Category: featured

BC stands for Bi-Crystalline graded

Published January 26, 2008 by lievenlb

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 :

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 $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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Bost-Connes for ringtheorists

Published January 23, 2008 by lievenlb

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 following (( 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 ))

(1) : $X_n^* X_n = 1, \forall n \in \mathbb{N}_+$

(2) : $X_n X_m = X_{nm}, \forall m,n \in \mathbb{N}_+$

(3) : $X_n X_m^* = X_m^* X_n, \text{whenever } (m,n)=1$

(4) : $Y_{\gamma} Y_{\mu} = Y_{\gamma+\mu}, \forall \gamma,mu \in \mathbb{Q}/\mathbb{Z}$

(5) : $Y_{\gamma}X_n = X_n Y_{n \gamma},~\forall n \in \mathbb{N}_+, \gamma \in \mathbb{Q}/\mathbb{Z}$

(6) : $X_n Y_{\lambda} X_n^* = \frac{1}{n} \sum_{n \delta = \gamma} Y_{\delta},~\forall n \in \mathbb{N}_+, \gamma \in \mathbb{Q}/\mathbb{Z}$

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+ \ldots + \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}} + \ldots + 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…

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the Bost-Connes Hecke algebra

Published January 19, 2008 by lievenlb

As before, $\Gamma $ is the subgroup of the rational linear group $GL_2(\mathbb{Q}) $ consisting of the matrices

$\begin{bmatrix} 1 & b \\ 0 & a \end{bmatrix} $ with $a \in \mathbb{Q}_+ $ and $\Gamma_0 $ the subgroup of all matrices $\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} $ with $n \in \mathbb{N} $. Last time, we have seen that the double coset space $\Gamma_0 \backslash \Gamma / \Gamma_0 $ can be identified with the set of all rational points in the fractal comb consisting of all couples $~(a,b) $ with $a=\frac{m}{n} \in \mathbb{Q}_+ $ and $b \in [0,\frac{1}{n}) \cap \mathbb{Q} $

The blue spikes are at the positive natural numbers $a={ 1,2,3,\ldots } $. Over $a=1 $ they correspond to the matrices $\begin{bmatrix} 1 & \gamma \\ 0 & 1 \end{bmatrix} $ with $\gamma \in [0,1) \cap \mathbb{Q} $ and as matrix-multiplication of such matrices corresponds to addition of the $\gamma $ we see that these cosets can be identified with the additive group $\mathbb{Q}/\mathbb{Z} $ (which will reappear at a later stage as the multiplicative group of all roots of unity).

The Bost-Connes Hecke algebra $\mathcal{H} = \mathcal{H}(\Gamma,\Gamma_0) $ is the convolution algebra of all comlex valued functions with finite support on the double coset space $\Gamma_0 \backslash \Gamma / \Gamma_0 $. That is, as a vector space the algebra has as basis the functions $e_X $ with $X \in \Gamma_0 \backslash \Gamma / \Gamma_0 $ (that is, $X $ is a point of the fractal comb) and such that $e_X(X)=1 $ and $e_X(Y)=0 $ for all other double cosets $Y \not= X $. The algebra product on $\mathcal{H} $ is the convolution-product meaning that if $f,f’ $ are complex functions with finite support on the Bost-Connes space, then they can also be interpreted as $\Gamma_0 $-bi-invariant functions on the group $\Gamma $ (for this just means that the function is constant on double cosets) and then $f \ast f’ $ is the function defined for all $\gamma \in \Gamma $ by

$f \ast f'(\gamma) = \sum_{\mu \in \Gamma/ \Gamma_0} f(\mu) f'(\mu^{-1} \gamma) $

Last time we have seen that the coset-space $\Gamma / \Gamma_0 $ can be represented by all rational points $~(a,b) $ with $b<1 $. At first sight, the sum above seems to be infinite, but, f and f’ are non-zero only at finitely many double cosets and we have see last time that $\Gamma_0 $ acts on one-sided cosets with finite orbits. Therefore, $f \ast f $ is a well-defined $\Gamma_0 $-bi-invariant function with finite support on the fractal comb $\Gamma_0 \backslash \Gamma / \Gamma_0 $. Further, observe that the unit element of $\mathcal{H} $ is the function corresponding to the identity matrix in $\Gamma $.

Looking at fractal-comb picture it is obvious that the Bost-Connes Hecke algebra $\mathcal{H} $ is a huge object. Today, we will prove the surprising result that it can be generated by the functions corresponding to the tiny portion of the comb, shown below.

That is, we will show that $\mathcal{H} $ is generated by the functions $e(\gamma) $ corresponding to the double-coset $X_{\gamma} = \begin{bmatrix} 1 & \gamma \\ 0 & 1 \end{bmatrix} $ (the rational points of the blue line-segment over 1, or equivalently, the elements of the group $\mathbb{Q}/\mathbb{Z} $), together with the functions $\phi_n $ corresponding to the double-coset $X_n = \begin{bmatrix} 1 & 0 \\ 0 & n \end{bmatrix} $ for all $ n \in \mathbb{N}_+ $ (the blue dots to the right in the picture) and the functions $\phi_n^* $ corresponding to the double cosets $X_{1/n} = \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{n} \end{bmatrix} $ (the red dots to the left).

Take a point in the fractal comb $X = \begin{bmatrix} 1 & \gamma \\ 0 & \frac{m}{n} \end{bmatrix} $ with $~(m,n)=1 $ and $\gamma \in [0,\frac{1}{n}) \cap \mathbb{Q} \subset [0,1) \cap \mathbb{Q} $. Note that as $\gamma < \frac{1}{n} $ we have that $n \gamma < 1 $ and hence $e(n \gamma) $ is one of the (supposedly) generating functions described above.

Because $X = \begin{bmatrix} 1 & \gamma \\ 0 & \frac{m}{n} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & m \end{bmatrix} \begin{bmatrix} 1 & n \gamma \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & \frac{1}{n} \end{bmatrix} = X_m X_{n \gamma} X_{1/n} $ we are aiming for a relation in the Hecke algebra $\phi_m \ast e(n \gamma) \ast \phi^*_n = e_X $. This is ‘almost’ true, except from a coefficient.

Let us prove first the equality of functions $e_X \ast \phi_n = n \phi_m \ast e(n \gamma) $. To do this we have to show that they have the same value for all points $Y \in \Gamma_0 \backslash \Gamma / \Gamma_0 $ in the fractal comb. Let us first study the function on the right hand side.

$\phi_m \ast e(n \gamma) = \sum_{g \in \Gamma/\Gamma_0} \phi_m(g) e(n \gamma)(g^{-1}Y) $. Because $X_m \Gamma_0 $ is already a double coset (over $m $ we have a comb-spike of length one, so all rational points on it determine at the same time a one-sided and a double coset. Therefore, $\phi_m(g) $ is zero unless $g = X_m $ and then the value is one.

Next, let us consider the function on the left-hand side. $e_X \ast \phi_n(Y) = \sum_{g \in \Gamma / \Gamma_0} e_X(g) \phi_m( g^{-1} Y) $. We have to be a bit careful here as the double cosets over $a=\frac{m}{n} $ are different from the left cosets. Recall from last time that the left-cosets over a are given by all rational points of the form $~(a,b) $ with $ b < 1 $ whereas the double-cosets over a are represented by the rational points of the form $~(a,b) $ with $b < \frac{1}{n} $ and hence the $\Gamma_0 $-orbits over a all consist of precisely n elements g.
That is, $e_X(g) $ is zero for all $ g \in \Gamma/\Gamma_0 $ except when g is one of the following matrices

$ g \in { \begin{bmatrix} 1 & \gamma \\ 0 & \frac{m}{n} \end{bmatrix}, \begin{bmatrix} 1 & \gamma+\frac{1}{n} \\ 0 & \frac{m}{n} \end{bmatrix}, \begin{bmatrix} 1 & \gamma + \frac{2}{n} \\ 0 & \frac{m}{n} \end{bmatrix}, \ldots, \begin{bmatrix} 1 & \gamma + \frac{n-1}{n} \\ 0 & \frac{m}{n} \end{bmatrix} } $

Further, $\phi_n(g^{-1}Y) $ is zero unless $g^{-1}Y \in \Gamma_0 \begin{bmatrix} 1 & 0 \\ 0 & n \end{bmatrix} \Gamma_0 $, or equivalently, that $Y \in \Gamma_0 g \Gamma_0 \begin{bmatrix} 1 & 0 \\ 0 & n \end{bmatrix} \Gamma_0 = \Gamma_0 g \begin{bmatrix} 1 & 0 \\ 0 & n \end{bmatrix} \Gamma_0 $ and for each of the choices for g we have that

$ \begin{bmatrix} 1 & \gamma + \frac{k}{n} \\ 0 & \frac{m}{n} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & n \end{bmatrix} = \begin{bmatrix} 1 & n \gamma + k \\ 0 & m \end{bmatrix} \sim \begin{bmatrix} 1 & n\gamma \\ 0 & m \end{bmatrix} $

Therefore, the function $e_X \ast \phi_n $ is zero at every point of the fractal comb unless at $\begin{bmatrix} 1 & n \gamma \\ 0 & m \end{bmatrix} $ where it is equal to $n $. This proves the claimed identity of functions and as one verifies easily that $\phi_n^* \ast \phi_n = 1 $, it follows that all base vectors $e_X $ of $\mathcal{H} $ can be expressed in the claimed generators

$ e_X = n \phi_m \ast e(n \gamma) \ast \phi_n^* $

Bost and Connes use slightly different generators, namely with $\mu_n = \frac{1}{\sqrt{n}} \phi_n $ and $\mu_n^* = \sqrt{n} \phi_n^* $ in order to have all relations among the generators being defined over $\mathbb{Q} $ (as we will see another time). This will be important later on to have an action of the cyclotomic Galois group $Gal(\mathbb{Q}^{cycl}/\mathbb{Q}) $ on certain representations of $\mathcal{H} $.

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