means that $u=e^{i\phi}$ is $\nabla^A$-parallel hence $\Delta_A$-harmonic. On the other hand, if the magnetic field $B=dA$ is non-zero then $\lambda_1(\Omega,A)>0$.

It then remains to examine the case when $A$ is closed but not exact. The situation was clarified in [@Sh] for closed manifolds and in [@HHHO] for Neumann boundary conditions.

\[shikegawa\]The following statements are equivalent:

a\) $\lambda_1(\Omega,A)=0$;

b\) $dA=0$ and $\Phi^A_c\in\bf Z$ for any closed curve $c$ in $\Omega$.

Thus, the first eigenvalue vanishes if and only if $A$ is a closed form whose flux around every closed curve is an integer; equivalently, if $A$ has non-integral flux around at least one closed loop, then $\lambda_1(\Omega,A)>0$.

Proof of the lower bound
------------------------

From now on we assume that $\Omega$ is a Riemannian cylinder. Fix a first eigenfunction $u$ associated to $\lambda_1(\Omega, A)$ and fix a level curve $$\Sigma_r=\{\psi=r\}, \quad\text{where $r\in [0,1]$.}$$ As $\psi$ has no critical points, $\Sigma_r$ is isometric to ${{\bf S}^{1}}(\frac{L_r}{2\pi})$, where $L_r$ is the length of $\Sigma_r$. The restriction of $A$ to $\Sigma_r$ is a closed $1$-form denoted by $\tilde A$; we use the restriction of $u$ to $\Sigma_r$ as a test-function for the first eigenvalue $\lambda_1(\Sigma_r,\tilde A)$ and obtain: $$\label{level}
\lambda_1(\Sigma_r,\tilde A)\int_{\Sigma_r}{\lvert{u}\rvert}^2\leq\int_{\Sigma_r}{\lvert{\nabla^{\tilde A}u}\rvert}^2.$$ By the estimate 