## Friday, July 30, 2010

### $\mathbb{Q}$-lattices and commensurability

My supervisor, Professor Kim, helped clear up some of my confusion on the homomorphism in $\mathbb{Q}$ lattices. Lets go over it again: A $\mathbb{Q}$-lattice is a normal lattice $\Lambda \subset \mathbb{R}^n$ together with a homomorphism $\phi : \mathbb{Q}^n / \mathbb{Z}^n \to \mathbb{Q}\Lambda / \Lambda$. Lets break down that notation a bit.

$\mathbb{Q}^n / \mathbb{Z}^n$ looked a bit frightening to me at first, but its really not so bad. Consider the far more neighbourly quotient $\mathbb{R} / \mathbb{Z}$, which of course is just a circle. $\mathbb{Q} / \mathbb{Z}$ clearly is a subset of the circle, but only including the points on the circle that go back to the origin when multiplied by an integer. $(\mathbb{Q} / \mathbb{Z})^2$ is a subset of the torus $(\mathbb{R} / \mathbb{Z})^2$, and so on, and that gives me a reasonable way to think about the more general looking $\mathbb{Q}^n / \mathbb{Z}^n$. Not that bad after all, right?

Now for $\mathbb{Q}\Lambda / \Lambda$. $\mathbb{Q}\Lambda$ turns out to be exactly what it sounds like. For a lattice $\Lambda$, we can consider it as the $\mathbb{Z}$-span of basis vectors $\{ v_1,...,v_n \}$. For $\mathbb{Q}\Lambda$, just take the $\mathbb{Q}$-span. But what about the quotient? Its really not that different from $\mathbb{Q}^n / \mathbb{Z}^n$. Consider that $\Lambda$ is like a "tilted" version of $\mathbb{Z}^n$. From there its not hard to imagine $\mathbb{Q}\Lambda / \Lambda$ as being a "tilted" version of the "hyper-rational-torus" $\mathbb{Q}^n / \mathbb{Z}^n$ (I call it "hyper" because it may be more than 2 dimensions, and "rational" because it only includes the points that get back to the origin when multiplied by an integer.) In fact, its not so hard to see that there's an isomorphism between $\mathbb{Q}^n / \mathbb{Z}^n$ and $\mathbb{Q}\Lambda / \Lambda$ that comes just from figuring out how much $\Lambda$ is titled from $\mathbb{Z}^n$, id est, expressing the basis for $\Lambda$ in terms of the standard basis of $\mathbb{Z}^n$.

So why then does the the definition of $\mathbb{Q}$-lattices include a homomorphism and not a full, invertible isomorphism? In fact, why do we even need a definition for $\mathbb{Q}$-lattices at all? It just seems to be a lattice with an obvious homomorphism attached to it. (this question is really quite moot. commensurability and the commensurability classes yield plenty results, among them is the topic of my 4th year project!)

Finally, we also have a notion of what it means for 2 $\mathbb{Q}$-lattices to be commensurable: $\mathbb{Q} \Lambda_1 = \mathbb{Q} \Lambda_2$ and $\phi_1 - \phi_2\, =\, 0\, mod\, \Lambda_1 + \Lambda_2$. What does this mean? As best I can tell [I'm really not too sure], the homomorphism labels some of our tilted-torus points as coming from our nice hyper-rational-torus. The statement that $\phi_1 - \phi_2\, =\, 0\, mod\, \Lambda_1 + \Lambda_2$ says that they have the same points labelled in the tilted-torus. It's pretty easy to see that if the homomorphisms are isomorphism, then 2 $\mathbb{Q}-lattices are only commensurable if they are in fact, equal. Its also not too hard to see that commensurability is an equivalence relation (Connes has a proof of this fact in Non-commutative geometry, Quantum Fields and motives, it was from that prove that I gathered the "same label" idea.) We denote all the commensurability classes of$\mathbb{Q}$-lattices in$\mathbb{R}^n$as$\mathcal{L}_n$. So whats next? Apparently, the Bost-Connes quantum statistical mechanical system a C*-algebra somehow generated from the quotient$\mathcal{L}_n / \mathbb{R}^{*}_{+}$(I don't recognize the notation$\mathbb{R}^{*}_{+}$,$\mathbb{R}^{*}$I thought was the real numbers excluding 0, and$\mathbb{R}_{+}$the positive reals. so is$\mathbb{R}^{*}_{+}$all the strictly positive reals?) So I suppose my next step is to 1) understand how to form that C*-algebra [which implicitly involves understanding what a C*-algebra is, I think I'll probably spend some time reading the first chapter to Averson's Invitation to C*-algebras] and 2) understand how that C*-algebra is a quantum statistical mechanical system. After that point, I can figure out how this is the same as the Hecke algebra mentioned in Connes' and Bost's work and what it has to do with$Gal(\mathbb{Q}^{cycl} / \mathbb{Q})\$.