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

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
$$pi :Ato B(H),$$
by
$$pi (a)xi = axi , quadforall ain A,quad forall xi in H,$$
(here $$axi$$ 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 $$aneq 0$$, 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.

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 $$|ppm q|=1$$, hence the parallelogram law fails.

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
$$tag2varphi(x^*x)=varphi((x^*x)^2)^{1/2},qquad xin A.$$
So, for each positive $$ain A$$,
$$tag3varphi(a)^2=varphi(a^2).$$
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) $$mathbb C$$.