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Double commutant theorem: For a unital $C^*$-subalgebra $M \subset B(H)$, one has $$\overline M^\text{SOT}=\overline M^\text{WOT}=M^{''}$$$$\smash{\overline M}^\text{SOT}=\smash{\overline M}^\text{WOT}=M''.$$

My question: For a $C^*$-subalgebra $M \subset B(H)$ but don't assume $M$ contains identity operator $1$, does $$\overline M^\text{SOT}=\overline M^\text{WOT}=M^{''}?$$

Thanks so much for your time and your answers.$$\smash{\overline M}^\text{SOT}=\smash{\overline M}^\text{WOT}=M''?$$

Double commutant theorem: For a unital $C^*$-subalgebra $M \subset B(H)$, one has $$\overline M^\text{SOT}=\overline M^\text{WOT}=M^{''}$$

My question: For a $C^*$-subalgebra $M \subset B(H)$ but don't assume $M$ contains identity operator $1$, does $$\overline M^\text{SOT}=\overline M^\text{WOT}=M^{''}?$$

Thanks so much for your time and your answers.

Double commutant theorem: For a unital $C^*$-subalgebra $M \subset B(H)$, one has $$\smash{\overline M}^\text{SOT}=\smash{\overline M}^\text{WOT}=M''.$$

My question: For a $C^*$-subalgebra $M \subset B(H)$ but don't assume $M$ contains identity operator $1$, does $$\smash{\overline M}^\text{SOT}=\smash{\overline M}^\text{WOT}=M''?$$

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user62498
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Double commutant theorem when $C^*$-subalgebra does not contain identity operator $1$

Double commutant theorem: For a unital $C^*$-subalgebra $M \subset B(H)$, one has $$\overline M^\text{SOT}=\overline M^\text{WOT}=M^{''}$$

My question: For a $C^*$-subalgebra $M \subset B(H)$ but don't assume $M$ contains identity operator $1$, does $$\overline M^\text{SOT}=\overline M^\text{WOT}=M^{''}?$$

Thanks so much for your time and your answers.