\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.
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