Given a Banach algebra $\mathcal{A}$, the collection of invertible elements in $\mathcal A$, $G(\mathcal{A})$ is a group. I wonder whether there is a measurement for how far $\mathcal{A}$ is from $G(\mathcal{A})$.
It seems to me that such an measurement can show how 'nice' a Banach algebra is algebraically.
Thanks!
On
The basic example of a Banach algebra is the algebra $\mathcal C(X)$ of continuous ($\mathbb R$- or $\mathbb C$-valued) functions on a compact space $X$, w.r.t. the sup norm.
A function is invertible in this Banach algebra precisely if it is nowhere zero. So as long as $X$ has more than one point, there will be elements in $\mathcal C(X)$ that are neither zero nor units. So the gap between $\mathcal A$ and $G(\mathcal A)$ is some kind of measure of how non-trivial the space underlying $\mathcal A$ is. (Here I am invoking the idea that $\mathcal C(X)$ is a basic model for a general Banach algebra --- at least a commutative one; this is in the spirit of the Gelfand--Naimark theorem, for example.)
There is a recent paper of Dales and Feinstein:
where they construct some "exotic" examples of uniform algebras satisfying the property in the title.
Note that $G(C(X))$ is not always dense in $C(X)$; those $X$ with this property are apparently the compact Hausdorff spaces with covering dimension $0$ or $1$. (The phrase to look up is "topological stable rank"; there is an article of B. Nica 0911.2945 with more background and related concepts.)
In a different direction: it may be worth noting that if $A$ is a commutative unital Banach algebra with a unique maximal ideal $M$ (in which case $M=Rad(A)$, the Jacobson radical of $A$) then $G(A)=\{ \lambda 1 + r : \lambda\in{\mathbb C}\setminus\{0\}, r\in M\}$, which is a "large" subset of $A$. Such algebras arise by adjoining units to so-called radical commutative Banach algebras. There might be some examples in the book of Bonsall and Duncan.