Two different C*-norms on the same $*$-algebra?

Question

Let $A$ be a $*$-algebra that admits at least one $C^*$-norm.

If $A$ is unital, it is well known that

$$\Vert x \Vert = r_A(xx^*)^{1/2}$$

where $r_A$ denotes the spectral radius thus in that case the norm is uniquely determined.

But what if $A$ does not have a unit? Is the norm then still unique?

Answer 1

(I'm assuming C*-norm means that it is a submultiplicative norm which satisfies the C*-identity, and not that the *-algebra is complete with respect to the norm; if it is complete, see the comment at the bottom).

To add to @MaoWao's comment, this isn't even true in the unital case. Take two unital C*-algebras, and form their algebraic tensor product, which is still a unital *-algebra. If one of the algebras is non-nuclear, there could be 2 C*-norms. For example, the reduced group C*-algebra of the free group $\mathbb{F}_2$, which is non-nuclear (since $\mathbb{F}_2$ is non-amenable), so there exists a C*-algebra such that there are atleast two C*-norms on the algebraic tensor product of it with that C*-algebra (I wrote tensor with itself initially, but to be more correct, $A$ non-nuclear means there exists $B$ such that $A \otimes_{\max} B \neq A \otimes_{\min} B$). To add to this, the algebra $\mathcal{B}(\mathcal{H}) \odot \mathcal{B}(\mathcal{H})$ (algebraic tensor) has a continuum of C*-norms ("A continuum of C*-norms on $\mathcal{B}(\mathcal{H})\otimes \mathcal{B}(\mathcal{H}) $ and related tensor products" by Ozawa and Pisier).

It is however true that if a *-algebra IS a C*-algebra (that is, it has a C*-norm, and it is complete with respect to it), then the norm is unique by your argument. The nonunital case follows from embedding into the unitization.

Edit: I suppose I missed a more obvious example for $\mathbb{F}_2$: one can form the reduced group C*-algebra and the full group C*-algebra, which are two different closures of the group ring $\mathbb{C} \mathbb{F}_2$ since $\mathbb{F}_2$ is non-amenable (the former is the closure with respect to the left regular representation, and the latter is the closure with respect to the universal representation).

Answer 2

It is not the case that the spectral radius defines a norm on a general unital *-algebra, and unital *-algebras can have infinitely many different norms.

For example, consider the commutative unital *-algebra of complex *-polynomials, i.e. the commutative polynomial algebra $\newcommand{\C}{\mathbb{C}}A = \C[z,z^*]$ given the involution defined by extending $\alpha z \mapsto \overline{\alpha}z^*$ in the obvious way.

The element $z - \lambda$ has no inverse for any $\lambda \in \C$ because $1$ has degree $0$ as a polynomial, while $z - \lambda$ has degree 1, and multiplying by an element of $A$ can only add to the degree. Therefore $z$ has infinite spectral radius, so spectral radius is not even a seminorm on $A$.

Furthermore, every element of $A$ defines a continuous function $\C \rightarrow \C$. Given any infinite compact set $K \subseteq \C$, we can define the norm $$ \| a \|_{K} = \sup \{ |a(z)| \mid z \in \C \} $$ We need $K$ to be infinite for this to be a norm - for finite $K$ it is only a seminorm. These norms are C$^*$-norms, and embed $A$ as a dense subset of $C(K)$ for each such $K$ (by Stone-Weierstrass). Following this reasoning, it is not hard to show that $\|a \|_{K_1} = \|a\|_{K_2}$ for all $a \in A$ iff $K_1 = K_2$.