Towards the end of the Bost-Connes for ringtheorists post I freaked-out because I realized that the commutation morphisms with the $X_n^*$ were given by non-unital algebra maps. I failed to notice the obvious, that algebras such as $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$ have plenty of idempotents and that this mysterious ‘non-unital’ morphism was nothing else but multiplication with an idempotent…

Here a sketch of a ringtheoretic framework in which the Bost-Connes Hecke algebra $\mathcal{H}$ is a motivating example (the details should be worked out by an eager 20-something). Start with a suitable semi-group $S$, by which I mean that one must be able to invert the elements of $S$ and obtain a group $G$ of which all elements have a canonical form $g=s_1s_2^{-1}$. Probably semi-groupies have a name for these things, so if you know please drop a comment.

The next ingredient is a suitable ring $R$. Here, suitable means that we have a semi-group morphism
$\phi~:~S \rightarrow End(R)$ where $End(R)$ is the semi-group of all ring-endomorphisms of $R$ satisfying the following two (usually strong) conditions :

1. Every $\phi(s)$ has a right-inverse, meaning that there is an ring-endomorphism $\psi(s)$ such that $\phi(s) \circ \psi(s) = id_R$ (this implies that all $\phi(s)$ are in fact epi-morphisms (surjective)), and

2. The composition $\psi(s) \circ \phi(s)$ usually is NOT the identity morphism $id_R$ (because it is zero on the kernel of the epimorphism $\phi(s)$) but we require that there is an idempotent $E_s \in R$ (that is, $E_s^2 = E_s$) such that $\psi(s) \circ \phi(s) = id_R E_s$

The point of the first condition is that the $S$-semi-group graded ring $A = \oplus_{s \in S} X_s R$ is crystalline graded (crystalline group graded rings were introduced by Fred Van Oystaeyen and Erna Nauwelaarts) meaning that for every $s \in S$ we have in the ring $A$ the equality $X_s R = R X_s$ where this is a free right $R$-module of rank one. One verifies that this is equivalent to the existence of an epimorphism $\phi(s)$ such that for all $r \in R$ we have $r X_s = X_s \phi(s)(r)$.

The point of the second condition is that this semi-graded ring $A$ can be naturally embedded in a $G$-graded ring $B = \oplus_{g=s_1s_2^{-1} \in G} X_{s_1} R X_{s_2}^*$ which is bi-crystalline graded meaning that for all $r \in R$ we have that $r X_s^*= X_s^* \psi(s)(r) E_s$.

It is clear from the construction that under the given conditions (and probably some minor extra ones making everything stand) the group graded ring $B$ is determined fully by the semi-group graded ring $A$.

what does this general ringtheoretic mumbo-jumbo have to do with the BC- (or Bost-Connes) algebra $\mathcal{H}$?

In this particular case, the semi-group $S$ is the multiplicative semi-group of positive integers $\mathbb{N}^+_{\times}$ and the corresponding group $G$ is the multiplicative group $\mathbb{Q}^+_{\times}$ of all positive rational numbers.

The ring $R$ is the rational group-ring $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$ of the torsion-group $\mathbb{Q}/\mathbb{Z}$. Recall that the elements of $\mathbb{Q}/\mathbb{Z}$ are the rational numbers $0 \leq \lambda < 1$ and the group-law is ordinary addition and forgetting the integral part (so merely focussing on the ‘after the comma’ part). The group-ring is then

$\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] = \oplus_{0 \leq \lambda < 1} \mathbb{Q} Y_{\lambda}$ with multiplication linearly induced by the multiplication on the base-elements $Y_{\lambda}.Y_{\mu} = Y_{\lambda+\mu}$.

The epimorphism determined by the semi-group map $\phi~:~\mathbb{N}^+_{\times} \rightarrow End(\mathbb{Q}[\mathbb{Q}/\mathbb{Z}])$ are given by the algebra maps defined by linearly extending the map on the base elements $\phi(n)(Y_{\lambda}) = Y_{n \lambda}$ (observe that this is indeed an epimorphism as every base element $Y_{\lambda} = \phi(n)(Y_{\frac{\lambda}{n}})$.

The right-inverses $\psi(n)$ are the ring morphisms defined by linearly extending the map on the base elements $\psi(n)(Y_{\lambda}) = \frac{1}{n}(Y_{\frac{\lambda}{n}} + Y_{\frac{\lambda+1}{n}} + \ldots + Y_{\frac{\lambda+n-1}{n}})$ (check that these are indeed ring maps, that is that $\psi(n)(Y_{\lambda}).\psi(n)(Y_{\mu}) = \psi(n)(Y_{\lambda+\mu})$.

These are indeed right-inverses satisfying the idempotent condition for clearly $\phi(n) \circ \psi(n) (Y_{\lambda}) = \frac{1}{n}(Y_{\lambda}+\ldots+Y_{\lambda})=Y_{\lambda}$ and

$\begin{eqnarray} \psi(n) \circ \phi(n) (Y_{\lambda}) =& \psi(n)(Y_{n \lambda}) = \frac{1}{n}(Y_{\lambda} + Y_{\lambda+\frac{1}{n}} + \ldots + Y_{\lambda+\frac{n-1}{n}}) \\ =& Y_{\lambda}.(\frac{1}{n}(Y_0 + Y_{\frac{1}{n}} + \ldots + Y_{\frac{n-1}{n}})) = Y_{\lambda} E_n \end{eqnarray}$

and one verifies that $E_n = \frac{1}{n}(Y_0 + Y_{\frac{1}{n}} + \ldots + Y_{\frac{n-1}{n}})$ is indeed an idempotent in $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$. In the previous posts in this series we have already seen that with these definitions we have indeed that the BC-algebra is the bi-crystalline graded ring

$B = \mathcal{H} = \oplus_{\frac{m}{n} \in \mathbb{Q}^+_{\times}} X_m \mathbb{Q}[\mathbb{Q}/\mathbb{Z}] X_n^*$

and hence is naturally constructed from the skew semi-group graded algebra $A = \oplus_{m \in \mathbb{N}^+_{\times}} X_m \mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$.

This (probably) explains why the BC-algebra $\mathcal{H}$ is itself usually called and denoted in $C^*$-algebra papers the skew semigroup-algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \bowtie \mathbb{N}^+_{\times}$ as this subalgebra (our crystalline semi-group graded algebra $A$) determines the Hecke algebra completely.

Finally, the bi-crystalline idempotents-condition works well in the settings of von Neumann regular algebras (such as all limits of finite dimensional semi-simples, for example $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$) because such algebras excel at idempotents galore

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