# When can $*$-algebras be turned into $C^*$-algebas?

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.

• 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. – s.harp Sep 23 '15 at 12:38
• @s.harp: What you ask for is only true for self-adjoint elements. In general, $\|A\| = \sqrt{\sup(|\text{spec}(A^{\ast}A)|)}$ – Prahlad Vaidyanathan Sep 24 '15 at 2:05
• @PrahladVaidyanathan thanks for the correction – s.harp Sep 24 '15 at 8:19
• 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. – 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 $$\|\pi(x)\| \leq C_x \qquad (1)$$

Proof:

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 $$\mu \circ \pi_u = \varphi$$

• 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. – MaoWao Oct 3 '15 at 5:07