Friday, July 02, 2010

"Holomorphic Functional Calculus"

Writing up the Connes-Renault notes, which I mentioned in a previous post, leads to a number of interesting digressions. For instance, the notion of "holomorphic closure" is discussed at some length in these early notes. But what exactly is the relationship between "holomorphic closure", "inverse closure", "complete holomorphic closure" (= holomorphic closure when tensored with any matrix algebra), and so on? I was aware that there had been some progress in this area but had not really sorted it out in my mind. Here's a summary (all these results are pretty old, so perhaps everyone knows this but me...)

Theorem 1. Let G be a group which has a complete metric topology for which multiplication is jointly continuous. Then inversion is continuous (i.e., G is a topological group).

There is a short, slick proof in Pfister, "Continuity of the inverse", Proc AMS 95(1985) 312-314, though the result is older.

For the rest of this note I follow Schweitzer, Larry B. “A Short Proof that $M_{n}(A)$ is local if $A$ is local and Fr\'echet.” International Journal of Mathematics 3 (1992): 581-589. Suppose that B is a C* or Banach algebra and that A is a dense subalgebra which is a Fr\'echet algebra under some topology stronger than the topology of B.

Theorem 2. A is holomorphically closed in B iff it is inverse closed.

For the proof, one notes that inverse closure implies that the invertibles are open in A, so their topology (in A) can be given by a complete metric. By Theorem 1, inversion is continuous. This means that the Cauchy integral formula for the holomorphic calculus converges in A.

Theorem 3. A is inverse closed in B iff every irreducible A module is a submodule of some B module.

This involves some topological-algebraic manipulation with maximal left ideals. An irreducible A module is of the form $A/m$, where $m$ is some maximal left ideal. Let M be the closure of m in B. Using density and inverse closure one sees that $m=M\cap A$. But then
$A/m = A/A\cap M = (A+M)/M \subseteq B/M $ is contained in the B-module $B/M$.

Corollary. $A$ is inverse (or holomorphically) closed in $B$ iff every matrix algebra $M_n(A)$ is inverse closed in $M_n(B)$.

This follows since an irreducible $M_n(A)$-module is just the direct sum of $n$ copies of some irreducible $A$-module.

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