Let $\mathcal{A}$ be a $C^*$-algebra generated by a single element $a \in \mathcal{A}$. Suppose that it is also generated by another element $b \neq a$. Consider a subalgebra $\tilde{\mathcal{A}}$ of a matrix algebra $M_2(\mathcal{A})$ generated by an element of the form

$$

\begin{pmatrix} a & 0 \\ 0 & b\end{pmatrix}.

$$

It is clear that if the map $a \mapsto b$ induces an automorphism, then $\tilde{\mathcal{A}} \cong \mathcal{A}$. Is this a necessary condition?

When is it true that $\tilde{\mathcal{A}} \cong\mathcal{A} \oplus \mathcal{A}$? What are the other options for $\tilde{\mathcal{A}}$?

A bit more generally, let $\mathcal{A}$ be a $C^*$-algebra generated by its elements $a$ and $\{c_i\}_{i\in \mathbb{N}}$. Suppose that $\mathcal{A}$ is also generated by $b\neq a$ and $\{c_i\}_{i\in \mathbb{N}}$. Now let $\tilde{\mathcal{A}}$ be a subalgebra of $M_2(\mathcal{A})$ generated by elements

$$

\begin{pmatrix} a & 0 \\ 0 & b\end{pmatrix} \quad \text{and} \quad \begin{pmatrix} c_i & 0 \\ 0 & c_i\end{pmatrix}, \quad i \in \mathbb{N}.

$$

What can we say about $\tilde{\mathcal{A}}$ then?

My hypothesis is that, at least for $\mathcal{A}$ commutative or $\mathcal{A} \subset M_n(\mathbb{C})$ where all the possible automorphisms and subalgebras are known, it is true that either $\tilde{\mathcal{A}} \cong\mathcal{A} \oplus \mathcal{A}$ or $\tilde{\mathcal{A}} \cong\mathcal{A}$, although my proof lacks details at the moment. Even in that case I can't quite tell how can we differ definitively the two cases by looking at the generators.

## Best Answer

This is more of a long comment addressing the commutative case and where no additional generators $\{c_i\}$ are involved. Note that as both $a$ and $b$ generates $A$, we must have $\sigma(a) \simeq \text{Spec}(A) \simeq \sigma(b)$. Since $a$ and $b$ are both normal, so is $x = \begin{pmatrix} a & 0\\0 & b\end{pmatrix}$, and one easily sees that $\sigma(x) = \sigma(a) \cup \sigma(b)$. As $\tilde{A}$ is generated by $x$, we have $\tilde{A} \simeq A$ iff $\sigma(a) \cup \sigma(b) \simeq \sigma(a)$. Similarly, $\tilde{A} \simeq A \oplus A$ iff $\sigma(a) \cup \sigma(b) \simeq \sigma(a) \times \{1, 2\}$.

Itâ€™s not hard to see that $a \mapsto b$ inducing an automorphism is equivalent to $\sigma(a) = \sigma(b)$. Of course, if $\sigma(a) = \sigma(b)$, then $\sigma(a) \cup \sigma(b) = \sigma(a)$, so $\tilde{A} \simeq A$. The converse is not true. As an example, choose $a = x$ and $b = 1 + x$ in $C[0, 1]$. Both clearly generates $C[0, 1]$. We have $\sigma(a) \cup \sigma(b) = [0, 2] \simeq [0, 1] = \sigma(a)$, so $\tilde{A} \simeq A$, but $\sigma(a) = [0, 1] \neq [1, 2] = \sigma(b)$, so $a \mapsto b$ does not induce an automorphism. Perhaps surprisingly, this also tells you the answer is true if $A$ is finite-dimensional. Since then $\sigma(a)$ and $\sigma(b)$ are both finite, and also since $|\sigma(a)| = |\sigma(b)|$ as $\sigma(a) \simeq \sigma(b)$, so if $a \mapsto b$ does not induce an automorphism, i.e., $\sigma(a) \neq \sigma(b)$, then $|\sigma(a) \cup \sigma(b)| > |\sigma(a)|$, so we must have $\sigma(a) \cup \sigma(a)$ is not homeomorphic to $\sigma(a)$, i.e., $\tilde{A}$ is not isomorphic to $A$.

It also can the case that neither $\tilde{A} \simeq A$ nor $\tilde{A} \simeq A \oplus A$. This can even happen when $A$ is both commutative and finite-dimensional. For example, in $A = \mathbb{C}^2$, let $a = (1, 2)$ and $b = (2, 3)$. Then both $a$ and $b$ generates $A$. But $\sigma(a) \cup \sigma(b) = \{1, 2, 3\}$ is not homeomorphic to either $\sigma(a) = \{1, 2\}$ or $\sigma(a) \times \{1, 2\} = \{1, 2\}^2$, so $\tilde{A}$ is not isomorphic to either $A$ or $A \oplus A$.

The following addresses the case where $A = M_n(\mathbb{C})$ and where we may have additional generators $\{c_i\}_{i \in I}$. Assume, WLOG, that all generators ($a$, $b$, and $c_i$) are contractions. (Replace $a$ by $\frac{a}{\max\{\|a\|,\|b\|\}}$, $b$ by $\frac{b}{\max\{\|a\|,\|b\|\}}$, and $c_i$ by $\frac{c_i}{\|c_i\|}$ if necessary, which will not alter the algebra $\tilde{A}$.) Let $B$ be the universal $C^\ast$ algebra generated by contractions $x_0$ and $\{x_i\}_{i \in I}$. Then we have representations,

$$\begin{split} &\pi_1: B \to M_n(\mathbb{C}), \; \pi_1(x_0) = a, \; \pi_1(x_i) = c_i \, \text{for} \, i \in I\\ &\pi_2: B \to M_n(\mathbb{C}), \; \pi_2(x_0) = b, \; \pi_2(x_i) = c_i \, \text{for} \, i \in I\\ &\pi: B \to M_{2n}(\mathbb{C}), \; \pi(x_0) = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}, \; \pi(x_i) = \begin{pmatrix} c_i & 0 \\ 0 & c_i \end{pmatrix} \, \text{for} \, i \in I \end{split}$$

Clearly, $\pi_1$ and $\pi_2$ are irreducible representations, $\pi = \pi_1 \oplus \pi_2$, and $\pi(B) = \tilde{A}$. Since every finite-dimensional $C^\ast$ algebra is a von Neumann algebra, applying standard representation theoretic techniques and using the double commutant theorem, we see that we indeed have $\tilde{A} = \pi(B) \simeq A$, when $\pi_1$ and $\pi_2$ are unitarily equivalent; or $\tilde{A} = \pi(B) \simeq A \oplus A$, when $\pi_1$ and $\pi_2$ are not unitarily equivalent. As all automorphisms of $M_n(\mathbb{C})$ are inner, $\pi_1$ and $\pi_2$ are unitarily equivalent iff there exists an automorphism of $M_n(\mathbb{C})$ that sends $a$ to $b$ and $c_i$ to itself for all $i$. Thus, in case $A = M_n(\mathbb{C})$, we indeed have:

$\tilde{A} \simeq A$ iff $a \mapsto b$ and $c_i \mapsto c_i$ extends to an automorphism of $A$. If $\tilde{A}$ is not isomorphic to $A$, then it must be the case that $\tilde{A} \simeq A \oplus A$.

One final comment: By counting the number of irreducible representations in $\pi_1$, $\pi_2$ and the number of shared irreducible representations (as in the commutative case, where we have counted $|\sigma(a)|$ and $|\sigma(b)|$), one may see that the assertion $\tilde{A} \simeq A$ iff $a \mapsto b$ and $c_i \mapsto c_i$ extends to an automorphism of $A$ is generally true for any finite-dimensional $A$. Though, of course, as we have seen, this needs not be true for infinite-dimensional $A$, and the second assertion that either $\tilde{A} \simeq A$ or $\tilde{A} \simeq A \oplus A$ needs not be true for general finite-dimensional $A$ without assuming $A = M_n(\mathbb{C})$.