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Let $B$ and $F$ be compact Hausdorff spaces.

Let $E\to B$ be a fiber bundle with fibre $F$ and structure group $\mathrm{Homeo}(F)$, the group of homeomorphisms of $F$.

I think this induces a fiber bundle $E'$ over $B$ with fiber $C(F,\mathbb C)$, the C*-algebra of continuous functions on $F$, and with structure group $\mathrm{Aut}(C(F,\mathbb C))\cong\mathrm{Homeo}(F)$, the group of *-automorphisms of $C(F,\mathbb C)$.

(To be more explicit about what happens here: my idea is: take a covering of $B$ which trivialises $E$. The transition functions give me a cocycle with values in the structure group $\mathrm{Homeo}(F)$. But, since $\mathrm{Homeo}(F)\cong\mathrm{Aut}(C(F,\mathbb C))$, I get a cocycle with values in $\mathrm{Aut}(C(F,\mathbb C))$, which I'd like to use to glue my new bundle.)

Let $\Gamma(B,E')$ denote the continuous sections of $E'$. I think pointwise operations turn this into a C*-algebra. Since the fiber $C(F,\mathbb C)$ is commutative, $\Gamma(B,E')$ is commutative as well.

Question: What is the spectrum of $\Gamma(B,E')$?

Example: If $E\cong B\times F$ is the trivial bundle, then $E'\cong B\times C(F,\mathbb C)$ and thus $$\Gamma(B,E')\cong C(B,C(F,\mathbb C))\cong C(B\times F,\mathbb C).$$ This suggests that the spectrum of $\Gamma(B,E')$ is actually $E$.

Edit: I posted this question on MO where it was solved in a comment by Anton Deitmar.

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by induce do you mean just take functions into $\mathbb{C}$ fiberwise? or rather, can you be more explicit in that step, i am missing what exactly you do there. It seems like there could be some dumb things you could do. – Sean Tilson Feb 16 '11 at 23:53
@Sean Tilson: I added some clarification. – Rasmus Feb 17 '11 at 7:10
No progress?${}{}$ – t.b. Oct 10 '11 at 21:18
Hey, t. I just added a tag about fiber-bundles tag because the bundes tag was deleted, it seems. Actually I posted the question on MO here. Anton Deitmar made a comment which solves the question. I should have linked to the MO thread earlier. – Rasmus Oct 10 '11 at 21:26
@Rasmus, why don't you answer your question? It is hang in the top of unanswered question of functional analysis, though it already answered. – Norbert Jun 5 '12 at 10:31
up vote 2 down vote accepted

[As requested by Norbert:]

I posted this question on MO where it was solved in a comment by Anton Deitmar.

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