Which $\ast$-algebras are $C^\ast$-algebras?

Question

It's well-known that the norm on a $C^\ast$-algebra is uniquely determined by the underlying $\ast$-algebra by the spectral radius formula. Therefore there should be a way to axiomatize $C^\ast$-algebras directly in terms of the $\ast$-algebra structure, without explicitly talking about a norm.

Question 1: How does one do this? That is, which $\ast$-algebras are $C^\ast$-algebras?

Question 2: How does one axiomatize those $\ast$-algebras which embed into a $C^\ast$-algebra (equivalently, embed into their $C^\ast$-enveloping algebra)?

Some possibilities:

• Perhaps a $\ast$-algebra is a $C^\ast$-algebra iff the spectral radius is a complete, submultiplicative norm?

• Perhaps a $\ast$-algebra embeds into a $C^\ast$-algebra iff every element has finite spectral radius?

If the first guess above (or something like it) is correct, it would still be nice to break it down into more manageable chunks.

EDIT: I'm currently fascinated by the following observation. Let that if $A$ be any algebra over $\mathbb C$, and $a \in A$. Let $B$ be the subalgebra of $A$ generated by $a$, and let $C$ be the subalgebra of $A$ obtained from $B$ by closing under those inverses which exist in $A$, so that $C \cong \mathbb C[a][\{(a-\lambda)^{-1} \mid \lambda \not \in Spec(a)\}]$. Writing a general element $c \in C$ as a rational function $c = \phi(a)$, we have $Spec(c) = \phi(Spec(a))$. It follows that he spectral radius is a homogenous, subadditive, submultiplicative, power-multiplicative function on $C$. If we assume that the spectral radius in $A$ of any nonzero element of $C$ is finite and nonzero, it follows that the spectral radius is in fact a submultiplicative, power-multiplicative norm on $C$. So it seems natural to stipulate that (if $A$ is a $\ast$-algebra, and maybe assuming that $a$ is normal?), every "Cauchy sequence" in $C$ should have a unique "limit" in $A$ with respect to the spectral radius. I wonder how far this condition is from guaranteeing that $A$ is a $C^\ast$-algebra?

Answer 1

TL;DR: A $*$-algebra embeds in a $C^*$-algebra iff it is archimedean with no non-trivial infinitesimals. In this case the $*$-algebra is a $C^*$-algebra iff a certain norm is complete. I do explicitly talk about a norm here, but it is explicitly constructed from a natural order structure associated with the $*$-operation.

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Let $A$ be a complex $*$-algebra. In what follows we assume $A$ is unital and regard accordingly $\mathbb{C}$ as subalgebra of $A$. This is not a real restriction, as one can always unitalize $A$.

We define $$ A_+=\left\{\sum_{i=1}^n x_i^*x_i\mid n\in\mathbb{N},~x_1,\ldots,x_n\in A\right\}$$ and note that it is a convex cone in $A$, thus it defines on it a partial order by $$ x\leq y \quad \Longleftrightarrow \quad y-x\in A_+. $$ Next we define for $x\in A$, $$ \|x\|=\sqrt{\inf\{\alpha\in \mathbb{R}_+\mid x^*x\leq \alpha\}}\in [0,\infty] $$ (using the conventions $\inf\emptyset=\infty$ and $\sqrt{\infty}=\infty$).

Claim: $A$ is embeddable in a $C^*$-algebra iff it is archimedean with no non-trivial infinitesimals iff $\|\cdot\|$ is a norm on $A$ and in this case, $A$ is a $C^*$-algebra iff this norm is complete.

Let me elaborate. If $-1\in A_+$ then $\|\cdot\|=0$ identically, thus we assume from now on that $-1\notin A_+$. $A$ is said to be archimedean if $\|\cdot\|$ attains only finite values. We define the set of infinitesimal elements in $A$ to be $A_i=\{x\mid \|x\|=0\}$. The claim above follows from the following two facts. For proofs see section 2.1 here.

Fact 1: $A$ has a unital $*$-representation into a $C^*$-algebra iff it is archimedean and $A_i$ is a two-sided ideal which is in the kernel of every such a representation.

Fact 2: If $A$ is archimedean then $\|\cdot\|$ is a seminorm on $A$ and the corresponding norm on $A/A_i$ satisfies the $C^*$-property. In particular, $A/A_i$ $*$-embeds into the $C^*$-algebra obtained by completion.

Answer 2

Given an algebra $A$, one can ask whether it has a unit. If one exists, one then shows it is unique: $1_A = 1_A1_A' = 1_A'$. Thus being unital is a property of an algebra and not extra structure. Either an algebra has a unit or it does not.

Similarly, given a complex $*$-algebra $A$, one can ask whether there exists a C${}^*$-norm on $A$, i.e., a norm on $A$ which satisfies the following properties:

• $A$ is complete in this norm,
• $\|ab\|\leq \|a\|\cdot\|b\|$ for all $a,b\in A$, and
• $\|a^*a\|=\|a\|^2$ for all $a\in A$.

If one exists, then one shows that it is unique; the norm is determined by the spectral radius as the OP point out: $$ \|a\| = \|a^*a\|^{1/2} = r(a^*a)^{1/2} \qquad\forall\, a\in A. $$ (The spectral radius only equals the norm for normal elements, and the C${}^*$-axiom does the rest.) Thus being a C${}^*$-algebra is a property of a complex $*$-algebra and not extra structure. Either there exists a C${}^*$-norm or there does not.

So how might one go about determining whether a complex $*$-algebra admits a C${}^*$-norm? When it is unital (or after one unitizes), as the OP suggests in the first bullet point, it suffices to look at the spectral radius and ask whether it gives a C${}^*$-norm. The following exact statement was pointed out to me by Andre Henriques.

A unital complex $*$-algebra $A$ is a C${}^*$-algebra if and only if the function $\|\cdot\|: A \to [0,\infty]$ given by $$ \|a\|^2 := \sup\left\{ |\lambda| : a^*a - \lambda 1_A \text{ is not invertible}\right\} $$ is a C${}^*$-norm on $A$.

Another way is to find a faithful $*$-representation $\pi$ from $A$ into another C${}^*$-algebra $B$ whose image is norm closed. Then $\|a\|:= \|\pi(a)\|_B$ works.

Often in my own work, I will have a finite dimensional unital complex $*$-algebra and need to know if it is a C${}^*$-algebra. Here are a couple conditions that work in the finite dimensional setting.

Suppose $A$ is a finite dimensional unital complex $*$-algebra. The following are equivalent.

1. $A$ is a C${}^*$-algebra.

2. For all $a\in A$, $a^*a=0$ implies $a=0$.

3. $A$ is $*$-isomorphic to a unital $*$-subalgebra of $M_n(\mathbb{C})$ for some $n\in \mathbb{N}$.

4. There exists a linear functional $\varphi:A \to \mathbb{C}$ such that $\varphi(a^*a) \in (0,\infty)$ for all $a\neq 0$, i.e., $A$ has a faithful positive linear functional.

An outline for a proof of 2 implies 1 can be found as Exercise 3.1.27 in these notes I wrote for a course on quantum algebra. A proof of 3 implies 1 can be found as Theorem 3.2.1 in Vaughan Jones' notes on von Neumann algebras (Wayback Machine). That 4 implies 3 follows by performing the GNS construction for $(A,\varphi)$.

Answer 3

A good way of figuring out whether a definition is good or not, is to see how it survives generalisations.

The notion of $C^*$-algebra admits a couple of interesting generalisations:

• Real $C^*$-algebra.
• Super $C^*$-algebra, and more generally: $C^*$-algebra internal to a rigid $C^*$-tensor category.

The notion of real $C^*$-algebra is particularly revealing: the algebraic definition of a real $C^*$-algebra involves a very weird-looking condition, involving $xx^*+yy^*$.

Which leads me to conclude that the "main definition" is:

"Embeddable into $B(H)$ as a norm-closed subset."

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What I mean by "main definition" is the following. Imagine that you have two competing definitions:

• On the one side, you have the naïve generalisation of the notion of complex C*-algebra to the real setting.
  
• And on the other side, you have "embeddable into $B(H)$".

It turns out that the above two definitions are not equivalent. Which one should win? Which one should be the official definition of real $C^*$-algebra? Well... the one that wins is "embeddable into $B(H)$", and the loser definition (namely the first one) has to accept the humiliation of having a weird-looking axiom being added to it.