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Let $V$ be some vector space and $P$ a projection onto a subvectorspace $C \subset V$. Let $\mathcal L$ be a set of unitary operators on $V$ satisfying $AP = PAP$ for all $A \in \mathcal L$. In this paper, the authors prove that $\mathcal L$ is a group. The step in their proof which confuses me a bit is when they show that this set contains inverses. Suppose $A \in \mathcal L$. They show $$(PA^\dagger P)(PAP) = PA^\dagger P^2AP = PA^\dagger (PAP) = PA^\dagger A P = P,$$ which shows that $PA^\dagger P$ is the inverse of $PAP$ on the subspace $C$. They claim that because of this, $P = (PAP)(PA^\dagger P)$, i.e. that $PAP$ and $PA^\dagger P$ commute, and use that fact to complete the proof.

I don't fully understand what they mean when they say that $PA^\dagger P$ is the inverse of $PAP$ on a subspace. I see that $P$ acts as the identity on the subspace, so I can see what they are getting at but I don't see why that guarantees that we can flip around the product. An explanation or proof of this fact would be appreciated!

Sergio Escobar
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More generally, let $M$ be a semi-group (i.e., a set with a multiplication that is associative; no neutral or inverse is postulated) and $P\in M$ an element with $P^2=P$. Define $$\mathcal L=\{\,A\in M\mid AP=PAP\,\}.$$ Then

  • If $A,B\in\mathcal L$ then $ABP=APBP=PAPBP=PABP$, i.e., also $AB\in\mathcal L$
  • $P\in\mathcal L$.

Further, for $A,B\in\mathcal L$, define the equivalence relation $$A\sim B\iff AP=BP$$ and set $$ \mathcal G=\mathcal L/{\sim}$$ and for $A\in \mathcal L$ let $[A]$ denote the equivalence class of $A$ in $\mathcal G$. If $A\sim A'$ and $B\sim B'$, then $ A'B'P=A'PB'P=APBP=ABP$, i.e., $A'B'\sim AB$. Thus multiplication $[A][B]:=[AB]$ in $\mathcal G$ is well-defined and turns $\mathcal G$ into a semi-group.

In this, $[P]$ is right-neutral: $$ P^2=P\implies APP=AP\implies AP\sim A\implies [A][P]=[A].$$ and also left-neutral: $$ PA\sim PAP=AP\sim A\implies [P][A]=[A].$$

Moreover, you already know that every element of $\mathcal G$ has a left-inverse, namely for arbitrary $[A]\in\mathcal G$, we have $[A^\dagger][A]=[P]$.

The rest is an abstract and general argument: a semi-group with left-neutrals and left-inverses is a group.

Hagen von Eitzen
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    The assumption $AP=PAP$ is equivalent to $A(C)\subset C,$ i.e. $C$ is an invariant subspace of $A.$ If $A(C)=C,$ then $A^{-1}(C)=C,$ hence $A^{-1}\in \mathcal{L}.$ Assume $A(C)\subsetneq C.$ What is the inverse of $A$ in $\mathcal{L}$ ? – Ryszard Szwarc Mar 26 '22 at 18:51
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    How do we know that $[A^\dagger]$ is in $\mathcal{G}$? If we knew this, then we must have known that $A^\dagger$ was in $\mathcal{L}$, right? –  Apr 19 '22 at 23:45