## 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.