BC stands for Bi-Crystalline graded

By lieven

Noncommutative geometry and the Riemann zeta function

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 , by which I mean that one must be able to invert the elements of and obtain a group 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 . 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 satisfying the following two (usually strong) conditions :

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
The composition $\psi(s) \circ \phi(s)$ usually is NOT the identity morphism (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, ) such that $\psi(s) \circ \phi(s) = id_R E_s$

The point of the first condition is that the -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 the equality X_s R = R X_s where this is a free right -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 can be naturally embedded in a -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 is determined fully by the semi-group graded ring .

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 is the multiplicative semi-group of positive integers $\mathbb{N}^+_{\times}$ and the corresponding group is the multiplicative group $\mathbb{Q}^+_{\times}$ of all positive rational numbers.

The ring 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 $\psi(n)(Y_{\lambda}) = \frac{1}{n}(Y_{\frac{\lambda}{n}} + Y_{\frac{\lambda+1}{n}} + \hdots + Y_{\frac{\lambda+n-1}{n}})$ (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 $\phi(n) \circ \psi(n) (Y_{\lambda}) = \frac{1}{n}(Y_{\lambda}+\hdots+Y_{\lambda})=Y_{\lambda}$ and

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

and one verifies that $E_n = \frac{1}{n}(Y_0 + Y_{\frac{1}{n}} + \hdots + Y_{\frac{n-1}{n}})$ 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 ) 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…

Previous in series Next in series

Connes, crystalline, graded, noncommutative, semi-group, simples

4 comments
Posted in geometry
Written on Sat, 26 January 2008 at 5:26 pm
Tags: Connes, crystalline, graded, noncommutative, semi-group, simples
If you liked this post, then consider subscribing to our full RSS feed.

4 Responses to “BC stands for Bi-Crystalline graded”

Doug Says:
January 27th, 2008 at 2:08 am
Hi Lieven,

When you write about idempotents and semirings, I think of Tropical Algebras [TA], specifically the Max-Plus Algebra, used by applications mathematicians such as engineers.

I do not know if these latter algebras are bi-crystalline or crystalline?

Is it possible that NCG and TA are related?
lieven Says:
January 28th, 2008 at 12:51 pm
Doug, i dont knw a thing about max-plus etc. algebras but i googled this site which gives me plenty to read up on my ignorance. maybe one day ill have a response… there are plenty of connections between markov-chains and NCG. here the buzz-term is the ‘Cuntz-Krieger algebra’ (more on this probably later in this series).
abc on adelic Bost-Connes | neverendingbooks Says:
February 2nd, 2008 at 2:52 pm
[...] function the Bost-Connes coset space the Bost-Connes Hecke algebra Bost-Connes for ringtheorists BC stands for Bi-Crystalline graded adeles and ideles Chinese remainders and adele classesabc on adelic [...]
KMS, Gibbs & zeta function | neverendingbooks Says:
February 21st, 2008 at 2:32 pm
[...] function the Bost-Connes coset space the Bost-Connes Hecke algebra Bost-Connes for ringtheorists BC stands for Bi-Crystalline graded adeles and ideles Chinese remainders and adele classes abc on adelic Bost-Connes “God given [...]