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.

Posted on September 4, 2020 by Dan
Categories: Research
Tags: algebras, automorphisms, directly infinite

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Automorphisms of Infinite Dimensional Algebras — Part 2

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