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

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