Let $A$ be a (not necessarily unital) complex $*$-algebra, i.e. an algebra over $\mathbb{C}$ together with an involution $*: A \to A$.

There exists at most one norm on $A$ turning $A$ into a $C^*$-algebra with respect to this norm. For instance, if $A = \mathcal{C}(X)$ where $X$ is a compact Hausdorff space, then the only norm on $A$ turning $A$ into a $C^*$-algebra is $\|f\| = \sup_{x \in X} |f(x)|$.

However, there could still exist several or no norms on $A$ such that the completion of $A$ with respect to this norm becomes a $C^*$-algebra. For instance, if $A$ and $B$ are $C^*$-algebras and $A \otimes B$ is their algebraic tensor product, then this can be turned into a $*$-algebra in an obvious way, but in general there exists several norms on the tensor product that completes it into a $C^*$-algebra.

I was wondering if there was any characterization of the $*$-algebras $A$ with the property that such a norm exists, or at least some necessary or sufficient conditions.

  • $\begingroup$ One free constraint that you get is that the spectrum of each operator must be bounded, as for a C*-star algebra you have $||A||=\sup(|\text{spec}(A)|)$. The spectrum is the set of complex numbers $\lambda$ so that $\lambda -A$ is not invertible if you adjoin a unit to the algebra, which is always possible and depends only on the algebraic structure the algebra. $\endgroup$ – s.harp Sep 23 '15 at 12:38
  • $\begingroup$ @s.harp: What you ask for is only true for self-adjoint elements. In general, $\|A\| = \sqrt{\sup(|\text{spec}(A^{\ast}A)|)}$ $\endgroup$ – Prahlad Vaidyanathan Sep 24 '15 at 2:05
  • $\begingroup$ @PrahladVaidyanathan thanks for the correction $\endgroup$ – s.harp Sep 24 '15 at 8:19
  • 1
    $\begingroup$ If we are allowed to take the completion after defining the norm, then the completion may in general have way more invertible elements than the original algebra (as s.harp and I discovered in a recent discussion). Thus, boundedness of spectra is not even a necessary condition. $\endgroup$ – Josse van Dobben de Bruyn Nov 18 '15 at 3:14

The question you ask can be rephrased as follows: We wish to know when there exists a $C^{\ast}$ algebra $B$ and a $\ast$-homomorphism $\pi_u : A \to B$ with the universal property that if $C$ is any other $C^{\ast}$-algebra, and $\varphi : A\to C$ any $\ast$-homomorphism, $\exists$ a unique $\ast$-homomorphism $\mu : B\to C$ such that $$ \mu \circ \pi_u = \varphi $$ The pair $(B,\pi_u)$ is called the universal $C^{\ast}$-envelope of $A$.

We have the following theorem, giving a condition that is sufficient for such an envelope to exist, but perhaps it is hard to check.

If $A$ is a $\ast$-algebra generated by a set $S$ (so every element of $A$ is a polynomial in $S\cup S^{\ast}$), then $A$ admits a universal $C^{\ast}$-envelope iff for every $x\in S, \exists C_x \geq 0$ such that for any $\ast$-representation $\pi : A\to B(H)$ on a Hilbert space, one has $$ \|\pi(x)\| \leq C_x \qquad (1) $$


Suppose $A$ has a $C^{\ast}$-envelope $(B,\pi_u)$, then for any $\ast$-represention $\pi :A\to B(H)$, there is a $\ast$-homomorphism $\mu :B\to B(H)$ such that $\mu \circ \pi_u = \pi$. Then, it follows that $C_x := \|\pi_u(x)\|$ works since $\mu$ is contractive.

Conversely, suppose such a $C_x$ exists for each $x\in S$, then such a $C_x$ also exists for any $x\in A$ (since $A$ is merely polynomial expressions in $S\cup S^{\ast}$). For each $x\in A$, let $C_x \geq 0$ be minimal with the property $(1)$.

Then fix $x\in A$: For each $n\in \mathbb{N}, \exists$ as $\ast$-representation $\pi_n : A\to B(H_n)$ such that $$ \|\pi_n(x)\| \geq C_x - 1/n $$ Take $\pi_x := \oplus \pi_n$ and $H_x := \oplus H_n$, then $\pi : A\to B(H_x)$ is a $\ast$-representation such that $$ \|\pi_x(x)\| = C_x $$ Now set $$ H = \oplus_{x\in A} H_x \text{ and } \pi_u := \oplus_{x\in A} \pi_x $$ Then $\pi_u : A\to B(H)$ is a $\ast$-representation such that $$ \|\pi_u(x)\| = C_x \quad\forall x\in A $$ Let $B$ denote the closure of $\pi_u(A)$ in $B(H)$, then we claim that $(B,\pi_u)$ is the universal $C^{\ast}$-envelope of $A$ : To see this, suppose $\varphi :A\to C$ is a $\ast$-homomorphism into a $C^{\ast}$-algebra $C$, then we may assume that $C \subset B(K)$ for some Hilbert space $K$. Hence $$ \|\varphi(x)| \leq C_x = \|\pi_u(x)\| \quad\forall x\in A $$ and so $\exists$ a $\ast$-homomorphism $\mu : B\to C$ such that $$ \mu \circ \pi_u = \varphi $$

  • 1
    $\begingroup$ One may note that the criterion is not too hard to check in many interesting cases, namely, partial isometries and projections satisfy property (1), and some interesting $C^\ast$-algebras arise as universal $C^\ast$-algebras generated by these (Cuntz algebras, group algebras of discrete groups, ...). However, if I understood the OP right, he is really interested in the question when $\pi_u$ is injective. $\endgroup$ – MaoWao Oct 3 '15 at 5:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.