0
$\begingroup$

If $A$ is a Banach algebra then let $\Omega (A)$ denote the character space of $A$. Apparently there exist abelian Banach algebras such that $\Omega (A) = \varnothing$. Also apparently, if $A$ is a (commutative) $C^\ast$ algebra then $\Omega (A) \neq \varnothing$. I am trying to prove this but got stuck.

Let $A \neq 0$ be a commutative $C^\ast$ algebra. Then there exists $a \in A$ with $\|a\|\neq 0$. By the Hahn Banach theorem there exists a continuous linear functional $\tau : A \to \mathbb C$ such that $|\tau (a)|=\|a\|$ and $\|\tau\|\le 1$. That's how it's stated on Wikipedia but of course $|\tau (a)|=\|a\|$ and $\|\tau\|\le 1$ implies $\|\tau\|=1$. Hence $\Omega(A) \neq \varnothing$.

To me this seems to be a correct proof. But in Murphy's book there is something quite different. Could someone help me and enlighten me about what is wrong with my proof above? I am including the text given in the book:

enter image description here

$\endgroup$
1
$\begingroup$

Note that your Hahn-Banach functional $\tau$ will in general not be multiplicative, hence not a character.

$\endgroup$
  • $\begingroup$ Thank you. If it does not follow from Hahn-Banach then why does $r(a) = \|a\|$ imply the existence of a character? $\endgroup$ – Student Jul 3 '14 at 11:53
  • $\begingroup$ Note that for any complex Banach algebra and any $x \in A$ we have $\sigma(x) = \{\gamma(x) \mid \gamma \in \Omega(A)\}$, now use that your $a$ has real spectrum (as it is hermitian) and hence $r(x) = \max\sigma(x)$ or $r(x) = -\min\sigma(x)$. $\endgroup$ – martini Jul 3 '14 at 12:05
  • $\begingroup$ To see the statement about the spectrum, let $\lambda \in \sigma(x)$. Then $x-\lambda$ is not invertible in $\hat A$, hence there is an maximal ideal $I$ of $\hat A$ containing $x-\lambda$. Now restrict the morphism $\hat A \to \hat A/I \cong \mathbb C$ to $A$. $\endgroup$ – martini Jul 3 '14 at 12:19
  • $\begingroup$ Ok. I'm not sure I understand, can you tell me if this is the argument you have in mind: If $r(a) = \max \sigma (a) > 0$ then $\{\gamma (x) \mid \gamma \in \Omega (A)\}\neq \varnothing$ and therefore $\Omega (A) \neq \varnothing$? $\endgroup$ – Student Jul 4 '14 at 9:33
  • $\begingroup$ Exactly.${}{}{}$ $\endgroup$ – martini Jul 4 '14 at 10:23

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.