# Elliptic boundary condition eigenvalue problem

In Chapter 1, section 1.5 of "The Dirac Spectrum" by Nicolas Ginoux, different elliptic boundary conditions for Dirac operators are introduced. On page 24, there is the following theorem

Theorem 1.5.2 Let $$(M^n, g)$$ be a compact Riemannian spin manifold with non-empty boundary $$\partial M$$. Let $$B$$ be an elliptic boundary condition for $$D$$. The the following eigenvalue problem $$D\phi = \lambda \phi \text{ on } M, \text{ } \text{ } B(\phi|_{\partial M}) = 0 \text { on } \partial M.$$ has a discrete spectrum with finite dimensional eigenspaces in $$C^{\infty}(\Sigma M)$$ [smooth sections of a fixed spinor bundle over $$M$$], unless the spectrum is $$\mathbb{C}$$ itself. Moreover, if $$B$$ is self-adjoint, then the Dirac spectrum is real.

Here, $$\Sigma M$$ is the spinor bundle associated with a fixed spin structure over $$M$$. $$D$$ is the Dirac operator defined by composition of the Clifford multiplication and the compatible connection. My question is the following:

• Why is the self-adjoint condition of $$B$$ is needed here so that the Dirac spectrum is real? Certainly if certain $$(\lambda, \phi_\lambda)$$ is a solution, necessarily by the above theorem $$\phi_\lambda \in C^{\infty}(\Sigma M)$$. And using the $$L^2-$$formally self-adjoint property of $$D$$, can we just conclude that $$\lambda ||\phi_\lambda||^2 = \langle D \phi_\lambda , \phi_\lambda \rangle = \langle \phi_\lambda, D\phi_\lambda \rangle = \overline{\lambda} ||\phi_{\lambda}||^2,$$ which implies that $$\lambda$$ has to be real? What part is wrong in my argument?