11
$\begingroup$

Let $\mathcal{A}$ be a non (necessarily) unital commutative Banach algebra, and let $$ M_{\mathcal{A}} = \{ \phi:\mathcal{A} \to \mathbb{C} : \phi \mbox{ is multiplicative and not trivial}\} $$ and $$ \mathrm{Max}(\mathcal{A})=\{ I \lhd \mathcal{A} : I \mbox{ maximal} \}.$$ If $\mathcal{A}$ is unital, it is well known that there is a bijection between $M_{\mathcal{A}}$ and $\mathrm{Max}(\mathcal{A})$ sending each functional to its kernel (the inverse is given by the quotient and the Gelfand-Mazur theorem).

My question is,

is this still a bijection in the non-unital case?

I'm aware that if $\mathcal{A}$ is a commutative C*-algebra it is still a bijection. Also that the restriction gives a bijection from $M_{\tilde{\mathcal{A}}} \setminus \{ \pi:\tilde{\mathcal{A}} \to \mathbb{C} \}$ to $M_{\mathcal{A}}$; but this fact don't seem enough to conclude the result. I haven't been able to find a source for this.

Thanks in advance.

$\endgroup$
  • 3
    $\begingroup$ The correspondence fails when $\mathcal A$ is non-commutative, even if it is a C$^*$-algebra (canonical example: $B(H)$ has no characters but it has one unique---so maximal---ideal). $\endgroup$ – Martin Argerami Jul 13 '16 at 1:24
  • $\begingroup$ I know about the non-commutative case, I'm asking about the non-unital but commutative case. I realized that I've failed to describe that in my original question, so I've edited it to reflect this. $\endgroup$ – sjvega Jul 13 '16 at 15:14
7
$\begingroup$

The kernels of nonzero homomorphisms to $\mathbb C$ are modular ideals, terminology that might help you find more references.

Without any further restriction on the algebras, using the zero product is a way to provide trivial counterexamples. E.g., take $\mathbb C$ with the $0$ product, which has maximal ideal $\{0\}$ and no nonzero homomorphisms to $\mathbb C$.

Googling led me to the following maybe more interesting example, Example 1.3 in this Feinstein and Somerset article: Take $C[0,1]$ with its usual Banach space structure but with multiplication $(f\diamond g)(t) = f(t)g(t)t$. Then the ideal of functions vanishing at $0$ is maximal but not the kernel of a nonzero homomorphism to $\mathbb C$.

$\endgroup$
5
$\begingroup$

Linear-multiplicative functionals (aka characters) on complex Banach algebras are automatically continuous, so their kernels are closed. (You will find a slick proof of this fact on p. 181 of Allan's and Dales' Introduction to Banach Spaces and Algebras.) However, in the non-unital case it may well happen that a maximal ideal is dense.

The right notion to look at is the notion of a maximal modular ideal. Such ideals are bijectively associated to (kernels of) characters.

You may also be interested in this thread.

$\endgroup$

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.