The following question is motivated by the definition of spectral triples in noncommutative geometry. This question was split in the following parts:
- First: Could somebody give diverse examples of operators on Hilbert spaces, having compact resolvent?
Now, suppose one has certain algebra $A$ acting as operators on a Hilbert Space $X$. Certain self-adjoint operator $D$ on $X$ is imposed to satisfy, in particular, following axioms in order to be a deemed (generalized) Dirac operator, in an abstract sense:
- $[D,a]$ is bounded for each $a\in A$ and
$(D^2+1)^{-1/2}$ is a compact operator.
- Second: could somebody explain with an example, why conditions 1. and 2. tend to contradict each other?
I'm not sure what you mean by "tend to contradict each other", but here is one relevant point. An easy way to satisfy 1 would be for $D$ to be a bounded operator and $A$ to consist of other bounded operators. However, this would make 2 nearly impossible. For if $D$ is bounded, then so is $T := (D^2 + 1)^{1/2}$. If $T^{-1}$ is compact then it is a homeomorphism, which means the closed unit ball of $H$ is compact. But this happens iff $H$ is finite dimensional.
So in this sense, boundedness properties like 1 tend to be in tension with compact resolvent properties like 2.