on the eigenvalues of a circle done in Section \[sectioncircle\] below we see : $$\lambda_1(\Sigma_r,\tilde A)=\dfrac{4\pi^2}{L_r^2}d(\Phi^{\tilde A},{\bf Z})^2,$$ where $\Phi^{\tilde A}$ is the flux of $\tilde A$ across $\Sigma_r$. Now note that $\Phi^{\tilde A}=\Phi^{A}$, because $\tilde A$ is the restriction of $A$ to $\Sigma_r$; moreover $L_r\leq L$ by the definition of $L$. Therefore: $$\label{llower}
\lambda_1(\Sigma_r,\tilde A)\geq \dfrac{4\pi^2}{L^2}d(\Phi^{ A},{\bf Z})^2$$ for all $r$. Let $X$ be a unit vector tangent to $\Sigma_r$. Then: $$\begin{aligned}
\nabla^{\tilde A}_{X}u&=\nabla_{X}u-i\tilde A(X)u\\
&=\nabla_{X}u-iA(X)u\\
&=\nabla^A_{X}u.
\end{aligned}$$ The consequence is that: $$\label{energy}
{\lvert{\nabla^{\tilde A}u}\rvert}^2={\lvert{\nabla^{\tilde A}_{X}u}\rvert}^2={\lvert{\nabla^{A}_{X}u}\rvert}^2\leq {\lvert{\nabla^{A}u}\rvert}^2.$$ [$\bullet\quad$]{}[*Note that equality holds in iff $\nabla^A_{N}u=0$ where $N$ is a unit vector normal to the level curve $\Sigma_r$ (we could take $N=\nabla\psi/{\lvert{\nabla\psi}\rvert}$).*]{}

For any fixed level curve $\Sigma_r=\{\psi=r\}$ we then have, taking into account , and : $$\dfrac{4\pi^2}{L^2}d(\Phi^{ A},{\bf Z})^2\int_{\psi=r}{\lvert{u}\rvert}^2\leq \int_{\psi=r}{\lvert{\nabla^Au}\rvert}^2.$$ Assume that $B_1\leq{\lvert{\nabla\psi}\rvert}\leq B_2$ for positive constants $B_1,B_2$. Then the above inequality implies: $$\dfrac{4\pi^2}{L^2}d(\Phi^{ A},{\bf Z})^2\cdot B_1\int_{\psi=r}\dfrac{{\lvert{u}\rvert}^2}{{\lvert{\nabla\psi}\rvert}}\leq B_2\int_{\psi=r}\dfrac{{\lvert{\nabla^Au}\rvert}^2}{{\lvert{\nabla\psi}\rvert}}.$$ We now integrate both sides from $r=0$ to $r=1$ and use the coarea formula. Conclude that $$\dfrac{4\pi^2}{L^2}d(\Phi^{ A},{\bf Z})^2\cdot B_1\int_{\Omega}{{\lvert{u}\rvert}^2}\leq B_2\int_{\Omega}{\lvert{\nabla^Au}\rvert}^2.$$ As $u$ is a first eigenfunction, one has: $$\int_{\Omega}{\lvert{\nabla^Au}\rvert}^2=\lambda_1(\Omega,A)\int_{\Omega}{\lvert{u}\rvert}^2.$$ Recalling that $K=\frac{B_2}{B_1}$ we finally obtain the estimate .

Proof of the equality case {#equalitycase}
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If the cylinder $\Omega$ is a 