A common lifting of a positive element?

Question

Suppose $A$ is a C$^{*}$-algebra and that we have finite-dimensional irreducible representations $\varphi_{1},\ldots,\varphi_{n}$ of $A$. It follows from the question C$^{*}$-algebra acting irreducibly on the finite-dimensional space $\mathbb{C}^{n}$ must be $M_{n}(\mathbb{C})$ that each $\varphi_{i}$ is surjective. Suppose we are given positive elements $b_{i}\in\varphi_{i}(A)$ for $1\leq i\leq n$. Due to surjectivity, we can find postive elements $a_{1},\ldots,a_{n}\in A$ such that $\varphi_{i}(a_{i})=b_{i}$. My question is:

Is it possible to find a common positive lifting of the $b_{i}'s$? I.e., does there exist a positive $z\in A$ such that $\varphi_{i}(z)=b_{i}$ for all $1\leq i\leq n$?

I am reading an old paper of Fell's and, if this is not possible in general, I was wondering how he accomplishes what I've underlined:

Answer 1

As Martin points out, the question obviously has a negative answer if two of the representations are equivalent.

However, Fell considers non-equivalent irreducible representations, and the question has a positive answer if all representations are pairwise non-equivalent.

Note that by Corollary 4.1.10 in Dixmier's book, two finite dimensional irreducible representations are equivalent if and only if they have the same kernel. Hence the positive answer follows from the following more general result. The proof uses basic theory of primitive ideal spaces, see for instance Sections 2.9, 2.10 and 2.11 in Dixmier's book.

Lemma. Let $A, B_1,\dots, B_n$ be $C^\ast$-algebras with $B_i$ simple for $i=1,\dots, n$, and suppose that $\pi_i \colon A \to B_i$ are surjective $\ast$-homomorphisms for $i=1,\dots, n$ such that $\mathrm{ker}\, \pi_i \neq \mathrm{ker} \, \pi_j$ for $i\neq j$. Then the $\ast$-homomorphism $\pi := \bigoplus_{i=1}^n \pi_i \colon A \to \bigoplus_{i=1}^n B_i$ is surjective.

Proof. Let $I_i = \mathrm{ker}\, \pi_i$ for $i=1,\dots, n$ which is a maximal two-sided, closed ideal in $A$. Hence $\{ I_i\}$ are distinct, closed (one-point) subset of the primitive ideal space $\mathrm{Prim}\, A$ for $i=1,\dots, n$, and the two-sided, closed ideal $I_i$ corresponds to the open subset $U_i := \mathrm{Prim}\, A \setminus \{I_i\}$ in $\mathrm{Prim}\, A$.

Let $I : = \bigcap_{i=1}^n I_i$. The open subset in $\mathrm{Prim}\, A$ corresponding to $I$ is $\bigcap_{i=1} U_i = \mathrm{Prim}\, A \setminus\{ I_1,\dots, I_n\}$. Hence $A/I$ has a primitive ideal space (canonically) homeomorphic to $\{ I_1,\dots, I_n\} \subseteq \mathrm{Prim}\, A$ in the subspace topology. As each $\{I_i\}$ is closed in $\mathrm{Prim}\, A$, it follows that $\mathrm{Prim}\, A/I$ is a discrete $n$-point space. Hence we obtain canonical isomorphisms \begin{equation} A/I \cong \bigoplus_{i=1}^n (A/I)/(I_i/I) \cong \bigoplus_{i=1}^n A/I_i \cong \bigoplus_{i=1}^n B_i. \end{equation} That $\pi$ is surjective is an easy consequence. QED

Answer 2

No, it's not possible. Take $A=M_2(\mathbb C)$, $\varphi_1(a)=a$, $\varphi_2(a)=uau^*$, where $$ U=\begin{bmatrix} 0&1\\1&0\end{bmatrix}. $$ Both $\varphi_1$ and $\varphi_2$ are bijective.

Take $$b_1=I,\ \ b_2=\begin{bmatrix} 1&0\\0&2\end{bmatrix}.$$ Thus if $\varphi_1(z)=I$, then $z=I$, while $\varphi_2(z)=b_2$ implies $$z=\begin{bmatrix} 2&0\\0&1\end{bmatrix}.$$

As for what is Fell doing, I cannot tell without knowing at least what $A$ is, which your text does not say.