$L$. Let $A=H(t)\,dt$. Then, the eigenvalues of the magnetic Laplacian with potential $A$ are: $$\lambda_k(M,A)=\dfrac{4\pi^2}{L^2}(k-\Phi^A)^2, \quad k\in\bf Z$$ with associated eigenfunctions $$u_k(t)=e^{i\phi(t)}e^{\frac{2\pi i (k-\Phi^A)}{L}s(t)}, \quad k\in\bf Z.$$ where $\phi(t)=\int_0^tH(\tau)\,d\tau$ and $s(t)=\int_0^t\theta(\tau)\,d\tau$.

In particular, if the metric is the canonical one, that is, $g=dt^2$, and the potential $1$-form is harmonic, so that $A=\frac{2\pi \Phi^A}{L}dt$, then the eigenfunctions are simply : $$u_k(t)=e^{\frac{2\pi i k}{L}t}, \quad k\in\bf Z.$$

We remark that if the flux $\Phi^A$ is not congruent to $1/2$ modulo integers, then the eigenvalues are all simple. If the flux is congruent to $1/2$ modulo integers, then there are two consecutive integers $k,k+1$ such that $
\lambda_{k}=\lambda_{k+1}.
$ Consequently, the lowest eigenvalue has multiplicity two, and the first eigenspace is spanned by $$e^{i\phi(t)}e^{\frac{\pi i}{L}s(t)}, \, e^{i\phi(t)}e^{-\frac{\pi i}{L}s(t)}.$$ The following proposition is an easy consequence (for a proof, see also Appendix \[riemannian circle\]).

\[cyl\] Consider the Riemannian product $\Omega=[0,a]\times{{\bf S}^{1}}(\frac{L}{2\pi})$, and let $A$ be a closed $1-$form on $\Omega$. Then, the spectrum of $\Delta_A$ is given by $$\dfrac{\pi^2 h^2}{a^2}+\dfrac{4\pi^2}{L^2}(k-\Phi^A)^2, \quad h, k\in{\bf Z}, h\geq 0.$$ In particular, $$\lambda_1(\Omega,A)=\dfrac{4\pi^2}{L^2}d(\Phi^A,{\bf Z})^2.$$

### Fourth step : a calculus lemma

In this section, we state a technical lemma which will allow us to conclude. The proof is 