# A continuous field of C* algebra, $C(\mathbb T)\rtimes\mathbb Z_2$

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.

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.

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