that the range of $\psi$ is the interval $[0,1]$, and that $\psi=0$ on $\Sigma_1$ and $\psi=1$ on $\Sigma_2$. Note that the level curves of the function $\psi$ are all smooth, closed and connected; moreover they are all homotopic to each other so that the flux of a closed $1$-form $A$ across any of them is the same, and will be denoted by $\Phi^A$.

We say, briefly, that $\Omega$ is [*$K$-foliated by the level curves of $\psi$.*]{} We also denote by $d(\Phi^A,{\bf Z})$ the minimal distance between $\Phi^A$ and the set of integer $\bf Z$: $$d(\Phi^A,{\bf Z})^2=\min\Big\{(\Phi^A-k)^2: k\in\bf Z\Big\}.$$

Finally, we say that [*$\Omega$ is a Riemannian product*]{} if it is isometric to $[0,a]\times{{\bf S}^{1}}(R)$ for suitable positive constants $a,R$.

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a\) Let $(\Omega,g)$ be a Riemannian cylinder, and let $A$ be a closed $1$-form on $\Omega$. Assume that $\Omega$ is $K$-foliated by the level curves of the smooth function $\psi\in{\cal F}_{\Omega}$. Then: $$\label{cylinder}
\lambda_1(\Omega,A)\geq\dfrac{4\pi^2}{KL^2}\cdot d(\Phi^A,{\bf Z})^2,$$ where $L$ is the maximum length of a level curve of $\psi$ and $\Phi^A$ is the flux of $A$ across any of the boundary components of $\Omega$.

b\) Equality holds if and only if the cylinder $\Omega$ is a Riemannian product.

[$\bullet\quad$]{}It is clear 