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!

Awesome proof , I tried to do this for about 2 hrs but could not get it , your proof really helped me. Thanks. btw could you try to correct that typo in "Extra close brace or missing open brace"? that would be nice , then people won't have to check the math for understanding it.

ReplyDeleteI'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. :)

ReplyDelete