Let $R$ be a topological ring. Let $I(R)$ be the set of invertible elements of $R$. This is a group, and the multiplication is continuous. But the inversion map $$I(R) \to I(R),\, a \mapsto a^{-1}$$ does not have to be continuous. In particular, $I(R)$ is not a topological group. See here for such an example. (Does anybody know a more basic example? I do not demand that $I(R)$ is open in $R$.)
When $R$ is a Banach ring, one has the von Neumann series and the inversion map is at least continuous on the open ball $B_1(1)$ (correct?). My question is: When is the inversion map continuous on $I(R)$? For example, is it really a coincidence that this happens to be true for $R=\mathbb{R}$? Are there large classes of topological rings where this holds?
Edit: One can show that the map is continuous iff it is continuous at $1 \in I(R)$. (This is because it is an anti-homomorphism, and the multiplication maps with elements are continuous.) In particular, it is continuous if $R$ is a Banach ring.
By the way, the "correct" definition of the topological group of units of $R$ is $U(R) = \{(x,y) \in R^2 : xy=yx=1\}$ with multiplication map $(x,y)(x',y')=(xx',y'y)$ and inversion map $(x,y) \mapsto (y,x)$, which are clearly continuous. In general, there is a group isomorphism $I(R) \to U(R)$, and the question is when it is a homeomorphism as well.
