For a compact spin Riemannian manifold $(M^n,g)$ without boundary, $n \not\equiv 3\mod 4$, it is well-known that the Dirac operator associated with a fixed spin structure $S\rightarrow M$ has real, discrete spectrum and symmetric about zero. In the case $(M^n,g)$ has non-empty connected boundary, for the spectrum of the Dirac operator $D^S$ of a fixed spin structure $S$ to be real and discrete, one has to subjugate the eigenvalue problem to the Atiyah-Patodi-Singer(APS) condition. My first question is as follows

Question 1. What sort of reasonable (topological/geometrical) conditions do we have to impose on compact spin $(M^n,g,\partial M \neq 0)$ or $S$ so that the spectrum of aforementioned $D^S$ restricted to APS condition is also symmetric about zero, besides being real and discrete?

If such a task for question 1 is possible, temporarily we shall call such manifold CSymm.

Let $E \rightarrow (M^n,g,\partial M\neq 0)$ be any Hermitian bundle equipped with a compatible connection $\nabla^E$. It is not hard to see that the twisted bundle $S \otimes E$ is a Clifford bundle over $M$, on which one can have a globally-defined notion of associated generalized Dirac operator $D^{S\otimes E}$. At this point, we carefully consider a metric $g$ on $M$ so that near $\partial M$ looks like $\partial M \times [0,r)$. With such a choice of $g$, $D^{S\otimes E}$ has a unambiguous induced Dirac operator on $\partial M$, denoted by $D^{S\otimes E, \partial}$. We say $(M^n,g,\partial M \neq 0)$ is CitD iff it possesses a Hermitian bundle $(E,\nabla^E)$ ($E$ may depend on $S$) such that $D^{S\otimes E,\partial}$ is invertible.

For $n\geq 4$, we say $M^n$ is a good manifold if and only if $M$ is CSymm and there exists a smooth map $f: M \rightarrow \mathbb{S}^n$ such that $(M,f^{*} S_0)$ makes $M$ a CitD manifold, where $S_0$ is the spin structure on $\mathbb{S}^n$. Here is my last question

Question 2. What is an example of a good manifold with positive scalar curvature?

If these questions turn out to be straight-up trivial, I would just like a hint or a smart observation that would point me in a right direction so that I can go on on my own.

  • $\begingroup$ Welcome to mathoverflow! For question 1, if $\dim M$ is even, the spectrum is symmetric because $D$ anticommutes with the Clifford volume element. Have you tried to check wether the Clifford volume acts on the space of sections satisfying the APS conditions? $\endgroup$ – Sebastian Goette Dec 4 '18 at 15:35
  • $\begingroup$ I thought about this and from the conversation I had with other people, it seems like the Clifford volume form $\omega^{\mathbb{C}}_n$ should acts on sections of spin structure on the boundary of $M$ (assuming dimension of $M$ is even) like $\text{identity} \oplus -\text{identity}$. If this is true, then the form should commute with the spectral projection, which in-turn preserves the APS boundary condition. $\endgroup$ – M. L. Nguyen Dec 5 '18 at 19:03

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