# Do GNS projections corresponding to commuting subalgebras commute?

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 $$\pi(A)$$.

• 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. Commented Jun 5, 2022 at 20:58

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.

• 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
– Lau
Commented Jun 5, 2022 at 22:14
• In my example one does have that $A\cap B=\mathbb C1$.
– Ruy
Commented Jun 5, 2022 at 22:34
• Oh, yes you're right. Thanks for pointing that out! But we don't have $C= A\otimes B$ in your example.
– Lau
Commented Jun 5, 2022 at 22:42