irr. representation induced from pure, positive form on C*-algebra My question is about the result 2.8.5 in Dixmier's "C* algebras".  The result says: Let $A$ be a C*-algebra, $f$ a pure positive form on $A$ and $N$ the left ideal of those $x\in A$ such that $f(x^*x)=0$.  Then $A/N$, with the scalar product derived from $f(y^*x)$ is complete and therefore coincides with the space of the representation defined by $f$.
I do not see how this is the case, because the representation $\pi$ defined by $f$ uses the Hilbert space obtained from completing $\widetilde{A}/N$, where $\widetilde{A}$ is the unitization of $A$.  Call this Hilbert space $H_\pi$.  Essentially the result claims that $A/N=H_\pi$.  My problem is that I believe $H_\pi$ should have more elements.
I see that elements of $A/N$ can be embedded into $\widetilde{A}/N$ by taking a cosets $\bar{x}\in A/N$ and embedding it as $\overline{(x,0)}\in\widetilde{A}/N$, where $(x,\lambda)$ with $x\in A, \lambda\in\mathbb{C}$ is a typical element in $\widetilde{A}$.
Given any $x\in A$, then what about the coset $\overline{(x,5)}\in\widetilde{A}/N$?  From 2.4.4, where Dixmier explains the GNS construction, then certainly this element belongs in $H_\pi$, but if $H_\pi=A/N$, then I do not know what element in $A$ is mapped to it under the canonical projection. 
$(x,5)$ cannot be equivalent to any element whose second component isn't $5$, and in particular is equivalent to no element of the form $(y,0)$, which is something in $A$, i.e. if $z\in\widetilde{A}\backslash A$, then $z$ is not equivalent to any element in $A$ since every element of $N$ is in $A$.
EDIT: I thought something might change because in the GNS construction we mod out by the set $N_f$ of all $z\in\widetilde{A}$ such that $\widetilde{f}(z^*z)=0$, where $\widetilde{f}$ is the canonical extension of $f$ to $\widetilde{A}$, but even when I do this $A/N$ embeds into $\widetilde{A}/N_f$ and my problem remains.
EDIT 2: I think I figured this out.  You can do the GNS construction either through $\widetilde{A}/N_f$ or $A/N$.  It just so happens that the cyclic vector is easy to construct if you do the former.  Either way, you end up getting a representation $\pi$ of $A$ in the completion of $\widetilde{A}/N_f$ with a cyclic vector $\xi$ and a representation $\phi$ of $A$ in the completion of $A/N$ with a cyclic vector $\eta$ such that $f(x)=(\pi(x)\xi|\xi)=(\phi(x)\eta|\eta)$ for every $x\in A$, and hence there is a unique isomorphism between these two Hilbert spaces that maps $\pi$ to $\phi$ and $\xi$ to $\eta$.
 A: It's still so interesting how unintuitive the notion of infinity can be.
As noted in my edit, the GNS construction can either be done by modding out $A$ by the set $N=\{x\in A|f(x^*x)=0\}$ or by modding out the unitization $\widetilde{A}$ by the set $M=\{x\in\widetilde{A}|\widetilde{f}(x^*x)=0\}$, where $\widetilde{f}$ is the canonical extension of $f$ to the unitization.  I believe the advantage to the latter process is that the accompanying cyclic vector is the image of $1$ in the completion of the quotient group, whereas the cyclic vector in the first method is not as obvious.
Regardless, by construction, let $\pi$, $H_\pi$, and $\xi$ be the representation, Hilbert space and cyclic vector, respectively, from the GNS construction of the latter method and $\phi$, $H_\phi$ and $\eta$ the representation, Hilbert space, and cyclic vector from the former.  Then we have that
$f(x)=(\pi(x)\xi|\xi)=(\phi(x)\eta|\eta)$ for all $x\in A$ and hence there is a unique isomorphism $U$ from $H_\pi\to H_\phi$ that maps $\pi$ to $\phi$ and $\xi$ to $\eta$.
This implies that indeed when $f$ is a pure form then $A/N$ is already complete and hence equal to $H_\phi$.
