m_1}{n_1}) + \phi_a(\frac{m_2}{n_2}) = \frac{a_{n_1} n_2 m_1 + a_{n_2} n_1 m_2} {n_1 n_2}$

But recall our description of $\hat{\mathbb{Z}}$, "for all $m|n, a_n \equiv a_m (mod m)$". That means that we can write $a_{n_1 n_2} = a_{n_1} + q n_1 = a_{n_2} + r n_2$. Or in other words,
$a_{n_1} = a_{n_1 n_2} - q n_1$ and $a_{n_2} = a_{n_1 n_2} - r n_2$. So the above becomes:

$a_{n_1 n_2} \frac{n_2 m_1 + n_1 m_2}{n_1 n_2} - n_1 n_2 \frac {q m_1 + r m_2}{n_1 n_2}$

The negative term is an integer, so it's 0 in our quotient $\mathbb{Q}/\mathbb{Z}$. Now we're halfway done. If we can find an inverse of that map that keeps the ring structure of $\hat{\mathbb{Z}}$ intact, we've won the game.

To define said inverse, we'll start with an $f \in \text{End}(\mathbb{Q}/\mathbb{Z})$. Define an $a \in \prod_n \mathbb{Z}/n\mathbb{Z}$ by $a_k = k f(\frac{1}{k})$. Again its not hard to see that this preserves component-wise addition on $\hat{\mathbb{Z}}$. Moreover, if $a \in \hat{\mathbb{Z}}$, then it's the inverse map of the above. Now if $m|n$, we can write $n=mr$ for some $r$. Then $m r f(\frac{1}{n}) - m f ( \frac{1}{m})$ is obviously divisible by $m$. Hence $a \in \hat{\mathbb{Z}}$. And we've found our isomorphism.

I did this with a very naive approach, and its a bit long and messy. Please let me know if there are any errors!

Someone at physicsforums.com suggested a category-theoretic proof of the above. It's a lot shorter and cleaner then my naive approach, but I don't yet know enough category theory to understand it. Since I am working on a writeup of my category theory reading anyways, perhaps I'll do a bit more and incorporate that in as well.

Stay tuned for more algebra and the Bost-Connes system!

## Saturday, September 4, 2010

### Getting back into things...

Apologies for the hiatus in maths posts; I've had some personal issues (trying to find food, fending off homelessness, grieving the death of my friend.) The past few weeks have been pretty stressful. I'm now in Colorado and shouldn't be worried about those important distractions for the next three weeks (when I return to London I should also have a place to live, and student loan money should be coming too.) I'm glad I'm here and that I get a chance to see my friend's family. I will be resuming maths post from today (inshallah!) Here's the plan for the next few days:

Yesterday and Today: catching up on some algebra - modules, algebras, group rings, group algebras, c* group algebras.
Sunday: Etale groupoids & category Theory
Monday: C* algebra construction of the bost-connes system
Tuesday until ? : Hecke algebras and their corresponding construction, plus the equivalence of the two.

My Advisor and I had set a goal of understanding the construction of bost-connes system by the end of August. I'm ashamed to say I'm a bit late. But I think I'll be able to make up lost ground and start working on the symmetries of the system here in the next few days. I don't think that schedule is too ambitious. I should have time to catch up and even enjoy the miles of mountain bike singletrack just outside my parent's house!

Additionally, one might expect some non-project related blogs to come soon too, particularly ones on the UCL undergrad colloquium, my talks there, and some other reading.

Right. enough talk, lets get to work

## Thursday, August 12, 2010

### Q-lattices revisited. (1-dim case)

Yesterday, I found out a friend of mine back home in Colorado died and have since not had the focus to do much work. I haven't done that review of algebra topics like I said I would. But it's important to keep good habits, so I thought I'd review and write a bit about $\mathbb{Q}$-lattices a bit more.

Connes and Marcolli's monographs define a $\mathbb{Q}$-lattice as a lattice $\Lambda \in \mathbb{R}^n$ together with a homomorphism $\phi : \mathbb{Q}^n / \mathbb{Z}^n \to \mathbb{Q} \Lambda / \Lambda$.  We call two $\mathbb{Q}$-lattices $\Lambda_1$ and $\Lambda_2$ commensurable ($\Lambda_1 \sim \Lambda_2$) if:
1) $\mathbb{Q} \Lambda_1 =\mathbb{Q} \Lambda_2$ and
2) $\phi_1$ = $\phi_2$ mod $\Lambda_1 + \Lambda_2$.

Condition 1) means that there exists some $\mathbb{Q}$-span in $\mathbb{R}^n$ such that $\Lambda_1$ and $\Lambda_2$ both sit inside it.  For instance, both $\frac{1}{2} \mathbb{Z}$ and $\frac{1}{5} \mathbb{Z}$ sit inside $\mathbb{Q}$, so they meet condition 1.  But if we take $< (1,0), (0,\sqrt{2}) >$ and $<(\frac{1}{2},0), (0,0) >$ we cannot find any vectors in $\mathbb{R}^2$ that, when spanned over $\mathbb{Q}$, would contain both of those lattices.

Condition 2) is actually how I originally understood it.  The homomorphism must "label the same points".  More formally, Lets say $X =\mathbb{Q} \Lambda_1 =\mathbb{Q} \Lambda_2$.  For each $\Lambda_i$ we have a projection map $\pi_i : X/\Lambda_i \to X/(\Lambda_1 + \Lambda_2)$ by $r+\Lambda_i \mapsto r + (\Lambda_1 + \Lambda_2)$ (of course, $r$ may not be the best representative element in $\Lambda_1 + \Lambda_2$).  Condition two says that $\pi_1 \circ \phi_1 = \pi_2 \circ \phi_2$.

1-dim Q-lattices

We can describe the Bost-Connes system with $\mathbb{Q}$-lattices in $\mathbb{R}$, and 1-dim $\mathbb{Q}$-lattices can be described by $\hat{\mathbb{Z}} \times \mathbb{R}_{+}^{*}$, where $\hat{\mathbb{Z}} = \varprojlim_{n} \mathbb{Z} / n\mathbb{Z}$.  First note any lattice $\Lambda \in \mathbb{R}$ can be described by $\lambda \mathbb{Z}$, where $\lambda \in \mathbb{R}_{+}^{*}$.  So the only data we need is the homomorphism $\phi : \mathbb{Q} / \mathbb{Z} \to \mathbb{Q} / \lambda \mathbb{Z}$.  But apparently $\hat{\mathbb{Z}} \cong Hom (\mathbb{Q} / \mathbb{Z},\mathbb{Q} / \mathbb{Z})$, that is, $\hat{\mathbb{Z}}$ is isomorphic to the set of all endomorphisms of $\mathbb{Q} / \mathbb{Z}$ (I'm not yet sure how! Something to do with Pontryagin duality?), so all that remains is a choice of $\rho \in \hat{\mathbb{Z}}$ and we have $(\Lambda,\phi) \mapsto (\rho,\lambda)$.  One should note that the inverse of this map is well defined, so we can identify the set of all 1-dim $\mathbb{Q}$-lattices with $\hat{\mathbb{Z}} \times \mathbb{R}_{+}^{*}$.

I'll continue reading about 1-dim $\mathbb{Q}$-lattices for the next several days, though I do still plan to take some time to catch up on some algebra (algebras over a field, group rings, group algebras, C* algebras, their representations, group C* algebras, Pontryagin duels, etale groupoids.)

Thanks to Professor Kim and the folks at mathoverlow for helping me understand this.
Also, for the next few days I will all but abandon Marcolli's monograph;  that text goes straight into a description of the Bost-Connes system via Shimura varieties, which is a bit above my level.  I do intent to understand the Shimura variety and the Hecke algebra description of Bost-Connes eventually, however.

## Tuesday, August 10, 2010

### KMS states and symmetries.

I'm still looking at Quantum Statistical Mechanical systems, which I'm treating as a C* algebra $\mathcal{A}$ together with a time evolution $\sigma_t$. Particularly, I'm interested in its equilibrium states, which we can characterize via the KMS condition (recall that a state on a QSM ($\mathcal{A}$,$\sigma_t$) is a positive [or zero] linear functional $\phi : \mathcal{A} \to \mathbb{C}$ with $\phi(1)=1$, or has operator norm 1 if $\mathcal{A}$ has no unit.)

A state $\phi$ satisfies the KMS condition at inverse temperature $\beta$ if for all $a,b \in \mathcal{A}$ we can find a function $F_{a,b} (z)$ with the following properties:
1. Its holomorphic on the strip $\{z \in \mathbb{C} : 0 < \Im(z) < \beta \}$

2. continuous on boundary and bounded

3. $F_{a,b}(t) = \phi(a \sigma_t(b))$ and $F_{a,b}(t+i\beta) = \phi(\sigma_t(b) a)$

4. for all $t \in \mathbb{R}$
We call such states $KMS_{\beta}$ states. We also have $KMS_{\infty}$ states, but rather than define them on half-plane (we call those ground states), we use a stronger condition (For reasons I neither know nor understand at this point): A state $\phi$ is a $KMS_{\infty}$ state if we have $KMS_{\beta}$ states $phi_{\beta}$ such that, for sufficiently large $\beta$, $\phi(a) = \lim_{\beta \to \infty} \phi_\beta (a)$ for all $a \in \mathcal{A}$. Mathematically, all I should need is the definitions, just to understand them and go from there. But I'm not quite satisfied with them. I'm wishing I took a course in statistical thermodynamics/mechanics at this point, as I have little intuition for how these equilibrium states relate to our phase transitions of the system (indeed, I'm still not sure what "phase transition" means for a C* dynamical system.) The $KMS_{\beta}$ states (we denote the set of them by $\Sigma_{\beta}$) are the equilibrium states [states that we can actually measure] for "temperatures" of $1/\beta$. At 0 temperature we have the $KMS_{\infty}$ states. Now, if $\phi_1$ and $\phi_2$ are two $KMS_{\beta}$ states, its not too hard to see that any linear combination of the two will also be, and from that its not too hard to see that $\Sigma_{\beta}$ is a convex set (there's a path between any two $KMS_{\beta}$ states). Connes goes further, and states that its compact under the weak topology (pointwise convergence?) and that its a Choquet Simplex. And I have no idea what that is. But for this blog's purposes, I don't think I need to. The important bit is that $KMS_{\beta}$ states are a convex set. Its not too hard to see that $\Sigma_{\infty}$ is a convex set too. Convex sets have extremal points (points that aren't in a line segment joining two other points.) We'll denote the set of extremal $KMS_{\beta}$ states by $\xi_{\beta}$. Connes has a proposition stating that the extremal $KMS_{\beta}$ states are the same as the states that have a factorial GNS representation (while I have an idea what a GNS representation is - the image of the state under a homomorphism to a Hilbert space, where the homomorphism has certain properties and is constructed in a certain way - I have no idea what it means for one to be factorial.) R. Haag in Local Quantum Physics has a similar statement, that $\xi_{\beta}$ is the same as the set of $KMS_{\beta}$ states that are primary. But I don't know what that means either.

So what's the structure and intuition behind $\xi_{\beta}$? How should I think of it? If I'm thinking just in terms of a physical thermodynamic system, what do the extremal equilibrium states represent? My guess is states of matter, or pure phase states (considering that a change in a state of matter is a "phase transition"). So does that mean there are "impure phase states", id est, states in a mixture of phases (so that $\Sigma_{\beta}$ is non-trivial). Back to the QSM: If we're working in a system with only one phase transition (I'm not sure what a phase transition means in this context) does that mean that there are only two extremal $KMS_{\beta}$ states?

Apart from the equilibrium states, we're also interested in symmetries of the QSM system. Rather frustratingly, I cannot talk much about the physics of QSM symmetries (but I resolve to be able to soon! as I do with statistical thermodynamics!), so I must stick to the mathematics. We're interested in both automorphisms (symmetries of the whole system) and endomorphisms (symmetries of only a part of the system....I think.) In either case, they must be compatible [commute] with our time evolution, id est, if g is our auto/endo-morphism, then $g \sigma_t = \sigma_t g$ for all t. for automorphism g, we say $g \in Aut(\mathcal{A}, \sigma_t)$.

A subgroup $G \subset Aut(\mathcal{A}, \sigma_t)$ we call a group of symmetries by automorphisms. It acts on KMS states via pullback, id est, $g^{*}(\phi) (a) = \phi ( g (a) )$. Connes included a lemma stating that inner automorphisms (from a unitary state $u \in \mathcal{A}$, inner automorphisms acts by $a \mapsto u a u^{-1}$) act trivially on KMS states. Actually, the lemma stated something else, but that was in the proof. The proof essentially uses an analytic continuation of the map $t \mapsto \sigma_t(a)$ to the whole complex plane (I don't know how that's possible) and uses the fact that KMS states have the property that $\phi(b \sigma_{i \beta} (a)) = \phi (ab)$. It then follows that $\phi(u a u^{-1}) = \phi(a)$, and the action is trivial.

An endomorphism $\rho$ also has its pullback, $\rho^{*} (\phi) (a) = \frac{\phi(\rho(a))}{\phi(e)}$, assuming that $\phi(e) \neq 1$.
$e=\phi(1)$ (assuming $\mathcal{A}$ has unity, of course) and its not too hard to see that e is an idempotent ($e^2 = e$, because $e^2 = \phi(1)^2 = \phi (1^2) = e$). Inner endomorphisms also act trivially, by a similar proof.

Things I don't know this post:

Choquet Simplex
factorial GNS representation
primary KMS state.
The analytic continuation used in the proof of the triviality of the action of a group of inner automorphism.
Connes also talks a bit about multiplier algebras and essential ideas - I may look into those tomorrow, but as best I can tell I won't be needing those ideas any time soon. Marcolli doesn't mention them at all, for instance.
Connes also mentions that we can pushforward KSM states: I largely understood what I read there [or I think I did], but I choose not to write about it, largely because of time, and largely because I don't yet see how it relates.

## Monday, August 9, 2010

### melancholy mondays

Today started out with so much progress. I started early, made ample progress with programming and business contacts, bided on freelance projects, and was anxious to start a day of mathematics work. But when the time to do maths rolled around, I procrastinated, and procrastinated, and procrastinated. (Actually, for the past 3-4 days I've not done much maths work, for the same reason.)

I finally started doing some reading about 2 hours ago. It was vague, unfocused, distracted reading. More about states on a QMS, representations, KMS states, and quite a few things I don't understand (list at the end.) But I must keep good habits, so I read anyways, and am now writing something barely coherent about it.

Here's what I plan for this week:
1) KMS states, the structure of their sets, symmetries on QMS. 3-4 more hours and 1 more blog post.
2) Some algebra: group rings, group algebras, algebras over a field, et cetera. Just to get my definitions, examples, and intuition straight. 3-4 hours. 1 blog post should be enough, I think.
3) QMS from 1-dim Q lattices and the Bost-Connes system. However long it takes!

What I don't understand today:
Connes has a proposition (3.8) about $KMS_\beta$ states...they're extremal iff the GNS representation is factorial. I have an idea of what a GNS rep is, I have no idea what it means for one to be factorial

Same prob mentions that the set of $KMS_\beta$ states is a convex compact Choquet simplex. I've never seen "Choquet Simplex" before. I'll have to do some googling when I'm more awake.

"phase transition" is mentioned often. While I have an idea what this means for a classical thermodynamics system (changing states in matter) I have no idea what it means for a C* dynamical system. I did manage to find some stuff in R. Haag's Local Quantum Physics (around page 213) that I'll have to look at tomorrow.

I can barely keep my eyes open. Goodnight!

## Wednesday, August 4, 2010

### Quantum Statistical Mechanics, C* algebra setup, and the KMS Condition

(random tidbit: Connes & Marcolli's "Non-commutative geometry, quantum fields, and motives" is actually a lot more readable than Marcolli's "Lectures on Arithmetic Non-commutative geometry". Connes goes more in-depth on several topics, and explains more of the things that I have never seen before. I'm reading both side-by-side.)

I'm moving on from the group $C^{*}$-algebra $C^{*} (\mathcal{L}_1/\mathbb{R}^{*}_{+})$ and am continuing through the monographs of Connes and Marcolli. My goal is to get a general, high level understanding of the construction of the Bost-Connes system (1-dim $\mathbb{Q}$-lattice systems) and their use in the explicit class field theory of $\mathbb{Q}$. As I develop that "big picture", I'm keeping a list of all the things I need to explore in-depth. But not until I have that big picture will I then start chasing down things like group C* algebras. (Of course, I'll need to pick up a quick idea of these things as I go on.)

So that brings me to the section on Quantum Statistical Mechanics (QSM). QSM, as best I can tell, is actually a lot easier to work with than Classical Stat Mech, as I deal with operator and C* algebras, rather than trying to work out some delicate approximation here and there. In classical mechanics the observables are functions on the phase space. But in the quantum world, observables are treated as operators, and they form a C* algebra. So in fact, we can think of a QSM system as just a C* dynamical system, id est, a C* algebra $\mathcal{A}$ together with a time evolution $\sigma$. By "time evolution", I simple mean a family of automorphisms of $\mathcal{A}$ that can be expressed by a single parameter. A "state" on such a system is then defined as a linear functional $phi : \mathcal{A} \to \mathbb{C}$ with $\phi(1)=1$ and $\phi(a^{*}a) \geq 0$ (the first condition is changed to stating that the operator norm of $\phi$ is 1, if $\mathcal{A}$ doesn't have a unit.)

According to the monographs [and what follows from physical intuition] there's one particular class of states we want to look at: the equilibrium states, states that are invariant with respect to time evolution. According to QSM, the equilibrium states are the ones that meet the "KMS" condition at a specific "inverse temperature" $\beta$ of the system. I can't say where the KMS condition comes from or how we got it, but I can read it from the monographs and make some sense of it: A state $\phi$ satisfies the KMS condition at inverse temperature $\beta$ if for all $a,b \in \mathcal{A}$ we can find a function $F_{a,b} (z)$ with the following properties:
1. Its holomorphic on the strip $\{z \in \mathbb{C} : 0 < \Im(z) < \beta \}$

2. continuous on boundary and bounded

3. $F_{a,b}(t) = \phi(a \sigma_t(b))$ and $F_{a,b}(t+i\beta) = \phi(\sigma_t(b) a)$

4. for all $t \in \mathbb{R}$
(holomorphic and bounded.....constant?)As best as I understand things, this condition should ensure that the state is invariant with respect to the time evolution over all $a \in \mathcal{A}$.Connes had a great example of the KMS condition in the finite case, using a finite dimensional Hilbert space. I had planned to write about it, but I'm all out of time!May write more about KMS next time...or may move on! Things I don't understand from today's work:Connes makes several references to representations of C* algebras. I've seen this before while I was collecting literature on group C* algebras and C* algebras in general. More and more I am convinced that I need to spend a couple of weeks studying these things.

I'm not sure how the KMS condition implies the equilibrium (that states are invariant with respect to the time evolution.) Connes mentions that it's a consequence of the Liouville Theorem. I suspect its just some complex analysis that wouldn't take me too long to think about [I can kinda see it...I think...] but we're not worrying about it right now.

Additionally....I don't know where the KMS condition comes from. I have a couple books on "operator algebras and QSM" that I use to figure that out later, as well.

Its probably worth noting that $C^{*} (\mathcal{L}_1/\mathbb{R}^{*}_{+})$ is a C* dynamical system, though some work is required to get the time evolution. I think $C^{*} (\mathcal{L}_1/\mathbb{R}^{*}_{+})$ is the Bost-Connes system.

## Sunday, August 1, 2010

### the group C* algebra for $\mathcal{L}_n$

Not much progress this weekend. But I had originally written that Bost-Connes type systems where the C* algebra generated by $\mathcal{L}_n / \mathbb{R}^{*}_{+}$ (n=1 for Bost-Connes, n=2 for imaginary quadratic fields.) I got that from the notation $C^{*} (\mathcal{L}_n)$. I've learned a bit more about that symbol.

Its not the C* algebra generated by $\mathcal{L}_n$...its the group C* algebra of $\mathcal{L}_n$, which is apparently the C*-enveloping algebra of $L^1 (\mathcal{L}_n)$ From what my googling and textbook-glancing today has told me, its more related to harmonic analysis and the Haar measure stuff then to the material covered in Averson's Invitation to C* Algebras. The Haar measure is a Borel measure that we can find on any [locally] compact topological group G, according to a theorem in Functional/Harmonic analysis. The notation $L^1 (G)$ is then pretty obvious, its exactly what one would expect from the Lebesque spaces (I think, I need to look into this stuff more.) I have no idea what a C*-enveloping algebra is yet. I'll be looking into all these things from tomorrow onwards.

But its starting to look like $C^{*} (\mathcal{L}_n)$ is a pretty complex object. Its not even obvious to me that $\mathcal{L}_n$ is [locally?] compact! Then again, I'm not worried about that right now. I'm just trying to get the "big picture".

## Friday, July 30, 2010

### random aside

I enjoyed [probably because I actually understood] the following post on group theory from Terry Tao. I thought I'd share a link:

http://terrytao.wordpress.com/2009/10/19/grothendiecks-definition-of-a-group/

### $\mathbb{Q}$-lattices and commensurability

My supervisor, Professor Kim, helped clear up some of my confusion on the homomorphism in $\mathbb{Q}$ lattices. Lets go over it again: A $\mathbb{Q}$-lattice is a normal lattice $\Lambda \subset \mathbb{R}^n$ together with a homomorphism $\phi : \mathbb{Q}^n / \mathbb{Z}^n \to \mathbb{Q}\Lambda / \Lambda$. Lets break down that notation a bit.

$\mathbb{Q}^n / \mathbb{Z}^n$ looked a bit frightening to me at first, but its really not so bad. Consider the far more neighbourly quotient $\mathbb{R} / \mathbb{Z}$, which of course is just a circle. $\mathbb{Q} / \mathbb{Z}$ clearly is a subset of the circle, but only including the points on the circle that go back to the origin when multiplied by an integer. $(\mathbb{Q} / \mathbb{Z})^2$ is a subset of the torus $(\mathbb{R} / \mathbb{Z})^2$, and so on, and that gives me a reasonable way to think about the more general looking $\mathbb{Q}^n / \mathbb{Z}^n$. Not that bad after all, right?

Now for $\mathbb{Q}\Lambda / \Lambda$. $\mathbb{Q}\Lambda$ turns out to be exactly what it sounds like. For a lattice $\Lambda$, we can consider it as the $\mathbb{Z}$-span of basis vectors $\{ v_1,...,v_n \}$. For $\mathbb{Q}\Lambda$, just take the $\mathbb{Q}$-span. But what about the quotient? Its really not that different from $\mathbb{Q}^n / \mathbb{Z}^n$. Consider that $\Lambda$ is like a "tilted" version of $\mathbb{Z}^n$. From there its not hard to imagine $\mathbb{Q}\Lambda / \Lambda$ as being a "tilted" version of the "hyper-rational-torus" $\mathbb{Q}^n / \mathbb{Z}^n$ (I call it "hyper" because it may be more than 2 dimensions, and "rational" because it only includes the points that get back to the origin when multiplied by an integer.) In fact, its not so hard to see that there's an isomorphism between $\mathbb{Q}^n / \mathbb{Z}^n$ and $\mathbb{Q}\Lambda / \Lambda$ that comes just from figuring out how much $\Lambda$ is titled from $\mathbb{Z}^n$, id est, expressing the basis for $\Lambda$ in terms of the standard basis of $\mathbb{Z}^n$.

So why then does the the definition of $\mathbb{Q}$-lattices include a homomorphism and not a full, invertible isomorphism? In fact, why do we even need a definition for $\mathbb{Q}$-lattices at all? It just seems to be a lattice with an obvious homomorphism attached to it. (this question is really quite moot. commensurability and the commensurability classes yield plenty results, among them is the topic of my 4th year project!)

## Saturday, January 30, 2010

### First Post

I've decided that I should make a diary of my progress in mathematics (both in my modules as a student at UCL and in my independent study and readings.) Currently, my UCL courses are: [examinable] Functional Analysis, Algebraic Topology, Algebraic Number Theory, Nuclear and Particle Physics; [non-examinable] Galois Theory, and Probability (from measure theory.) Just spent the day working through Algebraic Topology, particularly work with exact-sequences (and some diagram chasing!) and the homology of the "subdivided simplical circle". (This course is AT is "Via homology of simplical complexes" and doesn't assume ant general topology.)

Speaking of general topology, myself and some other UCL students have started an undergrad colloquium at the maths department. A friend an I gave a short talk and problem class on general topology to start it out (our department doesn't have a course in general topology, and my friend and I had done substantial reading on it - from Munkres and Dugundji.) We covered the definition of a topology, continuity, Hausdorff spaces, connectedness, and the quotient space (as best as we could given only an hour!) I typed up our notes from the talk and they're online:

Right, so....time for sleep...more maths tomorrow!