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Let $A$ be a $C^*$-algebra and $\pi_1 , \pi_2$ two *-representations on a same Hilbert space $\mathcal{H}$ so that the inclusions $\pi_1(A)^{''}\subseteq \pi_2(A)^{''}$ of the associated von Neumann algebras makes sense ( ' stands for commutant, the double commutant coincides with the "weak, strong, ultraweak = $\sigma$-weak, ultrastrong= $\sigma$-strong" closure).

If the inclusion holds, can one say anything then on $\pi_1(A)$ and $\pi_2(A)$ ? is there also an inclusion? Is that a purely topological question (the fact the one has $C^*$-algebras seems irrelevant)? Examples, counter examples?


Simple question, simple remarks:

  • the norm = uniform topology on $\mathcal{B}(\mathcal{H})$ is finer than the bunch of other topologies mentionned above, this means that an open in the "... " topologies is necessarily open in the norm but a "..." closed subset is also closed in the norm topology

  • for $\pi$ a *-representation and $A$ a $C^*$-algebra, $\pi(A)$ is also a $C^*$-algebra, namely it is complete, norm closed. Similarly, if $A$ were a von Neumann algebra and $\pi$ normal then the image is a von Neumann algebra. 3.12 p.136 in Takesaki (normal: In a von Neumann algebra, a bounded increasing net of positive elements has a supremum, a limit, cf. monotone closed. A morphism $\phi$ between two von Neumann algebras is normal if it preserves the limit, $\phi(\mathrm{lim} ...) = \mathrm{lim} \phi(...)$

so if $A$ were are von Neumann algebra, $\pi_2$ normal then it would work haha

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I don't think you can say anything. For example, let $A_1=K(H)\subset B(H)$ and $A_2\subset B(H)$ be an image of the Cuntz algebra $O_2$ via an irreducible representation; this representation is faithful, because $O_2$ is simple.

The intersection $A_1\cap A_2$ is $0$, because $O_2$ has no compact operators. Now form $A=A_1\oplus A_2$, and let $\pi_j:A\to B(H)$ be given by $\pi_1(a\oplus b)=a$, $\pi_2(a\oplus b)=b$.

Then $\pi(A_1)''=\pi_2(A_2)''=B(H)$, while $\pi_1(A)\cap\pi_2(A)=A_1\cap A_2=\{0\}$.

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