## 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.