\[lemma2.3\] Assuming $g\in\tilde{\Pi}$, the domain $D(g_0,g_1)$ is a core for $H(g)$.

We begin by noting that the potentials $\wp(x+i\alpha/2)$ and $\wp(x-\pi/2r-i\alpha/2)$ yield bounded self-adjoint operators, so that they do not influence domain issues. Therefore, we need only prove the core property for one value of $g_2$ and $g_3$. We choose $g_2=g_3=1$ (say), the point being that this entails $g\in\Pi_r$. (This is easily checked, cf. (\[defJN\]), (\[Pi\]) and (\[Pir\]).) Hence we can now use ${{\cal I}}(g)$.

Assume $\phi\in{{\cal H}}$ satisfies $$\begin{gathered}
(\phi,(H(g)+i)\psi)=0,\qquad \forall \, \psi\in D(g_0,g_1).\end{gathered}$$ By virtue of (\[Ran\]), this implies in particular $$\begin{gathered}
(\phi,(H(g)+i){{\cal I}}(g) f)=0,\qquad \forall \, f\in{{\cal H}}.\end{gathered}$$ Thanks to (\[eiB\]) and (\[cIad\]), this inner product can be rewritten as $$\begin{gathered}
((H(g')-i){{\cal I}}(g')\phi,f)=0,\qquad \forall \, f\in{{\cal H}}.\end{gathered}$$ This implies that the vector ${{\cal I}}(g')\phi$, which belongs to $D(g_0^{'},g_1^{'})$ by (\[Ran\]), is an eigenvector with eigenvalue $i$ for the symmetric operator $H(g')$. Thus we must have ${{\cal I}}(g')\phi=0$. By (\[KerI\]), this entails $\phi=0$. Therefore ${\rm Ran}\, (H(g)+i)$ is dense in ${{\cal H}}$. Likewise, ${\rm Ran}\, (H(g)-i)$ is dense in ${{\cal H}}$. Hence the core property results.

We would like to point out that the dense subspace $$\begin{gathered}
\label{cC}
{{\cal C}}\equiv C_0^{\infty}((0,\omega_1))\subset D(g_0,g_1),\end{gathered}$$ is already a core for $H(g)$ 