# Characterization of certain subalgebras of $M_2(\mathcal{A})$ where $\mathcal{A}$ is a $C^*$-algebra

c-star-algebrasoa.operator-algebras

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.

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 $$A = M_n(\mathbb{C})$$.