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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...)

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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).

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  • $\begingroup$ Do you have a proof (or reference) in mind for the fact that, if every self-adjoint element of $\mathcal{A}$ has finite spectrum, then $\mathcal{A}$ is finite-dimensional? $\endgroup$ – Mike F Apr 5 at 18:54
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    $\begingroup$ There is a proof here. $\endgroup$ – Martin Argerami Apr 5 at 19:36
  • $\begingroup$ I added a proof of this fact specifically for the $C^*$-algebra case since (surprising nobody) things can be done more simply there than in the Banach algebra context. $\endgroup$ – Mike F Apr 6 at 1:26
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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.


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.

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  • $\begingroup$ +1, but the argument when you have infinitely many commuting projections doesn't come from my answer. It comes from using them to produce infinitely many pairwise orthogonal projections, and so you can embed $\ell^\infty$. $\endgroup$ – Martin Argerami Apr 6 at 1:49
  • $\begingroup$ Another comment is that you don't need unital. If $A$ is not unital, just work on the unitization; the spectra do not change, as they are actually defined on the unitization. $\endgroup$ – Martin Argerami Apr 6 at 2:43
  • $\begingroup$ @MartinArgerami: Thanks I think I had in mind that, once one has an infinite-dimensional commutative algebra, it's (probably?) not hard to make an element of infinite spectrum or, equivalently, given an infinite, locally compact Hausdorff space $X$, make a function $f \in C_0(X)$ with infinite range. I agree that one can use the projections in Claim 3 above to directly make an element of infinite spectrum though, because you can make infinitely many pairwise orthogonal projections. I think you can only embed $c_0(\mathbb{N})$ though, rather than necsesarily $\ell^\infty$. $\endgroup$ – Mike F Apr 6 at 3:48
  • $\begingroup$ @MartinArgerami: Re: your other comment, I removed the unitality assumption from the final claim (but left it in some of the intermediate claims for convenience). $\endgroup$ – Mike F Apr 6 at 4:14
  • $\begingroup$ Yes, you are right about $c_0$. I would simply use $\sum_n2^{-n}p_n$. $\endgroup$ – Martin Argerami Apr 6 at 4:37

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