# 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. http://arxiv.org/abs/1306.1580.

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.