Consider a finite rank complex bundle $E$ over $S^1$ with connection $\nabla$. Let $Q_0, Q_1 \in C^\infty(S^1, E)$ be pseudo-differential operators. $Q_0$ is defined by the symbol $\sigma_0(x, \xi) = i \xi$, and $Q_1$ by $\sigma_1(x,\xi) = 2H(\xi)-1$ where $H$ is the Heaviside step function. Both of these have discrete spectra.

Now take $Q_t$ to be defined by $\sigma_t = t \sigma_0 + (1-t) \sigma_1$. Does $Q_t$ have a discrete spectrum?

Less rigorously, is there a ``continuous" way to connect $Q_0$ and $Q_1$ through operators with discrete spectra? Also, are there well known theorems about when such operator have discrete spectra, or well known classes of operators with discrete spectra? (Compact operators do, but that is insufficient for my purposes). Apologies for being vague, I'm a physicist.


Lukasz Fidkowski.

  • $\begingroup$ Post-edit, the question is much better. Oh, and welcome to MO. $\endgroup$ – Willie Wong Dec 9 '10 at 1:56
  • $\begingroup$ (I deleted some comments that no longer applied.) $\endgroup$ – Willie Wong Dec 9 '10 at 1:57
  • $\begingroup$ Compact operators don't usually have discrete spectrum near 0. You probably mean operators $D$ with compact resolvent $(D-\lambda)^{-1}$? The first place I look for answers to these kinds of questions is Kato's book on perturbation theory. Physicists sometimes prefer Reed-Simon. $\endgroup$ – Paul Dec 9 '10 at 3:22

You should have a look at p. 267, 23.35.2 of Dieudonné's Treatise on Analysis, part 7, as well as Lawson-Michelson, Spin Geometry, p. 196, Thm. 5.8, and check if your symbols are elliptic (they seem to be) and notice that your base manifold is compact. This might help.


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