# Q-lattices and commensurability, any insight into the definition and intuition?

I've been coming across $\mathbb{Q}$-lattices in $\mathbb{R}^n$ in my reading, and I'm having trouble understanding the definitions. Connes and Marcolli define it as a lattice $\Lambda \in \mathbb{R}^n$ together with a homomorphism $\phi : \mathbb{Q}^n / \mathbb{Z}^n \to \mathbb{Q} \Lambda / \Lambda$. Moreover, two $\mathbb{Q}$-lattices $\Lambda_1$ and $\Lambda_2$ are commensurable iff 1) $\mathbb{Q} \Lambda_1 = \mathbb{Q}\Lambda_2$ and 2) $\phi_1 = \phi_2$ mod $\Lambda_1 + \Lambda_2$.

I think I understand condition 1): the lattices must be rational multiples of each other to be commensurable. I don't even understand the notation for condition 2). The best I can gather is that the homomorphism $\phi$ labels which positions in $\mathbb{Q} \Lambda / \Lambda$ come from your more normal "discrete hyper-torus" $\mathbb{Q}^n / \mathbb{Z}^n$. Condition 2) then says that the same points are labelled. Is this anywhere near the right ballpark? Can anyone recommend any literature on the subject?

I'm a pretty young mathematician (not even in a PhD program...yet) so please forgive me if this question seems basic.

Thanks.

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Can you, please, give a reference to the paper where this is defined? As for your notational question: after identification $\mathbb{Q}\Lambda_1\simeq \mathbb{Q}\Lambda_2,$ you can mod out this group by the sublattice $\Lambda_1+\Lambda_2$ that contains $\Lambda_1$ and $\Lambda_2.$ The requirement is that compositions of $\phi_1$ and $\phi_2$ with the projection become equal. – Victor Protsak Aug 10 '10 at 18:04
Connes & Marcolli, Noncommutative Geometry, Quantum Fields, and Motives. Marcolli, Lectures on Arithmetic Noncommutative Geometry. Various other papers from those two. alainconnes.org/docs/Qlattices.pdf is a short one. Thanks for the comments on the notation, very helpful. – mebassett Aug 10 '10 at 19:34
For condition 1, the lattices don't have to be rational multiples of each other, e.g. take $\Lambda_1=\langle(1,0),(0,1)\rangle$, $\Lambda_2=\langle(\frac{1}{2},0),(0,2)\rangle$. They just have to be contained in the same $n$-dimensional $\mathbb{Q}$-vector space. Equivalently, there exist integers $N$ and $M$ so that $M\Lambda_1\subset\Lambda_2$ and $N\Lambda_2\subset\Lambda_1$. – Kevin Ventullo Aug 10 '10 at 21:24

The condition $\mathbb Q\Lambda_1=\mathbb Q\Lambda_2=:X$ means that we have $\Lambda_1,\Lambda_2\subseteq X$ and then we have $\Lambda_1,\Lambda_2\subseteq \Lambda_1+\Lambda_2\subseteq X$. This means that we have quotient maps
$$X/\Lambda_1\rightarrow X/(\Lambda_1+\Lambda_2){\rm\quad and\quad }X/\Lambda_2\rightarrow X/(\Lambda_1+\Lambda_2).$$ Condition 2) then means that the two composites $\mathbb Q^n/\mathbb Z^n\rightarrow X/\Lambda_i \rightarrow X/(\Lambda_1+\Lambda_2)$ are equal.
As for your interpretation of condition 2) it seems to be more or less OK, though the way you have phrased it $\phi$ could very well be $0$. To understand what 2) means in concrete terms it is convenient to use some more advanced notions (which may or may not be unfamiliar to you). First the source and target of $\phi$ are torsion groups. We then have that for each prime $p$ the $p$-torsion of $\mathbb Q^n/\mathbb Z^n$ is taken into the $p$-torsion of $\mathbb Q\Lambda/\Lambda$. Any map of these $p$-torsion groups corresponds exactly to a $\mathbb Z_p$-module map (where $\mathbb Z_p$ is the ring of $p$-adic integers) $\phi_p\colon\mathbb Z_p^n\rightarrow \Lambda\bigotimes \mathbb Z_p$. This is very analoguous to the situation when we instead would start with a continuous map $\phi\colon\mathbb R^n/\mathbb Z^n\rightarrow \mathbb R\Lambda/L$ which would correspond to a map $\phi_{\mathbb Z}\colon\mathbb Z^n\rightarrow\Lambda$.