Processing math: 1%

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.

No comments:

Post a Comment