$g_2$ and $g_3$, we get equality of spectra (and even equality of operators) under the transformations $$\begin{gathered}
\label{triv}
g_2\to 1-g_2,\qquad g_3\to 1-g_3.\end{gathered}$$ (A caveat is in order: these maps can take $g$ out of $\Pi_r$.) Less trivially, consider the permutation $$\begin{gathered}
\label{refl}
(g_0,g_1,g_2,g_3)\to (g_1,g_0,g_3,g_2).\end{gathered}$$ Defining the ‘mirror’ unitary $$\begin{gathered}
\label{mir}
(Mf)(x)\equiv f(\pi/2r-x),\qquad x\in[0,\pi/2r],\qquad f\in{{\cal H}},\end{gathered}$$ we clearly get $$\begin{gathered}
M{D(g_0,g_1)}=D(g_1,g_0).\end{gathered}$$ This implies $$\begin{gathered}
MH(g_0,g_1,g_2,g_3)M=H(g_1,g_0,g_3,g_2),\end{gathered}$$ so that (\[refl\]) yields operators with equal spectra; in this case the two operators are different for generic $g$.

Next, we note that the matrix $J_N$ (\[defJN\]) can be viewed as the reflection with respect to the unit vector $(1,-1,1,-1)/2$: $$\begin{gathered}
J_N={\bf 1}_4-2u\otimes u,\qquad u\equiv (1,-1,1,-1)/2.\end{gathered}$$ Thus the self-dual couplings are given by $$\begin{gathered}
g'=g \ \Leftrightarrow \ g_0+g_2=g_1+g_3,\end{gathered}$$ and on this 3-dimensional submanifold of $\Pi_r$ the HS operator ${{\cal I}}(g)$ is self-adjoint. The maps (\[triv\]) and (\[refl\]) just considered do not involve $J_N$. But it is clear that even when we start from a self-dual $g$, the maps (\[triv\]) give rise to new couplings $\hat{g}$ that are generically not self-dual. When also $\hat{g}\in\Pi_r$, we can act with $J_N$ to obtain $\hat{g}'\in\Pi_r$ for which $H(\hat{g}')$ has the same spectrum as $H(g)$ (by virtue of Theorem \[theorem2.6\]).

