# Question about Nate and Taka

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?

"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.