Timeline for Double commutant theorem when $C^*$-subalgebra does not contain identity operator $1$
Current License: CC BY-SA 4.0
7 events
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| Aug 14, 2020 at 2:01 | comment | added | Ruy | 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 $$M''=\smash{\overline M}^{\text{SOT}}\oplus \mathbb{C}I_{K^\perp}.$$ | |
| Aug 13, 2020 at 23:42 | comment | added | Yemon Choi | 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 $I_2$ ... | |
| Aug 12, 2020 at 16:34 | review | Close votes | |||
| Aug 28, 2020 at 3:02 | |||||
| Aug 12, 2020 at 16:16 | comment | added | user62498 | LSpice, thanks your edit | |
| Aug 12, 2020 at 16:10 | comment | added | LSpice |
$M''$ 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.
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| Aug 12, 2020 at 16:10 | history | edited | LSpice | CC BY-SA 4.0 |
TeX fixes
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| Aug 12, 2020 at 16:07 | history | asked | user62498 | CC BY-SA 4.0 |