# Spectrum of a first-order elliptic differential operator

Suppose that I have a first-order elliptic differential operator $$A: \mathrm{dom}(A) \subset L^2(E) \to L^2(E)$$, where $$(E,h^E) \to M$$ is a hermitian vector bundle and $$M$$ is a compact manifold.

I know also that $$A$$ is $$\omega$$-bisectorial with $$\omega < \frac{\pi}{2}$$. That is to say, the spectrum $$\sigma(A)$$ is in a bisector of angle $$\omega$$ in the complex plane, and outside of this bisector, we have resolvent bounds: there exists $$C > 0$$ with

$$|\zeta| ||(\zeta - A)^{-1}|| \leq C,$$

I want to be able to assert that $$A$$ has spectrum that goes to infinity on both sides of the imaginary axis. This should be true because you can see at the symbol level that:

$$\lambda \in \sigma( \mathrm{sym}_A(x,\xi) ) \iff -\lambda \in \sigma( \mathrm{sym}_A(x,-\xi)).$$

I have been told that this claim is true in the folklore, but I can't seem to find a reference.

Any ideas?

Consider the 0th order operator $$F:=A(1+A^*A)^{-1/2}$$. The spectrum of the operator $$F$$ is contained in the unit disc. The symbol mapping is a $$*$$-homomorphism, and therefore the spectrum of $$F$$ is contained in the spectrum of its symbol, call it $$a$$. In fact, the spectrum of $$a$$ coincides with the essential spectrum of $$F$$. The symbol argument you allude to shows that the spectrum of $$a$$ is symmetric around the imaginary axis. This shows that $$F$$ has essential spectrum on both sides of the imaginary axis. Now, returning to $$A$$, since $$A$$ is bisectorial its spectrum is not all of $$\mathbb{C}$$ and therefore it must be discrete. So the essential spectrum of $$F$$ consists of all limit points of the image of the spectrum of $$A$$ under the homeomorphism $$z\mapsto z(1+|z|^2)^{-1/2}$$ of the complex plane onto the open unit disc. Since $$F$$ has spectrum on both sides, $$A$$ has accumulation points on both sides.