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Let A be a $\mathcal{C}$* algebra. We define state , say $\phi$ on A ( linear functional on A) such that f is positive and $\phi$( 1)= 1 .

I'm trying to prove the following : If A is isomorphic to C iff order of state space is 1.

The usual part is trivial as if A is isomorphic then there is only one map , the identity function.

But how to I conclude the converse part? . As $\mathbb{C}$ is multiplicative , from there does it follow that state is homomorphism ? & What should I do to show its injective?

Leo
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1 Answers1

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Suppose there exists only one state $\varphi$ on the unital $C^*$-algebra $A$. Then $\varphi$ must be faithful. If not, then there is some nonzero positive element $a\in A$ with $\varphi(a)=0$. But every positive element achieves its norm by some state, a contradiction.

If now $a\in A$ is positive, $\|a\|=1$, then $\varphi(a)=1$, so $\varphi(1-a)=0$ and thus $a=1$. Thus every positive element is of the form $\lambda\cdot1$ for some $\lambda\in[0,\infty)$. Now every element of $A$ is in the span of the positive elements, hence lies in the span of $1$, and therefore $A=\mathbb C\cdot 1\cong\mathbb C$.

Aweygan
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  • I understood it. Then from this fact can I conclude that there's always exists a non trivial unitary element in a unital $\mathcal{C}$* algebra other than $\mathbb{C}$ . – Leo May 17 '20 at 00:11
  • And there always exists a non trivial self adjoint element in a $\mathcal{C}$* algebra other than $\mathbb{C}$ – Leo May 17 '20 at 00:20
  • Yes, if $A$ is a unital $C^*$-algebra, and every self-adjoint element is of the form $\lambda1$ for some $\lambda\in\mathbb R$, then $A\cong\mathbb C$. – Aweygan May 17 '20 at 00:41