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I would like to ask specialists in C*-algebras if the following variant of the Stone-Weierstrass theorem is true.

Suppose $A$ is a C*-algebra and $C$ is its center. Since $C$ is a commutative C*-algebra, there exists a compact space $T$ such that $C$ is isomorphic to the algebra $C(T)$ of continuous functions on $T$. Does this mean that there exists a C*-algebra $B$ such that

1) $A$ is isomorphic to a closed subalgebra in the algebra $C(T,B)$ of continuous mappings $f:T\to B$ with the pointwise algebraic operations (and the topology of uniform convergence on $T$), and

2) this isomorphism turns $C$ into the algebra of scalar mappings, i.e. the mappings of the form $f(x)=\lambda(x)\cdot 1_B$, where $1_B$ is the identity in $B$, and $\lambda(x)$ $\in$ $\mathbb{C}$ for all $x\in T$.

EDIT 21-03-12: All the C*-algebras here are supposed to be unital, excuse me for not mentioning this from the very beginning!

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3 Answers

up vote 1 down vote accepted

EDIT 20-03-12 It seems from the recent answers of Douglas Somerset and Ulrich Pennig that what I claim below is false, and so this answer should be "dis-accepted".


I think (although I admit I don't know the details) that the answer to both questions is yes, by a theorem of Dauns and Hoffman. According to the version quoted in the article

T. Becker, A few remarks on the Dauns-Hofmann theorems for $C^\ast$-algebras. Archiv der Mathematik 43 (1984) no. 3, 265-269 [Math Review]

$A$ can be realized as the algebra of continuous sections of some kind of continuous $C^\ast$-algebra-bundle with base space $T$.

However, since I am not a specialist, I may have misread or misunderstood.

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I am also under the impression that one should think of a C* algebra as some kind of continuous bundle over Spec of its center. It is not clear to me whether this implies the statement in the original question, however? –  Theo Johnson-Freyd Dec 21 '11 at 23:43
    
Theo: firstly, the devil is in the details (Dauns and Hofmann were openly trying to get bundle representations for C*-algebras and Banach algebras, but my understanding is that more than "soft" or "categorical" methods are needed en route). –  Yemon Choi Dec 22 '11 at 0:04
    
... Secondly, it's not clear to me either, if this bundle realization is enough to answer the original questions. But I thought I would mention the crucial result and give a link in case it helps. (Certainly both 1 and 2 are true for certain kinds of C*-algebra, but right now I'm not sure if this needs certain conditions on the topology of the primitive ideal space) –  Yemon Choi Dec 22 '11 at 0:06
    
Excuse me, I actually meant one question: 1) and 2) are just two conditions of one statement (I have now put "and" between them). But I would be satisfied if B is replaced by a C*-algebra bundle. –  Sergei Akbarov Dec 22 '11 at 0:23
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The statement of the Dauns-Hofmann theorem is actually too weak to get bundles of $C^*$-algebras. For completeness, let me state it:

Let $A$ be a $C^*$-algebra. For each $P \in Prim(A)$, let $\pi_P \colon A \to A/P$ be the quotient map. Then there is an isomorphism $\phi$ of $C_b(Prim(A))$ onto the center $ZM(A)$ of the multiplier algebra $M(A)$ such that for all $f \in C_b(Prim(A))$ and $a \in A$ $$ \pi_P(\phi(f)a) = f(P)\pi_P(a) $$ for every $P \in Prim(A)$. Usually one writes $f \cdot a = \phi(f)a$.

So, the best you could hope for is some kind of sheaf of $C^*$-algebras over the primitive ideal space. Getting local triviality in general is kind of hopeless, I think. A reading recommendation for these matters would be the book "Morita Equivalence and Continuous-Trace $C^*$-algebras" by Raeburn and Williams.

For continuous trace $C^*$-algebras things are quite different. These are all Morita equivalent (or stably isomorphic) to sections in a bundle of compact operators!

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Ulrich, the Yemon Choi reference led me to a book by M.Dupre and R.Gillette "Banach bundles, Banach modules and automorphisms of C*-algebras". As far as I understand, from their Theorem 2.4 (at p.40, see also the discussion at pp.38-39) it follows that if C is a closed subalgebra in the center of a (unital) C*-algebra A, then A is isomorphic to the algebra of sections of a C*-algebra bundle over the spectrum T of C. I didn't understand, whether your words contradict to what they write... Just in case: everywhere I speak about unital algebras, of course. –  Sergei Akbarov Mar 21 '12 at 9:21
    
I think there is no contradiction. It is just that the term "bundle" for me always implies some sort of local triviality, whereas the term "bundle" used by Dupre does not! In fact, Banach bundles as defined by Fell and others avoid local triviality. I would call this a field of C*-algebras instead of a bundle. –  Ulrich Pennig Mar 21 '12 at 15:34
    
btw. for the definition of a Banach bundle see http://books.google.de/books?id=nCCNodGa6UkC&lpg=PR6&dq=C*%20algebra%20bundle%2‌​0Fell&hl=de&pg=PA1#v=onepage&q=C*%20algebra%20bundle%20Fell&f=false –  Ulrich Pennig Mar 21 '12 at 15:35
    
Ulrich, if there's no contradiction, I won't change anything here, because what Dupre and Gillette write seems to be sufficient for me (and it was Yemon Choi, who gave me the first reference). Thank you for the comments anyway! –  Sergei Akbarov Mar 21 '12 at 17:18
    
Sure. No problem! –  Ulrich Pennig Mar 21 '12 at 18:25
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Three thoughts on this. The first is that $A$ probably has to be assumed unital to guarantee that $T$ is compact.

Assuming then that $A$ is unital, each point $t\in T$ corresponds to a maximal ideal $M_t$ of $C$ which generates a closed two-sided ideal $G_t$ in $A$. The ideals $\{G_t: t\in T\}$ are called the Glimm ideals (after James Glimm who used them in the case when $A$ is a von Neumann algebra). For each element $a\in A$, the mapping $t\mapsto \Vert a+G_t\Vert$ is upper semi-continuous but not in general continuous. Indeed these norm funcions are all continuous if and only if the 'complete regularisation' map from the primitive ideal space of $A$ with the hull kernel topology to $T$ is an open map (R-Y Lee, 1970s). The second thought, therefore, is that a necessary condition for the answer to the question to be yes is that the complete regularisation map should be open.

Even when the complete regularisation map is open, I expect that one can find examples where the answer to the question is no, although no such example comes to mind just now [in fact, see E. Kirchberg, S.Wassermann, Operations on continuous bundles of C*-algebras, Math. Ann. 303 (1995), 677-697]. The third thought, however, is that Blanchard showed that if $A$ is separable and exact and the complete regularisation map is open then such a $B$ can be found (E. Blanchard, Subtriviality of continuous fields of nuclear C*-algebras, J. Reine Angew. Math. 489 (1997), 133-149).

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Of course, I am speaking about unital algebras (excuse me for not clarifying this from the very beginning). Douglas, how is this connected with what M.Dupre and R.Gillette write in "Banach bundles, Banach modules and automorphisms of C*-algebras"? –  Sergei Akbarov Mar 21 '12 at 9:24
    
One can do a lot in this area without all the baggage of Banach bundles. The simplest approach is that of the $C_0(X)$-algebra where one simply assumes a continuous map $\phi$ from the primitive ideal space of $A$ to a locally compact Hausdorff space $X$. Then $A$ 'fibres' as an algebra of upper semi-continuous cross-sections over $X$ with the cross-sections taking values in the fibre algebras $A_x$, where $A_x=A/J_x$ (where $J_x$ is the kernel of the primitive ideals of $A$ which $\phi$ maps onto the point $x\in X$). The cross-sections are continuous iff $\phi$ is open. –  Douglas Somerset Mar 21 '12 at 21:49
    
Douglas, I hope it will be proper if I contact you (and Ulrich) by e-mail, because the more I read about this, the more questions occur to me. :) –  Sergei Akbarov Mar 28 '12 at 19:24
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