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Let $\omega$ be a state on a $C^*$-algebra $C$ and let $A$ and $B$ be commuting unital subalgebras of $C$ (e.g. if $C$ is a $C^*$-tensor product of $A$ and $B$) and let $(\pi,H,\Omega)$ be the GNS-representation of $\omega$. Let $P_A$ be the orthogonal projection onto $[\pi(A)\Omega]$ and let $P_B$ be the projection onto $[\pi(A)\Omega]$ (where $[V]$ is the closed linear hull of a subset $V \subset H$).

Is it true that $P_A$ and $P_B$ commute? Does it help if $\omega$ is pure?

I'm interested in $P_A$ because $(P_A(\pi \upharpoonright A),P_A H,\Omega)$ is unitarily equivalent to the GNS representation of the restricted state $\omega \upharpoonright A$. One can see that $P_A$ commutes with $\pi(A)$.

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – William
    Commented Jun 5, 2022 at 20:58

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The answer is no.

Consider $C={\mathbb C}^3$, with $$ A= \{(\lambda , \lambda , \mu ): \lambda , \mu \in {\mathbb C}\}, $$ and $$ B= \{(\lambda , \mu , \mu ): \lambda , \mu \in {\mathbb C}\}. $$ Taking $$ \varphi (\lambda , \mu , \nu ) = (\lambda + \mu +\nu )/3, $$ we have that $H={\mathbb C}^3$, with coordinatewise multiplication for the representation $\pi $, while the cyclic vector is $\Omega =(1,1,1)/\sqrt 3$.

We then have that $$ [\pi (A)\Omega ] = \{(\lambda , \lambda , \mu ): \lambda , \mu \in {\mathbb C}\}, $$ and $$ [\pi (B)\Omega ] = \{(\lambda , \mu , \mu ): \lambda , \mu \in {\mathbb C}\}, $$ while $$ P_A =\pmatrix {1/2 & 1/2 & 0 \cr 1/2 & 1/2 & 0 \cr 0 & 0 & 1}, $$ and $$ P_B =\pmatrix { 1& 0 & 0 \cr 0 & 1/2 & 1/2 \cr 0 & 1/2 & 1/2 }, $$ which are not commuting matrices.

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  • $\begingroup$ Nice counterexample. Thanks! In my case $C$ is a $C^*$-tensor product of $A$ and $B$, so one needs at least that $A\cap B = \mathbb C 1$. I guess, I should just post that as a new question $\endgroup$
    – Lau
    Commented Jun 5, 2022 at 22:14
  • $\begingroup$ In my example one does have that $A\cap B=\mathbb C1$. $\endgroup$
    – Ruy
    Commented Jun 5, 2022 at 22:34
  • $\begingroup$ Oh, yes you're right. Thanks for pointing that out! But we don't have $C= A\otimes B$ in your example. $\endgroup$
    – Lau
    Commented Jun 5, 2022 at 22:42

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