Every representation of a $C^*$-algebra lifts to a representation of double dual $A^{**}$

Question

Let $A$ be a $ C^*$-algebra and $\pi: A \to B(H)$ be a representation of $A$.

Does there exist a representation $\tilde{\pi}$ of $A^{\ast \ast}$ such that $\tilde{\pi} \vert_{A}=\pi$

Any ideas/ references?

Answer 1

The double dual $A^{**}$ is realized as a von Neumann algebra via the Universal Representation. That is, $A^{**}=\pi_U(A)''$, where $\pi_U$ is the universal representation. The "universality" in the universal representation comes from the fact that is satisfies a universal property. Said property is the following:

Given a representation $\pi:A\to B(H)$, there exists a normal representation $\tilde\pi:A^{**}\to B(H)$ such that $$ \tilde\pi\circ\pi_U=\pi.$$

This is done for instance on Lemma III.2.2 and Theorem III.2.4 of Takesaki's Theory of Operator Algebras I.