An algebra with a strictly positive state + condition ? to become a C$^*$-algebra.

Question

Let $(A,*)$ be an involutive $\mathbb{C}$-algebra i.e. a $\mathbb{C}$-algebra with a semi-linear anti-automorphism i.e. identically

1. $(x+y)^*=x^*+y^*$
2. $(\alpha.x)^*=\overline{\alpha}.x^*$
3. $(xy)^*=y^*x^*$

A strictly positive (or faithful) state $\varphi\in A^*$ is such that for all $x\in A\setminus \{0\}$, $\varphi(x^*x)> 0$. One can prove easily that $g(x,y):=\varphi(x^*y)$ is a non-degenerate hermitian form. It can be noted that, due to the fact that $g(a.x,y)=g(x,a^*y)$ and with finite dimensionality, one gets that $A$ is semi-simple. So, one can consider $A$ as simple.

My question is the following.

Q) We suppose $A$ to be finite dimensional. Under which additional condition(s) is the Hilbert space $(A,g)$ a C$^*$-algebra ?

Late edit I insist on the fact that I am in search of conditions for $(A,g)$ to be a C$^*$-algebra, not only $A$. This means that it should be a C$^*$-algebra for the precise norm $||x||=\sqrt{g(x,x)}$.

Late remark The fact that $A$ be a star-algebra of finite dimension, sum of matrix algebras is by no means sufficient to imply that the projectors on the blocks are *-invariant nor $A\simeq \mathbb{C}$ as shows the following counterexample. Take $B=\mathbb{C}^{n\times n}$ (algebra of complex square matrices of dimension $n>0$) and $A=B\oplus B$ with the anti-automorphism $(X,Y)^\star=(Y^*,X^*)$. Then $(A,\star)$ is easily checked to be a star algebra. It is of finite dimension, sum of matrix algebras but $dim_\mathbb{C}=2n^2\not= 1$. Indeed, the existence of a strictly positive state is crucial as there is none over $A$.

Answer 1

The following statements are equivalent:

• $A$ is a C$^*$-algebra with the norm induced by $\varphi$

• $\dim A=1$, i.e., $A=\mathbb C$

Proof. if $A=\mathbb C$, then the only state is the identity, which induces the norm.

Conversely, if $A$ is a C$^*$-algebra, its norm satisfies the C$^*$-identity $$\tag1\|x\|^2=\|x^*x\|.$$ In terms of $\varphi$, this is $$\tag2\varphi(x^*x)=\varphi((x^*x)^2)^{1/2},\qquad x\in A.$$ So, for each positive $a\in A$, $$\tag3\varphi(a)^2=\varphi(a^2).$$ As $\varphi$ is a ucp map, the equality $(3)$ says that $a$ is in the multiplicative domain of $\varphi$. Thus the multiplicative domain of $\varphi$ is all of $A$, since a C$^*$-algebra is spanned by its positive elements. So $\varphi$ is a faithful representation of $A$ into (thus onto) $\mathbb C$.

Answer 2

No Hilbert space is a $C^*$-algebra in any natural way since the structure of Hilbert spaces (a vector space with an inner product) is quite different from that of a C*-algebra (a Banach space equipped with an involutive algebra structure).

Nevertheless from the data provided by the OP one may give $A$ the structure of a $C^*$-algebra in a very natural way. The first step is to consider the hermitian form $g$ as an inner product on $A$, so that $$ H:= (A,g) $$ becomes a Hilbert space (completeness follows from finite dimensionality). One may then define a map $$ \pi :A\to B(H), $$ by $$ \pi (a)\xi = a\xi , \quad\forall a\in A,\quad \forall \xi \in H, $$ (here $a\xi $ is nothing other than the product of $a$ and $\xi $).

One may easily show that $\pi $ is a *-homomorphism. It is moreover injective since, for $a\neq 0$, one has that $\pi (a)a^*=aa^*\neq 0$, as a consequence of $\varphi (aa^*)>0$.

Finally, identifying $A$ with its image within $B(H)$ via $\pi $, we have that $A$ becomes a $C^*$-algebra.

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Later edit: The late edit indeed makes the question trivial and, as noted my Martin, the only example is the complex numbers.

A perhaps more elementary proof of this fact is that every finite dimensional $C^*$-algebra is the direct sum of matrix algebras, so unless $\text{dim}(A)=1$, there exists a pair of mutually orthogonal projections $p$ and $q$. It follows that $\|p\pm q\|=1$, hence the parallelogram law fails.