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?

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    $\begingroup$ Following @NikWeaver's comment, I assume you mean "submultiplicative norm" $\endgroup$
    – Yemon Choi
    Sep 27, 2019 at 15:12
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    $\begingroup$ I'm not able to spend time right now looking up precise statements, but if your library has a copy of Theodore Palmer's books on Banach algebras, then Volume 2 on star-algebras has a comprehensive look at various species of star-algebras and which ones admit a $C^*$-norm. $\endgroup$
    – Yemon Choi
    Sep 27, 2019 at 15:14
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    $\begingroup$ Just to address the second bullet point: you can have commutative (non-unital) Banach algebras where every element has spectral radius zero. These often admit a trivial involution given by complex conjugation of functions/measures. $\endgroup$
    – Yemon Choi
    Sep 27, 2019 at 15:16
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    $\begingroup$ The spectral radius of the $2 \times 2$ matrix with a 1 in the top right corner and zeroes elsewhere is 0, but it's operator norm is 1. However, if $A$ is a C$^*$-algebra and $a \in A$, the C$^*$-norm of $a$ is the square root of the spectral radius of $a^*a$. $\endgroup$ Sep 27, 2019 at 20:15
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    $\begingroup$ It should also be noted that completing a (star) algebra -- or indeed, taking a complete (star) algebra and completing it with respect to a weaker norm -- can drastically alter the shape of the spectrum. The "well-known fact" that you open with is surprisingly delicate, because the spectrum can shrink drastically once you add in the "ghost" elements of completion. Think about ${\mathbb C}[z,z^{-1}] \subset C({\mathbb T})$. $\endgroup$
    – Yemon Choi
    Sep 28, 2019 at 0:09

3 Answers 3


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.

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.

  • $\begingroup$ Thanks, this is even better than I hoped! So basically, the natural cone one gets from the $\ast$ structure seems to work better than the spectral radius formula -- one need only assume that $-1$ is not nonnegative (which I think would mean the cone is the whole algebra? and anyway it's a natural "reality" condition) to get a canonically associated $[0,\infty]$-valued, submultiplicative, power-multiplicative seminorm. Then the $C^\ast$ condition just asks for this intrinsic seminorm to be complete. All very tidy! $\endgroup$
    – Tim Campion
    Jan 28, 2022 at 21:48
  • $\begingroup$ If I'm not mistaken, one can define the $C^\ast$ uniformity directly in terms of the postive cone without mentioning the norm at all. Then one can simply say: a $\ast$-algebra has a natural (possibly non-separated) uniform structure, and it is a $C^\ast$-algebra if and only if it is complete (and in particular separated) with respect to this uniformity. Of course, one will probably still talk about the associated norm when proving the equivalence. (Also, thanks for including the condensed version of Cimpric's argument in your paper, that was very helpful!) $\endgroup$
    – Tim Campion
    Jan 28, 2022 at 21:54
  • $\begingroup$ I'm having a lot of fun with this! Here is an alternate argument, which unfortunately leans into the hypotheses a bit earlier on. Let $A$ be a $\ast$-algebra over $\mathbb C$, and suppose that $-1$ is not in the positive cone, and that for every $a \in A$, $0 < a^\ast a \leq \alpha$ for some $\alpha \in \mathbb R$. Then by the Hahn-Banach theorem, there exists a state $\rho$ on $A$. By the GNS construction, $A$ now embeds into a $C^\ast$-algebra constructed from $\rho$. (Compare Andre's point below that the "correct" definition of a $C^\ast$ algebra is about acting on a Hilbert space.) $\endgroup$
    – Tim Campion
    Jan 29, 2022 at 1:05
  • $\begingroup$ @TimCampion I am glad that you like it. I think this should be better known. In your last argument, I suppose what you mean by $a^*a>0$, is $\exists \epsilon>0,~a^*a\geq \epsilon$, right? How do you use Hahn Banach without pre-assuming the existence of a norm? $\endgroup$
    – Uri Bader
    Jan 29, 2022 at 8:40
  • $\begingroup$ I'm using the version of Hahn-Banach for a cone rather than a norm -- see e.g. Thm 2.1 here. This does require passing back and forth between real and complex statements. But if $-1 \not \in A_+$, then for $a \in A_+$, if $\lambda a \in A_+$ it follows that $\lambda$ is real. So pick a basis of the $\mathbb R$-vector space $X_0$ spanned by $A_+$ whose generators are positive; then $X_0 \otimes_{\mathbb R} \mathbb C \subseteq A$. Extend this to get a real vector space $X_0 \subseteq X$ with $X \otimes_{\mathbb R} \mathbb C = A$. $\endgroup$
    – Tim Campion
    Jan 29, 2022 at 12:47

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 $(A,\varphi)$.

  • $\begingroup$ Thanks! I think I misunderstood the definition of the spectral radius -- I thought this modification where one looks at $a^\ast a$ was built into the definition. Now I suppose what I'm wondering is "how close" these conditions are to holding automatically. For example, when $\|-\|$ is defined as above on a $\ast$-algebra, and assuming that $\|a\|$ is always finite, do any of the (in)equalities $\|a\| = 0 \Rightarrow a = 0$, $\|a+b\| \leq \|a\|+\|b\|$, $\|ab \| \leq \|a\|\|b\|$, $\|a^\ast a\| = \|a\|^2$ hold autormatically? Also, it would be nice to see a proof/reference... Or is it trivial? $\endgroup$
    – Tim Campion
    Sep 27, 2019 at 17:58
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    $\begingroup$ How does spectral radius enter into this characterization? Your three bullet points are the usual axioms for C*-algebras. $\endgroup$
    – Nik Weaver
    Sep 27, 2019 at 18:22
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    $\begingroup$ You've just listed the axioms of a C*-algebra and added that the norm comes from spectral radius. $\endgroup$
    – Nik Weaver
    Sep 27, 2019 at 18:24
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    $\begingroup$ @TimCampion My suggestion is to either learn about Banach algebras (I know, I know, but some of us in functional analysis have tried to learn about category theory) or consult Palmer as I suggested above. Your hopes for things to hold automatically read to me as if I, as a functional analyst, assumed every monoidal category is symmetric, closed, and compact $\endgroup$
    – Yemon Choi
    Sep 28, 2019 at 0:02
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    $\begingroup$ @TimCampion - I added more discussion to hopefully clarify what is going on. The proof is: $A$ is a C* algebra if and only if $A$ has a C*-norm (which is determined by spectral radius) if and only if the spectral radius defines a C*-norm on $A$. As for whether some of these properties hold automatically, that's a very interesting question for which I don't know the answer and would need to think more. $\endgroup$ Sep 29, 2019 at 10:15

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:

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."

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.


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