the metric $g$ in normal coordinates. Since $N={\dfrac{{\partial}}{{\partial}r}}$ one sees that $g_{11}=1$ everywhere; for any fixed $r=r_0$ we have that $F(r_0,\cdot)$ maps $\Sigma_1$ diffeomorphically onto the level set $\{\psi=r_0\}$ so that ${\dfrac{{\partial}}{{\partial}r}}$ and ${\dfrac{{\partial}}{{\partial}t}}$ will be mapped onto orthogonal vectors, and indeed $g_{12}=0$. Setting $\theta(r,t)^2={\langle{{\dfrac{{\partial}}{{\partial}t}}},{{\dfrac{{\partial}}{{\partial}t}}}\rangle}$ one sees that the metric takes the form . Finally note that $
\theta(0,t)=1
$ for all $t$, because $F(0,\cdot)$ is the identity.

### Second step : Gauge invariance

Let $\Omega$ be any Riemannian cylinder and $A=f(r,t)\,dr+h(r,t)\,dt$ a closed $1$-form on $\Omega$. Then, there exists a smooth function $\phi$ on $\Omega$ such that $$A+d\phi=H(t)\,dt$$ for a smooth function $H(t)$ depending only on t. Hence, by gauge invariance, we can assume from the start that $A=H(t)\,dt$.

Consider the function $
\phi(r,t)=-\int_0^rf(x,t)\,dx.
$ Then: $$A+d\phi=\tilde h(r,t)\,dt$$ for some smooth function $\tilde h(r,t)$. As $A$ is closed, also $A+d\phi$ is closed, which implies that ${\dfrac{{\partial}\tilde h}{\bdr}}=0$, that is, $
\tilde h(t,r)
$ does not depend on $r$; if we set $H(t)\doteq\tilde h(t,0)$ we get the assertion.

[$\bullet\quad$]{}We point out the following consequence. If $u=u(r,t)$ is an eigenfunction, we know from [**F4**]{} above that $\nabla^A_Nu=0$, where $N={\dfrac{{\partial}}{\bdr}}$. As $
\nabla^A_Nu={\dfrac{{\partial}u}{\bdr}}-iA({\dfrac{{\partial}}{\bdr}})u
$ and $A=H(t)\,dt$ we obtain $A({\dfrac{{\partial}}{\bdr}})=0$ hence $
{\dfrac{{\partial}u}{\bdr}}=0
