The proof relies on a delicate argument involving the nodal line of a first eigenfunction; in particular, the conclusion does not follow from a specific comparison argument, or from an explicit lower bound.

In this paper we give a geometric lower bound of $\lambda_1(\Omega,A)$ when $\Omega$ is, more generally, a [*Riemannian cylinder*]{}, that is, a domain $(\Omega,g)$ diffeomorphic to $[0,1]\times{{\bf S}^{1}}$ endowed with a Riemannian metric $g$, and when $A$ is a closed potential $1$-form : hence, the magnetic field $B$ associated to $A$ is equal to $0$. The lower bound will depend on the geometry of $\Omega$ and, in an explicit way, on the flux of the potential $A$.

Let us write $\partial\Omega=\Sigma_1\cup\Sigma_2$ where $$\Sigma_1=\{0\}\times{{\bf S}^{1}}, \quad \Sigma_2=\{1\}\times{{\bf S}^{1}}.$$ We will need to foliate the cylinder by the (regular) level curves of a smooth function $\psi$ and then we introduce the following family of functions.

$$\begin{aligned}{\cal F}_{\Omega}=\{\psi:\Omega\to{{\bf R}}: \quad &\text{\it $\psi$ is constant on each boundary component}\\
&\text{\it and has no critical points inside $\Omega$}\}
\end{aligned}$$

As $\Omega$ is a cylinder, we see that ${\cal F}_{\Omega}$ is not empty. If $\psi\in{\cal F}_{\Omega}$, we set: $$K=K_{\Omega,\psi}=\dfrac{\sup_{\Omega}{\lvert{\nabla\psi}\rvert}}{\inf_{\Omega}{\lvert{\nabla\psi}\rvert}}.$$ It is clear that, in the definition of the constant $K$, we can assume 