Let $M$ be a von Neumann algebra acting on a Hilbert space $H$, and let $\tau$ be a faithful tracial state on $M$.
What is the relation between the GNS representation of $(M,\tau)$ and the original action of $M$ on $H$?
More precisely, let $M_\tau$ denote the Hilbert space obtained by completing $M$ with the inner product $\langle x,y\rangle_\tau=\tau(y^*x)$ (since $\tau$ is faithful, this inner product is nondegenerate), and $M$ acts faithfully (again since $\tau$ is a faithful state) on $M_\tau$ by left multiplication. Let $\varphi_\tau:M\to B(M_\tau)$ denote this representation.
My question is: is $\varphi_\tau(M)$ a von Neumann algebra on $M_\tau$? If so, $\tau$ becomes a vector state under the identification $M\simeq\varphi_\tau(M)$: $\tau(u)=\langle u(1_M),1_M\rangle_\tau$.
In case $M$ admits a cyclic vector $x$ for which $\tau(u)=\langle u(x),x\rangle$, the result is true, but the last comment above becomes useless. In fact, in this case the actions of $M$ on $H$ and on $M_\tau$ are unitarily equivalent (see Murphy, Theorem 5.1.4).
The result is that $\varphi_\tau(M)$ is a von Neumann algebra if and only if $\tau $ is normal.
If $\tau$ is normal, then so is $\varphi_\tau$ and so $\varphi_\tau(M)$ is a von Neumann algebra.
Conversely, suppose that $\varphi_\tau(M)$ is a von Neumann algebra. Let $\{x_j\}$ be a monotone net of selfadjoints in $M$ and let $x=\sup x_j\in M$. As $\{\varphi_\tau(x_j)\}$ is a monotone bounded net of selfadjoints in $\varphi_\tau(M)$, since $\varphi_\tau$ is surjective there exists $y\in M$ such that $\varphi_\tau(y)=\sup\varphi_\tau(x_j)$. Since $\varphi_\tau$ is a $*$-isomorphism and $\varphi_\tau(y-x_j)\geq0$, we get that $y-x_j\geq0$ for all $j$. So $y$ is an upper bound for the net, and thus $x\leq y$ (because $x$ is the supremum of the net in $M$). But then $$\varphi_\tau(x_j)\leq\varphi_\tau(x)\leq\varphi_\tau(y),$$ which implies $\varphi_\tau(x)=\varphi_\tau(y)$ since $\varphi_\tau(y)$ is the supremum of the net $\{\varphi_\tau(x_j)\}$ in $\varphi_\tau(M)$. Then $y-x$ is a positive element of $M$ with $\varphi_\tau(y-x)=0$, so by faithfulness $y-x=0$. In other words, $$ \sup\varphi_\tau(x_j)=\varphi_\tau(y)=\varphi_\tau(x)=\varphi_\tau(\sup x_j). $$ So $\varphi_\tau$ is normal. And then $$ \sup\tau(x_j)=\sup\langle\varphi_\tau(x_j)\,\hat 1_M,\hat 1_M\rangle =\langle\sup\varphi_\tau(x_j)\hat1_M,\hat1_M\rangle =\langle\varphi_\tau(\sup x_j)\hat1_M,\hat1_M\rangle=\tau(\sup x_j). $$ (the second equality is due to the fact that $\sup\varphi_\tau(x_j)=\lim_{SOT}\varphi_\tau(x_j)$). So $\tau$ is normal.