any UCL folks who have taken or are about to take Measure Theory could probably benefit from following Tao's notes on the subject. good stuff!
http://terrytao.wordpress.com/2010/08/31/course-announcement-245a-real-analysis/
http://terrytao.wordpress.com/2010/09/04/245a-prologue-the-problem-of-measure/
http://terrytao.wordpress.com/2010/09/09/245a-notes-1-lebesgue-measure/
http://terrytao.wordpress.com/2010/09/19/245a-notes-2-the-lebesgue-integral/
http://terrytao.wordpress.com/2010/09/25/245a-notes-3-integration-on-abstract-measure-spaces-and-the-convergence-theorems/
And the category link:
http://terrytao.wordpress.com/category/teaching/245a-real-analysis/
Thursday, September 30, 2010
Thursday, September 16, 2010
Groupoids and Equivelance Relations
So in working out the details and the structure of the groupoids of commensurability relations on 1dQLs, I realized I could generalize the method a bit more.
Given a set S with an equivalence relation ~, we can form a groupoid consisting of all ordered pairs of equivalent elements, i.e., all pairs $(s_1,s_2)$ where $s_1 \sim s_2$. The composition is then $(s_1,s_2) \circ (s_3,s_4) = (s_1,s_4)$ defined when $s_2 = s_3$.
It's such an obvious example of a groupoid that I'm embarrassed that I missed it. I'm sure it's in a textbook somewhere. I would have saved a lot of time if I had found and read that textbook a week ago.
Edit
Oh! Look! There it is! It's on wikipedia. fml.
Given a set S with an equivalence relation ~, we can form a groupoid consisting of all ordered pairs of equivalent elements, i.e., all pairs $(s_1,s_2)$ where $s_1 \sim s_2$. The composition is then $(s_1,s_2) \circ (s_3,s_4) = (s_1,s_4)$ defined when $s_2 = s_3$.
It's such an obvious example of a groupoid that I'm embarrassed that I missed it. I'm sure it's in a textbook somewhere. I would have saved a lot of time if I had found and read that textbook a week ago.
Edit
Oh! Look! There it is! It's on wikipedia. fml.
The Étale Groupoids of the Bost-Connes System
PDF version of the past 3 posts.
Recall that a 1-dimensional $ {\mathbb{Q}}$-lattice (henceforth called a 1dQL) can be denoted by $ {(\frac{1}{\lambda} \mathbb{Z}, \frac{1}{\lambda} \rho)}$ with $ {\lambda}$ a positive real and $ {\rho \in End(\mathbb{Q}/\mathbb{Z})=\hat{\mathbb{Z}}}$. Let $ {\mathcal{Q}_1}$ denote the set of commensurability relations of such lattices. Then $ {\mathcal{Q}_1}$ consists of ordered pairs $ {(\Lambda_1,\Lambda_2)}$ of commensurable 1dQLs. There's a natural composition of elements, namely:
We can actually give a much more clear description of this groupoid. Let $ {\mathcal{G}_1 = \{(r,\rho,\lambda) : r \in \mathbb{Q}^{*}_{+}, \, \rho \in \hat{\mathbb{Z}}, \, \lambda \in \mathbb{R}^{*}_{+} \; such \; that \; r \rho \in \hat{\mathbb{Z}} \}}$ be the groupoid with the composition
Proposition. The map $ {\phi : \mathcal{G}_1 \rightarrow \mathcal{Q}_1}$ by
Proof: First note that the two 1dQLs, $ {\left( \frac{1}{r \lambda} \mathbb{Z}, \frac{1}{\lambda} \rho \right)}$ and $ {\left( \frac{1}{\lambda} \mathbb{Z}, \frac{1}{\lambda} \rho \right)}$ are commensurable. Indeed, $ {\mathbb{Q} \frac{1}{r \lambda} \mathbb{Z} = \frac{1}{\lambda} \mathbb{Q} \mathbb{Z}}$.
Second, note that every pair of commensurable 1dQLs is of this form. To see this, let $ {(\frac{1}{\lambda_1} \mathbb{Z}, \frac{1}{\lambda_1} \rho_1)}$ and $ {(\frac{1}{\lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2)}$ be another pair of commensurable 1dQLs. $ {\frac{1}{\lambda_2}}$ then must be a rational multiple of $ {\frac{1}{\lambda_1}}$, id est, $ {\lambda_1 = r \lambda_2}$ for a strictly positive rational $ {r=\frac{a}{b}}$ (see previous posts on $ {\mathbb{Q}}$-lattices). Additionally, we must have $ {\frac{1}{\lambda_1} \rho_1 = \frac{1}{\lambda_2} \rho_2 \mod (\frac{1}{\lambda_1} \mathbb{Z} + \frac{1}{\lambda_2} \mathbb{Z})}$ Factoring out by $ {\lambda_2}$, this yields $ {\frac{1}{r} \rho_1 = \rho_2 \mod \frac{1}{a} \mathbb{Z}}$, and then $ {a\rho_2 - b \rho_1 = 0}$ and finally $ {\rho_1 = r \rho_2}$, so that $ {\frac{1}{\lambda_1}\rho_1=\frac{1}{r \lambda_2} r \rho_2 = \frac{1}{\lambda_2}\rho_2}$. Hence the map is surjective.
Finally, note that the map preserves the groupoid composition and inversion. The composition
Proposition. $ {\mathcal{G}_1}$ is an Étale groupoid. (I do not yet know the significance of this statement.)
Proof: Since $ {\mathcal{G}_1}$ has the product topology, $ {\mathbb{Q}_{+}^{*}}$ has the subspace topology, multiplication in all spaces is continuous, and division in $ {\mathbb{Q}_{+}^{*}}$ is continuous, the groupoid operations are homeomorphisms. $ \Box$
But we're not done yet. All the action in the Bost-Connes system comes not from $ {\mathcal{G}_1}$, but from the commensurability relations of 1dQLs modulo scaling, id est, from $ {\mathcal{Q}_1/\mathbb{R}_{+}^{*}}$. The unit set $ {(\mathcal{Q}_1/\mathbb{R}_{+}^{*})^0}$ is simply $ {\hat{\mathbb{Z}}}$ in this case. As you might expect, we can model this groupoid with an Étale groupoid similar to $ {\mathcal{G}_1}$: Let
Proposition. The map $ {\gamma : \mathcal{U}_1 \rightarrow \mathcal{Q}_1/\mathbb{R}_{+}^{*}}$ by
Proof: The proof of this is similar to the last one, so I won't repeat myself. Oddly enough, Connes & Marcolli give a short proof of the first isomorphism, and a long proof of this one. $ \Box$
Since the topology of the groupoids largely comes from $ {\mathbb{R}}$, they are also locally compact (at least, I think that's the reason.)
1. Why This Matters
It's important to keep the larger goal of the project in mind when working out these smaller details. My goal is to describe the construction of the Bost-Connes system, as I did here. Connes & Marcolli give an explicit description in terms of generators and relations of the Bost-Connes algebra in Noncommutative Geometry, Quantum Fields, Motives [1]. Additionally, Marcolli gives the same description in Lectures on Arithmetic Noncommutative Geometry [2], Connes & Marcolli give it again in $ {\mathbb{Q}}$-Lattices: Quantum Statistical Mechanics and Galois Theory [3], and Bost & Connes give it in Hecke Algebras, Type III Factors and Phase Transitions with Spontaneous Symmetry Breaking in Number Theory [4]. However, only [1] and [4] provide a proof for the description, and the latter only in terms of Hecke Algebras. [1], however, starts from convolution algebras of these Étale groupoids, and shows the relation with the Hecke algebra description in the paper by Bost & Connes [4]. Hence, it's important that I understand and am able to describe these groupoids.
2. What's next
The next step is a careful study of the convolution and C* groupoid algebras of the above structures, leading to a proof of the explicit description of the Bost-Connes system (Prop. 3.23 in Noncommutative Geometry, Quantum Fields, Motives). Following that, I'll need to understand the relation with the Hecke Algebras given by Bost & Connes and the symmetries of the system. Then I can begin a study of the class field theory of $ {\mathbb{Q}}$ and learn how to apply the Bost-Connes system.
Eventually, I'll study the two other Quantum Statistical Mechanical systems, but first I need to finish the Bost-Connes system and its applications, and learn its corresponding sub algebras and Shimura varieties, which will be vital to generalizing the techniques used to apply it.
3. Sources
Noncommutative Geometry, Quantum Fields, Motives. Connes, Marcolli
Lectures on Arithmetic Noncommutative Geometry. Marcolli (the usual two)
A Homology Theory for Étale groupoids. Marius Crainic and Ieke Moerdijk
Abstract Algebra. Dummit, Foote (appendix on category theory.)
Category Theory. Steve Awodey
Wikipedia on Groupoids
Recall that a 1-dimensional $ {\mathbb{Q}}$-lattice (henceforth called a 1dQL) can be denoted by $ {(\frac{1}{\lambda} \mathbb{Z}, \frac{1}{\lambda} \rho)}$ with $ {\lambda}$ a positive real and $ {\rho \in End(\mathbb{Q}/\mathbb{Z})=\hat{\mathbb{Z}}}$. Let $ {\mathcal{Q}_1}$ denote the set of commensurability relations of such lattices. Then $ {\mathcal{Q}_1}$ consists of ordered pairs $ {(\Lambda_1,\Lambda_2)}$ of commensurable 1dQLs. There's a natural composition of elements, namely:
$ {(\Lambda_1,\Lambda_2) \circ (\Lambda_3, \Lambda_4) = (\Lambda_1,\Lambda_4)}$ defined for $ {\Lambda_2=\Lambda_3}$.
It's not too hard to see that each element has an inverse. Indeed $ {(\Lambda_1,\Lambda_2)^{-1} = (\Lambda_2,\Lambda_1)}$, so that $ {(\Lambda_2,\Lambda_1) \circ (\Lambda_1,\Lambda_4) = (\Lambda_3, \Lambda_4)}$. Hence we have a groupoid. The units $ {\mathcal{Q}_1^0}$ consists of the pairs $ {(\Lambda, \Lambda)}$, id est, just the set of 1dQLs. We can actually give a much more clear description of this groupoid. Let $ {\mathcal{G}_1 = \{(r,\rho,\lambda) : r \in \mathbb{Q}^{*}_{+}, \, \rho \in \hat{\mathbb{Z}}, \, \lambda \in \mathbb{R}^{*}_{+} \; such \; that \; r \rho \in \hat{\mathbb{Z}} \}}$ be the groupoid with the composition
$ {(r_1,\rho_1,\lambda_1) \circ (r_2,\rho_2,\lambda_2) = (r_1 r_2, \rho_2, \lambda_2)}$ defined for $ {r_2 \rho_2 = \rho_1}$, $ {r_2 \lambda_2 = \lambda_1}$.
and inverse elements $ \displaystyle (r,\rho,\lambda)^{-1} = \left(\frac{1}{r},r \rho, r \lambda \right) $
Proposition. The map $ {\phi : \mathcal{G}_1 \rightarrow \mathcal{Q}_1}$ by
$ \displaystyle (r,\rho,\lambda) \mapsto \left( \left( \frac{1}{r \lambda} \mathbb{Z}, \frac{1}{\lambda} \rho \right), \left( \frac{1}{\lambda} \mathbb{Z}, \frac{1}{\lambda} \rho \right) \right)$
is a groupoid isomorphism. (Connes, Marcolli; Noncommutative Geometry, Quantum Fields, and Motives. Lemma 3.21-2.) Proof: First note that the two 1dQLs, $ {\left( \frac{1}{r \lambda} \mathbb{Z}, \frac{1}{\lambda} \rho \right)}$ and $ {\left( \frac{1}{\lambda} \mathbb{Z}, \frac{1}{\lambda} \rho \right)}$ are commensurable. Indeed, $ {\mathbb{Q} \frac{1}{r \lambda} \mathbb{Z} = \frac{1}{\lambda} \mathbb{Q} \mathbb{Z}}$.
Second, note that every pair of commensurable 1dQLs is of this form. To see this, let $ {(\frac{1}{\lambda_1} \mathbb{Z}, \frac{1}{\lambda_1} \rho_1)}$ and $ {(\frac{1}{\lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2)}$ be another pair of commensurable 1dQLs. $ {\frac{1}{\lambda_2}}$ then must be a rational multiple of $ {\frac{1}{\lambda_1}}$, id est, $ {\lambda_1 = r \lambda_2}$ for a strictly positive rational $ {r=\frac{a}{b}}$ (see previous posts on $ {\mathbb{Q}}$-lattices). Additionally, we must have $ {\frac{1}{\lambda_1} \rho_1 = \frac{1}{\lambda_2} \rho_2 \mod (\frac{1}{\lambda_1} \mathbb{Z} + \frac{1}{\lambda_2} \mathbb{Z})}$ Factoring out by $ {\lambda_2}$, this yields $ {\frac{1}{r} \rho_1 = \rho_2 \mod \frac{1}{a} \mathbb{Z}}$, and then $ {a\rho_2 - b \rho_1 = 0}$ and finally $ {\rho_1 = r \rho_2}$, so that $ {\frac{1}{\lambda_1}\rho_1=\frac{1}{r \lambda_2} r \rho_2 = \frac{1}{\lambda_2}\rho_2}$. Hence the map is surjective.
Finally, note that the map preserves the groupoid composition and inversion. The composition
$ \displaystyle \left( (\frac{1}{r_1 \lambda_1} \mathbb{Z}, \frac{1}{\lambda_1} \rho_1),(\frac{1}{\lambda_1} \mathbb{Z}, \frac{1}{\lambda_1} \rho_1) \right) \circ \left( (\frac{1}{r_2 \lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2),(\frac{1}{\lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2) \right)$
is only defined in $ {\mathcal{Q}_1}$ if $ {\lambda_1=r_2 \lambda_2}$ and $ {\rho_1 = r_2 \rho_2}$. In that case, it equals $ \displaystyle \left( (\frac{1}{r_1 \lambda_1} \mathbb{Z}, \frac{1}{\lambda_1} \rho_1),(\frac{1}{\lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2) \right)=\left( (\frac{1}{r_1 r_2 \lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2),(\frac{1}{\lambda_2} \mathbb{Z}, \frac{1}{\lambda_2} \rho_2) \right) $
Which is exactly $ {\phi(r_1 r_2, \rho_2,\lambda_2)}$. Additionally, $ \displaystyle \phi(\frac{1}{r},r \rho, r \lambda ) = \left( (\frac{1}{\lambda} \mathbb{Z}, \frac{1}{\lambda} \rho), (\frac{1}{r \lambda} \mathbb{Z}, \frac{1}{\lambda} \rho) \right)$
Which is the inverse of $ {\left( (\frac{1}{r \lambda} \mathbb{Z}, \frac{1}{\lambda} \rho), (\frac{1}{\lambda} \mathbb{Z}, \frac{1}{\lambda} \rho) \right)}$ $ \Box$Proposition. $ {\mathcal{G}_1}$ is an Étale groupoid. (I do not yet know the significance of this statement.)
Proof: Since $ {\mathcal{G}_1}$ has the product topology, $ {\mathbb{Q}_{+}^{*}}$ has the subspace topology, multiplication in all spaces is continuous, and division in $ {\mathbb{Q}_{+}^{*}}$ is continuous, the groupoid operations are homeomorphisms. $ \Box$
But we're not done yet. All the action in the Bost-Connes system comes not from $ {\mathcal{G}_1}$, but from the commensurability relations of 1dQLs modulo scaling, id est, from $ {\mathcal{Q}_1/\mathbb{R}_{+}^{*}}$. The unit set $ {(\mathcal{Q}_1/\mathbb{R}_{+}^{*})^0}$ is simply $ {\hat{\mathbb{Z}}}$ in this case. As you might expect, we can model this groupoid with an Étale groupoid similar to $ {\mathcal{G}_1}$: Let
$ \displaystyle \mathcal{U}_1 = \{ (r,\rho) : r \in \mathbb{Q}_{+}^{*})^0, \rho \in \hat{\mathbb{Z}} \; such \; that \; r \rho \in \hat{\mathbb{Z}}\}$
With composition $ { (r_1,\rho_1) \circ (r_2,\rho_2) = (r_1 r_2,\rho_2)}$ defined when $ {\rho_1 = r_2 \rho_2}$.
.Proposition. The map $ {\gamma : \mathcal{U}_1 \rightarrow \mathcal{Q}_1/\mathbb{R}_{+}^{*}}$ by
$ \displaystyle (r,\rho) \mapsto \left( (\frac{1}{r}\mathbb{Z},\rho), (\mathbb{Z},\rho) \right)$
is a groupoid isomorphism. (Connes, Marcolli; Noncommutative Geometry, Quantum Fields, and Motives. Prop 3.22.)Proof: The proof of this is similar to the last one, so I won't repeat myself. Oddly enough, Connes & Marcolli give a short proof of the first isomorphism, and a long proof of this one. $ \Box$
Since the topology of the groupoids largely comes from $ {\mathbb{R}}$, they are also locally compact (at least, I think that's the reason.)
1. Why This Matters
It's important to keep the larger goal of the project in mind when working out these smaller details. My goal is to describe the construction of the Bost-Connes system, as I did here. Connes & Marcolli give an explicit description in terms of generators and relations of the Bost-Connes algebra in Noncommutative Geometry, Quantum Fields, Motives [1]. Additionally, Marcolli gives the same description in Lectures on Arithmetic Noncommutative Geometry [2], Connes & Marcolli give it again in $ {\mathbb{Q}}$-Lattices: Quantum Statistical Mechanics and Galois Theory [3], and Bost & Connes give it in Hecke Algebras, Type III Factors and Phase Transitions with Spontaneous Symmetry Breaking in Number Theory [4]. However, only [1] and [4] provide a proof for the description, and the latter only in terms of Hecke Algebras. [1], however, starts from convolution algebras of these Étale groupoids, and shows the relation with the Hecke algebra description in the paper by Bost & Connes [4]. Hence, it's important that I understand and am able to describe these groupoids.
2. What's next
The next step is a careful study of the convolution and C* groupoid algebras of the above structures, leading to a proof of the explicit description of the Bost-Connes system (Prop. 3.23 in Noncommutative Geometry, Quantum Fields, Motives). Following that, I'll need to understand the relation with the Hecke Algebras given by Bost & Connes and the symmetries of the system. Then I can begin a study of the class field theory of $ {\mathbb{Q}}$ and learn how to apply the Bost-Connes system.
Eventually, I'll study the two other Quantum Statistical Mechanical systems, but first I need to finish the Bost-Connes system and its applications, and learn its corresponding sub algebras and Shimura varieties, which will be vital to generalizing the techniques used to apply it.
3. Sources
Noncommutative Geometry, Quantum Fields, Motives. Connes, Marcolli
Lectures on Arithmetic Noncommutative Geometry. Marcolli (the usual two)
A Homology Theory for Étale groupoids. Marius Crainic and Ieke Moerdijk
Abstract Algebra. Dummit, Foote (appendix on category theory.)
Category Theory. Steve Awodey
Wikipedia on Groupoids
Wednesday, September 15, 2010
A Minor Problem with the Étale groupoids of Bost-Connes
EDIT: I was confused and delusional when writing this blog; the problem discussed below does not actually exist. I didn't know how to take the inverse in the groupoids mentioned. It's actually quite simple and will be discussed in a later post. This entry can be safely ignored.
I had hoped to move beyond my study of the groupoids mentioned in the construction of the Bost Connes system on Monday. Yet it wasn't til Tuesday that I had fully understood what an Étale groupoid was. Though I had thought I had understood the aforementioned groupoids and the proofs of their isomorphisms, I realized in the middle of my writeup last night that I was still confused about things. I worked out some details on paper, but there's one thing left that I'm hung up on. It has to do with the inversion of the elements in the groupoid. Here's the detail isolated all by itself:
Given a $\rho \in \hat{\mathbb{Z}}$ and a positive rational $r$ such that $r \rho \in \hat{\mathbb{Z}}$. I need that $\frac{1}{r} \rho \in \hat{\mathbb{Z}}$ as well. But I don't see how this could possible be true, at least not when I'm thinking of $\hat{\mathbb{Z}}$ as an inverse limit. However....
As I proved earlier, $\hat{\mathbb{Z}}$ is isomorphic to $End(\mathbb{Q}/\mathbb{Z})$, so we can think of $\rho$ as being a homomorphism $\rho : \mathbb{Q}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$. Now if $ r \rho$ is still such a homomorphism ($(r \rho )(x) =r \rho(x)$) then I can't see any reason why $\frac{1}{r} \rho$ wouldn't be. It would preserve addition and the identity on $\mathbb{Q}/\mathbb{Z}$ just fine (I think).
So that brings me to another question...is there any positive rational $r$ and homomorphism $\rho \in End(\mathbb{Q}/\mathbb{Z})$ such that $r \rho$ is not a homomorphism? If not, then why does Connes state that requirement?
Some thoughts immediately after writing this post
A homomorphism $\rho \in End(\mathbb{Q}/\mathbb{Z})$ must preserve torsion, id est, there's a smallest $n$ such that an $x \in \mathbb{Q}/\mathbb{Z}$ will have $n x = 0$. $\rho$ must preserve $n \rho(x) = 0$ if it's going to be a homomorphism. Multiplication by a natural number $k$ is an a homomorphism on $\mathbb{Q}/\mathbb{Z}$, but multiplication by $\frac{1}{k}$ is not. Hence I have my $r=k$ where $r \rho \in End(\mathbb{Q}/\mathbb{Z})$ but $\frac{1}{r} \rho$ is not. And thus I still have a problem inverting elements in the groupoids mentioned. I'll have to spell out this problem more explicitly in another post.
Tuesday, September 14, 2010
Groupoids
I'm continuing a series of posts working out some details regarding some details in the groupoids used to construct the Bost-Connes system.
Groupoids
A groupoid is essentially a group, but one where the binary operation is not defined for all pairs of elements. That is, a groupoid is set G such that for all $g \in G$, there exist an inverse $g^{-1}$, together with a partial function $ \ast : G \times G \to G$ ($\ast$ is not necessarily a binary operation; it needn't be defined for all ordered pairs.) and with the following properties:
While the above definition of a groupoid is familiar to one who has any experience with groups, it's sometimes more useful to think of it as a category: Let $G$ be a groupoid, and define $G^0$ to be the set of all elements of the form $g * g^{-1}$. Now let $x, y \in G^0$. If $f \in G$ is an element such that $y * f * x$ is defined, then we say $f \in G^1$ and $f: x \to y$.
In this case, it's not to hard to see that composition works: if we have $f:x \to y$ and $g:y \to z$, then $gf$ is defined and so is $z*gf*x$. Additionally, note that for every $x \in G^0$, the identity arrow $1_x$ is simply $x$. One oddity we have is that the inverse map $x * f^{-1} * y$ is defined (remember that since $x, y \in G^0$, we have $x=a*a^{-1}$ for some $a$, so that $x^{-1}=x$.) so that we have an inverse arrow $f^{-1} : y \to x$. Thus any groupoid can be considered as a category where all arrows have inverses.
Moreover, any category where all arrows have inverses is also a groupoid, so the two characterizations are actually equivalent definitions. Here's how: Given a groupoid $C$ in the category theoretic sense (id est, we have objects $C^0$ and arrows $C^{-1}$), we can denote all arrows $f: x \to y$ by $G(x,y)$. Let G be the disjoint union of all such $G(x,y)$. Then inverses are defined for all elements, and arrow composition because the partially defined groupoid operation.
Now lets take a groupoid $G$ (with objects, now called units $G^0$ and arrows/morphisms $G^{1}$). Recall the notation $G(x,y) = \{ f \in G^1 : f:x \to y. x,y \in G^0\}$. Let $x \in G^0$. Then $G(x,x)$ is a group (recall the example of a category of a group from earlier!) We call it the isotropy group at $x$ and denote it by $I_x$.
Before we continue, lets give an example. Let $GL(\mathbb{R})$ denote all the real invertible matrices under matrix multiplication. then $GL(\mathbb{R})^0$ is in bijective correspondence with the natural numbers, as for each natural number $n$ we have the corresponding $n \times n$ identity matrix. $G(n,m)$ is empty when $n \neq m$, but $G(n,n)$ is the general linear group $GL_n(\mathbb{R})$ of all $n \times n$ invertible matrices.
Finally, we say a groupoid $G$ is a topological groupoid when $G^0$ and $G^1$ are topological spaces, and all the corresponding maps on them (e.g. for $f \in G^1$ we have the domain $d(f)$, $d : G^1 \to G^0$, arrow composition $ \circ : G^1 \times G^1 \to G^1$, et cetera) must be continuous. Or equivalently, inversion and the groupoid operation must be continuous maps. A topological groupoid $G$ is said to be Étale if the domain map $d : G^1 \to G^0$ is a local homeomorphism (and thus all the other maps too), or equivalently, if inversion and the groupoid operation are local homeomorphism. Examples of these will come in the next post.
Sources
Noncommutative Geometry, Quantum Fields, Motives. Connes
Lectures on Arithmetic Noncommutative Geometry. Marcolli (the usual two)
A Homology Theory for Étale groupoids. Marius Crainic and Ieke Moerdijk
Abstract Algebra Dummit, Foote (appendix on category theory.)
Category Theory Steve Awodey
Groupoids
A groupoid is essentially a group, but one where the binary operation is not defined for all pairs of elements. That is, a groupoid is set G such that for all $g \in G$, there exist an inverse $g^{-1}$, together with a partial function $ \ast : G \times G \to G$ ($\ast$ is not necessarily a binary operation; it needn't be defined for all ordered pairs.) and with the following properties:
- Associativity: If $a * b$ and $b * c$ are defined, then $(a*b)*c=a*(b*c)$ is also defined
- Inverse: $a*a^{-1}=a^{-1}*a$ is always defined.
- Units: If $a*b$ is defined, then $a*b*b^{-1}=a$ and $a^{-1}*a*b=b$
While the above definition of a groupoid is familiar to one who has any experience with groups, it's sometimes more useful to think of it as a category: Let $G$ be a groupoid, and define $G^0$ to be the set of all elements of the form $g * g^{-1}$. Now let $x, y \in G^0$. If $f \in G$ is an element such that $y * f * x$ is defined, then we say $f \in G^1$ and $f: x \to y$.
In this case, it's not to hard to see that composition works: if we have $f:x \to y$ and $g:y \to z$, then $gf$ is defined and so is $z*gf*x$. Additionally, note that for every $x \in G^0$, the identity arrow $1_x$ is simply $x$. One oddity we have is that the inverse map $x * f^{-1} * y$ is defined (remember that since $x, y \in G^0$, we have $x=a*a^{-1}$ for some $a$, so that $x^{-1}=x$.) so that we have an inverse arrow $f^{-1} : y \to x$. Thus any groupoid can be considered as a category where all arrows have inverses.
Moreover, any category where all arrows have inverses is also a groupoid, so the two characterizations are actually equivalent definitions. Here's how: Given a groupoid $C$ in the category theoretic sense (id est, we have objects $C^0$ and arrows $C^{-1}$), we can denote all arrows $f: x \to y$ by $G(x,y)$. Let G be the disjoint union of all such $G(x,y)$. Then inverses are defined for all elements, and arrow composition because the partially defined groupoid operation.
Now lets take a groupoid $G$ (with objects, now called units $G^0$ and arrows/morphisms $G^{1}$). Recall the notation $G(x,y) = \{ f \in G^1 : f:x \to y. x,y \in G^0\}$. Let $x \in G^0$. Then $G(x,x)$ is a group (recall the example of a category of a group from earlier!) We call it the isotropy group at $x$ and denote it by $I_x$.
Before we continue, lets give an example. Let $GL(\mathbb{R})$ denote all the real invertible matrices under matrix multiplication. then $GL(\mathbb{R})^0$ is in bijective correspondence with the natural numbers, as for each natural number $n$ we have the corresponding $n \times n$ identity matrix. $G(n,m)$ is empty when $n \neq m$, but $G(n,n)$ is the general linear group $GL_n(\mathbb{R})$ of all $n \times n$ invertible matrices.
Finally, we say a groupoid $G$ is a topological groupoid when $G^0$ and $G^1$ are topological spaces, and all the corresponding maps on them (e.g. for $f \in G^1$ we have the domain $d(f)$, $d : G^1 \to G^0$, arrow composition $ \circ : G^1 \times G^1 \to G^1$, et cetera) must be continuous. Or equivalently, inversion and the groupoid operation must be continuous maps. A topological groupoid $G$ is said to be Étale if the domain map $d : G^1 \to G^0$ is a local homeomorphism (and thus all the other maps too), or equivalently, if inversion and the groupoid operation are local homeomorphism. Examples of these will come in the next post.
Sources
Noncommutative Geometry, Quantum Fields, Motives. Connes
Lectures on Arithmetic Noncommutative Geometry. Marcolli (the usual two)
A Homology Theory for Étale groupoids. Marius Crainic and Ieke Moerdijk
Abstract Algebra Dummit, Foote (appendix on category theory.)
Category Theory Steve Awodey
Categories and Functors
In my last post, when describing the setup for the Bost-Connes system, I displayed my lack of understanding of groupoids. In the next few posts I want to clear that up, describe Étale groupoids, and provide some more details on the proofs of the groupoid isomorphisms in Connes' monograph that I mentioned in the last construction. First let me talk about categories.
Categories and Functors
To describe a groupoid, it's good to have the language of category theory in your toolkit. Category theory abstracts the notion of functions, homomorphisms, et cetera and allow you to make statements about those relations. More formally, by a category (actually, a small category) $\mathcal{C}$ I mean a set objects $\mathcal{C}^0$, together with a set of "morphisms" or "arrows" between these objects $\mathcal{C}^1$. For each arrow $f \in \mathcal{C}^1$ we have the domain $d(f) \in \mathcal{C}^0$ and the range $r(f) \in \mathcal{C}^0$. We write $f: A \to B$ to denote that $A=d(f)$ and $B=r(f)$. The arrows are functions on the objects, and we can compose them. Formally, if $f : A\to B$ and $g: B \to C$, then there is an arrow $g \circ f : A \to C$. Additionally, for every $A \in \mathcal{C}^0$ there is an identity array $1_A : A \to A$ such that $f \circ 1_A = f = 1_B \circ f$. Composition is, of course, associative.
As an example, lets pretend we have a well-defined notion of the set of all groups, call it $\mathcal{G}^0$. Between any two groups there exist some homomorphism $\phi$. $\phi$ may be trivial, sending everything to to the identity, but it exists nevertheless. Collect all such $\phi$'s into the set $\mathcal{G}^1$. Then $\mathcal{G}$ is a category.
We can also look at a single group, $G$, and consider it as a category. The object set, $G^0$ contains only the identity element $e$. All the fun happens in the set $G^1$, which contains all the other elements of of $G$. Then "arrow" composition is not only associative, but also has an inverse and the identity is $1_e$. This example will make more sense after I discuss groupoids, but keep it in mind!
One can think of categories as being purely algebraic structures. And like any algebraic structure, we'll want some structure-preserving map between them. This brings us to functors. A functor $F: \mathcal{C} \to \mathcal{D}$ between categories $\mathcal{C}$ and $\mathcal{D}$ is a mapping of objects to objects and arrows to arrows such that it preserves all the expected notions. To be more explicit, a functor meets:
Sources
Noncommutative Geometry, Quantum Fields, Motives. Connes
Lectures on Arithmetic Noncommutative Geometry. Marcolli (the usual two)
A Homology Theory for Étale groupoids. Marius Crainic and Ieke Moerdijk
Abstract Algebra Dummit, Foote (appendix on category theory.)
Category Theory Steve Awodey
Categories and Functors
To describe a groupoid, it's good to have the language of category theory in your toolkit. Category theory abstracts the notion of functions, homomorphisms, et cetera and allow you to make statements about those relations. More formally, by a category (actually, a small category) $\mathcal{C}$ I mean a set objects $\mathcal{C}^0$, together with a set of "morphisms" or "arrows" between these objects $\mathcal{C}^1$. For each arrow $f \in \mathcal{C}^1$ we have the domain $d(f) \in \mathcal{C}^0$ and the range $r(f) \in \mathcal{C}^0$. We write $f: A \to B$ to denote that $A=d(f)$ and $B=r(f)$. The arrows are functions on the objects, and we can compose them. Formally, if $f : A\to B$ and $g: B \to C$, then there is an arrow $g \circ f : A \to C$. Additionally, for every $A \in \mathcal{C}^0$ there is an identity array $1_A : A \to A$ such that $f \circ 1_A = f = 1_B \circ f$. Composition is, of course, associative.
As an example, lets pretend we have a well-defined notion of the set of all groups, call it $\mathcal{G}^0$. Between any two groups there exist some homomorphism $\phi$. $\phi$ may be trivial, sending everything to to the identity, but it exists nevertheless. Collect all such $\phi$'s into the set $\mathcal{G}^1$. Then $\mathcal{G}$ is a category.
We can also look at a single group, $G$, and consider it as a category. The object set, $G^0$ contains only the identity element $e$. All the fun happens in the set $G^1$, which contains all the other elements of of $G$. Then "arrow" composition is not only associative, but also has an inverse and the identity is $1_e$. This example will make more sense after I discuss groupoids, but keep it in mind!
One can think of categories as being purely algebraic structures. And like any algebraic structure, we'll want some structure-preserving map between them. This brings us to functors. A functor $F: \mathcal{C} \to \mathcal{D}$ between categories $\mathcal{C}$ and $\mathcal{D}$ is a mapping of objects to objects and arrows to arrows such that it preserves all the expected notions. To be more explicit, a functor meets:
- $F(f:A \to B) = F(f) : F(A) \to F(B)$
- $F(1_A) = 1_{F(A)}$ and finally
- $F(g \circ f) = F(g) \circ F(f)$
Sources
Noncommutative Geometry, Quantum Fields, Motives. Connes
Lectures on Arithmetic Noncommutative Geometry. Marcolli (the usual two)
A Homology Theory for Étale groupoids. Marius Crainic and Ieke Moerdijk
Abstract Algebra Dummit, Foote (appendix on category theory.)
Category Theory Steve Awodey
Thursday, September 9, 2010
Constructing the Bost-Connes system. Attempt 1.
At the end of my last blog on lattices I described how the space of 1-dimensional $\mathbb{Q}$-lattices can be identified with $\hat{\mathbb{Z}} \times \mathbb{R}_{+}^{*}$. It's not too hard to see that if we mod out scaling by real factors, then we just end up with $\hat{\mathbb{Z}}$. Essentially, in one dimension, if we treat $\mathbb{Q}$-lattices that are multiples of each other as the same, all we have left is the homomorphism.
Now I want to describe a few more descriptions of 1 dimensional $\mathbb{Q}$-lattices (I'll henceforth use the symbol "1dQL" to mean 1 dimensional $\mathbb{Q}$-lattices), culminating in my describing a natural time evolution, and thus constructing the Quantum Statistical Mechanical system of Bost and Connes. First off, I want to describe the structure of the commensurability relation of 1dQL. But to do that, I first need to talk about groupoids.
Without delving too deep into category theory (which I am just now getting used to) let me suffice to say that a groupoid $\mathcal{G}$ is a collection of objects which will we call $\mathcal{G}^0$, and a collection of morphisms/mappings/functions (or just "arrows") $\mathcal{G}^1$ on $\mathcal{G}^0$ . We require every morphism in $\mathcal{G}^1$ to be invertible. Without category theory, you can think of a groupoid as being a lot like a normal group, except the group operation is not a binary operation. Id est, while every element has an inverse, not every pair of elements can be composed.
Now when I talk about the structure of the commensurability relation of 1dQL, its important to note that I don't mean equivalence classes. I'm actually talking about pairs $(\Lambda_1, \phi_1)$ and $(\Lambda_2, \phi_2)$ of 1dQLs that are commensurable. In Lemma 3.21 of Connes' monograph, he posits that this is isomorphic to the étale groupoid
$\mathcal{G}_1 = \{ (r, \rho, \lambda) : r \in \mathbb{Q}_{+}^{*}, \rho \in \hat{\mathbb{Z}}, \lambda \in \mathbb{R}_{+}^{*}$ such that $r \rho \in \hat{\mathbb{Z}}\}$
with the composition $(r_1,\rho_1,\lambda_1) \circ (r_2,\rho_2,\lambda_2) = (r_1 r_2,\rho_2,\lambda_2)$ if $r_2 \rho_2 = \rho_1$ and $r_2 \lambda_2 = \lambda_1$.
Étale groupoids are just groupoids with a specific topological structure (I'm still working out the details on that.) You can see this isomorphism from the map $l: (\rho,\lambda) \mapsto (\Lambda,\phi) $. In particular:
$(r,\rho,\lambda) \in \mathcal{G}_1 \mapsto (l(r \rho, r \lambda), l(\rho, \lambda))$
Finally, via Proposition 3.22, we can go even further. The following groupoid is isomorphic to the quotient $\mathcal{G}_1/\mathbb{R}_{+}^{*}$, hence we can also describe the commensurability classes of 1dQL up to scaling.
$\mathcal{U}_1 = \{(r,\rho) : r \in \mathbb{Q}_{+}^{*}, \rho \in \hat{\mathbb{Z}} $ such that $r \rho \in \hat{\mathbb{Z}} \}$
The isomorphism is via the map $\gamma : \mathcal{U}_1 \to \mathcal{G}_1/\mathbb{R}_{+}^{*}$ by $(r,\rho) \mapsto ((r^{-1}\mathbb{Z},\rho),(\mathbb{Z},\rho)) \in\mathcal{G}_1/\mathbb{R}_{+}^{*} $. The proof that this is an isomorphism isn't too hard (at least it doesn't look to hard) but it is yet still one of the details I've not finished.
Now to a locally compact groupoid (and I'm trusting Connes here when he says its locally compact) $\mathcal{G}$ we can construct the convolution algebra $\mathcal{A}_{c}(\mathcal{G})$. This convolution algebra is a space of functionals (into $\mathbb{C}$) on the groupoid. In the case of $\mathcal{U}_1$, the convolution product is
$f_1 * f_2 (L_1, L_2) = \sum_{L_1 \sim K \sim L_2} f_1(L_1, K) f_2(L, L_2)$
The involution is $f*(L_1, L_2) = \overline{f(L_1, L_2)}$. The symbols $L_1, L_2, K$ are 1dQLs and $\sim$ denotes commensurability.
(Note to readers: an algebra is essentially a vector space with a multiplication. The convolution product is our multiplication. Convolution is just a fancy word for an operation on two functions that produces a third. Involution is what makes C* algebras into C* algebras; its an abstraction of complex conjugation. Indeed, $\mathbb{C}$ is not only an algebra, it's a C* algebra.)
By completing the appropriate norm (which I do not fully understand, so I won't mention it here.) we can upgrade the convolution algebra $\mathcal{A}_{c}(\mathcal{U}_1)$ to the C* groupoid algebra $C^{*}(\mathcal{U}_1)$.
Connes has another description of the C* algebra in terms of generators and relations. However, I do not yet understand the proposition or the proof. It involves semigroup cross products, Morita equivalences, adeles, and notation I've not deciphered. I'll return to it soon.
The last thing left to do before we have a full Quantum Statistical Mechanical system is to define the time evolution (Lemma 3.24 in Connes' monograph). We can actually do this using data from an element in $\mathcal{G}_1/\mathbb{R}_{+}^{*}$. Recall that an element there is actually a pair of 1dQLs $L_1 = (\Lambda_1,\phi_1)$ and $L_2 = (\Lambda_2,\phi_2)$. Since they are commensurable, the lattices $\Lambda_1$ and $\Lambda_2$ differ by a scaling factor $r$. This $r$ happens to be the ratio of covolumes of two lattices.
Recall that the covolume of a lattice is the "area" of its fundamental parallelogram (or equivalent hyper-dimensional shape). And that the we can find this area by taking the determinate of the basis vectors of the lattice. In one dimension, we only have a single basis vector (some member of $\mathbb{R}$). Now when we take the ratio of the covolumes of lattices of commensurable 1 dimensional $\mathbb{Q}$-lattices, we simply get the scaling factor between the two, id est, $r$.
Now members of $C^{*}(\mathcal{G}_1/\mathbb{R}_{+}^{*})$ are simply functions on pairs of commensurable 1dQLs, e.g., $f(\Lambda_1, \Lambda_2)$. We define the time evolution as
$\sigma_{t}(f)(\Lambda_1, \Lambda_2) = r^{i t} f(\Lambda_1, \Lambda_2)$ where $r$ is the scaling factor.
Or in terms of the equivalent $C^{*}(\mathcal{U}_1)$ (functions $f(r,\rho)$ for $(r,\rho) \in \mathcal{U}_1$) we have
$\sigma_{t}(f)(r,\rho) = r^{i t} f(r,\rho)$.
And there you have it, folks, a bona fide Quantum Statistical Mechanical system. Bit embarrassing that it took me til September!
Details I've missed and Things I don't understand
Étale groupoids
How the groupoid inversion and operation works on $\mathcal{G}_1$ and $\mathcal{U}_1$. Plus how it's locally compact.
The proof of the isomorphism.
Proof of Proposition 3.22
All of Proposition 3.23 in Connes' monograph
Construction of C* group algebras and C* groupoid algebras.
Whats next
All of the above, plus the Hecke algebra description and its equivalence to the above. Then on to symetries of the system and the class field theory of $\mathbb{Q}$.
Now I want to describe a few more descriptions of 1 dimensional $\mathbb{Q}$-lattices (I'll henceforth use the symbol "1dQL" to mean 1 dimensional $\mathbb{Q}$-lattices), culminating in my describing a natural time evolution, and thus constructing the Quantum Statistical Mechanical system of Bost and Connes. First off, I want to describe the structure of the commensurability relation of 1dQL. But to do that, I first need to talk about groupoids.
Without delving too deep into category theory (which I am just now getting used to) let me suffice to say that a groupoid $\mathcal{G}$ is a collection of objects which will we call $\mathcal{G}^0$, and a collection of morphisms/mappings/functions (or just "arrows") $\mathcal{G}^1$ on $\mathcal{G}^0$ . We require every morphism in $\mathcal{G}^1$ to be invertible. Without category theory, you can think of a groupoid as being a lot like a normal group, except the group operation is not a binary operation. Id est, while every element has an inverse, not every pair of elements can be composed.
Now when I talk about the structure of the commensurability relation of 1dQL, its important to note that I don't mean equivalence classes. I'm actually talking about pairs $(\Lambda_1, \phi_1)$ and $(\Lambda_2, \phi_2)$ of 1dQLs that are commensurable. In Lemma 3.21 of Connes' monograph, he posits that this is isomorphic to the étale groupoid
$\mathcal{G}_1 = \{ (r, \rho, \lambda) : r \in \mathbb{Q}_{+}^{*}, \rho \in \hat{\mathbb{Z}}, \lambda \in \mathbb{R}_{+}^{*}$ such that $r \rho \in \hat{\mathbb{Z}}\}$
with the composition $(r_1,\rho_1,\lambda_1) \circ (r_2,\rho_2,\lambda_2) = (r_1 r_2,\rho_2,\lambda_2)$ if $r_2 \rho_2 = \rho_1$ and $r_2 \lambda_2 = \lambda_1$.
Étale groupoids are just groupoids with a specific topological structure (I'm still working out the details on that.) You can see this isomorphism from the map $l: (\rho,\lambda) \mapsto (\Lambda,\phi) $. In particular:
$(r,\rho,\lambda) \in \mathcal{G}_1 \mapsto (l(r \rho, r \lambda), l(\rho, \lambda))$
Finally, via Proposition 3.22, we can go even further. The following groupoid is isomorphic to the quotient $\mathcal{G}_1/\mathbb{R}_{+}^{*}$, hence we can also describe the commensurability classes of 1dQL up to scaling.
$\mathcal{U}_1 = \{(r,\rho) : r \in \mathbb{Q}_{+}^{*}, \rho \in \hat{\mathbb{Z}} $ such that $r \rho \in \hat{\mathbb{Z}} \}$
The isomorphism is via the map $\gamma : \mathcal{U}_1 \to \mathcal{G}_1/\mathbb{R}_{+}^{*}$ by $(r,\rho) \mapsto ((r^{-1}\mathbb{Z},\rho),(\mathbb{Z},\rho)) \in\mathcal{G}_1/\mathbb{R}_{+}^{*} $. The proof that this is an isomorphism isn't too hard (at least it doesn't look to hard) but it is yet still one of the details I've not finished.
Now to a locally compact groupoid (and I'm trusting Connes here when he says its locally compact) $\mathcal{G}$ we can construct the convolution algebra $\mathcal{A}_{c}(\mathcal{G})$. This convolution algebra is a space of functionals (into $\mathbb{C}$) on the groupoid. In the case of $\mathcal{U}_1$, the convolution product is
$f_1 * f_2 (L_1, L_2) = \sum_{L_1 \sim K \sim L_2} f_1(L_1, K) f_2(L, L_2)$
The involution is $f*(L_1, L_2) = \overline{f(L_1, L_2)}$. The symbols $L_1, L_2, K$ are 1dQLs and $\sim$ denotes commensurability.
(Note to readers: an algebra is essentially a vector space with a multiplication. The convolution product is our multiplication. Convolution is just a fancy word for an operation on two functions that produces a third. Involution is what makes C* algebras into C* algebras; its an abstraction of complex conjugation. Indeed, $\mathbb{C}$ is not only an algebra, it's a C* algebra.)
By completing the appropriate norm (which I do not fully understand, so I won't mention it here.) we can upgrade the convolution algebra $\mathcal{A}_{c}(\mathcal{U}_1)$ to the C* groupoid algebra $C^{*}(\mathcal{U}_1)$.
Connes has another description of the C* algebra in terms of generators and relations. However, I do not yet understand the proposition or the proof. It involves semigroup cross products, Morita equivalences, adeles, and notation I've not deciphered. I'll return to it soon.
The last thing left to do before we have a full Quantum Statistical Mechanical system is to define the time evolution (Lemma 3.24 in Connes' monograph). We can actually do this using data from an element in $\mathcal{G}_1/\mathbb{R}_{+}^{*}$. Recall that an element there is actually a pair of 1dQLs $L_1 = (\Lambda_1,\phi_1)$ and $L_2 = (\Lambda_2,\phi_2)$. Since they are commensurable, the lattices $\Lambda_1$ and $\Lambda_2$ differ by a scaling factor $r$. This $r$ happens to be the ratio of covolumes of two lattices.
Recall that the covolume of a lattice is the "area" of its fundamental parallelogram (or equivalent hyper-dimensional shape). And that the we can find this area by taking the determinate of the basis vectors of the lattice. In one dimension, we only have a single basis vector (some member of $\mathbb{R}$). Now when we take the ratio of the covolumes of lattices of commensurable 1 dimensional $\mathbb{Q}$-lattices, we simply get the scaling factor between the two, id est, $r$.
Now members of $C^{*}(\mathcal{G}_1/\mathbb{R}_{+}^{*})$ are simply functions on pairs of commensurable 1dQLs, e.g., $f(\Lambda_1, \Lambda_2)$. We define the time evolution as
$\sigma_{t}(f)(\Lambda_1, \Lambda_2) = r^{i t} f(\Lambda_1, \Lambda_2)$ where $r$ is the scaling factor.
Or in terms of the equivalent $C^{*}(\mathcal{U}_1)$ (functions $f(r,\rho)$ for $(r,\rho) \in \mathcal{U}_1$) we have
$\sigma_{t}(f)(r,\rho) = r^{i t} f(r,\rho)$.
And there you have it, folks, a bona fide Quantum Statistical Mechanical system. Bit embarrassing that it took me til September!
Details I've missed and Things I don't understand
Étale groupoids
How the groupoid inversion and operation works on $\mathcal{G}_1$ and $\mathcal{U}_1$. Plus how it's locally compact.
The proof of the isomorphism.
Proof of Proposition 3.22
All of Proposition 3.23 in Connes' monograph
Construction of C* group algebras and C* groupoid algebras.
Whats next
All of the above, plus the Hecke algebra description and its equivalence to the above. Then on to symetries of the system and the class field theory of $\mathbb{Q}$.
Tuesday, September 7, 2010
End(Q/Z) is isomorphic to \hat{Z}
So I've, more or less, finished my work set out in my last post: caught up on some algebra and category theory and wrapped up my study of the construction of the Bost Connes system (though I've not touched the Hecke algebra description yet.) I've been spending the past few days trying to write something coherent on my reading and notes, and today I got stuck on a rather small detail for an embarrassingly long time. To make matters worse, it was something I had discussed with my project supervisor just a month ago.
In any case, I think I've cleared it up, and I thought I'd post my work here for review.
Proposition $End(\mathbb{Q}/\mathbb{Z})$ is isomorphic to $\hat{\mathbb{Z}}$
First some background. $End(\mathbb{Q}/\mathbb{Z})$ is the set of all endomorphism of Q/Z. Id est, all homomorphisms from $\mathbb{Q}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$. $\mathbb{Q}/\mathbb{Z}$ is an Abelian [torsion] group, and hence a $\mathbb{Z}$-module, which makes $End(\mathbb{Q}/\mathbb{Z})$ a $\mathbb{Z}$-module as well (its group operation is addition of the mappings, e.g., if $\phi_1 \in End(\mathbb{Q}/\mathbb{Z})$ and $\phi_2 \in End(\mathbb{Q}/\mathbb{Z})$, then $(\phi_1 + \phi_2)(r) = \phi_1(r) + \phi_2(r)$) If you recall my earlier post on $\mathbb{Q}$-lattices, you'll know that End(Q/Z) is vital to describing the structure of the commensurability equivalence classes of 1 dimensional $\mathbb{Q}$-lattices.
$\hat{\mathbb{Z}}$ is a profinite ring (in case you want to google it.) It's the inverse limit, $\varprojlim \mathbb{Z}/n\mathbb{Z}$. The inverse system of groups and homomorphisms is so easy to work with that I can give an explicit description of the resulting profinite ring:
$\hat{\mathbb{Z}} = \{a=(a_1, a_2, \dots ) : \forall m|n, a_n \equiv a_m (mod m)\}$
These two rings are apparently isomorphic, so we can describe our 1 dimensional $\mathbb{Q}$-lattices with $\hat{\mathbb{Z}}$. The minor detail that had me stuck all day was the proof of that fact. So let me propose a proof now.
Given an $a \in \hat{\mathbb{Z}}$, I'll define a homomorphism $\phi_a : \mathbb{Q} / \mathbb{Z} \to \mathbb{Q} / \mathbb{Z}$ by $\phi_a (\frac{m}{n}) = a_n \frac{m}{n}$. The map $a \mapsto \phi_a$ is a homomorphism between $\hat{\mathbb{Z}}$ and $End(\mathbb{Q}/\mathbb{Z})$ (it's pretty clear that it respects the group operation, which is just component-wise modular arithmetic in the former ring, and addition of functions in the latter.) I still need to check that $\phi_a$ is an endomorphism of $\mathbb{Q}/\mathbb{Z}$, however. Let's do that:
$\phi_a ( \frac{m_1}{n_1} + \frac{m_2}{n_2}) = \phi_a (\frac{n_2 m_1 + n_1 m_2}{n_1 n_2}) = a_{n_1 n_2} \frac{n_2 m_1 + n_1 m_2}{n_1 n_2}$.
And
$\phi_a( \frac{m_1}{n_1}) + \phi_a(\frac{m_2}{n_2}) = \frac{a_{n_1} n_2 m_1 + a_{n_2} n_1 m_2} {n_1 n_2}$
But recall our description of $\hat{\mathbb{Z}}$, "for all $m|n, a_n \equiv a_m (mod m)$". That means that we can write $a_{n_1 n_2} = a_{n_1} + q n_1 = a_{n_2} + r n_2$. Or in other words,
$a_{n_1} = a_{n_1 n_2} - q n_1$ and $a_{n_2} = a_{n_1 n_2} - r n_2$. So the above becomes:
$a_{n_1 n_2} \frac{n_2 m_1 + n_1 m_2}{n_1 n_2} - n_1 n_2 \frac {q m_1 + r m_2}{n_1 n_2}$
The negative term is an integer, so it's 0 in our quotient $\mathbb{Q}/\mathbb{Z}$. Now we're halfway done. If we can find an inverse of that map that keeps the ring structure of $\hat{\mathbb{Z}}$ intact, we've won the game.
To define said inverse, we'll start with an $f \in \text{End}(\mathbb{Q}/\mathbb{Z})$. Define an $a \in \prod_n \mathbb{Z}/n\mathbb{Z}$ by $a_k = k f(\frac{1}{k})$. Again its not hard to see that this preserves component-wise addition on $\hat{\mathbb{Z}}$. Moreover, if $a \in \hat{\mathbb{Z}}$, then it's the inverse map of the above. Now if $m|n$, we can write $n=mr$ for some $r$. Then $m r f(\frac{1}{n}) - m f ( \frac{1}{m})$ is obviously divisible by $m$. Hence $a \in \hat{\mathbb{Z}}$. And we've found our isomorphism.
I did this with a very naive approach, and its a bit long and messy. Please let me know if there are any errors!
Additional Notes
Someone at physicsforums.com suggested a category-theoretic proof of the above. It's a lot shorter and cleaner then my naive approach, but I don't yet know enough category theory to understand it. Since I am working on a writeup of my category theory reading anyways, perhaps I'll do a bit more and incorporate that in as well.
Stay tuned for more algebra and the Bost-Connes system!
In any case, I think I've cleared it up, and I thought I'd post my work here for review.
Proposition $End(\mathbb{Q}/\mathbb{Z})$ is isomorphic to $\hat{\mathbb{Z}}$
First some background. $End(\mathbb{Q}/\mathbb{Z})$ is the set of all endomorphism of Q/Z. Id est, all homomorphisms from $\mathbb{Q}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$. $\mathbb{Q}/\mathbb{Z}$ is an Abelian [torsion] group, and hence a $\mathbb{Z}$-module, which makes $End(\mathbb{Q}/\mathbb{Z})$ a $\mathbb{Z}$-module as well (its group operation is addition of the mappings, e.g., if $\phi_1 \in End(\mathbb{Q}/\mathbb{Z})$ and $\phi_2 \in End(\mathbb{Q}/\mathbb{Z})$, then $(\phi_1 + \phi_2)(r) = \phi_1(r) + \phi_2(r)$) If you recall my earlier post on $\mathbb{Q}$-lattices, you'll know that End(Q/Z) is vital to describing the structure of the commensurability equivalence classes of 1 dimensional $\mathbb{Q}$-lattices.
$\hat{\mathbb{Z}}$ is a profinite ring (in case you want to google it.) It's the inverse limit, $\varprojlim \mathbb{Z}/n\mathbb{Z}$. The inverse system of groups and homomorphisms is so easy to work with that I can give an explicit description of the resulting profinite ring:
$\hat{\mathbb{Z}} = \{a=(a_1, a_2, \dots ) : \forall m|n, a_n \equiv a_m (mod m)\}$
These two rings are apparently isomorphic, so we can describe our 1 dimensional $\mathbb{Q}$-lattices with $\hat{\mathbb{Z}}$. The minor detail that had me stuck all day was the proof of that fact. So let me propose a proof now.
Given an $a \in \hat{\mathbb{Z}}$, I'll define a homomorphism $\phi_a : \mathbb{Q} / \mathbb{Z} \to \mathbb{Q} / \mathbb{Z}$ by $\phi_a (\frac{m}{n}) = a_n \frac{m}{n}$. The map $a \mapsto \phi_a$ is a homomorphism between $\hat{\mathbb{Z}}$ and $End(\mathbb{Q}/\mathbb{Z})$ (it's pretty clear that it respects the group operation, which is just component-wise modular arithmetic in the former ring, and addition of functions in the latter.) I still need to check that $\phi_a$ is an endomorphism of $\mathbb{Q}/\mathbb{Z}$, however. Let's do that:
$\phi_a ( \frac{m_1}{n_1} + \frac{m_2}{n_2}) = \phi_a (\frac{n_2 m_1 + n_1 m_2}{n_1 n_2}) = a_{n_1 n_2} \frac{n_2 m_1 + n_1 m_2}{n_1 n_2}$.
And
$\phi_a( \frac{m_1}{n_1}) + \phi_a(\frac{m_2}{n_2}) = \frac{a_{n_1} n_2 m_1 + a_{n_2} n_1 m_2} {n_1 n_2}$
But recall our description of $\hat{\mathbb{Z}}$, "for all $m|n, a_n \equiv a_m (mod m)$". That means that we can write $a_{n_1 n_2} = a_{n_1} + q n_1 = a_{n_2} + r n_2$. Or in other words,
$a_{n_1} = a_{n_1 n_2} - q n_1$ and $a_{n_2} = a_{n_1 n_2} - r n_2$. So the above becomes:
$a_{n_1 n_2} \frac{n_2 m_1 + n_1 m_2}{n_1 n_2} - n_1 n_2 \frac {q m_1 + r m_2}{n_1 n_2}$
The negative term is an integer, so it's 0 in our quotient $\mathbb{Q}/\mathbb{Z}$. Now we're halfway done. If we can find an inverse of that map that keeps the ring structure of $\hat{\mathbb{Z}}$ intact, we've won the game.
To define said inverse, we'll start with an $f \in \text{End}(\mathbb{Q}/\mathbb{Z})$. Define an $a \in \prod_n \mathbb{Z}/n\mathbb{Z}$ by $a_k = k f(\frac{1}{k})$. Again its not hard to see that this preserves component-wise addition on $\hat{\mathbb{Z}}$. Moreover, if $a \in \hat{\mathbb{Z}}$, then it's the inverse map of the above. Now if $m|n$, we can write $n=mr$ for some $r$. Then $m r f(\frac{1}{n}) - m f ( \frac{1}{m})$ is obviously divisible by $m$. Hence $a \in \hat{\mathbb{Z}}$. And we've found our isomorphism.
I did this with a very naive approach, and its a bit long and messy. Please let me know if there are any errors!
Additional Notes
Someone at physicsforums.com suggested a category-theoretic proof of the above. It's a lot shorter and cleaner then my naive approach, but I don't yet know enough category theory to understand it. Since I am working on a writeup of my category theory reading anyways, perhaps I'll do a bit more and incorporate that in as well.
Stay tuned for more algebra and the Bost-Connes system!
Labels:
algebra,
groups of homomorphisms,
inverse limits,
msci project
Saturday, September 4, 2010
Getting back into things...
Apologies for the hiatus in maths posts; I've had some personal issues (trying to find food, fending off homelessness, grieving the death of my friend.) The past few weeks have been pretty stressful. I'm now in Colorado and shouldn't be worried about those important distractions for the next three weeks (when I return to London I should also have a place to live, and student loan money should be coming too.) I'm glad I'm here and that I get a chance to see my friend's family. I will be resuming maths post from today (inshallah!) Here's the plan for the next few days:
Yesterday and Today: catching up on some algebra - modules, algebras, group rings, group algebras, c* group algebras.
Sunday: Etale groupoids & category Theory
Monday: C* algebra construction of the bost-connes system
Tuesday until ? : Hecke algebras and their corresponding construction, plus the equivalence of the two.
My Advisor and I had set a goal of understanding the construction of bost-connes system by the end of August. I'm ashamed to say I'm a bit late. But I think I'll be able to make up lost ground and start working on the symmetries of the system here in the next few days. I don't think that schedule is too ambitious. I should have time to catch up and even enjoy the miles of mountain bike singletrack just outside my parent's house!
Additionally, one might expect some non-project related blogs to come soon too, particularly ones on the UCL undergrad colloquium, my talks there, and some other reading.
Right. enough talk, lets get to work
Yesterday and Today: catching up on some algebra - modules, algebras, group rings, group algebras, c* group algebras.
Sunday: Etale groupoids & category Theory
Monday: C* algebra construction of the bost-connes system
Tuesday until ? : Hecke algebras and their corresponding construction, plus the equivalence of the two.
My Advisor and I had set a goal of understanding the construction of bost-connes system by the end of August. I'm ashamed to say I'm a bit late. But I think I'll be able to make up lost ground and start working on the symmetries of the system here in the next few days. I don't think that schedule is too ambitious. I should have time to catch up and even enjoy the miles of mountain bike singletrack just outside my parent's house!
Additionally, one might expect some non-project related blogs to come soon too, particularly ones on the UCL undergrad colloquium, my talks there, and some other reading.
Right. enough talk, lets get to work
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