In which C*-algebras one-sided-invertible elements are invertible?

Question

For matrices $AB=I$ is equivalent to $BA=I$, however this is no longer true for operators on infinite-dimensional spaces. I was wondering whether there was a characterisation of C*-algebras where similar phenomena takes place.

What would be sufficient conditions for a Banach algebra to make it work too?

Answer 1

For a $C^{\ast}$ algebra $A$, we say that a project $p\in A$ is infinite if it is Murray Von Neumann equivalent to a proper subprojection of itself. ie. $\exists v\in A$ such that $vv^{\ast} = p$, and $v^{\ast}v < p$.

If $A$ is unital with unit $1_A$, we say that $A$ is finite if $1_A$ is not infinite. If $A$ is non-unital, we say $A$ is finite, if its unitization is finite.

As it turns out the following are equivalent for a unital $C^{\ast}$ algebra $A$

1. $A$ is finite
2. All isometries are unitary.
3. All projections in $A$ are finite.
4. Every left invertible element is invertible (the condition you want).
5. Every right invertible element is invertible.

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For a Banach algebra, I don't know if there is any characterization for this. However, there is a sufficient condition. If the Banach algebra has left (or right) topological stable rank one [See this paper by Rieffel], then any left invertible element is invertible.