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Let $A$ be a unital Baer *-ring.

1- Assume that $\{p_i\}$ is a family of projections in $A$. Let $x$ be an isometry in A (I mean $x^*x=1$ where $1$ is the unit of $A$). True or false: $\inf (xp_ix^*)=x(\inf p_i)x^*$ !

2- Let $y$ be an element of $A$. Let us denote $[y]$ by the smallest projection with $[y]y=y$. Let $q$ be a projection in $A$ and assume that $qy=0$. Can we conclude that $q[y]=0$?!

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1 Answer 1

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A useful reference is Berberian's book "Baer *-rings".

1 - If $x^*x=1$, then $x$ is certainly a partial isometry, since $xx^*x=x$. By taking $e=1$ and $f=xx^*$ in Proposition 1.9 of Berberian, we obtain a *-isomorphism $\varphi:A\to fAf$, $a\mapsto xax^*$ that restricts to an order isomorphism between the projections of $A$ and $fAf$. Since order isomorphisms preserve infima, we obtain $\varphi(\inf_i p_i)=\inf_i \varphi(p_i)$, i.e., $x(\inf_i p_i)x^*=\inf_i(xp_ix^*)$.

2 - Yes, see Proposition 3.3 of Berberian.

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  • $\begingroup$ Dear Bert Lindenhovius,thanks a lot. It was great. $\endgroup$
    – ABB
    Commented Nov 3, 2016 at 6:07

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