Metric approach to limit operators II

Following on from my earlier post on the Spakula-Willett paper, let my try to summarize sections 5 and 6.  These parts produce, for their generalized notion of limit operator, an equivalent of how the classical limit operator theory looked prior to the Lindner-Seidel paper earlier this year.

Thus the main result of these parts is the following Theorem: A band dominated operator is Fredholm if and only if all its limit operators are invertible and there is a uniform bound on the norms of the inverses of all the limit operators.

This result works on $$\ell^p$$ for $$1<p<\infty$$.  I have ignored the extra words that one has to add in the case that there are coefficients in an infinite-dimensional Banach space (“P-Fredholm” rather than just “Fredholm”, and restrict attention to “rich” operators).

The proof that Fredholm implies that all limit operators are invertible closely follows the usual case.  In fact, by translating off a finite set one turns a compact operator into one with arbitrarily small norm, and therefore turns the Fredholmness of a given BDO into the invertibility “modulo arbitrarily small norm” of the limit operators.

The converse uses the formulation of property A in terms of good partitions of unity.  Specifically, property A is equivalent to the following: for all $$r,\epsilon>0$$ there exists a uniform partition of unity having

$d(x,y)\le r \Longrightarrow \sum_i |\phi_i(x)-\phi_i(y)|^p \le \epsilon^p.$

This is most familiar for $$p=1,2$$ but it works in the same way for any $$p<\infty$$.  Now the hypothesis that the limit operators are all uniformly invertible means that, for any uniform cover, the “localizations” of the given operator $$A$$ to the cover will be invertible.  (To be specific here, the map $$AQ_i$$, where $$Q_i$$ is the projection onto the $$i$$th space of the covering, will have a left inverse $$B$$ such that $$BAQ_i=Q_i$$.  The argument in the end will produce a left and a right parametrix for $$A$$ and then the usual yoga shows that they agree modulo compacts.  One must beware of saying “the restriction of $$A$$ to the sets of the cover is invertible” — shifts give counterexamples.)    Then the local inverses are carefully glued together using the good partition of unity that comes from property A.