$\mathbb{Z}$-invariant function on $\mathbb{R}^3$ manifold.

Question

Let $\phi_k : k\in \mathbb{Z}, \phi: \mathbb{R}^3\rightarrow\mathbb{R}^3$ be a family on mappings given by: $$\phi_k (x,y,z)=(x+k,e^k\cdot y, e^{-k}\cdot z).$$ I've proven the following statements:

• $\phi_k$ are diffeomorphisms.
• Mappings $\phi_k$ form a group of diffeomorphisms.
• Action of $\mathbb{Z}$ group through mappings $\phi_k$ on $\mathbb{R}^3$ is free and properly discontinuous.

Now I need to prove that on quotient manifold $\mathbb{R}^3/ \mathbb{Z}$ there exists an unbounded and smooth function $f:\mathbb{R}^3/ \mathbb{Z}\rightarrow \mathbb{R}$.

I know I need to find the $\mathbb{Z}$-invariant function (so $f:\mathbb{R}^3\rightarrow \mathbb{R}$ such that for every $k\in\mathbb{Z}$ we have $f(k(x))=f(x)$). I would be grateful for a hint how to find such function.

Answer 1

Just take $f(x,y,z) = yz$. This is unbounded, smooth, and $$f(\phi_k(x,y,z))= ({\rm e}^ky)({\rm e}^{-k}z) = yz,$$so it passes to the quotient as an unbounded smooth function.