Non-unital $C^\star$ subalgebra of a unital $C^\star$ algebra

Question

Straightforward techniques exist for unitizing a given non-unital $C^\star$ algebra, that is, viewing it as a $C^\star$ subalgebra (in fact, a closed ideal) of a unital $C^\star$ algebra. I am trying to see if something of a similar sort can be done in the opposite direction.

Given a unital $C^\star$ algebra $\mathcal{A}$, when does it contain a $C^\star$ subalgebra (i.e., a norm closed subalgebra) $\mathcal{B}$ that is not unital? How to find such subalgebras, when they exist?

Note: The subalgebra $\mathcal{B}$ might fail to contain the multiplicative identity of the full algebra $\mathcal{A}$, $\mathbb{1}_\mathcal{A}$, and still end up being unital. Some other element than $\mathbb{1}_\mathcal{A}$, might end up serving as $\mathbb{1}_\mathcal{B}$.

Partial solution:

1. Such subalgebras can never exist in finite dimensions. It is easy to see from several different arguments (eg, the Artin-Wedderburn theorem, compactness of the unit ball in finite dimensions), that every finite-dimensional $C^\star$ algebra is unital.
2. In the infinite dimensional commutative case, consider $\mathcal{A}\cong C(X)$ for some compact Hausdorff space $X$ such that $X$ contains a limit point $x_0$. Then $\hat{X}:= X \backslash \{x_0\}$ is a noncompact but locally compact Hausdorff space, and $C_0(\hat{X})$ is (isomorphic to) a closed proper ideal of $C(X)$, and a non-unital $C^\star$ subalgebra.
3. For $\mathcal{B(H)}$, of course, $\mathcal{K(H)}$ comes to the rescue (for infinite dimensional $\mathcal{H}$, $\mathcal{B(H)}$ is unital, while the closed subalgebra (ideal) of compact operators $\mathcal{K(H)}$ is non-unital).

How should I proceed in the general case? Otherwise, how to proceed for

a) infinite dimensional noncommutative $C^\star$ algebras?

b) infinite dimensional commutative $C^\star$ algebras where the spectrum does not contain a limit point?

I don't think GNS and step 3) from above will be of much help here, since I don't know if compactness passes on through $\star$-homomorphisms in a natural way (I don't think it does, but I don't know, really...)

Answer 1

Since you don't put other restrictions on your $\mathcal B$, this can always be done as long as $\mathcal A$ is infinite-dimensional.

It is enough to take $a\in\mathcal A$ selfadjoint such that $\sigma(a)$ is infinite. Now remove an accumulation point $t_0\in\sigma(a)$, and form $\mathcal B=C_0(\sigma(a)\setminus \{t_0\})$ (properly, brought back inside $C^*(a)$ via the Gelfand transform).

Answer 2

A key part of the story here is that a $C^*$-algebra whose self-adjoint elements all have finite spectrum must be finite-dimensional. Martin Argerami gave a reference which implies this result by way of several more general results in the context of Banach algebras. However, I decided to write up an account specifically for the $C^*$-algebra case in order to take advantage of the simplifications this context has to offer.

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Claim 1: Suppose $A$ is a unital $C^*$-algebra all of whose self-adjoint elements have one point spectrum. Then $A$ is one-dimensional.

Sketch: Use functional calculus to show every self-adjoint element is a multiple of $1_A$, then use the fact that every $C^*$-algebra is the span of its self-adjoint part.

Claim 2: Let $A$ be a $C^*$-algebra in which every self-adjoint element has a finite spectrum and let $p$ be a nonzero projection in $A$ such that the "corner algebra" $pAp$ is not one dimensional. Then, one can write $p=p_1+p_2$ where $p_1$ and $p_2$ are nonzero, orthogonal projections.

Sketch: Working inside $pAp$, whose unit is $p$, the previous claim gives a self-adjoint element whose spectrum is a finite set with at least two elements. Now use functional calculus.

Claim 3: Let $A$ be unital $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then there exist nonzero, pairwise orthogonal projections $p_1,\ldots,p_n$ summing to $1_A$ such that each of the corner algebras $p_iAp_i$ is one-dimensional.

Sketch: If $A$ is not one-dimensional, subdivide $1_A$ into two projections. If either of the new projections does not determine a one-dimensional corner, subdivide again. This process must eventually terminate and give a decomposition of the desired type. Otherwise, we would have an infinite collection of pairwise orthogonal projections in $A$ which could then be used to embed a copy of $c_0(\mathbb{N})$ into $A$ leading to self-adjoint elements of infinite spectrum, contrary to assumption.

Claim 4: Let $p$ and $q$ be nonzero, orthogonal projections in a $C^*$-algebra $A$ satisfying that $pAp$ and $qAq$ are one-dimensional. Then $qAp$ and $pAq$ are either both one-dimensional or both zero.

Proof: Since $(qAp)^* = pAq$, it suffices to look at $qAp$. Fix any nonzero $a \in qAp$. By the $C^*$-identity, $a^*a \in pAp$ and $aa^* \in qAq$ are nonzero. Since $pAp$ is one-dimensional, up to rescaling $a$, we may assume $a^*a=p$. But then $a$ is a partial isometry, so $aa^*$ is also a projection and, being a nonzero projection in $qAq$, is equal to $q$. The identities $a^*a=p$ and $aa^*=q$ show that $x \mapsto ax : pAp \to qAp$ and $x \mapsto a^* x : qAp \to pAp$ are mutually inverse bijections, and the result follows.

Claim 5: Let $A$ be a unital $C^*$-algebra and let $p_1,\ldots,p_n$ be (necessarily orthogonal) projections satisfying $1_A=p_1 + \ldots + p_n$. Then, $A$ is the internal direct sum of the spaces $p_i A p_j$ where $1 \leq i,j \leq n$.

Basically, each element of $A$ can be thought of as a "matrix of operators" with respect to this partition of the identity.

Final Claim: Let $A$ be a $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then, $A$ is finite-dimensional.

Proof: First unitize $A$ if necessary (which does not affect the spectra of any of its elements). Then combine Claims 3, 4 and 5.