Let $M$ be a von Neumann algebra. For any faithful normal state $\omega$ on $M$, according to the GNS construction, we have $\omega(x)=\langle x \Omega,\Omega\rangle$, where $\Omega$ is a cyclic vector.
For any two different faithful normal states, are the cyclic vectors in the above construction the same?
Best Answer
Yes, they both coincide with the unit of $M$, so in a sense they are indeed the same. However the two GNS representations take place in DIFFERENT Hilbert spaces, namely the completion of $M$ relative to two different inner products.
If the states are $\omega _1$ and $\omega _2$, and the corresponding GNS representations are $(\pi _1,H_1,\xi _1)$ and $(\pi _2,H_2,\xi _2)$, then both $H_1$ and $H_2$ contain a dense copy of $M$ and, for every $m$ on $M$, the diagram $\require{AMScd}$ \begin{CD} H_1 @<\iota_1<<M @>\iota_2>> H_2\\ @V\pi_1(m)VV @VL_mVV @VV\pi_2(m)V\\ H_1 @<\iota_1<<M @>\iota_2>> H_2\\ \end{CD} commutes, where $L_m$ is the operator of left-multiplication by $m$, and $\iota_1$ and $\iota_2$ are the inclusion maps.
This might give an impression that $π_1$ and $π_2$ are equivalent representations, and indeed they are (!) when restricted to the dense subspaces $\iota_1(M)$ and $\iota_2(M)$.
However, the extra vectors added in the completion process strongly depend on the states $\omega_1$ and $\omega_2$, so the representations $\pi_1$ and $\pi_2$ might very well be inequivalent!