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:

  1. $[D,a]$ is bounded for each $a\in A$ and
  2. $(D^2+1)^{-1/2}$ is a compact operator.

    • Second: could somebody explain with an example, why conditions 1. and 2. tend to contradict each other?
  • 1
    $\begingroup$ As I understand it, self-adjoint elliptic operators on a compact manifold always have compact resolvent. This includes, for example, the Laplace-Beltrami operator on a compact Riemannian manifold. What do you mean by $\lambda \not \in \mathbb{C}$? $\endgroup$ – Qiaochu Yuan Jun 21 '12 at 23:17
  • $\begingroup$ It tries to say "$D$ has compact resolvent". $\endgroup$ – c.p. Jun 21 '12 at 23:55

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.

  • $\begingroup$ I see. However, one doesn't have control on the boundedness of the elements in $A$. Anyway, very helpful answer. $\endgroup$ – c.p. Jun 23 '12 at 15:55

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