Question about the completion of the GNS representation

Question

Given an underlying C*-algebra $A$ and a state defining a Gelfand ideal $I$, the GNS representation $\pi(A)$ is said to be extended from $A/I$ to the entire Hilbert space $H$ which is the norm-completion. Is $\pi(A)$ not the same as $\pi(A)''$ because the former is norm-complete whereas the latter is weak-complete?

Answer 1

You might be mixing two things. Initially you have a representation $\pi:A\to B(A/I)$. Because $\pi$ is bounded: $$ \|\pi(a)(b+I)\|^2=\langle ab+I,ab+I\rangle=\phi(b^*a^*ab)\leq\|a\|^2\,\phi(b^*b)=\|a\|^2\,\|b+I\|^2, $$ the linear map $\pi$ extends canonically to the completion $H=\overline{A/I}$, and it is not hard to check that the extension is still a $*$-homomorphism.

Now that you have a representation $\pi:A\to B(H)$, you can ask about comparing $\pi(A)$ with $\pi(A)''=\overline{\pi(A)}^{\rm sot}$. Most of the time, they will be different, unless $A$ is a von Neumann algebra and $\phi$ is normal.