3
$\begingroup$

Consider the (additive) group $G$ of all homomorphisms $\phi: \mathbb{R}/\mathbb{Z} \rightarrow \mathbb{R}/\mathbb{Z}$.

I think there is an injective homomorphism $\mathbb{Z} \rightarrow G$ where an integer $q$ is sent to the map $[x] \mapsto [qx]$ for $[x] \in \mathbb{R}/\mathbb{Z}$.

I'm wondering if this injection is also a surjection? In other words are there endomorphisms of $\mathbb{R}/\mathbb{Z}$ that do not come from integers?

$\endgroup$

2 Answers 2

5
$\begingroup$

If you allow $\phi$ discontinuous (and assume the axiom of choice), there are such "exceptional" endomorphisms. For example, given a Hamel basis with generating set $B$ and $1\in B$, designate any $1\neq b\in B$ and map all elements of $B$ besides $b$ to $0$.

On the other hand, if you wish only to consider continuous endomorphisms, then you do have all of them. To show this, take a continuous $\phi:\mathbb R/\mathbb Z\to\mathbb R/\mathbb Z$, and take its pullback $\pi:\mathbb R\to\mathbb R/\mathbb Z$. Identify $\mathbb R/\mathbb Z$ with $[-1/2,1/2)$, and consider some $\epsilon>0$ for which $\pi((-\epsilon,\epsilon))\subset (-1/4,1/4)$. Now, consider a map $\tilde\pi:\mathbb R\to\mathbb R$ constructed by extending $\pi$ on $(-\epsilon,\epsilon)$ to all of $\mathbb R$ by $\tilde\pi(2x)=2\tilde\pi(x)$. This map is continuous since, for any $x$, there exists some $n$ for which $\tilde\pi(y)=2^n\pi(y/2^n)$ holds for all $y$ near $x$ (with $|y/2^n|<\epsilon$). It's clearly additive for similar reasons (dividing by a large power of $2$). So, $\tilde\pi:\mathbb R\to\mathbb R$ is a continuous homomorphism.

We claim that this implies $\tilde\pi(x)=\alpha x$ for some fixed constant $x$. If not, take some distinct $x$ and $y$, linearly independent over $\mathbb Q$, for which $\tilde\pi(x)=\alpha x$ and $\tilde\pi(y)=\beta y$ and $\alpha\neq \beta$. Since $\mathbb Zx+\mathbb Zy$ contains nonzero reals arbitrarily close to $0$, but $\alpha mx+\beta ny$ doesn't tend to $0$ as $mx+ny\to 0$ and $(m,n)$ grows (it looks like $(\beta-\alpha)ny$), this contradicts continuity at $0$.

So, $\tilde\pi(x)=\alpha x$ for some real $\alpha$, which implies that $\pi(x)=\alpha x$ for some real $\alpha$. The fact that $\pi(1)=0$ implies that $\alpha$ is an integer, which finishes the proof.

$\endgroup$
4
$\begingroup$

Yes, if you require the homomorphisms to be continuous (see the other answer otherwise).

Note for any (locally compact, abelian) group $G$ the group of continuous homomorphisms $\{ \phi : G \to \mathbb{R} / \mathbb{Z} \}$ is the Pontryagin Dual $G$, often written $\widehat{G}$.

Then you're asking for the pontryagin dual of the circle group, but this is well known to be $\mathbb{Z}$. Moreover, if you chase through the definitions, you'll find these characters are exactly the same copy of $\mathbb{Z}$ that you've found (though they're traditionally denoted $z \mapsto z^n$ instead of $[x] \mapsto [nx]$ since in this context we usually view the circle group as the unit circle in $\mathbb{C}$ under multiplication).


I hope this helps ^_^

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .