Let $\Gamma$ be a discrete group. Consider a conditional expectation $\Phi: B(l^{2}(\Gamma))\rightarrow l^{\infty}(\Gamma)$ defined by $$\Phi(T)=\sum_{g\in \Gamma}e_{g,g}Te_{g,g},$$ where $e_{g,g}$ is the rank-one projection onto $\delta_{g}\in l^{2}(\Gamma)$ (where $\{\delta_{t} : t\in \Gamma\}$ is the canonical orthonormal basis) and the sum is taken in the strong operator topology.

Then, can we verify that $\Phi(\lambda_{s}T\lambda_{s}^{*})=\lambda_{s}\Phi(T)\lambda_{s}^{*}$ ? (Here, the $\lambda_{s}$ denote the left regular representation: $\lambda_{s}(\delta_{t})=\delta_{st}$ for all $s, t\in \Gamma$.)


Note that $\Phi$ has norm one and $\Phi\circ \Phi = \Phi$. Then use Tomiyama's Theorem (Theorem II.6.10.2 from Blackadar's book Operator algebras; thanks Martin!). Actually, the proof of this theorem will show you how to prove the module property.

  • $\begingroup$ Thanks, but I think Martin's answer is more detailed, so I accpet his answer. $\endgroup$
    – Yan kai
    Oct 15 '14 at 14:31
  • $\begingroup$ @Tomek: I would include in your answer the fact that the theorem you quote is usually known as "Tomiyama's Theorem". $\endgroup$ Oct 16 '14 at 13:08
  • $\begingroup$ Thanks! Will edit accordingly. $\endgroup$ Oct 16 '14 at 13:12

You can also do this explictly: for any $h\in \Gamma$, $$ \Phi(\lambda_s T\lambda_s^*)\delta_h=\sum_g e_{gg}\lambda_s T\lambda_{s^{-1}} e_{gg}\delta_h=e_{hh}\lambda_s T\lambda_{s^{-1}}\delta_h=e_{hh}\lambda_s T\delta_{s^{-1}h}\\=\sum_ge_{hh}\lambda_s\langle T\delta_{s^{-1}h},\delta_g\rangle\delta_g=\sum_g \langle T\delta_{s^{-1}h},\delta_g\rangle e_{hh}\delta_{sg}=\langle T\delta_{s^{-1}h},\delta_{s^{-1}h}\rangle \delta_{h}. $$ And $$ \lambda_s\Phi(T)\lambda_{s^{-1}}\delta_h=\sum_g\lambda_se_{gg}Te_{gg}\delta_{s^{-1}h}=\lambda_s e_{s^{-1}h,s^{-1}h}T\delta_{s^{-1}h}=e_{h,s^{-1}h}T\delta_{s^{-1}h}\\ =\sum_g e_{h,s^{-1}h}\langle T\delta_{s^{-1}h},\delta_g\rangle\delta_g =\langle T\delta_{s^{-1}h},\delta_{s^{-1}h}\rangle\delta_h. $$


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