Crystalline | neverendingbooks

Posts Tagged ‘crystalline’

BC stands for Bi-Crystalline graded

Saturday, January 26th, 2008

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

Tags: Connes, crystalline, graded, noncommutative, semi-group, simples
Posted in geometry | 4 Comments »

Bost-Connes for ringtheorists

Wednesday, January 23rd, 2008

Noncommutative geometry and the Riemann zeta function

Over the last days I’ve been staring at the Bost-Connes algebra to find a ringtheoretic way into it. Ive had some chats about it with the resident graded-guru but all we came up with so far is that it seems to be an extension of Fred’s definition of a ‘crystalline’ graded algebra. Knowing that several excellent ringtheorists keep an eye on my stumblings here, let me launch an appeal for help :

What is the most elegant ringtheoretic framework in which the Bost-Connes Hecke algebra is a motivating example?

Let us review what we know so far and extend upon it with a couple of observations that may (or may not) be helpful to you. The algebra $\mathcal{H}$ is the algebra of $\mathbb{Q}$ -valued functions (under the convolution product) on the double coset-space $\Gamma_0 \backslash \Gamma / \Gamma_0$ where

$\Gamma = \{ \begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix}~:~a,b \in \mathbb{Q}, a > 0 \}$ and $\Gamma_0 = \{ \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}~:~n \in \mathbb{N}_+ \}$

We have seen that a $\mathbb{Q}$ -basis is given by the characteristic functions $X_{\gamma}$ (that is, such that $X_{\gamma}(\gamma') = \delta_{\gamma,\gamma'}$ ) with $\gamma$ a rational point represented by the couple ~(a,b) (the entries in the matrix definition of a representant of $\gamma$ in $\Gamma$ ) lying in the fractal comb

defined by the rule that $b < \frac{1}{n}$ if $a = \frac{m}{n}$ with $m,n \in \mathbb{N}, (m,n)=1$ . Last time we have seen that the algebra $\mathcal{H}$ is generated as a $\mathbb{Q}$ -algebra by the following elements (changing notation)

$\begin{cases}X_m=X_{\alpha_m} & \text{with } \alpha_m = \begin{bmatrix} 1 & 0 \\ 0 & m \end{bmatrix}~\forall m \in \mathbb{N}_+ \\ X_n^*=X_{\beta_n} & \text{with } \beta_n = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{n} \end{bmatrix}~\forall n \in \mathbb{N}_+ \\ Y_{\gamma} = X_{\gamma} & \text{with } \gamma = \begin{bmatrix} 1 & \gamma \\ 0 & 1 \end{bmatrix}~\forall \lambda \in \mathbb{Q}/\mathbb{Z} \end{cases}$

Using the tricks of last time (that is, figuring out what functions convolution products represent, knowing all double-cosets) it is not too difficult to prove the defining relations among these generators to be the following¹

$\begin{enumerate} \item{(1) : $X_n^* X_n = 1, \forall n \in \mathbb{N}_+$} \item{(2) : $X_n X_m = X_{nm}, \forall m,n \in \mathbb{N}_+$} \item{(3) : $X_n X_m^* = X_m^* X_n, \text{whenever } (m,n)=1$} \item{(4) : $Y_{\gamma} Y_{\mu} = Y_{\gamma+\mu}, \forall \gamma,mu \in \mathbb{Q}/\mathbb{Z}$} \item{(5) : $Y_{\gamma}X_n = X_n Y_{n \gamma},~\forall n \in \mathbb{N}_+, \gamma \in \mathbb{Q}/\mathbb{Z}$} \item{(6) : $X_n Y_{\lambda} X_n^* = \frac{1}{n} \sum_{n \delta = \gamma} Y_{\delta},~\forall n \in \mathbb{N}_+, \gamma \in \mathbb{Q}/\mathbb{Z}$} \end{enumerate}$

Simple as these equations may seem, they bring us into rather uncharted ringtheoretic territories. Here a few fairly obvious ringtheoretic ingredients of the Bost-Connes Hecke algebra $\mathcal{H}$

the group-algebra of $\mathbb{Q}/\mathbb{Z}$

The equations (4) can be rephrased by saying that the subalgebra generated by the $Y_{\gamma}$ is the rational groupalgebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$ of the (additive) group $\mathbb{Q}/\mathbb{Z}$ . Note however that $\mathbb{Q}/\mathbb{Z}$ is a torsion group (that is, for all $\gamma = \frac{m}{n}$ we have that $n.\gamma = (\gamma+\gamma+ \hdots + \gamma) = 0$ ). Hence, the groupalgebra has LOTS of zero-divisors. In fact, this group-algebra doesn’t have any good ringtheoretic properties except for the fact that it can be realized as a limit of finite groupalgebras (semi-simple algebras)

$\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] = \underset{\rightarrow}{lim}~\mathbb{Q}[\mathbb{Z}/n \mathbb{Z}]$

and hence is a quasi-free (or formally smooth) algebra, BUT far from being finitely generated…

the grading group $\mathbb{Q}^+_{\times}$

The multiplicative group of all positive rational numbers $\mathbb{Q}^+_{\times}$ is a torsion-free Abelian ordered group and it follows from the above defining relations that $\mathcal{H}$ is graded by this group if we give

$deg(Y_{\gamma})=1,~deg(X_m)=m,~deg(X_n^*) = \frac{1}{n}$

Now, graded algebras have been studied extensively in case the grading group is torsion-free abelian ordered AND finitely generated, HOWEVER $\mathbb{Q}^+_{\times}$ is infinitely generated and not much is known about such graded algebras. Still, the ordering should allow us to use some tricks such as taking leading coefficients etc.

the endomorphisms of $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$

We would like to view the equations (5) and (6) (the latter after multiplying both sides on the left with X_n^* and using (1)) as saying that X_n and X_n^* are normalizing elements. Unfortunately, the algebra morphisms they induce on the group algebra $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$ are NOT isomorphisms, BUT endomorphisms. One source of algebra morphisms on the group-algebra comes from group-morphisms from $\mathbb{Q}/\mathbb{Z}$ to itself. Now, it is known that

$Hom_{grp}(\mathbb{Q}/\mathbb{Z},\mathbb{Q}/\mathbb{Z}) \simeq \hat{\mathbb{Z}}$ , the profinite completion of $\mathbb{Z}$ . A class of group-morphisms of interest to us are the maps given by multiplication by n on $\mathbb{Q}/\mathbb{Z}$ . Observe that these maps are epimorphisms with a cyclic order n kernel. On the group-algebra level they give us the epimorphisms

$\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \longrightarrow^{\phi_n} \mathbb{Q}[\mathbb{Q}/\mathbb{Z}]$ such that $\phi_n(Y_{\lambda}) = Y_{n \lambda}$ whence equation (5) can be rewritten as $Y_{\lambda} X_n = X_n \phi_n(Y_{\lambda})$ , which looks good until you think that $\phi_n$ is not an automorphism…

There are even other (non-unital) algebra endomorphisms such as the map $\mathbb{Q}[\mathbb{Q}/\mathbb{Z}] \rightarrow^{\psi_n} R_n$ defined by $\psi_n(Y_{\lambda}) = \frac{1}{n}(Y_{\frac{\lambda}{n}} + Y_{\frac{\lambda + 1}{n}} + \hdots + Y_{\frac{\lambda + n-1}{n}})$ and then, we can rewrite equation (6) as $Y_{\lambda} X_n^* = X_n^* \psi_n(Y_{\lambda})$ , but again, note that $\psi_n$ is NOT an automorphism.

almost strongly graded, but not quite…

Recall from last time that the characteristic function X_a for any double-coset-class $a \in \Gamma_0 \backslash \Gamma / \Gamma_0$ represented by the matrix $a=\begin{bmatrix} 1 & \lambda \\ 0 & \frac{m}{n} \end{bmatrix}$ could be written in the Hecke algebra as $X_a = n X_m Y_{n \lambda} X_n^* = n Y_{\lambda} X_m X_n^*$ . That is, we can write the Bost-Connes Hecke algebra as

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

Hence, if only the morphisms $\phi_n$ and $\psi_m$ would be automorphisms, this would say that $\mathcal{H}$ is a strongly $\mathbb{Q}^+_{\times}$ -algebra with part of degree one the groupalgebra of $\mathbb{Q}/\mathbb{Z}$ .

However, they are not. But there is an extension of the notion of strongly graded algebras which Fred has dubbed crystalline graded algebras in which it is sufficient that the algebra maps are all epimorphisms. (maybe I’ll post about these algebras, another time). However, this is not the case for the $\psi_m$ …

So, what is the most elegant ringtheoretic framework in which the algebra $\mathcal{H}$ fits??? Surely, you can do better than generalized crystalline graded algebra…

Previous in series Next in series

if someone wants the details, tell me and I’ll include a ‘technical post’ or consult the Bost-Connes original paper but note that this scanned version needs 26.8Mb [↩]

Tags: Connes, crystalline, geometry, graded, groups, noncommutative, profinite, ringtheory
Posted in geometry | 5 Comments »

Posts Tagged ‘crystalline’

BC stands for Bi-Crystalline graded

Noncommutative geometry and the Riemann zeta function

Bost-Connes for ringtheorists

Noncommutative geometry and the Riemann zeta function

Recent Posts

Recent Comments