# Automorphisms of Infinite Dimensional Algebras–Part 3

In this post we’ll show that for any algebra with a faithful ideal isomorphic to $M_\infty(K)$, every automorphism is inner. Buckle up; this’ll be a long one. It’ll build on my previous two posts, so make sure to read them if you haven’t already: Part 1, Part 2

Let’s gather together some facts which will be useful:

We will be working with three different infinite matrix algebras; let $V$ denote an infinite-dimensional vector space with basis $\mathcal B = \{b_i : i \in \mathbb Z^+\}$.

• The column finite matrix algebra $\text{CFM}(K)$, which is isomorphic to the algebra of endomorphisms of the countable dimensional vector space $V$, $\text{End}_K(V)$.
• The row and column finite matrix algebra $B(K)$ which is isomorphic to the following sub-algebra of $\text{End}_K(V)$: Let $V_k = \bigoplus_{i = k}^\infty Kb_i$. Note that $V_k$ is the subspace consisting of the “tail” of $V$. Then $B(K)$ is isomorphic to $B(V) := \{\phi \in \text{End}_K(V) : \forall n, \exists m \text{ with } \phi(V_m) \subseteq V_n\}$.
• The non-unital $K$-algebra of infinite matrices with only finitely many nonzero entries, $M_\infty(K)$, which is ismorphic to the endomorphisms of $\mathcal B(V)$ which have finite dimensional image, call this ideal of $B(V)$ by $M_\infty(V)$

Its clear that $M_\infty(K)$ is a two-sided ideal of $B(K)$. Moreover $M_\infty(K)$ is faithful as an ideal, which means that it has nontrivial intersection with any other ideal. Naturally these facts still hold in $B(V)$ and $M_\infty(V)$.

The faithful-ness of $M_\infty(V)$ is what makes the following proof work,

Theorem: Let $\alpha$ be an automorphism of some algebra $A \subseteq B(V)$ which has a faithful ideal $I \simeq M_\infty(V)$ Then there is some $\Gamma \in B(V)$ such that $\alpha(x) = \Gamma^{-1} x \psi$. (We’ll denote such an inner automorphism by $\widehat{\Gamma}(x)$.)

We will prove this theorem via two lemmas. In the first that every automorphism of $M_\infty(V)$ is inner, and in the second we use the faithfulness of $M_\infty(K)$ to lift this inner automorphism to an automorphism of $A$.

Lemma: Let $\alpha$ be an automorphism of $M_\infty(V)$, then there exists some $\Gamma \in B(V)$ such that $\widehat \Gamma \circ \alpha = \text{Id}_{M_\infty(V)}$.

Proof: Construct a spanning set of $M_\infty(V)$ by $\{\epsilon_{ij} : i,j \in \mathbb Z\}$ such that for any $x = \sum x_i b_i \in V$

$\epsilon_{ii}(x) = x_i b_i$ and $\epsilon_{ij}(x) = x_j b_i$

For brevity we will denote $\epsilon_{ii}$ by $\epsilon_i$. A reflection on the left action of the matrix units of $M_\infty(K)$ shows that $\epsilon_{ij} \in M_\infty(V)$ is $e_{ij} \in M_\infty(K)$.

Let us examine the image of $\alpha$ on this set of matrix units of $M_\infty(V)$. First the main diagonal: Define an endomorphism $\phi_i \in \text{End}_k(V)$ by $\phi_i = \alpha \epsilon_i$ for each $i \in \mathbb Z^+$. Now $\alpha$ is an automorphism, which makes $\{\phi_i : i \in \mathbb Z^+\}$ a set of primitive orthogonal idempotents of $B(V)$. Then we must have another basis $\{g_{b_i} : i \in \mathbb Z^+\}$ of $A$ and $\text{im}(\phi_i) = K g_{b_i}$. Being an automorphism of $A$, $\alpha$ must, in particular, be a change of basis map which sends $b_i$ to $g_{b_i}$. To prevent endless subscripts, we will write $g_{b_i}$ merely as $g_i$. So define an endomorphism $\gamma$ such that $\gamma(b_i) = g_i$ for all $i \in \mathbb Z^+$.

This map $\gamma$ is clearly invertible. So let us show that $\widehat{\gamma}$ is an inner automorphism such that $\widehat{\gamma}(\phi_i) = \epsilon_i$. Let $u = \sum u_i b_i$ be some element of $V$. Then

$\phi_i \gamma(u) = \phi_i(\sum u_i g_i) = u_i g_i$, and

$\gamma \epsilon_i = \gamma(u_i b_i) = u_i g_i$,

giving the desired equality. Again for notation’s sake, define $\beta: A \rightarrow \text{End}_K(V)$ by $\beta(a) = \widehat{\gamma} \alpha(a)$. By definition of $\gamma$, $\beta(\epsilon_i) = \epsilon_i$, i.e. $\beta$ acts as the identity on the ‘main diagonal’ of $M_\infty(V)$.

Examining the action of $\beta$ on the ‘off-diagonal’ elements of $M_\infty(V)$, a technical argument then shows that there is some $s_{ij} \in K$ such that $\beta(\epsilon_{ij}) = s_{ij} \epsilon_{ij}$ for all $i,j \in \mathbb Z^+$. Moreover, $\{\beta(e_{ij}) : i,j \in \mathbb Z^+\}$ is a new set of matrix units which span $M_\infty(V)$. With this correspondence in mind, define

$\psi = \sum_{b_x \in \mathcal B} s_{1x} \epsilon_x$ which then gives $\gamma^{-1} = \sum_{b_x \in \mathcal B} s_{1x}^{-1} \epsilon_x$.

Calculation then shows that $\widehat{\psi^{-1}} \beta( \epsilon_{ij}) = \epsilon_{ij}$. Since $\beta = \widehat \gamma \alpha$, we have $\widehat{\psi^{-1}} \widehat \gamma \alpha(\epsilon_{ij}) = \epsilon_{ij}$ . Setting $\Gamma = \psi \circ \gamma$, we have our desired equality.

Claim: This endomorphism $\Gamma$ must be in $B(V)$.

Proof: Suppose that the matrix representation of $\Gamma$ has an infinite row (without loss of generality assume it is the first row, which will be denoted by $\Gamma_{1*}$). Then $\Gamma^{-1} x \Gamma = y \in M_\infty(K)$ for all $x \in M_\infty(K)$. In particular choose $x = e_{11}$. Then

$m \Gamma = \Gamma n$

Then $x \Gamma = \begin{pmatrix} \Gamma_{1*}\\ 0 \\0 \\\vdots \end{pmatrix}$, which is a matrix with an infinite first row (by assumption) and zeroes elsewhere. However, since $y \in M_\infty(K)$ we must have that $\Gamma y \in M_\infty(K)$ since $\Gamma$ is column-finite. This is a contradiction.

Hence this $\Gamma$ must be row and column finite.

$\blacksquare$

Lemma: Let $\phi, \psi \in \text{End}(V)$, then $\phi = \psi$ if and only if $\epsilon_i \phi \epsilon_j = \epsilon_i \psi \epsilon_j$ for all $i,j \in \mathbb Z^+$.

Proof: The forward direction of this statement is evident. So suppose that $\epsilon_i \phi \epsilon_j = \epsilon_i \psi \epsilon_j$. Note that for any basis element, $b_i$ we may write

$\phi(b_i) = \sum \lambda_{ij} b_j$

$\psi(b_i) = \sum \mu_{ij} b_j$

Then $\epsilon_i \phi \epsilon_j (u) = \epsilon_i \phi(u_j b_j) = \epsilon_i u_j \phi(b_j)= \epsilon_i (u_j \sum \lambda_{jk}b_k = \lambda_{ji} \epsilon_{ij}(u)$, and in a similar way $\epsilon_i \psi \epsilon_j(u) = \mu_{ji} \epsilon_{ij}(u)$. Thus $\lambda_{ji} = \mu_{ji}$ for all $i$ and $j$, and thus $\phi = \psi$.

$\blacksquare$

To finally prove the theorem let $\phi \in A$, then

$\epsilon_i \widehat \Gamma \alpha(\phi) \epsilon_j = \widehat \Gamma \alpha (\epsilon_i \phi \epsilon_j) = \widehat \Gamma \alpha (\lambda_{ji}\epsilon_{ij} = \lambda_{ji} \epsilon_{ij} = \epsilon_{i} \phi \epsilon_j.$

Hence by the previous lemma $\widehat \Gamma \alpha (\phi) = \phi$ for all $\phi \in A$. Then $\alpha = \widehat \Gamma^{-1}$.

$\blacksquare$

I find this result very important because we can often find a faithful ideal isomorphic to $M_\infty(K)$ within a directly infinite algebra. In an article which I recently submitted for publication, I used this characterization of automorphisms to classify a wide swath of algebras which have the property that

$A/M_\infty(K) \simeq K[x,x^{-1}],$

where $K[x,x^{-1}]$ is the ring of Laurent polynomials.

I’ve recently been working with a group here at Baylor on some set-theoretic aspects of abelian groups. I think that I will devote my next few posts to that.

–DB