# Additive maps modulo $1$ - what do they look like?

Recently, some convergence problems I have been considering led me to look at additive maps of the torus (which for me is the additive group $T := \mathbb{R}/\mathbb{Z}$).

A map $f:\ T \to T$ is additive if $f(\alpha + \beta) = f(\alpha) + f(\beta)$. Other names that seem to make sense would be $\mathbb{Z}$-linear map or group homomorphism, depending on how one wants to view things. The simplest examples are of course the linear maps $f(\alpha) = c \alpha$, $c \in \mathbb{Z}$.

If we were talking about additive maps $f:\ \mathbb{R} \to \mathbb{R}$, then the situation is fairly well understood. One can show that there exists a basis of $\mathbb{R}$ as a $\mathbb{Q}$-linear space. Let $\{\alpha\}_{i \in I}$ be such a basis, and choose arbitrary reals $\{\beta\}_{i \in I}$ : then there is a unique additive map $f$ such that $f(\alpha_i ) = \beta_i$ (more explicitly, $f(\sum_{i \in I} q_i \alpha_i) = \sum_{i \in I} q_i \beta_i$, where $q_i \in \mathbb{Q}$ and $q_i = 0$ except finitely many $i$). Moreover, any additive map arises in this way, so we have a complete characterisation (modulo the fact that a $\mathbb{Q}$-linear basis of $\mathbb{R}$ is not something easy to come by).

If we consider an additive map $f:\ \mathbb{R} \to \mathbb{R}$ such that $f(1) \in \mathbb{Z}$, then $f$ factors through the quotient with $\mathbb{Z}$, giving rise to the additive map $[f] :\ T \to T$ given by $[f]([\alpha]) = [f(\alpha)]$. Hence, there exists a fair number of additive maps of the torus.

Question: Does every additive map $T \to T$ arise in the way just described? Is there a general description of additive maps $T \to T$, like the one for $\mathbb{R}$? [Affirmative answer to the second question probably resolves both].

Note: if $f:\ T \to T$ is additive, and $\{\alpha\}_{i \in I}$ is a basis of $\mathbb{R}$ over $\mathbb{Q}$ as above, then clearly $f(\sum_i a_i \alpha_i) = \sum_i a_i f(\alpha_i)$ as long as $a_i$ are integers. However, it is not the case that one can select $\alpha_i$ such that each $\beta \in T$ is a unique integral combination of $\alpha_i$ (in other words, $T$ is not a free group). It is true that $f(\sum_i q_i \alpha_i) = \sum_i q_i f(\alpha_i) + Q$ with $q_i \in \mathbb{Q}$ and some $Q \in \mathbb{Q}$ dependent on $q_i$ (since we have the additional rational term anyway, we may understand rational multiplication $q_i f(\alpha_i)$ in any way we like.).

If we have $A<T$ and an additive map $f\colon A\to T$, then the set of additive maps $g\colon G\to T$ with $A\le G$ and $g|_A=f$ has a maximal element $m\colon M\to T$ by Zorn's lemma. Assume there exists $a\in T\setminus M$. The kernel of $\mathbb Z\to T/M$, $k\mapsto ka+M$ has the form $n\mathbb Z$. If $n>0$, select $b\in T$ with $nb=m(na)$. Then $h\colon\langle a\rangle + M\to T$, $ka+x\mapsto kb+m(x)$ defines an additive map contradicting maximality. If $n=0$, we get the same contradiction with $na+x\mapsto m(x)$. Therefore, $M=T$, i.e. any additive map on a subgroup can be extended to all of $T$.
Now for $n\in\mathbb N_0$ let $A_n$ be the kernel of $T\to T$, $x\mapsto 2^nx$. Given an additive map $f_{n-1}\colon A_{n-1}\to T$, we find $f_{n}\colon A_{n}\to T$ with $f_{n}|_{A_{n-1}}=f_{n-1}$ and $f_{n}(2^{-n}+\mathbb Z)\in[\frac14,\frac34]+\mathbb Z$ because we have the choice among two values that differ by $\frac12$. In the limit be get an additive map $f\colon \bigcup A_n\to T$ and from that an extension to $T\to T$, that does not come from an additive map $F\colon\mathbb R\to\mathbb R$ because there would have to be arbitrarily small rational numbers $x=2^{-n}$ with $|F(x)|\ge\frac14$.
• I don't understand this argument. The definition of $f_n$ given here does not seem to necessarily satisfy the requirement that its restriction to $A_{n-1}$ is equal to $f_{n-1}$. But it is true that, for each $f_{n-1}$ there are two possible $f_n$, so is the point rather that there are uncountably many possible $f$ using this construction, but only countably many values of $f(1)$ in a possible lift to ${\mathbb R}$? – Derek Holt May 6 '13 at 10:48