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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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    $\begingroup$ $M''$ is unchanged upon adjoining $1$, so it equals the strong and weak operator closures of $\mathbb C + M$. By the way, in terms of TeX, don't use $M^{''}$ $M^{''}$; instead use $M''$ $M''$ or $M^{\prime\prime}$ $M^{\prime\prime}$. I have edited accordingly. $\endgroup$
    – LSpice
    Commented Aug 12, 2020 at 16:10
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    $\begingroup$ LSpice, thanks your edit $\endgroup$
    – user62498
    Commented Aug 12, 2020 at 16:16
  • $\begingroup$ As @LSpice points out, $M''$ always contains the identity operator $I\in B(H)$. Now think about whether every ${\rm C}^*$-subalgebra of $M_2({\bf C})$ contains the identity matrix $I_2$ ... $\endgroup$
    – Yemon Choi
    Commented Aug 13, 2020 at 23:42
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    $\begingroup$ If $K$ is the essential space for $M$, namely $K=\overline{\text{span}}\{T(\xi): T\in M, \xi\in H\}$, then $H=K\oplus K^\perp$ and $M$ acts on $H$ like $M\oplus0$, while $$M''=\smash{\overline M}^{\text{SOT}}\oplus \mathbb{C}I_{K^\perp}.$$ $\endgroup$
    – Ruy
    Commented Aug 14, 2020 at 2:01

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