# neverendingbooks

The adelic interpretation of the Bost-Connes Hecke algebra $\mathcal{H}$ is based on three facts we’ve learned so far :

1. The diagonal embedding of the rational numbers $\delta~:~\mathbb{Q} \rightarrow \prod_p \mathbb{Q}_p$ has its image in the adele ring $\mathcal{A}$. ( details )

2. There is an exact sequence of semigroups $1 \rightarrow \mathcal{G} \rightarrow \mathcal{I} \cap \mathcal{R} \rightarrow \mathbb{N}^+_{\times} \rightarrow 1$ where $\mathcal{I}$ is the idele group, that is the units of $\mathcal{A}$, where $\mathcal{R} = \prod_p \mathbb{Z}_p$ and where $\mathcal{G}$ is the group (!) $\prod_p \mathbb{Z}_p^*$. ( details )

3. There is an isomorphism of additive groups $\mathbb{Q}/\mathbb{Z} \simeq \mathcal{A}/\mathcal{R}$. ( details )

Because $\mathcal{R}$ is a ring we have that $a\mathcal{R} \subset \mathcal{R}$ for any $a=(a_p)_p \in \mathcal{I} \cap \mathcal{R}$. Therefore, we have an induced ‘multiplication by $a$’ morphism on the additive group $\mathcal{A}/\mathcal{R} \rightarrow^{a.} \mathcal{A}/\mathcal{R}$ which is an epimorphism for all $a \in \mathcal{I} \cap \mathcal{R}$.

In fact, it is easy to see that the equation $a.x = y$ for $y \in \mathcal{A}/\mathcal{R}$ has precisely $n_a = \prod_p p^{d(a)}$ solutions. In particular, for any $a \in \mathcal{G} = \prod_p \mathbb{Z}_p^*$, multiplication by $a$ is an isomorphism on $\mathcal{A}/\mathcal{R} = \mathbb{Q}/\mathbb{Z}$.

But then, we can form the crystalline semigroup graded skew-group algebra $\mathbb{Q}(\mathbb{Q}/\mathbb{Z}) \bowtie (\mathcal{I} \cap \mathcal{R})$. It is the graded vectorspace $\oplus_{a \in \mathcal{I} \cap \mathcal{R}} X_a \mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$ with commutation relation
$Y_{\lambda}X_a = X_a Y_{a \lambda}$ for the base-vectors $Y_{\lambda}$ with $\lambda \in \mathbb{Q}/\mathbb{Z}$. Recall from last time we need to use approximation (or the Chinese remainder theorem) to determine the class of $a \lambda$ in $\mathbb{Q}/\mathbb{Z}$.

We can also extend it to a bi-crystalline graded algebra because multiplication by $a \in \mathcal{I} \cap \mathcal{R}$ has a left-inverse which determines the commutation relations $Y_{\lambda} X_a^* = X_a^* (\frac{1}{n_a})(\sum_{a.\mu = \lambda} Y_{\mu})$. Let us call this bi-crystalline graded algebra $\mathcal{H}_{big}$, then we have the following facts

1. For every $a \in \mathcal{G}$, the element $X_a$ is a unit in $\mathcal{H}_{big}$ and $X_a^{-1}=X_a^*$. Conjugation by $X_a$ induces on the subalgebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$ the map $Y_{\lambda} \rightarrow Y_{a \lambda}$.

2. Using the diagonal embedding $\delta$ restricted to $\mathbb{N}^+_{\times}$ we get an embedding of algebras $\mathcal{H} \subset \mathcal{H}_{big}$ and conjugation by $X_a$ for any $a \in \mathcal{G}$ sends $\mathcal{H}$ to itself. However, as the $X_a \notin \mathcal{H}$, the induced automorphisms are now outer!

Summarizing : the Bost-Connes Hecke algebra $\mathcal{H}$ encodes a lot of number-theoretic information :

• the additive structure is encoded in the sub-algebra which is the group-algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$
• the multiplicative structure in encoded in the epimorphisms given by multiplication with a positive natural number (the commutation relation with the $X_m$
• the automorphism group of $\mathbb{Q}/\mathbb{Z}$ extends to outer automorphisms of $\mathcal{H}$

That is, the Bost-Connes algebra can be seen as a giant mashup of number-theory of $\mathbb{Q}$. So, if one can prove something specific about this algebra, it is bound to have interesting number-theoretic consequences.

But how will we study $\mathcal{H}$? Well, the bi-crystalline structure of it tells us that $\mathcal{H}$ is a ‘good’-graded algebra with part of degree one the group-algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$. This group-algebra is a formally smooth algebra and we study such algebras by studying their finite dimensional representations.

Hence, we should study ‘good’-graded formally smooth algebras (such as $\mathcal{H}$) by looking at their graded representations. This will then lead us to Connes’ “fabulous states”…

### Similar Posts:

This site uses Akismet to reduce spam. Learn how your comment data is processed.