1
$\begingroup$

I am trying to understand the proof of the following theorem:

Let $A$ be a unital Banach algebra generated by $1$ and $a$. Then $A$ is abelian and the map $\widehat{a}:\Omega(A) \to \sigma (a), \tau \mapsto \tau (a)$ is a homeomorphism.

The beginning of the proof is: It is clear that $A$ is abelian and that $\widehat{a}$ is a continuous bijection.

But it is not clear to me. To show that it is injective wouldn't the characters have to separate points?

$\endgroup$
0
$\begingroup$

First, observe that an element $\tau$ of $\Omega(A)$ is necessarily continuous with operator norm $\|\tau\| \leq 1$. Indeed, let $x \in A$. Since $\tau(x - \tau(x)1) = 0$, the element $x - \tau(x)1$ lies in the maximal ideal $\ker\tau$, and hence cannot be invertible. In other words, $\tau(x)$ is in the spectrum of $x$, so that $|\tau(x)| \leq \|x\|$, and hence $\|\tau\| \leq 1$ — otherwise, one could use the Neumann identity to construct $(x - \tau(x) 1)^{-1}$. In fact, $\|\tau\|=1$, since $\tau(1) = 1$.

Not, let $\sigma$, $\tau \in \Omega(A)$ — a priori, then, $\sigma(1) = 1 = \tau(1)$ — and suppose that $\hat{a}(\sigma) = \hat{a}(\tau)$, which is just to say that $\sigma(a) = \tau(a)$. Because $\sigma$ and $\tau$ agree on the generators $1$ and $a$ of $A$, they agree on all polynomials in $1$ and $a$. Since such polynomials are dense in $A$, precisely by definition of $A$, it follows by continuity of $\sigma$ and $\tau$ that they agree everywhere, i.e., that $\sigma = \tau$.

$\endgroup$
  • $\begingroup$ For surjectivity is it okay to argue like this: Let $\lambda \in \sigma (a)$ be arbitrary and let $\tau \in \Omega(A)$ be arbitrary. Then $\tau' = {\lambda \over \tau (a)}\tau$ is a character and $\tau' (a) = \lambda$ hence $\widehat{a}$ is surjective. $\endgroup$ – user167889 Nov 21 '14 at 1:16
  • $\begingroup$ That construction gives you a character iff $\lambda = \tau(a)$, because you need $\tau^\prime(1)=1$. $\endgroup$ – Branimir Ćaćić Nov 21 '14 at 16:51
  • $\begingroup$ Suppose that $\lambda \in \sigma(a)$, so that $a - \lambda$ isn't invertible. Since $A$ is commutative, $\langle a-\lambda \rangle := A(a-\lambda) = (a-\lambda)A$ is a two-sided ideal, and since $a-\lambda$ isn't invertible, $\langle a-\lambda \rangle$ is proper, for otherwise, we could write $1 = b(a-\lambda) = (a-\lambda)b$ for some $b \in A$. Hence, by Zorn's lemma, $\langle a-\lambda \rangle$ is contained in a maximal two-sided ideal $\mathfrak{m}$, and one can then check that the canonical map $\tau : A \to A/\mathfrak{m} \cong \mathbb{C}$ is the desired character. $\endgroup$ – Branimir Ćaćić Nov 21 '14 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy