on functions (with Neumann boundary conditions if ${\partial}\Omega$ is not empty).

b If $A$ is a closed $1$-form, then $A$ is gauge equivalent to a unique (harmonic) $1$-form $\tilde A$ satisfying $$\twosystem
{d\tilde A=\delta\tilde A=0\quad\text{on}\quad \Omega}
{\tilde A(N)=0\quad\text{on}\quad {\partial}\Omega}$$ The form $\tilde A$ is often called the [Coulomb gauge]{} of $A$. Note that $\tilde A$ is the harmonic representative of $A$ for the absolute boundary conditions.

a This comes from the fact that $
\Delta_A e^{-i\phi}=e^{-i\phi} \Delta_{A+d\phi}
$ hence $\Delta_A$ and $\Delta_{A+d\phi}$ are unitarily equivalent.

b Consider a solution $\phi$ of the problem: $$\twosystem{\Delta\phi=\delta A \quad\text{on}\quad \Omega,}
{{\dfrac{{\partial}\phi}{\bdN}}=A(N) \quad\text{on}\quad {\partial}\Omega.}$$ Then one checks that $\tilde A=A-d\phi$ is a Coulomb gauge of $A$. As $\phi$ is unique up to an additive constant, $d\phi$, hence $\tilde A$, is unique.

We now focus on the first eigenvalue. Clearly, if $A=0$, then $\lambda_1(\Omega,A)=0$ simply because $\Delta_A$ reduces to the usual Laplacian, which has first eigenvalue equal to zero and first eigenspace spanned by the constant functions. If $A$ is exact, then $\Delta_{A}$ is unitarily equivalent to $\Delta$, hence, again, $\lambda_1(\Omega,A)=0$. In fact one checks easily from the definition of the connection that, if $A=d\phi$ for some real-valued function $\phi$ then $
\nabla^{A}e^{i\phi}=0,
$ which 