MathOverflow Asked by Sascha on February 20, 2021

I am looking for a function with the following property:

Let $v_1,v_2$ be two linearly independent vectors in $mathbb{R}^2.$

I am given a smooth function $g:(0,1) rightarrow (0,infty).$

I am trying to understand if there exists a smooth (non-constant) function $f:(0,1) times mathbb R^2 rightarrow mathbb R^2$ with the property that for all $n,m in mathbb Z$ such that $g(t)=n/m$, where $n/m$ is a reduced fraction, we have for all $x in mathbb R^2$

$$f(t,mv_1+x)=f(t,x)=f(t,nv_2+x)$$ and $n,m$ are the minimal periods of the function $f$, i.e. for all natural numbers $1le n_0 <n$ we do not have

$$f(t,nv_2+x)=f(t,x)$$ and the same for $m.$

Define $f(t,xmathbf v_1+ymathbf v_2)=e^{2pi i(x/g(t)+yg(t))}$. Then $f(t,mathbf x+jmathbf v_1)=e^{2pi ij/g(t)}f(t,mathbf x)$ and similarly $f(t,mathbf y+kmathbf v_2)=e^{2pi ikg(t)}f(t,mathbf x)$.

Suppose $t$ is such that $g(t)=frac nm$ (in lowest terms). Then $f(t,mathbf x+jmathbf v_1)=e^{2pi ijm/n}f(t,mathbf x)$ so that $f(t,cdot)$ is $jmathbf v_1$-periodic if and only if $n|jm$ if and only if $n|j$. Similarly $f(t,mathbf x+kmathbf v_2)=e^{2pi ikn/m}$, so that $f(t,cdot)$ is $kmathbf v_2$-periodic if and only if $m|k$.

Correct answer by Anthony Quas on February 20, 2021

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