## 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.