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?


1 Answer 1


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.


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