Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 10 / 4

The two algebras have the same identity, so an inverse in $A$ is also an inverse in $B$. Therefore $G(A)\subseteq G_B(A)$.

For strictness, take $B=C_{\mathbb{C}}(\mathbb{T})$, where $\mathbb{T}$ is the unit circle, and let $A$ be the uniform closure of the polynomials in the coordinate function $z$. This is a closed unital subalgebra of the Banach algebra $B$. The coordinate $z$ has inverse $\overline{z}$ in $B$. But $\overline{z}$ cannot belong to $A$: for every polynomial $p$,and the same identity holds for every element of $A$ by uniform convergence. It fails for $\overline{z}$, since $z\overline{z}=1$. Thus $z\in G_B(A)\setminus G(A)$. This algebra is the boundary realization of the disk algebra.

Because $A$ is a closed subalgebra of a Banach algebra, it is itself complete. If $\Vert a-1\Vert <1$, the Neumann series converges in $A$:Multiplying the partial sums by $a$ on either side gives $1-(1-a{)}^{N+1}$, which tends to $1$. Thus every such $a$ is invertible in $A$, and $1$ is an interior point of $G(A)$.

For $a\in G(A)$, any $b\in A$ with $\Vert a^{-1}\Vert \Vert b-a\Vert <1$ has the factorizationThe second factor has a Neumann series inverse in $A$, so $b\in G(A)$. Hence the group of invertible elements of a Banach algebra $G(A)$ is open in $A$.

For $a\in G_B(A)$ use its inverse in $B$ instead. The same small-perturbation argument proves invertibility of $b$ in $B$ whenever $b\in A$ and $\Vert a^{-1}{\Vert }_B\Vert b-a{\Vert }_B<1$. Therefore both $G(A)$ and $G_B(A)$ are open subsets of $A$; in the second argument the inverse need not lie in $A$.

Let $c=(1-ab{)}^{-1}\in A$. From the two inverse identities we have $cab=c-1=abc$. The candidate inverse isIndeedandBoth products have been checked, so this is valid in a noncommutative Banach algebra. It is the identity underlying the nonzero spectra of products in opposite orders result.

Suppose the inverse norms did not tend to infinity. Some subsequence would satisfy $\Vert a_n^{-1}\Vert \le C$. Since $a_n\to a$, for its large indices $\Vert a_n^{-1}(a-a_n)\Vert <1$. Thenis invertible in $A$ by the Neumann series, contradicting the hypothesis. Hence $\boxed{\Vert a_n^{-1}\Vert \to \infty }$, the inverse norm divergence at noninvertible boundary points property.

If $a$ were invertible in $B$, then continuity of inversion in a Banach algebra would give $a_n^{-1}\to a^{-1}$ in $B$. Explicitly, the same factorization around $a$ gives a Neumann series and the bound $\Vert a_n^{-1}-a^{-1}\Vert \le \Vert a_n^{-1}\Vert \Vert a-a_n\Vert \Vert a^{-1}\Vert$, with the inverse norms locally bounded in $B$. Since every $a_n^{-1}$ belongs to the closed subalgebra $A$, so would $a^{-1}$. That would make $a\in G(A)$, again a contradiction. Therefore $a\notin G_B(A)$.

Part (c) makes $G(A)$ relatively open in $G_B(A)$. It is also relatively closed: if a sequence in $G(A)$ converges to $a\in G_B(A)$, part (e) forbids $a\notin G(A)$. Sequences detect closure here because the topology comes from a norm.

Intersecting this clopen subset with any connected component $C$ of $G_B(A)$ gives a clopen subset of the connected space $C$. It is therefore either empty or all of $C$. Thus $G(A)$ is the union of those connected components of $G_B(A)$ that it meets. This is connected-component permanence of subalgebra invertibility; it does not require connected components to be path connected.