Let $A$ and $B$ be unital $C^*$-algebras and let $\omega$ be a state on a $C^*$-tensor product $A\otimes_\beta B$. Let $(\pi,H,\Omega)$ be the GNS-representation of $\omega$. Let $P_A$ be the projection onto $[\pi(A\otimes1)\Omega]$ and let $P_B$ be the projection onto $[\pi(1\otimes B)\Omega]$. Here $[V]$ denotes the closed linear hull of a subset $V \subset H$.
Is it true that $P_A$ and $P_B$ commute?
I'm interested in these projections because they induce the GNS-representations of the "reduced states" $\omega_A$ and $\omega_B$, e.g. $\omega_A(a) =\omega(a\otimes1)$.
Because of uniqueness one has that $(P_A(\pi( \,- \otimes 1)),P_A H,\Omega)$ equivalent to the GNS representation $\omega_A$.
An answer for von Neumann algebras would also be much appreciated, if that's easier.
Best Answer
I think the answer is again no. Take $A=B=\mathbb C^2$, so that $C=A\otimes B = \mathbb C^4$.
The goal is to reduce this to my answer to your previous question by choosing an appropriate state.
Let $e_1$ and $e_2$ be the canonical basis vectors in $\mathbb C^2$, and choose the state $\phi$ on $C$ such that $\phi(e_i\otimes e_j)=1/3$, for all $i,j$, except for $\phi(e_2\otimes e_1)=0$.
If $(\pi,H,\Omega)$ is the corresponding GNS representation we then have that $\pi(e_2\otimes e_1)\Omega =0$, and hence $$ f_1:= \pi (e_1\otimes e_1)\Omega , $$ $$ \qquad\ f_2:= \pi (e_1\otimes e_2)\Omega ,\quad \text{and} $$ $$ f_3:= \pi (e_2\otimes e_2)\Omega $$ span $H$.
It is not hard to see that these form a linearly independent set, and hence that $H$ is 3-dimensional.
Given any $a$ in $A$, say $a=(\lambda ,\mu )$, we have that $$ \pi (a\otimes 1) = \pi ((\lambda e_1 + \mu e_2)\otimes 1)= \lambda \pi (e_1\otimes 1) + \mu \pi (e_2\otimes 1), $$ so it is easy to see that $$ \pi (a\otimes 1)f_1 = \lambda f_1, \quad \pi (a\otimes 1)f_2 = \lambda f_2, \quad \text {and} \quad \pi (a\otimes 1)f_3 = \mu f_3, $$ whence the matrix of $\pi (a\otimes 1)$ in the basis $\{f_1, f_2, f_3\}$ is $$ \pmatrix{\lambda & 0 & 0 \cr 0 & \lambda & 0 \cr 0 & 0 & \mu }. $$ If instead $b = (\lambda ,\mu )\in B$, then $$ \pi (1\otimes b) = \pi (1\otimes (\lambda e_1 + \mu e_2)) =\lambda \pi (1\otimes e_1) + \mu \pi (1\otimes e_2), $$ so a similar analysis gives the matrix of $\pi (1\otimes b)$, as being $$ \pmatrix{\lambda & 0 & 0 \cr 0 & \mu & 0 \cr 0 & 0 & \mu }. $$
Identifying $H$ with ${\mathbb C}^3$ via the basis mentioned above we see that $\Omega$ is proportional to $(1,1,1)$, and $$ [\pi (A\otimes 1)\Omega ] = \{(\lambda , \lambda , \mu ): \lambda , \mu \in {\mathbb C}\}, $$ and $$ [\pi (1\otimes B)\Omega ] = \{(\lambda , \mu , \mu ): \lambda , \mu \in {\mathbb C}\}, $$ whence $$ P_A =\pmatrix {1/2 & 1/2 & 0 \cr 1/2 & 1/2 & 0 \cr 0 & 0 & 1}, $$ and $$ P_B =\pmatrix { 1& 0 & 0 \cr 0 & 1/2 & 1/2 \cr 0 & 1/2 & 1/2 }, $$ which do not commute.
I am still thinking about what happens when $\phi$ is a product state and I believe the answer is affirmative in that case.