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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. That 4 implies 3 follows by performing the GNS construction for $(A,\varphi)$.

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 $(A,\varphi)$.

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. That 4 implies 3 follows by performing the GNS construction for $(A,\varphi)$.

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. That 4 implies 3 follows by performing the GNS construction for $(A,\varphi)$.

Your first bullet point is essentially correct, but you need to give a slight modification since the spectral radius only equals the norm for normal elements. I first heard this exact characterization from Andre Henriques.

A unital complex *-algebra $A$ is a C${}^*$-algebra if and only if the function $\|\cdot\|: A \to [0,\infty]$ given by $$ \|a\|^2 := \sup\left\{ |\lambda| : a^*a - \lambda 1_A \text{ is not invertible}\right\} $$ gives a well-defined norm on $A$ which satisfies the following properties:

• $A$ is complete in this norm,

• this norm is sub-multiplicative $\|ab\| \leq \|a\|\cdot\|b\|$, and
• this norm satisfies the C${}^*$-axiom $\|a^*a\|=\|a\|^2$.

Edit: Here is another characterization which does not use the spectral radius.

A complex *-algebra is a C${}^*$-algebra if and only if it admits a faithful *-representation into $B(H)$ for some Hilbert space $H$ whose image is norm closed.

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. That 4 implies 3 follows by performing the GNS construction for $(A,\varphi)$.