Function $\mathbb{R} \to \mathbb{R}$ descending to a function $\mathbb{Z}/m\mathbb{Z} \to \mathbb{R}/\mathbb{Z}$

Question

Given a positive integer $m \in \mathbb{Z}_{>0}$, and a function $g: \mathbb{R} \to \mathbb{R}$ defined by

$x \mapsto g(x) := \frac{x^2}{2m} - \frac{x}{2}$ ,

how does this descend to a function from $\mathbb{Z}/m\mathbb{Z}$ to $\mathbb{R}/\mathbb{Z}$?

I can see that if $k \in \mathbb{Z}$,

$g(mk) = m\cdot\underbrace{\frac{k(k-1)}{2}}_{=\textrm{integer}} \in m\mathbb{Z} \subset \mathbb{Z}$ ,

so under the map $g$, elements of $m\mathbb{Z}$ are mapped into themselves, so it does make sense for $g$ to descend to a function -- call it $\widetilde{g}$ -- from $\mathbb{Z}/m\mathbb{Z}$ to $\mathbb{R}$. But what is the form of $\widetilde{g}$?

Answer 1

Let $g:\mathbb R\to\mathbb R$ be a function.

It always induces a function $G:\mathbb Z\to \mathbb R/\mathbb Z$ via restriction of the domain (note: I am denoting $g$ and its restriction to $\mathbb Z$ by the same letter, for simplicity) and composition $G=\pi\circ g$ with the natural projection $\pi: \mathbb R\to \mathbb R/\mathbb Z$. If $[x]$ is the class of $x\in\mathbb R$ in $\mathbb R/\mathbb Z$, then $\pi(x)=[x]$.

Now, $G$ induce a map $\overline{g}:\mathbb Z/m\mathbb Z\to \mathbb R/\mathbb Z$ if two elements $x$ and $y$ in $\mathbb Z$ that are equivalent modulo $m\mathbb Z$ have the same image by $G$ (equivalent, their images by $g$ are equivalent modulo $\mathbb Z$). This allows to define, $\overline{g}(\overline{x})=G(x)$ if $\overline{x}$ is the class of $x$ in $\mathbb Z/m\mathbb Z$.

Since, in $\mathbb R/\mathbb Z$, we have $[x]=[y] \iff x-y\in \mathbb Z$, we can formulate this as

For every $x$ and $y$ in $\mathbb Z$ such that $y=x+km$ for some $k\in\mathbb Z$, we have $g(y)-g(x) \in \mathbb Z$

A quick computation shows that $g(y)-g(x)=kx+\frac{k(k-1)}{2}m$ which is an integer. So $G$ indeed induces a map $\overline{g}$ from $\mathbb Z/m\mathbb Z$ into $\mathbb R/\mathbb Z$.

Note that $G$ does not induce a map from $\mathbb R/m\mathbb Z$ into $\mathbb R/\mathbb Z$. Indeed, if $x$ and $y$ are real number such that $y=x+km$ for some integer $k$, then $g(y)-g(x)=k(2x+(k-1)m)$ is not always an integer. For example, for $m=2$, $x=\sqrt{2}$ and $y=\sqrt{2}+4$, we have $g(y)-g(x)=2+2\sqrt{2}$, not an integer.