Let $\mathcal{A}$ be a commutative unital Banach algebra, $\mathcal{B} \subset \mathcal{A}$ a closed unital subalgebra, $\mathcal{I} \subset \mathcal{B}$ a closed ideal.

Is there in general a way to "identify" (soft question, perhaps) the maximal ideal spaces $\Sigma(\mathcal{B})$ and $\Sigma(\mathcal{A}/\mathcal{I})$, in terms of $\Sigma(\mathcal{A})$ and possibly some sort of other data? A couple of examples of the sort of thing I have in mind:

  • If $\mathfrak{X}$ is a Banach space and $M \subset \mathfrak{X}$ a closed subspace, then $M^* \simeq \mathfrak{X}^*/M^\perp$ and $(\mathfrak{X}/M)^* \simeq M^\perp$, where $$ M^\perp = \{f \in \mathfrak{X}^* \mid \forall m \in M: \, f(m) = 0\}. $$
  • In the special case where $\mathcal{A}$ is a $C^*$-algebra, the contravariant equivalence of categories with (compact Hausdorff spaces, continuous maps) implies that $C^*$-subalgebras of $\mathcal{A}$ correspond to quotients of $\Sigma(\mathcal{A})$, and quotients of $\mathcal{A}$ correspond to closed subspaces of $\Sigma(\mathcal{A})$.

Are there some sort of analogous relationships with commutative Banach algebras? An example: Viewing the disc algebra $A(\mathbb{D})$ as a closed subalgebra of $C(\mathbb{T})$, the above analogies might lead us to expect that $\Sigma(A(\mathbb{D}))$ is a quotient of $\Sigma(C(\mathbb{T}))$. But $\Sigma(A(\mathbb{D})) \simeq \overline{\mathbb{D}}$ while $\Sigma(C(\mathbb{T})) \simeq \mathbb{T}$, so it looks like in this case the relationship is a subspace rather than a quotient (and going the other direction).

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    $\begingroup$ There's some relevant stuff in Rickart's General theory of Banach algebras (the U of Iowa library has a copy). E.g.: "THEOREM (3.1.17). Let $\tau$ be a homomorphism of $\mathfrak B$ onto $\mathfrak A$ and let $\mathfrak K$ be the kernel of $\tau$. Then the dual mapping of $\Phi_{\mathfrak A^\infty}$ into $\Phi_{\mathfrak B^\infty}$ takes $\Phi_{\mathfrak A}$ homeomorphically onto $\mathcal h(\mathfrak K)$, the hull in $\Phi_{\mathfrak B}$ of the ideal $\mathfrak K$." $\endgroup$ – Jonas Meyer Jul 5 '12 at 16:11

Placeholder until I can come back with a more thought-out answer.

Take B to be the Jacobson radical of A to see that the natural map from max ideal space of A to that of B need not be injective.

Take B to be the disc algebra and A to be C(T), as you did, to see that said map need not be surjective.

As Jonas has mentioned in his comment, the natural map from max ideal space of A/I to that of A will be injective with closed range.

In the non-unital setting, note that one can have commutative Banach algebras with trivial Jacobson radical which quotient onto radical Banach algebras. The standard example is the Volterra algebra arising as a quotient of the convolution algebra L^1(R_+).


I'm not really sure what you're asking. $\Sigma$ is still a contravariant functor from commutative Banach algebras to compact Hausdorff spaces (it just isn't an equivalence), so from the sequence of morphisms $$B \to A \to A/I$$

you get a sequence of morphisms in the other direction $$\Sigma(A/I) \to \Sigma(A) \to \Sigma(B)$$

but I don't think there's much you can say anything in general about the corresponding morphism $\Sigma(A/I) \to \Sigma(B)$ without more information.

  • $\begingroup$ I voted up because this answer points out that you still get maps the other way, which was not pointed out in the question. However, I do think that more can be said, even if it is (admittedly) a soft question. $\endgroup$ – Jonas Meyer Jul 5 '12 at 16:20
  • $\begingroup$ +1 for noting that the Gelfand rep is a functor from CBAs to Spaces^op, and not just some theorem about the category of C^*-algebras, contrary to the impression one might get from some posts on MO... ;-) $\endgroup$ – user16299 Jul 5 '12 at 23:11
  • $\begingroup$ Contra @JonasMeyer, my gut feeling is that for general CBAs this is about as much as can be said, but I don't have immediate examples to hand. One major difference is that in any C^*-algebra (not just the commutative ones) all ideals have bounded approximate identities. Another is that for commutative C^*-algebras the Gelfand topology coincides with the hull-kernel topology. $\endgroup$ – user16299 Jul 5 '12 at 23:17

Thanks everyone. Here's what I understand so far:

  • Regarding the maximal ideal space of a quotient, one has (in the unital case) the identification $$ \Sigma(\mathcal{A}/I) \simeq \text{hull}(I) = \{\omega \in \Sigma(\mathcal{A}) \mid \omega = 0 \text{ on } I\}. $$
  • Regarding the maximal ideal space of a subalgebra, things aren't as clean. Denote by $E(\mathcal{B}) \subseteq \Sigma(\mathcal{B})$ the subspace of homomorphisms which are extendible to $\mathcal{A}$, and by $\sim$ the equivalence relation on $\Sigma(\mathcal{A})$ induced by restriction to $\mathcal{B}$. Then $$ E(\mathcal{B}) \simeq \Sigma(\mathcal{A})/\sim. $$ Some examples:

(1) $\mathcal{A} = C(\mathbb{T})$, $\mathcal{B} = A(\mathbb{D})$ shows that $E(\mathcal{B})$ can be a proper subspace of $\Sigma(\mathcal{B})$. The multiplicative linear functionals on $\mathcal{B}$ correspond to evaluation at points in the closed disc $\overline{\mathbb{D}}$, but only those corresponding to points in $\mathbb{T}$ are extendible to $\mathcal{A}$.

(2) $\mathcal{B}$ could be the Jacobson radical of $\mathcal{A}$, showing that $\sim$ can be nontrivial.

(3) Another (unital) example where $\sim$ is nontrivial is $\mathcal{A} = C(K)$ and $\mathcal{B} = C(K/\approx)$ where $\approx$ is a (nontrivial) equivalence relation on the compact Hausdorff space $K$. Then $\sim$ is the same as $\approx$, modulo the identification of $K$ with $\Sigma(C(K))$ and $K/\approx$ with $\Sigma(C(K/\approx))$. Forgive my sense of humor, but I can't pass up the opportunity to write $\sim \simeq \approx$ and have it almost mean something.

(4) Let $\mathcal{B} \subseteq A(\mathbb{D})$ be the subalgebra generated by $z^2$, i.e. the functions whose odd Taylor coefficients are all zero. Then $\sim$ is the antipodal equivalence on $\Sigma(\mathcal{A}) \simeq \mathbb{T}$, and $\Sigma(\mathcal{B})$ is the quotient of the disc under the antipodal map. In this example we have both that $E(\mathcal{B})$ is properly contained in $\Sigma(\mathcal{B})$, and that $\sim$ is nontrivial.

Rather elementary considerations, but I hadn't really thought through them before. Thanks for your patience.


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