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Consider a set of linearly independent vectors $\{x_1,\dots,x_n\}$ in some finite-dimensional Hilbert space $H$. For any subset $S \subset [n]$, let $P_S$ be the (orthogonal) projection (operator) onto the span of $x_S := \{x_i, \;i \in S\}$. Let us also write $P_j = P_{\{j\}}$.

We would like to study the collection of projections $\{P_S : \; S \subset [n]\}$. We also have some extra information which can be encoded in the form of a graph $G = ([n], E)$ such that for any $(i,j) \notin E$ \begin{align*} P_S^\perp P_i \perp P_S^\perp P_j, \quad S=[n]\setminus \{i,j\}. \end{align*} In other words, the residual errors after projecting $x_i$ on $x_S$ and $x_j$ on $x_S$ are orthogonal for any two nodes $i,j$ not connected with an edge.

My question is: Are there known algebraic techniques that help study these projections? Searching around, it seems that there is some connection to (finite-dimensional) von Neumann algebras, but I don't know much about them to see the link.

As a concrete question consider this: Fix $j \in [n]$ and $S \subset [n]\setminus\{j\}$ and consider $$ \mathcal{T}_j(S) := \{ T \subset [n]\setminus\{j\}:\; P_T P_j = P_S P_j\}. $$ I believe $\mathcal{T}_j(S)$ is a complete lattice (and the minimum and maximum elements can be read from the graph $G$ ...). Does this follow easily from a more general result?

EDIT: Concrete question 2: Consider $A,B,C \subset [n]$ such that $C$ separates $A$ and $B$ in graph $G$, i.e., there is no path in $G$ from $A$ to $B$ that does not share a node with $C$. Then, do we have: $$ P_C^\perp P_A \perp P_C^\perp P_B? $$

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I don't know about question 2, but question 1 can indeed be answered using a general result about the projection lattice $P$ (ordered by $p\leq q\Leftrightarrow p=pq$) of a von Neumann algebra $A$.

$Q=\{q\in P:pa=qa\}$ is a complete sublattice of $P$, for any $a\in A$ and $p\in P$

Proof: Let $[b]$ denote the range projection of any $b\in A$. If $R\subseteq Q$ then, for all $q\in R$, $[pa]=[qa]\leq q$ so $r=\bigwedge R$ satisfies $[qa]\leq r\leq q$ and hence $ra=rqa=r[qa]qa=[qa]qa=qa=pa$, i.e. $r\in Q$. So $Q$ is closed under taking infimums and, as $pa=qa\Leftrightarrow p^\perp a=q^\perp a$ (where $p^\perp=1-p$), the same applies to $Q^\perp=\{q^\perp:q\in Q\}$. But $p\leq q\Leftrightarrow q^\perp\leq p^\perp$ so this is saying $Q$ is closed under taking supremums, i.e. $Q$ is a complete sublattice of $P$. $\Box$

In fact, the above proof works more generally for any Baer *-ring $A$ (see Berberian's book "Baer *-rings"), or even Rickart *-ring $A$ (where $Q$ is a complete sublattice of $P$ means that $Q$ is closed under infimums and supremums whenever they exist).

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  • $\begingroup$ Thanks. This is so very nice! I assume range projection means projection onto the (1-dim.) subspace spanned by $\{b\}$ (?). Also, your set $Q$ is in fact closer to what I had in mind. I wasn't sure if it is natural to express everything in terms of projections. The way I wrote $Q$ is $\{q \in P:\; p[a] = q[a]\}$ which I guess is equivalent, but maybe not natural (?). $\endgroup$ – passerby51 Jan 25 '16 at 18:36
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    $\begingroup$ For any $b\in A$, $[b]$ is the projection onto the closed linear span of the vectors $bv$, for all $v$ in your Hilbert space. Equivalently, $[b]$ is the unique projection such that $ab=0\Leftrightarrow a[b]=0$, for all $a\in A$ (which is how $[b]$ has to be defined in a Baer/Rickart *-ring). For projection $p\in A$ we simply have $[p]=p$ so the way you have written $Q$ agrees with what I have above if $a$ is a projection. $\endgroup$ – Tristan Bice Jan 25 '16 at 20:06
  • $\begingroup$ OK, I see. I forgot that $b$ is itself an operator. So, basically $[b]$ is the projection onto the closure of range of $b$. $\endgroup$ – passerby51 Jan 25 '16 at 22:35
  • $\begingroup$ By the way, is this result connected to other results (lead to other interesting results) or is something one can derive on the side given general knowledge in the field? (For example, can one characterize $\wedge Q$ and $\vee Q$?) $\endgroup$ – passerby51 Jan 25 '16 at 22:58
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This is an expanded version of Tristan Bice's argument above, as far as I understand. Please feel free to correct. (For example, is it also true that $p \le q \iff p = qp$?)

Let $[b]$ be the range projection of any $b \in A$, i.e., projection onto the closure of the range of $b$. For any $q \in P$ and $a \in A$, we have $[qa] \le q$ (since the range of $qa$ is included in the range of $q$). Also note the identity (2) $b = [b] b, \; \forall b \in A$.

If $R \subset Q$, then for all $q \in R$, we have (1) $[pa] = [qa] \le q$, hence $[pa]$ is lower bound on $R$. Letting $r := \bigwedge R$, by definition of infimum, $ [pa] \le r \le q, \; \forall q \in R$, hence $[qa] \le r \le q, \forall q \in R$ by (1). Hence, \begin{align*} ra &= r qa & (\text{By} \; r \le q \iff r = rq) \\ & = r[qa] qa & (\text{By (2) with $b = qa$}) \\ & = [qa] qa & (\text{By}\; [qa] \le r \iff [qa] = r [qa] ?) \\ & = qa & (\text{By (2) with $b = qa$)}\\ &= pa, \end{align*} showing that $r \in Q$. So, $Q$ is closed under infimums.

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  • $\begingroup$ Yes, it is true that $p=pq$ is equivalent to $p=qp$, as projections are self-adjoint, so $p=pq$ implies $p=p^*=(pq)^*=q^*p^*=qp$. $\endgroup$ – Tristan Bice Feb 3 '16 at 16:34

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