# How far is a Banach algebra from being a multiplicative group?

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!

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I'm not sure I understand what you are looking for, but here are two threads that might be related: math.stackexchange.com/q/92079 and math.stackexchange.com/q/92354 – t.b. Dec 22 '11 at 3:09

There is a recent paper of Dales and Feinstein:

H.G. Dales and J. F. Feinstein. Banach function algebras with dense invertible group. Proceedings of the American Mathematical Society, 136 (2008), 1295-1304.

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.

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I originally posted this as a comment to your comment in Matt E's answer, but this seems to be the better place: The result in your second paragraph is a result of Vaserstein, Stable rank of rings and dimensionality of topological spaces, Funct. Anal. Appl. 5 (1971), 102-110 that $\operatorname{tsr}{C(X)} = [\dim{X}/2] + 1$, where $\dim{X}$ denotes the Lebesgue covering dimension (so density holds precisely for zero- and one-dimesional spaces. See also this article – t.b. Jan 4 '12 at 4:12
Yemon: Thanks for the nice answer, and thanks again for the correction. – Jonas Meyer Jan 4 '12 at 5:15

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.)

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On the other hand, in the $\mathbb C$-valued case, $G(C(X))$ is dense in $C(X)$, unlike what happens in $B(H)$, the algebra of bounded linear operators on an infinite dimensional Hilbert space, or in many commutative Banach algebras, like the continuous functions on the closed unit disk in $\mathbb C$ that are analytic in the interior. In the latter case, there are no zero divisors, yet the algebra is quite far from $G(A)$. – Jonas Meyer Dec 22 '11 at 4:04
It might be worth mentioning the Gelfand-Mazur theorem stating that a Banach division algebra is isomorphic to $\mathbb{C}$, the space of continuous functions on a point. – t.b. Dec 22 '11 at 4:07
@t.b. Dear t.b., Yes, that is a pertinent remark! Best wishes, – Matt E Dec 22 '11 at 4:26
@JonasMeyer: are you sure that $G(C(X))$ is always dense in $C(X)$ for arbitrary $X$? I think that's the definition of topological stable rank, and $C(X)$ does not always have tsr=1. – user16299 Jan 4 '12 at 3:56
@YemonChoi: No, I thought I knew what I was talking about there but wasn't thinking clearly, thank you for asking. I was thinking about the fact that invertible normal operators are dense in the normal operators of $B(H)$. (I was also vaguely thinking that because for $x\in C(X)$, $C^∗(1,x)\cong C(\sigma(x))$ we can reduce the problem to $X$ being a compact subset of $\mathbb C$. Indeed, but it is false in general for compact subsets of $\mathbb C$, such as the closed unit disk. Thank you again.) – Jonas Meyer Jan 4 '12 at 5:11