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Given a $C^*$-algebra,

$A=${$f:[0,1]\rightarrow M_2(\mathbb C)$ where $f(0),f(1) $ are diagonal }

which is isomorphic to $C(\mathbb T)\rtimes\mathbb Z_2$,

How can I determine its continuous field of $C^*$-algebras?

I know the center is given by

$Z(A)=${$f : f(e^{i\pi t})=f(e^{-i\pi t})$} where $t\in [0,1]$

The ideal in $Z(A)$ is given by {$f : f(e^{i\pi t})=f(e^{-i\pi t})$=0} where $t\in [0,1]$

But I don't know how to go further from here.

Thank you.

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As already mentioned, $$ (\spadesuit) \qquad C(\Bbb{T}) \rtimes_{\alpha} \Bbb{Z}_{2} \cong \{ f \in C([0,1] \to {\text{M}_{2}}(\Bbb{C})) \mid \text{$ f(0) $ and $ f(1) $ are diagonal} \}. $$ Hence, by the definition of a continuous field of $ C^{*} $-algebras, $ C(\Bbb{T}) \rtimes_{\alpha} \Bbb{Z}_{2} $ is a continuous field of $ C^{*} $-algebras over the compact Hausdorff space $ [0,1] $ with the following structure:

  • The fibers over $ 0 $ and $ 1 $ are the $ C^{*} $-subalgebra of $ {\text{M}_{2}}(\Bbb{C}) $ consisting of all diagonal matrices, which is isomorphic to $ \Bbb{C} \oplus \Bbb{C} $.
  • The fibers over $ (0,1) $ are $ {\text{M}_{2}}(\Bbb{C}) $.
  • The generating $ * $-subalgebra of cross-sections is simply the set on the right-hand side of the relation $ (\spadesuit) $.

This agrees with a $ 1976 $ result by Ru-Ying Lee, which states that a $ C^{*} $-algebra $ A $ is a continuous field over a locally compact Hausdorff space $ X $ if and only if there exists a continuous open map from the primitive-ideal space of $ A $, $ \text{Prim}(A) $, onto $ X $.

As $ (\Bbb{Z}_{2},\Bbb{T},\alpha) $ is a second-countable transformation group, certain results in the theory of transformation-group $ C^{*} $-algebras show that $ \text{Prim}(C(\Bbb{T}) \rtimes_{\alpha} \Bbb{Z}_{2}) $ is homeomorphic to a closed and bounded interval of $ \Bbb{R} $.

Of course, any $ C^{*} $-algebra is a continuous field over a single point, but this is uninteresting.

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  • $\begingroup$ It’s to be assumed that a continuous field of $ C^{*} $-algebras has no zero-fibers. $\endgroup$ – Berrick Caleb Fillmore May 6 '15 at 4:55

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