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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!

C. Dubussy
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reza
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    I guess that the OP asks for "a commutative Banach algebra without characters". If so, this is a reasonable question. – Tomasz Kania Oct 23 '14 at 21:49
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    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? –  Nov 30 '16 at 13:54
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    @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 :) – s.harp Nov 30 '16 at 14:08
  • @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! –  Nov 30 '16 at 14:10

2 Answers2

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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).

Community
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Tomasz Kania
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  • 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? – reza Oct 25 '14 at 14:42
  • Is this set closed under multiplication (convolution)? – Tomasz Kania Oct 25 '14 at 15:33
  • in my idea your answer is right, but unfortunately because i can not find your refrences, i don't find out your proof. – reza Oct 26 '14 at 21:27
  • You only need Dixmier's result. I will sketch it over the weekend for you. – Tomasz Kania Oct 27 '14 at 15:42
  • thanks for your reply and your time – reza Oct 27 '14 at 20:16
  • I wonder if there are any "easy" examples, like $L^p$ spaces or continuous functions. Of course these are all non-examples but something "common" would be nice. Alas, I don't know any "common" algebra that has empty character space : ( – Rudy the Reindeer Oct 29 '14 at 07:58
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    As far as I know from Garth Dales, this is the standard example of an algebra without maximal ideals. – Tomasz Kania Oct 29 '14 at 08:47
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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.

Amol Sasane
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