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When introducing central covers in this book the following is stated:

For every Non-degenerate representationt $\pi:A\rightarrow B(H)$ there exists a unique normal extension $\gamma:A^{**}\rightarrow B(H)$ such that $\gamma$ restricted to $A$ is $\pi$ and $im(\gamma)=\pi(A)^{''}$.

My first question is what exactly is a normal extension in this context. And my second question, if not obvious, why does this hold?

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"Normal" in this context means that $\gamma$ is continuous with respect to the weak$^\ast$ topologies on $A^{\ast\ast}$ and $B(H)$. The map $\gamma$ is constructed as follows.

Let $\mathscr{M}=\pi(A)^{\prime\prime}$, $\mathscr{M}_\ast$ the set of all normal linear functionals on $\mathscr{M}$, and $\tilde\pi$ the restriction of $\pi^\ast$ to $\mathscr{M}_\ast$ (as a map to $\mathscr{M})$. Then $\gamma=\tilde\pi^\ast\colon A^{\ast\ast}\to (\mathscr{M}_\ast)^\ast=\mathscr{M}$ is continuous w.r.t. the weak$^\ast$ topologies as adjoint of a bounded linear operator.

Let $\iota\colon A\to A^{\ast\ast}$ be the canonical injection. For $a\in A$ and $\omega\in \mathscr{M}_\ast$ we have $$ \langle\gamma(\iota(a)),\omega\rangle_{\mathscr{M},\mathscr{M}_\ast}=\langle \iota(a),\pi^\ast(\omega)\rangle_{A^{\ast\ast},A^\ast}=\langle\pi^\ast(\omega),a\rangle_{A^\ast,A}=\langle\omega,\pi(a)\rangle_{\mathscr{M}_\ast,\mathscr{M}}. $$ Thus $\gamma$ is an extension of $\pi$ in the sense that $\gamma\circ\iota=\pi$.

To see that the image of $\gamma$ is $\mathscr{M}$, first note that the image of the unit ball of $A^{\ast\ast}$ under $\gamma$ is weak$^\ast$ compact in $\mathscr{M}$. Moreover, it contains the image of the unit ball of $A$ under $\pi$, which is weak$^\ast$ dense in the unit ball of $\mathscr{M}$ by the bicommutant theorem. Thus $\gamma$ maps the unit ball of $A^{\ast\ast}$ onto the unit ball of $\mathscr{M}$. Hence $\gamma(A^{\ast\ast})=\mathscr{M}$.

Finally, $\gamma$ is the unique normal extension of $\pi$ to $A^{\ast\ast}$ since $\iota(A)$ is weak$^\ast$ dense in $A^{\ast\ast}$ by Goldstine's theorem.

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