Automorphisms of Infinite Dimensional Algebras–Part 3
In this post we’ll show that for any algebra with a faithful ideal isomorphic to , 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 denote an infinitedimensional vector space with basis .

 The column finite matrix algebra , which is isomorphic to the algebra of endomorphisms of the countable dimensional vector space , .
 The row and column finite matrix algebra which is isomorphic to the following subalgebra of : Let . Note that is the subspace consisting of the “tail” of . Then is isomorphic to .
 The nonunital algebra of infinite matrices with only finitely many nonzero entries, , which is ismorphic to the endomorphisms of which have finite dimensional image, call this ideal of by
Its clear that is a twosided ideal of . Moreover is faithful as an ideal, which means that it has nontrivial intersection with any other ideal. Naturally these facts still hold in and .
The faithfulness of is what makes the following proof work,
Theorem: Let be an automorphism of some algebra which has a faithful ideal Then there is some such that . (We’ll denote such an inner automorphism by .)
We will prove this theorem via two lemmas. In the first that every automorphism of is inner, and in the second we use the faithfulness of to lift this inner automorphism to an automorphism of .
Lemma: Let be an automorphism of , then there exists some such that $\widehat \Gamma \circ \alpha = \text{Id}_{M_\infty(V)}$.
Proof: Construct a spanning set of by such that for any
and
For brevity we will denote by . A reflection on the left action of the matrix units of shows that is .
Let us examine the image of on this set of matrix units of . First the main diagonal: Define an endomorphism by for each . Now is an automorphism, which makes a set of primitive orthogonal idempotents of . Then we must have another basis of and . Being an automorphism of , must, in particular, be a change of basis map which sends to . To prevent endless subscripts, we will write merely as . So define an endomorphism such that for all .
This map is clearly invertible. So let us show that is an inner automorphism such that . Let be some element of . Then
, and
,
giving the desired equality. Again for notation’s sake, define by . By definition of , , i.e. acts as the identity on the ‘main diagonal’ of .
Examining the action of on the ‘offdiagonal’ elements of , a technical argument then shows that there is some such that for all . Moreover, is a new set of matrix units which span . With this correspondence in mind, define
which then gives .
Calculation then shows that . Since , we have . Setting , we have our desired equality.
Claim: This endomorphism must be in .
Proof: Suppose that the matrix representation of has an infinite row (without loss of generality assume it is the first row, which will be denoted by ). Then for all . In particular choose . Then
Then , which is a matrix with an infinite first row (by assumption) and zeroes elsewhere. However, since we must have that since is columnfinite. This is a contradiction.
Hence this must be row and column finite.
Lemma: Let , then if and only if for all .
Proof: The forward direction of this statement is evident. So suppose that . Note that for any basis element, we may write
Then , and in a similar way . Thus for all and , and thus .
To finally prove the theorem let , then
Hence by the previous lemma for all . Then .
I find this result very important because we can often find a faithful ideal isomorphic to 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
where is the ring of Laurent polynomials.
I’ve recently been working with a group here at Baylor on some settheoretic aspects of abelian groups. I think that I will devote my next few posts to that.
–DB
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