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Let $\mathcal{A}$ be a $C^*$ -algebra and $\mathcal{B} \subset \mathcal{A}$ a $C^*$ -subalgebra, both having the same identity element.
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Proposition - Given a state $\phi$ of $\mathcal{A}$ , its restriction $\phi|_{\mathcal{B}}$ to $\mathcal{B}$ is also a state of $\mathcal{B}$ .
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Remark - Note that the requirement that the $C^*$ -algebras $\mathcal{A}$ and $\mathcal{B}$ have a (common) identity element is necessary.
For example, let $X$ be a compact space and $C(X)$ the $C^*$ -algebra of continuous functions $X \to \mathbb{C}$ . Pick a point $x_0 \in X$ and consider the $C^*$ -subalgebra of continuous functions $X \to \mathbb{C}$ which vanish at $x_0$ . Notice that this subalgebra never has the same identity element of
$C(X)$ (the constant function that equals $1$ ). In fact, this subalgebra may not have an identity at all.
Now the evaluation mapping at $x_0$ , i.e. the function $\mathrm{ev}_{x_0}: C(X) \to \mathbb{C}$
is a state of $C(X)$ . Of course, its restriction to the subalgebra in question is the zero mapping, therefore not being a state.
Let $\mathcal{A}$ be a $C^*$ -algebra and $\mathcal{B} \subset \mathcal{A}$ a $C^*$ -subalgebra (not necessarily unital).
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Theorem 1 - Every state $\phi$ of $\mathcal{B}$ admits an extension to a state $\widetilde{\phi}$ of $\mathcal{A}$ . Moreover, every pure state $\phi$ of $\mathcal{B}$ admits an extension to a pure state $\widetilde{\phi}$ of $\mathcal{A}$ .
Theorem 2 - The set of extensions of a state $\phi$ of $\mathcal{B}$ is a compact and convex subset of $S_{\mathcal{A}}$ , the state space of $\mathcal{A}$ endowed with the weak-* topology.
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