Explaining causal completion axiom in Haag-Kastler axioms?

Question

There are several variants of the Haag-Kastler axioms for algebraic quantum field theory. Usually one associates an algebra $\mathcal{A}(O)$ to each open region $O$ of spacetime. An often-suggested axiom is that the algebra associated to a region is the same as the algebra associated to its causal completion. To be precise: Let the causal complement $O'$ of a spacetime region $O$ be the set of points that are spacelike separated from all points in $O$. Then $O''=(O')'$ is called the causal completion, and the axiom in question axiom states $$\mathcal{A}(O'')=\mathcal{A}(O).$$ (For instance, see Haag, "Local Quantum Physics," Equation III.1.10.) Due to inconsistent terminology, I'll just call this the causal shadow axiom.

Can one provide a more elaborate explanation/intuition for the causal shadow axiom or show that it holds for free scalar field theory?

In some presentations -- including, it seems, the original Haag-Kastler paper -- the authors make an alternative statement: $$\mathcal{A}(D(O))=\mathcal{A}(O),$$ where $D(O)$ is the "domain of dependence" or "causal shadow" of $O$, i.e. the set of points $p$ such that every inextendible causal curve through $p$ intersects $O$. This weaker axiom makes more sense to me. (I think the causal completion axiom implies the causal shadow axiom.)

To highlight the difference between the causal shadow and causal completion, consider $O$ as the union of two small balls centered at the points $(1,0)$ and $(-1,0)$ in $(t,x)$ coordinates. Then $O''$ will approximately be the "causal diamond" with vertices at $(\pm 1,0), (0,\pm 1)$, but the causal shadow is much smaller.

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In response to comments: Another axiom is the "Haag duality" axiom: $$\mathcal{A}(O')=\mathcal{A}(O)',$$ where $\mathcal{A}'$ is the commutant. Because $\mathcal{A}''=\mathcal{A}$ for von Neumann algebras, you can iterate the duality axiom to see it implies the causal completion axiom. In some places (e.g. Haag and Araki textbooks), I only found the duality axiom stated for the case that $O$ is a causal diamond, but e.g. in "Introduction to Algebraic Quantum Field Theory" by Horuzhy (p21 and p48), Haag duality is discussed for arbitrary regions, thus implying the causal completion axiom for arbitrary regions. Rather than getting lost in a web of claims, I'd rather try to justify/refute the causal completion axiom, perhaps considering the case of a free theory.

Answer 1

I got a personnal proof for $ D(O)= O'' $ in Minkowski spacetime and I'm sure it holds in more general spacetimes...

Notations and conventions: $S\subset \mathcal M$ subset of the Minkowski spacetime, $\eta$ has signature (+,-,-,-) $$\textbf{causal cone}\qquad J(p) := \left\lbrace q\in\mathcal{M},\ (q-p)^2 \geq 0,\ \forall\ p\in S\right\rbrace,\quad J(S)= \bigcup_{p\in S} J(p)$$ $$\textbf{causal complement}\qquad S' := \left\lbrace q\in\mathcal{M},\ (q-p)^2 < 0,\ \forall\ p\in S\right\rbrace = \mathcal{M} \backslash J(S) $$

One wants to show (cf. comment, this is not the correct def. of $D(S)$) $$\textbf{domain of dependence} \qquad S'' \stackrel{!}{=} \left\lbrace q\in\mathcal{M},\ J(q) \subseteq J(S) \right\rbrace =: D(S) $$

This can be written as an equivalence $\ J(q) \subseteq J(S) \enspace \Longleftrightarrow\enspace q\in S''$ and note that each side can be understood as an implication $ A\ \Rightarrow\ B$ or with logical connectives $\neg A\ \text{or}\ B$. (i.e. $p\in J(q)\ \Rightarrow\ p\in J(S)$ and $ p\in S'\ \Rightarrow \ (q-p)^2 <0$)

$\underline{(\Rightarrow):}$ assuming that for all $ p\in \mathcal{M}$, either $ p\in \{q\}'\ \text{or}\ p\in J(S)$, one checks that $\ p\in S'\enspace\Rightarrow \enspace (q-p)^2 <0$.

$\underline{(\Leftarrow):}$ the contraposition of the last implication reads $\ (q-p)^2 \geq 0 \enspace\Rightarrow \enspace \exists\ s\in S,\ (p-s)\geq 0 $.

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Remark: the causal completion of two points $p,q \in \mathcal{M},\ q\in I^+(p)$ (chronological future) is indeed the double cone, but not the open one $I^+(p)\cap I^-(q)$ but actually the "causal" one $J^+(p)\cap J^-(q)$