Let $M$ be a vN algebra (represented GNS space with respect to state) in standard form. Under which condition we can say a subalgebra $B$ of $M$ is also in standard form? If there exist $\varphi$ preserving conditional expectation $E_{\varphi}: M\rightarrow B$, will $B$ be in standard form (on GNS space of M)?
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3$\begingroup$ Questions of the form "under what extra conditions does this desired result hold" are not ideal, and feel like a shopping list, because you are asking other people to invent things rather than answer a specific problem. Could you try to think of a more specific version of your question, such as: "if B has this condition, will it also be in standard form"? $\endgroup$– Yemon ChoiCommented Apr 17, 2019 at 11:47
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$\begingroup$ @YemonChoi now you feel specific right not a shopping list. $\endgroup$– user136400Commented Apr 17, 2019 at 11:53
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1$\begingroup$ Thanks, that is much better - I am not a specialist on vN algebras but the specific question should now be much easier for people to answer quickly $\endgroup$– Yemon ChoiCommented Apr 17, 2019 at 12:11
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$\begingroup$ Thanks to @ Yemon Choi $\endgroup$– user136400Commented Apr 17, 2019 at 12:32
1 Answer
I'm not an expert on von Neumann algebras either, but here are some easy observations. 1. Having a conditional expectation is not sufficient. Any state on $M$ can be regarded as a conditional expectation onto the one dimensional subalgebra $B$ consisting of the scalar multiples of the identity. If ${\rm dim}(M) > 1$ then $L^2(M)$ will be too big to be the Hilbert space of the standard representation of this $B$. 2. $L^2(M)$ will never be naturally the same as $L^2(B)$, if $B$ is properly contained in $M$. 3. On the other hand, you could get lucky and have the representation of $B$ be unitarily equivalent to its standard representation. For instance, if $M = B(H)$ then its standard rep is as $B(H)\otimes I$ on $H \otimes H$. If $H = K\otimes K$ and $B = B(K)\otimes I \subset M$ then restricting the standard rep of $M$ to $B$ gives you $B(K)\otimes I\otimes I \otimes I$, which if $K$ is infinite dimensional is unitarily equivalent to $B(K)\otimes I$.