## 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!

2. I'm glad you found it helpful! It's been so long ago I forgot exactly how I came up with it, but chances are that I got it from someone else, I should have linked to the referenced physics forum discussion. In any case, I fixed the $\LaTeX$ error. :)