Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a non-abelian C*-algebra $A$. I am wondering what are the possible abelian sub-C*-algebras of $A$. Let $K$ be the spectrum of $A$. Does $A$ contain an isomorphic copy (as a Banach space) of the space $C(K)$? (if $A$ is abelian and unital, then thay are of course isometrically indistinguishable).

If the answer for my question is negative, let $L$ be a compact metric space. Must $A$ contain $C(L)$?

share|cite|improve this question

Let $ (A,\| \cdot \|_{A}) $ be a non-commutative C$ ^{\ast} $-algebra. We shall first show that it is relatively easy to construct commutative C$ ^{\ast} $-subalgebras of $ A $.

Pick a non-zero element $ x \in A $. Then $ x^{*} x $ is a non-zero self-adjoint element of $ A $. Why $ x^{*} x $ should be non-zero can be seen easily from the C$ ^{\ast} $-identity $ \| x^{*} x \|_{A} = \| x \|_{A}^{2} $. If we had $ x^{*} x = 0_{A} $, then the C$ ^{\ast} $-identity would yield $ x = 0_{A} $, which is a contradiction.

Now, consider $$ C(x^{*} x) ~ \stackrel{\text{def}}{=} ~ \overline{ \{ p(x^{*} x) \in A ~|~ \text{$ p \in \mathbb{C}[X] $ and $ p $ has no constant term} \} }^{\| \cdot \|_{A}}. $$ This is a non-trivial commutative C$ ^{\ast} $-subalgebra of $ A $. By the Commutative Gelfand-Naimark Theorem, it is isometrically $ ^{\ast} $-isomorphic to $ {C_{0}}(X) $ for some locally compact (though not necessarily compact) Hausdorff space $ X $.

If $ A $ happens to be unital, then we can also consider $$ C(1_{A},x^{*} x) ~ \stackrel{\text{def}}{=} ~ \overline{ \{ p(x^{*} x) \in A ~|~ p \in \mathbb{C}[X] \} }^{\| \cdot \|_{A}}, $$ which is a unital commutative C$ ^{\ast} $-subalgebra of $ A $. This time, however, we can find a compact Hausdorff space $ K $ such that $ C(1_{A},x^{*} x) = C(K) $.

For general C$ ^{\ast} $-algebras, we do not have a good notion of ‘spectrum’. Ideally, we would like to define the spectrum in a way that allows us to recreate the success that Gelfand Duality has had in the commutative case, but up till now, there have been no clear breakthroughs. One possible approach is the topos of spectral presheaves, which is described in Andreas Döring’s notes entitled Some Steps Towards Noncommutative Gel’fand Duality, but it is still in its experimental stages.

However, we have the following definition of ‘spectrum’ for general C$ ^{\ast} $-algebras that is suited to the theory of C$ ^{\ast} $-algebras associated with locally compact topological groups.

Definition The spectrum of a C$ ^{\ast} $-algebra $ A $ is defined as the set $ \hat{A} $ of unitary equivalence classes of irreducible $ ^{\ast} $-representations of $ A $. We put on $ \hat{A} $ the pre-image topology that is induced by the natural map $ \text{k}: \hat{A} \to \text{Prim}(A) $, where $ \text{Prim}(A) $ is the set of primitive ideals of $ A $ equipped with the Jacobson topology.

It seems to be a difficult question of whether or not there is a Banach subspace of $ A $ that is isomorphic to $ C(\hat{A}) $. I do have the following example in mind, however. If $ A = K(\mathcal{H}) $, where $ K(\mathcal{H}) $ is the non-unital and non-commutative C$ ^{\ast} $-algebra of compact operators on a non-trivial Hilbert space $ \mathcal{H} $, then $ \hat{A} $ is just the one-point topological space. Hence, we simply have $ C(\hat{A}) \cong \mathbb{C} $, which is clearly isomorphic to any one-dimensional Banach subspace of $ A $.

In general, $ \hat{A} $ is not compact, so instead of $ C(\hat{A}) $, we should be looking at $ {C_{0}}(\hat{A}) $ or $ {C_{b}}(\hat{A}) $ instead, as the supremum norm on these function spaces is well-defined. However, I know very little about these objects, so I have practically nothing more to offer beyond this point. Any contribution from other members of the community would be welcome! :)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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