Let $A=\{f:[0,1]\to M_2(\mathbb{C}): $f continuous and $ f(0)=\begin{pmatrix} f_{11}(0) & 0 \\ 0 & f_{22}(0) \end{pmatrix}\}$ be a $C^*$-algebra with pointwise multiplication, involutions and norm $\|f\|=\sup\limits_{t\in [0,1]}\|f(t)\|_{op}$. I want to determine $\hat{A}=\{[\pi]:\pi$ is a irreducible representation of $A$}, (Here is $\pi\sim \rho :\iff$ there is an unitary operator $V:H_{\pi}\to H_{\rho}$ such that $V\pi(a)=\rho(a)V$ for all $a\in A$).

I want to use the following fact: If $I$ is a closed ideal in a $C^*$-algebra $A$, it is $$\hat{A}=\hat{A}_I\coprod \hat{A}_{A/I}=\widehat{I}\coprod \widehat{A/I},$$ where $\hat{A}_I=\{[\pi]\in \hat{A}: \pi(I)\neq 0\}$ and $\hat{A}_{A/I}=\{[\pi]\in \hat{A}: \pi(I)=0\}$.

Consider $I=\{f\in A: f(0)=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\}$. $I$ is a closed Ideal in $A$, because $I$ is the kernel of the evaluation-homomorphism $\epsilon_0:A\to M_2(\mathbb{C}), f\mapsto f(0)$. Therefore $I$ is a $C^*$-subalgebra of $A$. Then it follows that $I$ is isomorphic to $C_0((0,1],M_2(\mathbb{C}))$ as $C^*$-algebra. Maybe we get $\widehat{A/I}$ now, because we know what $A/I$ is, it is $A/I=A/ \ker(\epsilon_0)\cong Im(\epsilon_0)\cong \mathbb{C}^2$. But here I'm stuck, I'm not sure what $\widehat{A/I}$ should be and I really don't know what $\hat{I}$ is, I assume $\hat{I}$ is homeomorphic to $(0,1]$, but I can't prove it.

I need that because I want to describe the hull-kernel-topology on $\hat{A}$.

I'm not sure if the question is ok for mathoverflow, I'm sorry if not. But I appreciate your help.

edit: is $\widehat{A/I}$ a discrete set with 2 elements?

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    $\begingroup$ Indeed, $\widehat{C_0(X,\mathbb M_n)}$ is homeomorphic to $X$. Do you see a natural map? (This gives you both $\widehat{A/I}$ and $\hat{I}$. It is not hard to describe the topology on $\hat A=\hat I\sqcup\widehat{A/I}$.) $\endgroup$
    – Rasmus
    Sep 28, 2015 at 7:42
  • $\begingroup$ yes, thank you, I remember! therefore $\hat{I}$ must be homeomorphic to $(0,1]$. I know that $\widehat{C_0(X,M_2)}$ equals the Gelfand space of $C_0(X,M_2)$ and the Gelfand space of $C_0(X,M_2)$ is homeomorphic to $X$, the homeomorphism is the point-evaluation-map. But I still don't know what $\widehat{A/I}$ is. It must be a discrete set with 2 elements, but I don't see why... $\endgroup$ Sep 28, 2015 at 9:14
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    $\begingroup$ $\mathbb C^2=C_0(\{a,b\},\mathbb M_1)$. $\endgroup$
    – Rasmus
    Sep 28, 2015 at 9:29
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    $\begingroup$ You're welcome! If everthing's clear now, I encourage you to write a detailed answer! :) $\endgroup$
    – Rasmus
    Sep 28, 2015 at 9:35
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    $\begingroup$ ok, I will do it:) (in a few hours) $\endgroup$ Sep 28, 2015 at 11:17

1 Answer 1


Let $X$ be a compact Hausdorff space, $A$ a unital $C^*$-algebra, first you can prove that $$\gamma:X\times \hat{A}\to \widehat{C_(X,A)},\; (x,[\pi])\mapsto [\pi\circ ev_x]$$ is a bijection, where $ev_x: C(X,A)\to A,\; f\mapsto ev_x(f):=f(x)$ is the evaluation map. Then you can deduce the same for $\widehat{C_0(X,A)}$ and $X\times \hat{A}$, if $X$ is local compact, Hausdorff and if $A$ is an arbitrary $C^*$-algebra.

[Note: $\widehat{C_0(X,A)}$ and $\hat{A}$ are homeomorphic if $\widehat{C_0(X,A)}$ and $\hat{A}$ are endowed with the kern-hull-topology and $X\times \hat{A}$ takes the product topology. ]

Therefore it is $\hat{I}=\widehat{C_0( (0,1], M_2(\mathbb{C}) )}=(0,1]\times \widehat{M_2(\mathbb{C}) }$.
Next, note that all irreducible $*$-representations $\pi:M_2(\mathbb{C})\to L(H_{\pi})$ are equivalent to the irreducible $\ast$-representation $id:M_2(\mathbb{C})\to M_2(\mathbb{C})\cong L(\mathbb{C}^2)$, i.e. $ \widehat{M_2(\mathbb{C}) }=\{[id]\}$. So $\hat{I}$ is basically $(0,1]$.

To obtain $\hat{A}$ we also need $\widehat{A/I}$, because as sets we have the equality $$\hat{A}=\widehat{I}\coprod \widehat{A/I}.$$ It is $\widehat{A/I} =\widehat{ \mathbb{C}\oplus \mathbb{C} }=\hat{\mathbb{C}}\coprod \hat{\mathbb{C}}$, a discrete set with 2 elements.

In summary: $\hat{A}$ has the form $\{\pi_t|t\in (0,1]\}\cup \{\chi_1,\chi_2\}$ with $\pi_t(f)=f(t)$ and $\chi_i(f)=f_{ii}(0)$.

Thanks to Rasmus Bentmann for the help.

  • $\begingroup$ You're welcome! Some details concerning the topology are still missing. Can you work them out? $\endgroup$
    – Rasmus
    Sep 30, 2015 at 19:08
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    $\begingroup$ ok, I promise I will add some details the next days =). $\endgroup$ Sep 30, 2015 at 19:14

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