Thursday, August 12, 2010

Q-lattices revisited. (1-dim case)

Yesterday, I found out a friend of mine back home in Colorado died and have since not had the focus to do much work. I haven't done that review of algebra topics like I said I would. But it's important to keep good habits, so I thought I'd review and write a bit about $\mathbb{Q}$-lattices a bit more.

Connes and Marcolli's monographs define a $\mathbb{Q}$-lattice as a lattice $\Lambda \in \mathbb{R}^n$ together with a homomorphism $\phi : \mathbb{Q}^n / \mathbb{Z}^n \to \mathbb{Q} \Lambda / \Lambda$.  We call two $\mathbb{Q}$-lattices $\Lambda_1$ and $\Lambda_2$ commensurable ($\Lambda_1 \sim \Lambda_2$) if:
1) $\mathbb{Q} \Lambda_1 =\mathbb{Q} \Lambda_2$ and
2) $\phi_1$ = $\phi_2$ mod $\Lambda_1 + \Lambda_2$.

Condition 1) means that there exists some $\mathbb{Q}$-span in $\mathbb{R}^n$ such that $\Lambda_1$ and $\Lambda_2$ both sit inside it.  For instance, both $\frac{1}{2} \mathbb{Z}$ and $\frac{1}{5} \mathbb{Z}$ sit inside $\mathbb{Q}$, so they meet condition 1.  But if we take $< (1,0), (0,\sqrt{2}) >$ and $<(\frac{1}{2},0), (0,0) >$ we cannot find any vectors in $\mathbb{R}^2$ that, when spanned over $\mathbb{Q}$, would contain both of those lattices.

Condition 2) is actually how I originally understood it.  The homomorphism must "label the same points".  More formally, Lets say $X =\mathbb{Q} \Lambda_1 =\mathbb{Q} \Lambda_2$.  For each $\Lambda_i$ we have a projection map $\pi_i : X/\Lambda_i \to X/(\Lambda_1 + \Lambda_2)$ by $r+\Lambda_i \mapsto r + (\Lambda_1 + \Lambda_2)$ (of course, $r$ may not be the best representative element in $\Lambda_1 + \Lambda_2$).  Condition two says that $\pi_1 \circ \phi_1 = \pi_2 \circ \phi_2$.

1-dim Q-lattices 

We can describe the Bost-Connes system with $\mathbb{Q}$-lattices in $\mathbb{R}$, and 1-dim $\mathbb{Q}$-lattices can be described by $\hat{\mathbb{Z}} \times \mathbb{R}_{+}^{*}$, where $\hat{\mathbb{Z}} = \varprojlim_{n} \mathbb{Z} / n\mathbb{Z}$.  First note any lattice $\Lambda \in \mathbb{R}$ can be described by $\lambda \mathbb{Z}$, where $\lambda \in \mathbb{R}_{+}^{*}$.  So the only data we need is the homomorphism $\phi : \mathbb{Q} / \mathbb{Z} \to \mathbb{Q} / \lambda \mathbb{Z}$.  But apparently $\hat{\mathbb{Z}} \cong Hom (\mathbb{Q} / \mathbb{Z},\mathbb{Q} / \mathbb{Z})$, that is, $\hat{\mathbb{Z}}$ is isomorphic to the set of all endomorphisms of $\mathbb{Q} / \mathbb{Z}$ (I'm not yet sure how! Something to do with Pontryagin duality?), so all that remains is a choice of $\rho \in \hat{\mathbb{Z}}$ and we have $(\Lambda,\phi) \mapsto (\rho,\lambda)$.  One should note that the inverse of this map is well defined, so we can identify the set of all 1-dim $\mathbb{Q}$-lattices with $\hat{\mathbb{Z}} \times \mathbb{R}_{+}^{*}$.

I'll continue reading about 1-dim $\mathbb{Q}$-lattices for the next several days, though I do still plan to take some time to catch up on some algebra (algebras over a field, group rings, group algebras, C* algebras, their representations, group C* algebras, Pontryagin duels, etale groupoids.)

Thanks to Professor Kim and the folks at mathoverlow for helping me understand this.
Also, for the next few days I will all but abandon Marcolli's monograph;  that text goes straight into a description of the Bost-Connes system via Shimura varieties, which is a bit above my level.  I do intent to understand the Shimura variety and the Hecke algebra description of Bost-Connes eventually, however.

No comments:

Post a Comment