by parts and the coupling condition  we show first that $A_{\eta, \tau}$ is symmetric:

\[lemma\_symmetric\] Let $\eta, \tau \in \mathbb{R}$. Then the operator $A_{\eta, \tau}$ defined by  is symmetric.

Let $f, g  \in {\mathrm{dom}\,}A_{\eta, \tau}$. Employing integration by parts in $\Omega_\pm$ we get first $$\begin{split}
      (&A_{\eta, \tau} f, g )_{L^2(\mathbb{R}^3)^4}
              - (f, A_{\eta, \tau} g)_{L^2(\mathbb{R}^3)^4} \\
      &= (-i c \alpha \cdot \nu f_+, g_+ )_{L^2(\Sigma)^4}   -( -i c \alpha \cdot \nu f_-, g_- )_{L^2(\Sigma)^4} \\
      &= \frac{1}{2} \big( -i c \alpha \cdot \nu (f_+ - f_-), g_+ + g_- \big)_{L^2(\Sigma)^4}
      - \frac{1}{2} \big( f_+ + f_-, -i c \alpha \cdot \nu (g_+ - g_-)\big)_{L^2(\Sigma)^4}.
    \end{split}$$ Using the coupling condition  for $f$ and $g$, we conclude that the last term is $$\begin{split}
      \frac{1}{2} \big( -i c \alpha \cdot \nu (f_+ &- f_-), g_+ + g_- \big)_{L^2(\Sigma)^4} 
          - \frac{1}{2} \big(f_+ + f_-, -i c \alpha \cdot \nu (g_+ - g_-)\big)_{L^2(\Sigma)^4} \\
      &= \left( \frac{1}{4} (\eta I_4 + \tau \beta) (f_+ + f_-), g_+ + g_- \right)_{L^2(\Sigma)^4} \\
      &\qquad \qquad-  \left( f_+ + f_-, \frac{1}{4} (\eta I_4 + \tau \beta) (g_+ + g_-) \right)_{L^2(\Sigma)^4} = 0.
    \end{split}$$ Since this is true for any $f, g \in {\mathrm{dom}\,}A_{\eta, \tau}$, the operator $A_{\eta, \tau}$ is 