3
$\begingroup$

Can one give an example of an abelian Banach algebra with empty character space? Such algebra must be necessarily non-unital.

I couldn't find any examples of such algebras.

Thanks!

$\endgroup$
4
  • 2
    $\begingroup$ I guess that the OP asks for "a commutative Banach algebra without characters". If so, this is a reasonable question. $\endgroup$ Oct 23, 2014 at 21:49
  • 2
    $\begingroup$ the edit history of this question is very strange, I'm not a CA export but it looks as if the question has been changed to a new question, in place of the OP. Can someone with greater experience please assist? $\endgroup$
    – user284001
    Nov 30, 2016 at 13:54
  • 1
    $\begingroup$ @Bacon it appears the OP has decided to edit the question to an entirely different question about complex analysis two years after it was originally asked :) $\endgroup$
    – s.harp
    Nov 30, 2016 at 14:08
  • $\begingroup$ @s.harp That was my initial thought, but couldn't be absolutely clear as I'm not a CA expert. I was worried that if I tried to revert as C.Dubussy did I would be in trouble! Many thanks! $\endgroup$
    – user284001
    Nov 30, 2016 at 14:10

2 Answers 2

6
$\begingroup$

The Volterra algebra $V$ is an example of a commutative Banach algebra without maximal ideals (hence with empty character space). See also Definition 4.7.38 in

H. G. Dales, Banach algebras and automatic continuity, London Math. Soc. Monographs, Volume 24, Clarendon Press, Oxford, 2000.

Okay, let me prove this claim. This relies on three facts:

The algebra $V$ has a bounded approximate identity, e.g. $(n\cdot \mathbf{1}_{\big[0, \tfrac{1}{n}\big]})_{n=1}^\infty$, hence by the Cohen factorisation theorem, $V = V^2$. Consequently, all maximal ideals of $V$ (if exist) are closed.

Now apply a result of Dixmier which tells you that no prime ideal of $V$ is closed. Of course, maximal ideals are prime so, the conclusion follows. You will find the proof of Dixmier's result in the above-mentioned book by Dales (Theorem 4.7.58).

$\endgroup$
7
  • $\begingroup$ Thanks for your answer But I think there is a maximal ideal in the Volterra algebra For example, suppose set M={f \in L^1([0,1]) : f(1)=0 }. Is not a maximal ideal in the Volterra algebra? $\endgroup$
    – reza
    Oct 25, 2014 at 14:42
  • $\begingroup$ Is this set closed under multiplication (convolution)? $\endgroup$ Oct 25, 2014 at 15:33
  • $\begingroup$ in my idea your answer is right, but unfortunately because i can not find your refrences, i don't find out your proof. $\endgroup$
    – reza
    Oct 26, 2014 at 21:27
  • $\begingroup$ You only need Dixmier's result. I will sketch it over the weekend for you. $\endgroup$ Oct 27, 2014 at 15:42
  • 1
    $\begingroup$ As far as I know from Garth Dales, this is the standard example of an algebra without maximal ideals. $\endgroup$ Oct 29, 2014 at 8:47
0
$\begingroup$

A related comment: I think I have an example of a commutative unital complex algebra $A$ without any nontrivial complex homomorphisms. Take $A$ to be all rational functions with complex coefficients, that is $$ A= \{[p/q]: p,q \text{ complex polynomials and }q \text{ not identically } 0\}. $$ Here $[p/q]$ denotes the equivalence class of $(p,q)$ under the relation $(p,q)\sim(r,s)$ if $ps=qr$. Then $A$ is a commutative unital complex algebra (which is also a field, and the only maximal ideal is $0$). But there is no nontrivial complex homomorphism $\varphi:A \rightarrow \mathbb{C}$, because if $\varphi(z)=\alpha$, then $$ \varphi\Big(\frac{1}{z-\alpha}\Big)=\frac{1}{\varphi(z-\alpha)}=\frac{1}{0}, $$ a contradiction.

$\endgroup$
2
  • $\begingroup$ what is the norm ?... $\endgroup$
    – reuns
    Nov 30, 2016 at 14:49
  • 1
    $\begingroup$ This is an infinite-dimensional field, hence it is not a Banach algebra under any norm. $\endgroup$ Nov 30, 2016 at 22:22

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.