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?



Which $\ast$-algebras are $C^\ast$-algebras?
**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](https://arxiv.org/pdf/2001.10185.pdf).


**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.


