# Automorphisms of Infinite Dimensional Algebras — Part 2

This post is devoted to determining the set of automorphisms of the Toeplitz-Jacobson $K$-algebra $A = \langle x, y \;|\; xy = 1 \rangle$.

There are two general approaches to finding the automorphisms of these algebras. One could take the “outside-in” approach and look for bijective endomorphisms of $A$. This might be an interesting approach since one can, in theory, leverage the decomposition of the regular module $_A A \simeq _A A \oplus \text{ann}_l(y)$

and use the fact that $\text{End}(_AA) \simeq A$ as a $K$-algebra to show that $\text{End}(_AA) \simeq \begin{pmatrix} \text{End}(_AA) &\text{Hom}_K(\text{ann}_l(y), \text{End}(_AA))\\ \text{Hom}_K(\text{End}(_AA), \text{ann}_l(y)) &\text{End}(\text{ann}_l(y)) \end{pmatrix}$

This approach, however, has two significant drawbacks. The first is that we still need to classify the invertible endomorphisms of $\text{End}(_AA)$, and then we now must classify the invertible endomorphisms, $f: \text{ann}_l(y) \rightarrow \text{ann}_l(y)$.

Ew…

This leaves the “inside-out approach” inspired by this paper of Courtemanche and Dugas. It is easiest to fix an embedding of the Toeplitz-Jacobson algebra into the row and column matrices; specifically, I’ll be using the embedding from my last post here. The inside-out approach depends on the fact that the algebra $M_\infty(K)$ (the infinite matrix algebra with only finitely many nonzero entries) is a faithful ideal of the algebra $A$.

Definition: An ideal $I \subseteq A$ is called faithful if whenever there is some element $x \in A$ such that $xI = \{0\}$ or $Ix = \{0\}$ then $x = 0$. In other words, $I$ is faithful if it is faithful as a left and a right module.

This can be seen to be a strengthening of an essential ideal, that is, an ideal $I$ which has nontrivial intersection with any other nonzero ideal. In the theory of C*-algebras, which inspires much of my current research, these two are indistinguishable, but it is not hard to see that the principal ideal $(x)$ of $K[x]/(x^2)$ is an essential ideal without being faithful.

What does being faithful give us? For one, it allows the ideal to control much of the structure of its ambient algebra; faithfulness is to algebras as density is to analysis. If we can show that something must be true on a faithful ideal, then it must be true in the algebra itself. The classification of automorphisms is a perfect example.

Recall that $M_\infty(K)$ is the unique minimal ideal of the Toeplitz-Jacobson algebra $A$. Moreover, it is a faithful ideal since for every nonzero element $a \in A$, there is some $m \in M_\infty(K)$, such that $am \neq 0$.

To see an example of the power of a faithful ideal isomorphic to $M_\infty(K)$ we finish this post with the following result. Recall that $B(K)$ denotes the $K$-algebra of infinite matrices (indexed by $\mathbb{Z}^+$)

Lemma: Suppose that the algebra $A$ has an ideal $I$ which is isomorphic to $M_\infty(K)$, and let $\iota$ denote that composition of the embedding which takes $e_{ij} \in M_\infty(K)$ and maps it to $e_{ij} \in B(K)$ with the isomorphism $I \simeq M_\infty(K)$ and $i$ be the standard inclusion map of the ideal $I$ into $A$. Then there exists a unique algebra homomorphism $\phi: A \rightarrow B(K)$ which makes the following diagram commute. Furthermore $\phi$ is injective if, and only if, $I$ is a faithful ideal of $A$.

Proof:

Let $E_{ij}$ denote the matrix units which generate $I$ as a indeal of $A$. Define $\phi$ by $\phi(a) = (a_{ij})$ where $(a_{ij})$ is the matrix whose entries are defined $a_{ij} := E_{ii}aE_{jj}$ (note also that $a_{ij} \in K$). Because any decomposition of an algebra by a complete set of orthogonal idempotents is a direct sum, it is simple to show that $(a_{ij}) \in B(K)$ for any $a \in A$. Commutativity of the diagram is similarly straightforward. So all that remains to show is uniqueness. Suppose that there exists some $\psi$ which also completes the commutative diagram. Consider for all choices of $i, j \in \mathbb Z^+$, the $(i,j)$ entry of $\psi(a)$. Then $e_{ii} \psi(a) e_{jj} = \psi(E_{ii}) \psi(a) \psi(E_{jj}) = \psi(E_{ii}aE_{jj}) = E_{ii} a E_{jj}$. Because the $(i,j)$ entry of $\psi$ is equal to the $(i,j)$ entry of $\phi$, it follows that $\psi = \phi$. This completes the proof of the first statement.

Now suppose that $\phi$ is injective, but that $I$ is not a faithful ideal. Then there is some nonzero $a \in A$ such that $aI = 0$ or $Ia = 0$. In this case $\phi(a) = (E_{ii} a E_{jj}) = 0$, contradicting the injectivity of $\phi$. Now suppose that $I$ is faithful. If $\phi$ were not injective, then there would be some nonzero $a \in A$ such that $E_{ii} a E_{jj} = 0$ for all $i,j \in \mathbb Z^+$. In particular this means that $E_{ii} a = 0$ and $a E_{jj} = 0$. By noting that $E_{ij} = E_{ii} E_{ij} E_{jj}$, one has that $a E_{ij} = E_{ij} a = 0$. Thus $a$ is a nonzero element of $A$ such that $a I = 0$ and $Ia = 0$, contradicting that $I$ is faithful. $\square$

The presence of a faithful ideal isomorphic to $M_\infty(K)$ allows us to create a unique matrix embedding of a countable dimensional algebra. This allows us to work within the algebra of row and column finite matrices to analyze the automorphisms. Tune in next time for that proof.