Dense domains, symmetric operators and spectral triples

A cautionary paper appeared on the arXiv last week

Forsyth, I., B. Mesland, and A. Rennie. Dense Domains, Symmetric Operators and Spectral Triples. ArXiv e-print, June 6, 2013.

The paper is about the definition of unbounded Fredholm modules and how certain, initially plausible, weakenings of the definition do not work as expected. (Some well-known authors are called out for insufficient attention to these points; I was initially worried that Analytic K-Homology appears in the bibliography, but I don’t think Nigel and I are among the criminals here.)

Let’s recall that the basic elements of an unbounded Fredholm module, a.k.a. spectral triple, over a \(C^*\)-algebra \(A\) are a Hilbert space \(H\), an unbounded self-adjoint operator \(D\) on \(H\), and an action of \(A\) on \(H\) such that, for some dense subalgebra \(\mathcal A\) of \(A\),

  1. Each \(a\in\mathcal A\) maps the domain of \(D\) into itself, and the commutator \([a,D]\) (defined on the domain of \(D\) by assumption) is norm bounded and so extends uniquely to a bounded operator on \(H\).
  2. For each \(a\in\mathcal A\) the “localized resolvent” \( a (1+D^2)^{-1/2}\) is compact.

Under these circumstances the bounded transform \(F = d(1+D^2)^{-1/2}\) is well-defined and defines a Kasparov cycle for \(A\) (in particular, it commutes modulo compact operators with all elements of \(A\)).

It suffices in fact to assume that there is some core \(X\) for \(D\) such that, for every \(a\in\mathcal A\), \(a\cdot X \subseteq \mbox{dom}(D) \) and that \([D,a]\) is bounded on \(X\).  (Recall that a core for an unbounded closed operator \(D\) is a subset \(X\)  of its domain such that \(D\) is the closure of its restriction to \(X\).)

But it is tempting to go further and to claim that it suffices that there should be some dense subspace \(X\) of the domain of \(D\), not necessarily a core, that satisfies the same properties as those listed.  And this is false.  In fact, under these circumstances the bounded transform  need not commute compactly with members of \(A\).  Moreover, the example is a simple one: we consider the operator \( i d/dx\) on the interval, with boundary conditions \( f(0)=f(1) \), and take \(X\) to be the subspace of the domain consisting of functions that satisfy \(f(0)=f(1)=0\).  Restricted to this domain, \(D\) commutes boundedly with all smooth functions on the interval; but its bounded transform does not commute compactly with functions \(f\) for which \(f(0)\neq f(1)\).

The paper contains some more sophisticated counterexamples too (anything that involves explicit computations with Bessel functions is sophisticated in my book), but this is the basic one.


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