Riemannian product then it is obvious that we can take $K=1$ and then we have equality by Proposition \[cyl\] below. Now assume that we do have equality: we have to show that $\Omega$ is a Riemannian product. Going back to the proof, we must have the following facts.

[**F1.**]{} [*All level curves of $\psi$ have the same length $L$*]{}.

[**F2.**]{} [*${\lvert{\nabla\psi}\rvert}$ must be constant and, by renormalization, we can assume that it is everywhere equal to $1$.* ]{}Then, $\psi:\Omega\to [0,a]$ for some $a>0$ and we set $$N\doteq\nabla\psi.$$

[**F3.**]{} [*The eigenfunction $u$ on $\Omega$ restricts to an eigenfunction of the magnetic Laplacian of each level set $\Sigma_r=\{\psi=r\}$, with potential given by the restriction of $A$ to $\Sigma_r$.*]{}

[**F4.**]{} [*One has $\nabla^A_Nu=0$ identically on $\Omega$.* ]{}

### First step: description of the metric

$\Omega$ is isometric to the product $[0,a]\times {{\bf S}^{1}}(\frac{L}{2\pi})$ with metric $$\label{metric}
g={{\begin{pmatrix}}{1&0\\}{0&\theta^2(r,t)\\}{\end{pmatrix}}}, \quad (r,t)\in [0,a]\times [0,L]$$ where $\theta(r,t)$ is positive and periodic of period $L$ in the variable $t$. Moreover $\theta(0,t)=1$ for all $t$.

We first show that the integral curves of $N$ are geodesics; for this it is enough to show that $
\nabla_NN=0
$ on $\Omega$. Let $e_1(x)$ be a vector tangent to the level curve 