29.3 The Index Group
Definition 29.3.1 (Identity Component).label Let $A$ be a unital Banach algebra, then the connected component $G_{0}(A)$ of $G(A)$ containing $1$ is the identity component of $G(A)$, and:
- (1)
$G_{0}(A)$ is an open, closed, and normal subgroup of $G(A)$.
- (2)
The cosets of $G_{0}(A)$ are the connected components of $G(A)$.
Proof, [Theorem 2.4, Zhu93]. (1): Since $G_{0}(A)$ is connected, it is open and closed. By Proposition 5.14.2, $G_{0}(A)$ coincides with the path component of $G(A)$ containing $1$.
Let $x, y \in G_{0}(A)$, then there exists paths $f, g \in C([0, 1]; G_{0}(A))$ such that $f(0) = g(0) = 1$, $f(1) = x$, and $g(1) = y$. The concatenation of $f$ and $xg$ then is a path from $1$ to $xy$, so $xy \in G_{0}(A)$. In addition, $t \mapsto f(t)^{-1}$ is a path from $1$ to $x^{-1}$, so $x^{-1}\in G_{0}(A)$ as well. Therefore $G_{0}(A)$ is a subgroup of $G(A)$.
Finally, let $x \in G(A)$, then $x^{-1}G_{0}(A)x$ is a connected subset of $G(A)$ containing $1$, so $x^{-1}G_{0}(A)x \subset G_{0}(A)$ and $G_{0}(A) \subset xG_{0}(A)x^{-1}$. Since the above holds for all $x \in G(A)$, $x^{-1}G_{0}(A)x = G_{0}(A)$.
(2): For each $x \in G(A)$, $xG_{0}(A)$ is connected, closed, and open, so it is a connected component by Lemma 5.12.5.$\square$
Proposition 29.3.2.label Let $A$ be a unital Banach algebra and $x \in A$. If there exists $\theta \in [0, 2\pi)$ such that
then $x \in G_{0}(A)$.
Proof. Since there exists a branch of the logarithm $\ell: \complex \setminus e^{i\theta}[0, \infty) \to \complex$, there exists $y \in A$ such that $x = \exp(y)$. In which case, $x \in G_{0}(A)$ by Proposition 29.5.3.$\square$
Definition 29.3.3 (Index Group).label Let $A$ be a unital Banach algebra, then the index group $I(A)$ is the quotient $G(A)/G_{0}(A)$, which is a discrete group.
Proof. By Definition 29.3.1, the components of $G(A)$ are the cosets of $G_{0}(A)$, so the quotient group is discrete.$\square$
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