$c_{k}^{({\alpha})}\lvert{\Omega}\rangle=0,\quad{\alpha}=S,{\bar{S}},\quad\omega>0,\quad\left(c_{k}^{({\alpha})}\right)^{\dagger}=c_{-k}^{({\alpha})}\
,$ (102)

which yields a mode expansion of the generalized free field on the entire
Minkowski spacetime in terms of the $\\{c_{k}^{({\alpha})}\\}$

${{\mathcal{O}}}(x)=\sum_{{\alpha}=S,{\bar{S}}}\int\frac{d^{2}k}{(2\pi)^{2}}w_{k}^{({\alpha})}(x)c_{k}^{({\alpha})}=\sum_{{\alpha}=S,{\bar{S}}}\int\frac{d^{2}k}{(2\pi)^{2}}u_{k}^{({\alpha})}(x)a_{k}^{({\alpha})}\
,$ (103)

where the ‘extended’ Rindler mode functions are explicitly given by

$\displaystyle
u^{(S)}_{k}(x)=\frac{N_{k}}{\sinh\pi\omega}\begin{cases}\sinh\pi\omega~{}(-x^{-})^{-\bar{q}_{+}}(x^{+})^{-q_{-}},&x^{-}<0,~{}x^{+}>0\;(R)\cr-i\sin\pi
q_{-}~{}(x^{-})^{-\bar{q}_{+}}(x^{+})^{-q_{-}},&x^{-}>0,~{}x^{+}>0\;(F)\cr
0,&x^{-}>0,~{}x^{+}<0\;(L)\cr
i\sin\pi{\bar{q}_{+}}~{}(-x^{-})^{-\bar{q}_{+}}(-x^{+})^{-q_{-}},&x^{-}<0,~{}x^{+}<0\;(P)\cr\end{cases},$
(104) $\displaystyle
u^{(\bar{S})}_{k}(x)=\frac{N_{k}}{\sinh\pi\omega}\begin{cases}0,&x^{-}<0,~{}x^{+}>0\;(R)\cr-i\sin
q_{+}~{}(x^{-})^{-q_{+}}(x^{+})^{-\bar{q}_{-}},&x^{-}>0,~{}x^{+}>0\;(F)\cr\sinh\pi\omega~{}(x^{-})^{-q_{+}}(-x^{+})^{-\bar{q}_{-}},&x^{-}>0,~{}x^{+}<0\;(L)\cr
i\sin\pi{\bar{q}_{-}}~{}(-x^{-})^{-q_{+}}(-x^{+})^{-\bar{q}_{-}},&x^{-}<0,~{}x^{+}<0\;(P)\cr\end{cases}.$
(105)

The mode expansion in terms of the Rindler oscillators in (103),
$\\{a_{k}^{({\alpha})}\\}$, is obtained via the Bogoliubov transform

$\displaystyle
c^{({\alpha})}_{k}={1\over\sqrt{2\sinh\pi|\omega|}}\left(e^{{\pi|\omega|\over
2}}a^{({\alpha})}_{k}-e^{-{\pi|\omega|\over 2}}a^{(\bar{\alpha})}_{-k}\right)\
.$ (106)

The inverse Bogoliubov transform is

$\displaystyle
a^{({\alpha})}_{k}={1\over\sqrt{2\sinh\pi|\omega|}}\left(e^{{\pi|\omega|\over
2}}c^{({\alpha})}_{k}+e^{-{\pi|\omega|\over 2}}c^{(\bar{\alpha})}_{-k}\right)\
.$ (107)

###  A.2 Deriving the change of basis

The expansions (98) and (103) are for the same operator, ${{\mathcal{O}}}(x),$
thus, equating these two expressions allows us to express the Minkowski and
Rindler oscillators in terms of each other. Explicitly, we may write

$b_{p}=\sum_{{\alpha}=S,{\bar{S}}}\int{d^{2}k\over(2\pi)^{2}}A_{kp}^{({\alpha})}a_{k}^{({\alpha})},\quad
a_{k}^{({\alpha})}=\int{d^{2}p\over(2\pi)^{2}}B_{pk}^{({\alpha})}b_{p}\ .$
(108)

Substituting the first definition into (98) and equating with (103), or
alternatively the second definition into (103) and equating with (98), one
obtains

$u_{k}^{({\alpha})}=\int{d^{2}p\over(2\pi)^{2}}A_{kp}^{({\alpha})}f_{p}e^{ip\cdot
x},\quad A_{kp}^{({\alpha})}={1\over f_{p}}\int d^{2}x\,e^{-ip\cdot
x}u_{k}^{({\alpha})}(x),\quad f_{p}e^{ip\cdot
x}=\int{d^{2}k\over(2\pi)^{2}}B_{pk}^{({\alpha})}u_{k}^{({\alpha})}\ ,$ (109)

and, with
${{\delta}}_{kk^{\prime}}=(2\pi)^{2}{{\delta}}^{(2)}(k-k^{\prime})={{\frac{1}{2}}}(2\pi)^{2}{{\delta}}(k^{\pm}-k^{\prime\pm})$,

$\int{d^{2}p\over(2\pi)^{2}}A_{kp}^{({\alpha})}B_{pk^{\prime}}^{({\beta})}={{\delta}}_{kk^{\prime}}{{\delta}}_{{\alpha}{\beta}},\qquad\sum_{\alpha}\int{d^{2}k\over(2\pi)^{2}}B_{pk}^{({\alpha})}A_{kp^{\prime}}^{({\alpha})}={{\delta}}_{pp^{\prime}}\
.$ (110)

Note that

$A^{(\bar{S})}_{kp}=A^{(S)}_{-k,-p}=A^{(S)*}_{kp},\qquad
B^{(\bar{S})}_{pk}=B^{(S)}_{-p,-k}=B^{(S)*}_{pk}\ .$ (111)

The explicit expressions for $A$ and $B$ can be computed using the middle
equation of (109) and equations (110). One finds

$\displaystyle
A^{(S)}_{kp}={\pi\epsilon(\omega)\over\sqrt{2\sinh\pi|\omega|}}e^{-i{\theta}(k^{+},k^{-})}\begin{cases}e^{\frac{\pi}{2}\omega}(p^{+})^{-ik^{+}-\frac{1}{2}}(p^{-})^{ik^{-}-\frac{1}{2}},\quad&p^{\pm}>0\cr-e^{-\frac{\pi}{2}\omega}(-p^{+})^{-ik^{+}-\frac{1}{2}}(-p^{-})^{ik^{-}-\frac{1}{2}},\quad&p^{\pm}<0\end{cases},$
(112) $\displaystyle
B_{pk}^{(S)}=\frac{\pi}{\sqrt{2\sinh(\pi|\omega|)}}e^{i{\theta}(k^{+},k^{-})}(p^{+})^{ik^{+}-\frac{1}{2}}(p^{-})^{-ik^{-}-\frac{1}{2}}\begin{cases}e^{\frac{\pi}{2}\omega}&p^{\pm}>0\cr
e^{-\frac{\pi}{2}\omega}&p^{\pm}<0\end{cases}\ ,$ (113)

where

$\displaystyle
e^{i{\theta}(k^{+},k^{-})}=\frac{\left|\Gamma\left(\frac{\Delta}{2}-ik^{+}\right)\right|}{\Gamma\left(\frac{\Delta}{2}+ik^{+}\right)}\frac{\left|\Gamma\left(\frac{\Delta}{2}+ik^{-}\right)\right|}{\Gamma\left(\frac{\Delta}{2}-ik^{-}\right)}={|{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})|\over{{\Gamma}}(q_{+}){{\Gamma}}(\bar{q}_{-})},$
(114)
$\displaystyle{\theta}(-k^{+},-k^{-})=-{\theta}(k^{+},k^{-}),\qquad{\theta}(-k^{-},-k^{+})={\theta}(k^{+},k^{-})\
.$ (115)

###  A.3 Shifted Rindler wedges

One can easily consider a Rindler decomposition about a point
$(x^{+},x^{-})=(x^{+}_{0},x^{-}_{0})$ that is not the origin. The
corresponding mode functions are given by

$u^{({\alpha},x_{0})}_{k}(x)=u^{({\alpha})}_{k}(x-x_{0}),\quad
w^{({\alpha},x_{0})}_{k}(x)=w^{({\alpha})}_{k}(x-x_{0}),$ (116)

where $u^{({\alpha})}_{k},w^{({\alpha})}_{k}$ are mode functions for the
Rindler decomposition about the origin, given in (104)–(105) and (101). From
(109) we thus find that

$A_{kp}^{({\alpha},x_{0})}={1\over f(p)}\int dx\,e^{-ip\cdot
x}u_{k}^{({\alpha})}(x-x_{0})=e^{-ip\cdot x_{0}}A_{kp}^{({\alpha})},\quad
B_{kp}^{({\alpha})}=e^{ip\cdot x_{0}}B_{kp}^{({\alpha})}\ ,$ (117)

where $A_{kp}^{({\alpha})},~{}B_{kp}^{({\alpha})}$ are the basis
transformation matrices for the Rindler decomposition the origin. They are
given explicitly in (112) and (113).

###  A.4 Equivalence of ${\mathcal{X}}_{S}$ and ${\mathcal{Y}}_{\hat{S}}$

For this particular choice of spatial subregion on the $x^{0}=0$ slice,
$S:x^{1}>0,$ the entanglement wedge and casual domain subalgebras,
${\mathcal{X}}_{S}$ and ${\mathcal{Y}}_{\hat{S}},$ are equal in the vacuum
representation. Recall that ${\mathcal{Y}}_{\hat{S}}$ is generated by
$\\{{{\mathcal{O}}}(x),x\in\hat{S}\\}.$ Using (100), one has

$a^{(S)}_{k}={1\over
N_{k}}\int_{x\in\hat{S}}d^{2}x\left(-x^{-}\right)^{\bar{q}_{+}-1}\left(x^{+}\right)^{q_{-}-1}{{\mathcal{O}}}(x)\
,$ (118)

so, equivalently one can say that ${\mathcal{Y}}_{\hat{S}}$ is generated by
$\\{a^{(S)}_{k},~{}k\in\mathbb{R}^{2}\\}.$

With the definition of ${\mathcal{X}}_{S}$ given in (32)–(33) one has that
${\mathcal{X}}_{S}$ is generated by
$\\{{{\mathcal{O}}}(s;x^{1}),~{}s\in\mathbb{R},~{}x^{1}\in S\\}.$ The modular
flow for a Rindler region in the vacuum is a boost. Recalling that $\omega$ is
the quantum number associated to boosts, one has that

${{\Delta}}_{0}^{is}a^{(S)}_{k}{{\Delta}}_{0}^{-is}=e^{-i\omega s}a^{(S)}_{k}\
,$ (119)

which leads to

${{\mathcal{O}}}(s;x^{1})=\int{d^{2}k\over(2\pi)^{2}}N_{k}\left(x^{1}\right)^{-\bar{q}_{+}}\left(x^{1}\right)^{-q_{-}}e^{-i\omega
s}a^{(S)}_{k}={{\mathcal{O}}}\left(x^{-}=-x^{1}e^{-s},~{}x^{-}=x^{1}e^{s}\right)\
.$ (120)

For any $s\in\mathbb{R},~{}x^{1}>0,$ there is exactly one $x_{*}\in\hat{S}$
such that ${{\mathcal{O}}}(s;x^{1})={{\mathcal{O}}}(x_{*})$. Thus we find
$\\{{{\mathcal{O}}}(s;x^{1}),~{}s\in\mathbb{R},~{}x^{1}\in
S\\}=\\{{{\mathcal{O}}}(x),x\in\hat{S}\\},$ so ${\mathcal{X}}_{S}$ and
${\mathcal{Y}}_{\hat{S}}$ are generated by the same set of operators and thus
are equivalent. This is consistent with the coincidence of the entanglement
wedge and causal domain for this situation.

##  Appendix B Interval decomposition of the generalized free field

In this appendix we derive the change of basis from interval/complement
oscillators $\\{a^{(I)}_{k},~{}a^{(O)}_{k}\\}$ to Minkowski oscillators,
$\\{b_{p}\\}.$ We first discuss the localized mode expansion of the
generalized free field and then derive the basis change. We also explicitly
demonstrate the additivity anomaly
${\mathcal{Y}}_{\hat{O}}={{\mathcal{M}}}_{\hat{O}_{L}}\lor{{\mathcal{M}}}_{\hat{O}_{R}}\subset{\mathcal{X}}_{O}$
in the final subsection.

###  B.1 Local mode expansion

We are interested in decomposing the generalized free field in terms of mode
expansions associated to the regions $I:|x^{1}|<a$ and $O=\bar{I}:|x^{1}|>a$
on the $x^{0}=0$ slice. Consider the conformal transformation (with $b,c>0$)

$x^{-}=a~{}{y^{-}+b\over b-y^{-}},\qquad x^{+}=a~{}{y^{+}-c\over y^{+}+c}\ ,$
(121)

which smoothly maps the right Rindler wedge $\hat{S}$ of the $y-$plane to the
diamond region $\hat{I}:|x^{\pm}|<a$ and the region
$\\{0<y^{-}<b,~{}-c<y^{+}<0\\}$ to the (shifted) Rindler region
$\hat{O}_{L}:x^{-}>a,~{}x^{+}<-a$. The inverse transformation is

$y^{-}=b~{}{x^{-}-a\over x^{-}+a},\qquad y^{+}=c~{}{x^{+}+a\over a-x^{+}},$
(122)

and we can compute the conformal factor by the change in the metric

$ds^{2}=-dy^{+}dy^{-}=-\Omega^{2}dx^{+}dx^{-},\qquad\Omega^{2}={4a^{2}bc\over(x^{-}+a)^{2}(a-x^{+})^{2}}\
.$ (123)

Recall that ${{\mathcal{O}}}(x)$ is the vacuum representation of a scalar
primary of dimension ${{\Delta}},$ thus, under a conformal transformation we
have

${{\mathcal{O}}}(y)\to{{\mathcal{O}}}(x)=\Omega^{{{\Delta}}}{{\mathcal{O}}}(y),\quad\Omega(x)=\left({4a^{2}bc\over(x^{-}+a)^{2}(a-x^{+})^{2}}\right)^{1\over
2}\ ,$ (124)

with $y=y(x).$ Using the Rindler mode expansion (100) in the $y-$plane and the
transformation (124) we can obtain mode expansions for the generalized free
field in $\hat{I}$ and $\hat{O}_{L}.$

One finds (setting $b=c=1$)

$\displaystyle{{\mathcal{O}}}(x\in\hat{I})=\int{d^{2}k\over(2\pi)^{2}}u^{(I)}_{k}(x)~{}a^{(I)}_{k}\
,$ (125)
$\displaystyle{{\mathcal{O}}}(x\in\hat{O}_{L})=\int{d^{2}k\over(2\pi)^{2}}u^{(O)}_{k}(x)~{}a^{(O)}_{k}\
,$ (126) $\displaystyle
u^{(I)}_{k}(x)=N_{k}(2a)^{{{\Delta}}}(a-x^{-})^{-\bar{q}_{+}}(x^{-}+a)^{-q_{+}}(a-x^{+})^{-\bar{q}_{-}}(x^{+}+a)^{-q_{-}},$
(127) $\displaystyle
u^{(O)}_{k}(x)=N_{k}(2a)^{{{\Delta}}}(x^{-}-a)^{-q_{+}}(x^{-}+a)^{-\bar{q}_{+}}(a-x^{+})^{-q_{-}}(-x^{+}-a)^{-\bar{q}_{-}}\
.$ (128)

The mode functions $u_{k}^{(I,O)}$ can be analytically continued to the entire
Minkowski plane in the lower (upper) half $x^{\pm}$ planes for positive
(negative) $\omega$ to the (normalized) functions

$\displaystyle w_{k}^{(I)}(x)$ $\displaystyle={e^{\pi|\omega|\over
2}N_{k}(2a)^{{{\Delta}}}\over\sqrt{2\sinh\pi|\omega|}}(a-x^{-}+i\epsilon(\omega)\epsilon)^{-\bar{q}_{+}}(x^{-}+a-i\epsilon(\omega)\epsilon)^{-q_{+}}(a-x^{+}+i\epsilon(\omega)\epsilon)^{-\bar{q}_{-}}(x^{+}+a-i\epsilon(\omega)\epsilon)^{-q_{-}}$
(129) $\displaystyle w_{k}^{(O)}(x)$ $\displaystyle={e^{\pi|\omega|\over
2}N_{k}(2a)^{{{\Delta}}}\over\sqrt{2\sinh\pi|\omega|}}(x^{-}-a-i\epsilon(\omega)\epsilon)^{-q_{+}}(x^{-}+a-i\epsilon(\omega)\epsilon)^{-\bar{q}_{+}}(a-x^{+}+i\epsilon(\omega)\epsilon)^{-q_{-}}(-x^{+}-a+i\epsilon(\omega)\epsilon)^{-\bar{q}_{-}}\
,$

which are supported on purely positive (negative) frequency Minkowski modes
for positive (negative) $\omega.$ One immediate property of these functions is

$\left(w^{(I)}_{k}(x)\right)^{*}=w^{(I)}_{-k}(x),\qquad\left(w^{(O)}_{k}(x)\right)^{*}=w^{(O)}_{-k}(x)\
.$ (130)

One can associate oscillators $\\{c_{k}^{({\alpha})}\\}$ to the mode functions
(129) with

$c_{k}^{({\alpha})}\lvert{\Omega}\rangle=0,\quad{\alpha}=I,O,\quad\omega>0,\quad\left(c_{k}^{({\alpha})}\right)^{\dagger}=c_{-k}^{({\alpha})}\
,$ (131)

which yields a mode expansion of the generalized free field on the entire
Minkowski spacetime in terms of the $\\{c_{k}^{({\alpha})}\\}$

${{\mathcal{O}}}(x)=\sum_{{\alpha}=I,O}\int\frac{d^{2}k}{(2\pi)^{2}}w_{k}^{({\alpha})}(x)c_{k}^{({\alpha})}=\sum_{{\alpha}=I,O}\int\frac{d^{2}k}{(2\pi)^{2}}u_{k}^{({\alpha})}(x)a_{k}^{({\alpha})}\
,$ (132)

where the ‘extended’ interval/complement mode functions are explicitly given
by

$\displaystyle u^{(I)}_{k}(x)$
$\displaystyle={N_{k}(2a)^{{{\Delta}}}\over\sinh\pi\omega}$ (133)
$\displaystyle\cdot\begin{cases}-i\sin\pi{{\Delta}}\left(x^{-}-a\right)^{-\bar{q}_{+}}\left(x^{-}+a\right)^{-q_{+}}\left(x^{+}-a\right)^{-\bar{q}_{-}}\left(x^{+}+a\right)^{-q_{-}},&x\in
F_{M}\\\ -i\sin\pi
q_{-}\left(x^{-}-a\right)^{-\bar{q}_{+}}\left(x^{-}+a\right)^{-q_{+}}\left(a-x^{+}\right)^{-\bar{q}_{-}}\left(x^{+}+a\right)^{-q_{-}},&x\in
F_{L}\\\ -i\sin\pi
q_{+}\left(a-x^{-}\right)^{-\bar{q}_{+}}\left(x^{-}+a\right)^{-q_{+}}\left(x^{+}-a\right)^{-\bar{q}_{-}}\left(x^{+}+a\right)^{-q_{-}},&x\in
F_{R}\\\ 0,&x\in\hat{O}_{L}\\\ 0,&x\in\hat{O}_{R}\\\
\sinh\pi\omega\left(a-x^{-}\right)^{-\bar{q}_{+}}\left(x^{-}+a\right)^{-q_{+}}\left(a-x^{+}\right)^{-\bar{q}_{-}}\left(x^{+}+a\right)^{-q_{-}},&x\in\hat{I}\\\
i\sin\pi\bar{q}_{+}\left(a-x^{-}\right)^{-\bar{q}_{+}}\left(x^{-}+a\right)^{-q_{+}}\left(a-x^{+}\right)^{-\bar{q}_{-}}\left(-x^{+}-a\right)^{-q_{-}},&x\in
P_{L}\\\
i\sin\pi\bar{q}_{-}\left(a-x^{-}\right)^{-\bar{q}_{+}}\left(-x^{-}-a\right)^{-q_{+}}\left(a-x^{+}\right)^{-\bar{q}_{-}}\left(x^{+}+a\right)^{-q_{-}},&x\in
P_{R}\\\
i\sin\pi{{\Delta}}\left(a-x^{-}\right)^{-\bar{q}_{+}}\left(-x^{-}-a\right)^{-q_{+}}\left(a-x^{+}\right)^{-\bar{q}_{-}}\left(-x^{+}-a\right)^{-q_{-}},&x\in
P_{M}\\\ \end{cases}\ ,$ $\displaystyle u^{(O)}_{k}(x)$
$\displaystyle={N_{k}(2a)^{{{\Delta}}}\over\sinh\pi\omega}$ (134)
$\displaystyle\cdot\begin{cases}-i\sin\pi\left({{\Delta}}+i\omega\right)\left(x^{-}-a\right)^{-q_{+}}\left(x^{-}+a\right)^{-\bar{q}_{+}}\left(x^{+}-a\right)^{-q_{-}}\left(x^{+}+a\right)^{-\bar{q}_{-}},&x\in
F_{M}\\\ -i\sin\pi
q_{+}\left(x^{-}-a\right)^{-q_{+}}\left(x^{-}+a\right)^{-\bar{q}_{+}}\left(a-x^{+}\right)^{-q_{-}}\left(x^{+}+a\right)^{-\bar{q}_{-}},&x\in
F_{L}\\\ -i\sin\pi
q_{-}\left(a-x^{-}\right)^{-q_{+}}\left(x^{-}+a\right)^{-\bar{q}_{+}}\left(x^{+}-a\right)^{-q_{-}}\left(x^{+}+a\right)^{-\bar{q}_{-}},&x\in
F_{R}\\\
\sinh\pi\omega\left(x^{-}-a\right)^{-q_{+}}\left(x^{-}+a\right)^{-\bar{q}_{+}}\left(a-x^{+}\right)^{-q_{-}}\left(-x^{+}-a\right)^{-\bar{q}_{-}},&x\in\hat{O}_{L}\\\
\sinh\pi\omega\left(a-x^{-}\right)^{-q_{+}}\left(-x^{-}-a\right)^{-\bar{q}_{+}}\left(x^{+}-a\right)^{-q_{-}}\left(x^{+}+a\right)^{-\bar{q}_{-}},&x\in\hat{O}_{R}\\\
0,&x\in\hat{I}\\\
i\sin\pi\bar{q}_{-}\left(a-x^{-}\right)^{-q_{+}}\left(x^{-}+a\right)^{-\bar{q}_{+}}\left(a-x^{+}\right)^{-q_{-}}\left(-x^{+}-a\right)^{-\bar{q}_{-}},&x\in
P_{L}\\\
i\sin\pi\bar{q}_{+}\left(a-x^{-}\right)^{-q_{+}}\left(-x^{-}-a\right)^{-\bar{q}_{+}}\left(a-x^{+}\right)^{-q_{-}}\left(x^{+}+a\right)^{-\bar{q}_{-}},&x\in
P_{R}\\\
i\sin\pi\left({{\Delta}}-i\omega\right)\left(a-x^{-}\right)^{-q_{+}}\left(-x^{-}-a\right)^{-\bar{q}_{+}}\left(a-x^{+}\right)^{-q_{-}}\left(-x^{+}-a\right)^{-\bar{q}_{-}},&x\in
P_{M}\\\ \end{cases}\ .$

As in the Rindler case, the expansion of ${{\mathcal{O}}}(x)$ in terms of
$\\{a^{(I)}_{k},~{}a^{(O)}_{k}\\}$ is obtained from that in terms of
$\\{c^{(I)}_{k},~{}c^{(O)}_{k}\\}$ via the Bogoliubov transform (106) (with
${\alpha}=I,O$).

###  B.2 Derivation of the basis change

Equating the mode expansions (98) and (132) one can relate the
interval/complement oscillators $\\{a^{(I)}_{k},~{}a^{(O)}_{k}\\}$ to the
Minkowski oscillators $\\{b_{p}\\}.$ Defining the change of basis ‘matrices’
$P$ and $Q$ by

$b_{p}=\int{d^{2}k\over(2\pi)^{2}}\,\left(Q^{(I)}_{kp}c^{(I)}_{k}+Q^{(O)}_{kp}c^{(O)}_{k}\right)=\int{d^{2}k\over(2\pi)^{2}}\,\left(P^{(I)}_{kp}a^{(I)}_{k}+P^{(O)}_{kp}a^{(O)}_{k}\right)\
,$ (135)

we have

$Q^{(I)}_{kp}=f_{p}^{-1}\int d^{2}x\,e^{-ip\cdot x}w^{(I)}_{k}(x),\qquad
Q^{(O)}_{kp}=f_{p}^{-1}\int d^{2}x\,e^{-ip\cdot x}w^{(O)}_{k}(x)\ .$ (136)

Since $w^{(I,O)}_{k}(x)$ is analytic in lower (upper) $x^{\pm}$-plane for
$\omega>0$ ($\omega<0$), we have

$\displaystyle
Q^{(I)}_{kp}={\theta}(\omega){\theta}(p^{+}){\theta}(p^{-})A(k,p)+{\theta}(-\omega){\theta}(-p^{+}){\theta}(-p^{-})B(k,p)$
(137) $\displaystyle
Q^{(O)}_{kp}={\theta}(\omega){\theta}(p^{+}){\theta}(p^{-})\tilde{A}(k,p)+{\theta}(-\omega){\theta}(-p^{+}){\theta}(-p^{-})\tilde{B}(k,p)\
.$ (138)

Note that, from (136) and (130), we have

$(Q^{(I)}_{kp})^{*}=Q^{(I)}_{(-k)(-p)},\quad(Q^{(O)}_{kp})^{*}=Q^{(O)}_{(-k)(-p)},\quad
B(-k,-p)=A^{*}(k,p),\quad\tilde{B}(-k,-p)=\tilde{A}^{*}(k,p)\ .$ (139)

From the Bogoliubov transform (106), we have

$\displaystyle P^{(I)}_{kp}={e^{{\pi|\omega|\over
2}}\over\sqrt{2\sinh\pi|\omega|}}Q^{(I)}_{kp}-{e^{-{\pi|\omega|\over
2}}\over\sqrt{2\sinh\pi|\omega|}}Q^{(O)}_{(-k)p},$ (140) $\displaystyle
P^{(O)}_{kp}={e^{{\pi|\omega|\over
2}}\over\sqrt{2\sinh\pi|\omega|}}Q^{(O)}_{kp}-{e^{-{\pi|\omega|\over
2}}\over\sqrt{2\sinh\pi|\omega|}}Q^{(I)}_{(-k)p}\ .$ (141)

Using the explicit form of the mode functions (129) we have

$\displaystyle A(k,p)={1\over 2f_{p}}{e^{\pi|\omega|\over
2}\over\sqrt{2\sinh\pi|\omega|}}N_{k}(2a)^{{{\Delta}}}I(p^{+};\bar{q}_{+},q_{+})I(p^{-};\bar{q}_{-},q_{-})$
(142) $\displaystyle\tilde{A}(k,p)={1\over 2f_{p}}{e^{\pi|\omega|\over
2}\over\sqrt{2\sinh\pi|\omega|}}N_{k}(2a)^{{{\Delta}}}\tilde{I}(p^{+};\bar{q}_{+},q_{+})(\tilde{I}(p^{-};\bar{q}_{-},q_{-}))^{*}$
(143)

where

$\displaystyle I(p^{+};\bar{q}_{+},q_{+})=\int
dx\,e^{ip^{+}x}\,(a+i\epsilon-x)^{-\bar{q}_{+}}(x+a-i\epsilon)^{-q_{+}},$
(144) $\displaystyle\tilde{I}(p^{+};\bar{q}_{+},q_{+})=\int
dx\,e^{ip^{+}x}\,(x+a-i\epsilon)^{-\bar{q}_{+}}(x-a-i\epsilon)^{-q_{+}}\ .$
(145)

The above integrals can be evaluated to give

$\displaystyle I(p^{+};\bar{q}_{+},q_{+})={\theta}(p^{+}){2\pi e^{-\pi
k^{+}}\over{{\Gamma}}({{\Delta}})}e^{-ip^{+}a}(p^{+})^{{{\Delta}}-1}{}_{1}F_{1}\left(\bar{q}_{+};{{\Delta}};2iap^{+}\right)$
(146)
$\displaystyle\tilde{I}(p^{+};\bar{q}_{+},q_{+})={\theta}(p^{+})e^{-ip^{+}a}(p^{+})^{{{\Delta}}-1}{2\pi
e^{i\pi{{\Delta}}\over
2}\over{{\Gamma}}({{\Delta}})}{}_{1}F_{1}\left(q_{+};{{\Delta}};2iap^{+}\right)$
(147)

which gives

$\displaystyle A(k,p)=C\kappa^{\nu}e^{-{\pi\over
2}\omega}e^{-ip^{0}a}{}_{1}F_{1}\left(\bar{q}_{+};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left(\bar{q}_{-};{{\Delta}};2iap^{-}\right),$
(148) $\displaystyle\tilde{A}(k,p)=C^{*}\kappa^{\nu}e^{\pi\omega\over
2}e^{-ip^{1}a}{}_{1}F_{1}\left(q_{+};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left(\bar{q}_{-};{{\Delta}};-2iap^{-}\right),$
(149) $\displaystyle C={\pi\over
2^{{{\Delta}}-1}({{\Gamma}}({{\Delta}}))^{2}}\left|\Gamma\left(q_{+}\right)\Gamma\left(q_{-}\right)\right|(2a)^{{{\Delta}}}\
.$ (150)

With the identity

${}_{1}F_{1}(\bar{q}_{-};{{\Delta}};-2iap^{-})=e^{-2iap^{-}}{}_{1}F_{1}(q_{-};{{\Delta}};2iap^{-})$
(151)

we find

$\tilde{A}(k,p)=A(-k,p)\ .$ (152)

Using the expression for $P$ in terms of $Q$ and $Q$ in terms of
$A,\tilde{A},B,\tilde{B}$ one obtains the explicit expression for the change
of basis from interval/complement oscillators to Minkowski oscillators

$\displaystyle P^{(I)}_{kp}$
$\displaystyle={\epsilon(\omega)\epsilon(p^{0})\theta(p^{+}p^{-})\over\sqrt{2\sinh\pi|\omega|}}C\kappa^{\nu}e^{-ip^{0}a}{}_{1}F_{1}\left({\bar{q}_{+}};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left({\bar{q}_{-}};{{\Delta}};2iap^{-}\right)$
(153) $\displaystyle P^{(O)}_{kp}$
$\displaystyle={\epsilon(\omega)\epsilon(p^{0})\theta(p^{+}p^{-})e^{\pi\epsilon(p^{0})\omega}\over\sqrt{2\sinh\pi|\omega|}}C^{*}\kappa^{\nu}e^{-ip^{0}a}{}_{1}F_{1}\left({q_{+}};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left({q_{-}};{{\Delta}};2iap^{-}\right)\
.$

The inverse transformation from Minkowski to interval/complement oscillators
such that

$a^{({\alpha})}_{k}=\int{d^{2}p\over(2\pi)^{2}}R^{({\alpha})}_{pk}b_{p}$ (154)

is

$\displaystyle R^{(I)}_{pk}$
$\displaystyle={\theta(p^{+}p^{-})\over\sqrt{2\sinh\pi|\omega|}}C^{*}\kappa^{\nu}e^{-ip^{0}a}{}_{1}F_{1}\left({\bar{q}_{+}};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left({\bar{q}_{-}};{{\Delta}};2iap^{-}\right)$
(155) $\displaystyle R^{(O)}_{pk}$
$\displaystyle={\theta(p^{+}p^{-})e^{\pi\epsilon(p^{0})\omega}\over\sqrt{2\sinh\pi|\omega|}}C\kappa^{\nu}e^{-ip^{0}a}{}_{1}F_{1}\left({q_{+}};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left({q_{-}};{{\Delta}};2iap^{-}\right)\
.$

###  B.3 Relations between ${\mathcal{X}}_{I,O}$ and
${\mathcal{Y}}_{\hat{I},\hat{O}}$

For the subregion $\hat{I},$ which is conformally equivalent to a Rindler
wedge, the modular flow with respect to the vacuum is local. We will see that
this leads to the equivalence of ${\mathcal{X}}_{I}$ and
${\mathcal{Y}}_{\hat{I}}.$ Recall that ${\mathcal{Y}}_{\hat{I}}$ is defined as
the algebra generated by $\\{{{\mathcal{O}}}(x),~{}x\in\hat{I}\\}.$ From the
definition of ${\mathcal{X}}_{I}$ given in (32)–(33) one has that
${\mathcal{X}}_{I}$ is generated by
$\\{{{\mathcal{O}}}(s;x^{1}),~{}s\in\mathbb{R},~{}x^{1}\in I\\}.$ Since
$\omega$ was the modular frequency for the Rindler wedge (i.e. quantum number
for the boost symmetry) before the conformal transformation it still has this
interpretation after the conformal transformation. In particular, we have

${{\Delta}}_{0(I)}^{is}a^{(I)}_{k}{{\Delta}}_{0(I)}^{-is}=e^{-i\omega
s}a^{(I)}_{k}\ ,$ (156)

which leads to

$\displaystyle{{\mathcal{O}}}(s;x^{1})$
$\displaystyle=\int{d^{2}k\over(2\pi)^{2}}N_{k}(2a)^{{{\Delta}}}\left(\Omega_{s}(-x^{1},x^{1})\right)^{{{\Delta}}}\left(a-x^{-}_{s}\right)^{-\bar{q}_{+}}\left(x^{-}_{s}+a\right)^{-q_{+}}\left(a-x^{+}_{s}\right)^{-\bar{q}_{-}}\left(x^{+}_{s}+a\right)^{-q_{-}}a^{(I)}_{k}$
(157)
$\displaystyle=\left(\Omega_{s}(-x^{1},x^{1})\right)^{{{\Delta}}}{{\mathcal{O}}}\left(x^{-}_{s},x^{+}_{s}\right)\
.$

In the expression above, $x^{-}_{s}$ is evaluated at $x^{-}=-x^{1},$ i.e.
$x^{-}_{s}\to x^{-}_{s}(-x^{1})$ and $x^{+}_{s}$ is evaluated at
$x^{+}=x^{1},$ i.e. $x^{+}_{s}\to x^{+}_{s}(x^{1})$. We see that, as expected,
the modular flow acts as a conformal transformation with

$x_{s}^{-}(x^{-})=-a~{}{a(1-e^{s})-(1+e^{s})x^{-}\over
a(1+e^{s})-(1-e^{s})x^{-}},\qquad
x_{s}^{+}(x^{+})=a~{}{a(1-e^{-s})+(1+e^{-s})x^{+}\over
a(1+e^{-s})+(1-e^{-s})x^{+}}\ ,$ (158)

and conformal factor

$ds^{2}=-dx^{-}_{s}dx^{+}_{s}=-\Omega_{s}^{2}dx^{-}dx^{+},~{}\left(\Omega_{s}(x^{-},x^{+})\right)^{2}=\left({(2a)^{2}\over(a(1+e^{s})-(1-e^{s})x^{-})(a(1+e^{-s})+(1-e^{-s})x^{+})}\right)^{2}\
.$ (159)

This is in exact agreement with the computation of modular flow for an
interval in Casini:2011kv . We see that for each $s\in\mathbb{R},~{}x^{1}\in
I$ there is exactly one $x_{*}\in\hat{I}$ such that
${{\mathcal{O}}}(s;x^{1})=\Omega_{s}^{{{\Delta}}}{{\mathcal{O}}}(x_{*}).$ We
then have $\\{{{\mathcal{O}}}(s;x^{1}),~{}s\in\mathbb{R},~{}x^{1}\in
I\\}=\\{{{\mathcal{O}}}(x),~{}x\in\hat{I}\\},$ so ${\mathcal{X}}_{I}$ and
${\mathcal{Y}}_{\hat{I}}$ are generated by the same operators and thus
coincide.

Now we turn to the complementary region $O.$ The algebra
${\mathcal{Y}}_{\hat{O}}$ is generated by
$\\{{{\mathcal{O}}}(x),~{}x\in\hat{O}\\},$ while ${\mathcal{X}}_{O}$ is, via
(32)–(33), generated by
$\\{{{\mathcal{O}}}(s;x^{1}),~{}s\in\mathbb{R},x^{1}\in O\\}.$ Before the
conformal transformation, the oscillators $\\{a^{(O)}_{k}\\}$ were associated
to the left Rindler wedge and $\omega$ was the modular frequency, thus
$\omega$ still has this interpretation after the conformal transformation
though it is not associated to any spacetime symmetry of the region $\hat{O}.$
We therefore have

${{\Delta}}_{0(O)}^{is}a^{(O)}_{k}{{\Delta}}_{0(O)}^{-is}=e^{-i\omega
s}a^{(O)}_{k}\ ,$ (160)

which leads to

$\displaystyle{{\mathcal{O}}}(s;x^{1})$
$\displaystyle=\int{d^{2}k\over(2\pi)^{2}}N_{k}(2a)^{{{\Delta}}}\left|x^{-}-a\right|^{-q_{+}}\left|x^{-}+a\right|^{-\bar{q}_{+}}\left|a-x^{+}\right|^{-q_{-}}\left|-x^{+}-a\right|^{-\bar{q}_{-}}e^{-i\omega
s}a^{(O)}_{k}\ .$ (161)

This expansion of ${{\mathcal{O}}}(s;x^{1})$ can be inverted to solve for
$\\{a^{(O)}_{k}\\}$ as

$\displaystyle a^{(O)}_{k}=\int_{x^{1}\in
O}dx^{1}ds~{}f_{k}(x^{1},s){{\mathcal{O}}}(s;x^{1}),$ (162) $\displaystyle
f_{k}(x^{1},s)={1\over N_{k}}\left({(x^{1})^{2}-a^{2}\over
2a}\right)^{{{\Delta}}-1}\left({x^{1}-a\over x^{1}+a}\right)^{-iq}e^{i\omega
s}\ .$ (163)

We therefore have that $\\{{{\mathcal{O}}}(s;x^{1}),~{}s\in\mathbb{R},x^{1}\in
O\\}$ and $\\{a^{(O)}_{k},~{}k\in\mathbb{R}^{2}\\}$ generate the same algebra.
Thus ${\mathcal{X}}_{O}$ can be viewed as being generated by the
$\\{a^{(O)}_{k}\\}.$

We will now prove that $\\{{{\mathcal{O}}}(x),~{}x\in\hat{O}\\}$ does not
generate the same algebra as $\\{a^{(O)}_{k}\\}.$ From (132)–(134) it is clear
that all ${{\mathcal{O}}}(x\in\hat{O})$ can be expressed in terms of the
$a^{(O)}_{k},$ so clearly ${\mathcal{Y}}_{\hat{O}}\subseteq{\mathcal{X}}_{O}.$
However, we will see that the $\\{a^{(O)}_{k}\\}$ cannot be expressed in terms
of the $\\{{{\mathcal{O}}}(x\in\hat{O})\\},$ so we have a proper inclusion
${\mathcal{Y}}_{\hat{O}}\subset{\mathcal{X}}_{O}.$

In order for it be possible to express the $\\{a^{(O)}_{k}\\}$ in terms of the
$\\{{{\mathcal{O}}}(x\in\hat{O})\\}$ we would need to find a set of functions
$\\{f_{k}(x),k\in\mathbb{R}^{2}\\}$ supported for $x\in\hat{O}$ such that

$a^{(O)}_{k}=\int_{x\in\hat{O}}d^{2}x~{}f_{k}(x){{\mathcal{O}}}(x),\quad\forall
k\ .$ (164)

Consider the mode function
$u^{(O)}_{k}(x\in\hat{O}=\hat{O}_{L}\cup\hat{O}_{R}).$ We may view
$u^{(O)}_{k}(x\in\hat{O})$ as a matrix taking us from the basis of vectors
$\\{a^{(O)}_{k},k\in\mathbb{R}^{2}\\}$ to the set of vectors
$\\{{{\mathcal{O}}}(x),x\in\hat{O}\\}.$ An expression such as (164) would
require $f_{k}(x)$ to be the left inverse of $u^{(O)}_{k}(x\in\hat{O}).$
Namely, (164) requires the set of functions
$\\{f_{k}(x),k\in\mathbb{R}^{2}\\}$ to satisfy

$\int_{x\in\hat{O}}d^{2}xf_{k}(x)u^{(O)}_{k^{\prime}}(x)=(2\pi)^{2}\delta^{(2)}(k-k^{\prime})\
.$ (165)

Viewed as a matrix, $u^{(O)}_{k}(x\in\hat{O}),$ only has a left inverse (i.e.
the $f_{k}(x)$ only exist) if the nullspace of $u^{(O)}_{k}(x\in\hat{O})$ is
trivial. We will see that this is not the case and hence such $f_{k}(x)$ do
not exist.

The most general vector in the domain of $u^{(O)}_{k}(x\in\hat{O})$ is an
arbitrary function of the two momenta, $\hat{g}(k^{+},k^{-}).$ For
convenience, we will remove some non-zero factors from $\hat{g}(k^{+},k^{-})$
and write

$\hat{g}(k^{+},k^{-})={1\over N_{k}(2a)^{{{\Delta}}}}g(k^{+},k^{-})\ .$ (166)

We have a complete basis for functions on $\mathbb{R}^{2}$ given by the
Fourier modes. i.e. for any function $g(k^{+},k^{-})$ we may write

$g(k^{+},k^{-})=\int
d^{2}y~{}e^{ik^{+}y^{-}+ik^{-}y^{+}}\tilde{g}(y^{-},y^{+})\ ,$ (167)

where $y^{-}$ is the conjugate variable to $k^{+}$ and $y^{+}$ the conjugate
variable to $k^{-}.$

Let us now consider $u^{(O)}_{k}(x\in\hat{O})$ acting on the most general
vector $\hat{g}(k^{+},k^{-}).$ The result is, of course, some function of $x$
on $\hat{O}.$ Naming the result of the multiplication of
$\hat{g}(k^{+},k^{-})$ by this matrix $r_{g}(x),$ we have (for
$\hat{O}_{L}:x^{-}>a,x^{+}<-a$)

$\displaystyle r_{g}(x\in\hat{O}_{L})$
$\displaystyle=\int{d^{2}k\over(2\pi)^{2}}\tilde{u}^{(O)}_{k}(x\in
O_{L})~{}\hat{g}(k^{+},k^{-})$ (168) $\displaystyle=\int
d^{2}y\int{d^{2}k\over(2\pi)^{2}}(x^{-}-a)^{-q_{+}}(x^{-}+a)^{-\bar{q}_{+}}(a-x^{+})^{-q_{-}}(-x^{+}-a)^{-\bar{q}_{-}}e^{ik^{+}y^{-}+ik^{-}y^{+}}\tilde{g}(y^{-},y^{+})$
$\displaystyle=2\left(((x^{-})^{2}-a^{2})((x^{+})^{2}-a^{2})\right)^{-{{{\Delta}}\over
2}}$ $\displaystyle\cdot\int
d^{2}y~{}\tilde{g}(y^{-},y^{+})\delta\left(y^{-}-\log\left(x^{-}-a\over
x^{-}+a\right)\right)\delta\left(y^{+}-\log\left(a-x^{+}\over-x^{+}-a\right)\right)$
$\displaystyle=2\left(((x^{-})^{2}-a^{2})((x^{+})^{2}-a^{2})\right)^{-{{{\Delta}}\over
2}}\tilde{g}\left(\log\left(x^{-}-a\over
x^{-}+a\right),\log\left(a-x^{+}\over-x^{+}-a\right)\right)\ .$

and for $\hat{O}_{R}:x^{-}<-a,x^{+}>a$

$\displaystyle r_{g}(x\in\hat{O}_{R})$
$\displaystyle=\int{d^{2}k\over(2\pi)^{2}}\tilde{u}^{(O)}_{k}(x\in
O_{R})~{}\hat{g}(k^{+},k^{-})$ (169) $\displaystyle=\int
d^{2}y\int{d^{2}k\over(2\pi)^{2}}(a-x^{-})^{-q_{+}}(-x^{-}-a)^{-\bar{q}_{+}}(x^{+}-a)^{-q_{-}}(x^{+}+a)^{-\bar{q}_{-}}e^{ik^{+}y^{-}+ik^{-}y^{+}}\tilde{g}(y^{-},y^{+})$
$\displaystyle=2\left(((x^{-})^{2}-a^{2})((x^{+})^{2}-a^{2})\right)^{-{{{\Delta}}\over
2}}$ $\displaystyle\cdot\int
d^{2}y~{}\tilde{g}(y^{-},y^{+})\delta\left(y^{-}-\log\left(a-x^{-}\over-x^{-}-a\right)\right)\delta\left(y^{+}-\log\left(x^{+}-a\over
x^{+}+a\right)\right)$
$\displaystyle=2\left(((x^{-})^{2}-a^{2})((x^{+})^{2}-a^{2})\right)^{-{{{\Delta}}\over
2}}\tilde{g}\left(\log\left(a-x^{-}\over-x^{-}-a\right),\log\left(x^{+}-a\over
x^{+}+a\right)\right)\ .$

Equations (168)–(169) completely describe the function obtained by the
multiplication of $u^{(O)}_{k}(x\in\hat{O})$ on the vector $\hat{g}(k).$
Notice in (168) that the first and second arguments of $\tilde{g}$ on the
final line range over $(-\infty,0)$ and $(0,\infty)$ respectively while $x$
ranges over all of $\hat{O}_{L}.$ Similarly, in (169) the first and second
arguments of $\tilde{g}$ on the final line range over $(0,\infty)$ and
$(-\infty,0)$ respectively while $x$ ranges over all of $\hat{O}_{R}.$

Thus, any function $\hat{g}(k)$ such that its (rescaled) Fourier transform
$g(y^{-},y^{+})$ has support only for $y^{-}y^{+}>0$ gives
$r_{g}(x\in\hat{O})=0.$ Such functions of $k$ need not be identically zero,
thus we have found a family of non-trivial functions of $k$ that are
annihilated by $u^{(O)}_{k}(x\in\hat{O}).$ Thus the null-space of
$u^{(O)}_{k}(x\in\hat{O})$ is non-trivial so it is not left-invertible,
showing that $a^{(O)}_{k}$ cannot be written in terms of position-space
operators supported only in $\hat{O}.$ Thus, $\\{a^{(O)}_{k}\\}$ and
$\\{{{\mathcal{O}}}(x\in\hat{O})\\}$ do not generate the same algebra and we
have ${\mathcal{X}}_{O}\supset{\mathcal{Y}}_{\hat{O}}.$

####  B.3.1 Lack of invertibility and the additivity anomaly

The lack of invertibility of the mode expansion is crucial for
${\mathcal{X}}_{O}$ to describe the entanglement wedge, rather than the causal
wedge of the union of the two Rindler regions $\hat{O}_{L}$ and $\hat{O}_{R}$.
One can see immediately that invertibility of the mode expansion precludes the
possibility for ${\mathcal{X}}_{O}$ to describe the entanglement wedge.
Suppose that it was possible to find a set of $f_{k}(x)$ such that (164) was
satisfied. We could then split each $f_{k}(x)$ into a piece supported only on
$\hat{O}_{L}$ and a piece supported only on $\hat{O}_{R},$ i.e. we could write
$f_{k}(x)=f^{L}_{k}(x)+f^{R}_{k}(x).$ Using the Rindler mode expansions on
$\hat{O}_{L}$ and $\hat{O}_{R}$ separately, (164) could be written as

$a^{(O)}_{k}=\int{d^{2}k^{\prime}\over(2\pi)^{2}}M^{k^{\prime}}_{k}a^{(S,(0,a))}_{k^{\prime}}+\int{d^{2}k^{\prime}\over(2\pi)^{2}}N^{k^{\prime}}_{k}a^{({\bar{S}},(0,-a))}_{k^{\prime}}\
,$ (170)

for some functions $M^{k^{\prime}}_{k},~{}N^{k^{\prime}}_{k}.$ Since
${\mathcal{X}}_{O}$ is generated by $\\{a^{(O)}_{k}\\}$, (170) would imply
that all elements of ${\mathcal{X}}_{O}$ can be written in terms of
$\\{a^{(S,(0,a))}_{k},~{}a^{({\bar{S}},(0,-a))}_{k}\\},$ the generators of the
algebras ${\mathcal{Y}}_{\hat{O}_{R}}={\mathcal{X}}_{O_{R}}$ and
${\mathcal{Y}}_{\hat{O}_{L}}={\mathcal{X}}_{O_{L}},$ respectively. Namely,
(170) would imply additivity:
${\mathcal{X}}_{O}={\mathcal{X}}_{O_{L}}\vee{\mathcal{X}}_{O_{R}}.$ However,
from AdS-Rindler reconstruction and additivity of the bulk free field theory
we know that ${\mathcal{X}}_{O_{L}}\vee{\mathcal{X}}_{O_{R}}$ only describes
the causal wedge of $O_{L}\cup O_{R}.$ Thus, for ${\mathcal{X}}_{O}$ to
describe the entanglement wedge, additivity of the boundary GFF theory must be
violated and thus there cannot exist any expression like (164).

##  Appendix C RT surface without entropy: Rindler

We consider a bulk scalar field $\phi$, which has the expansion

$\displaystyle\phi(X)=\int{d^{2}p\over(2\pi)^{2}}v_{p}(X)b_{p}=\int{d^{2}k\over(2\pi)^{2}}\left(\tilde{v}^{(S)}_{k}(X)a^{(S)}_{k}+\tilde{v}^{(\bar{S})}_{k}(X)a^{(\bar{S})}_{k}\right)\
,$ (171) $\displaystyle v_{p}(X)=\theta(-p^{2})\sqrt{\pi}z^{d\over
2}J_{{{\Delta}}-1}(pz)e^{ip\cdot x},\qquad p=\sqrt{-p^{2}},$ (172)

as in (67). The expansion in terms of Rindler oscillators is obtained by using
the basis change from Rindler to Minkowski oscillators as in (108). The
expansion coefficients are then

$\displaystyle\tilde{v}^{(S)}_{k}(X)=\int{d^{2}p\over(2\pi)^{2}}v_{p}(X)A_{kp}^{(S)},\qquad\tilde{v}^{(\bar{S})}_{k}(X)=\int{d^{2}p\over(2\pi)^{2}}v_{p}(X)A_{kp}^{(\bar{S})}\
.$ (173)

We would like to show that

$\tilde{v}^{(S)}_{k}(X)=0,\quad\forall z\;\text{and}\;x^{\pm}\in\hat{\bar{S}}\
.$ (174)

From (111) we have

$\tilde{v}^{(\bar{S})}_{-k}(z,x^{\mu})=\tilde{v}^{(S)}_{k}(z,-x^{\mu})$ (175)

thus, if (174) is true it implies that $\tilde{v}^{(\bar{S})}_{k}(X)=0$ for
$x\in\hat{S}$.

More explicitly, from (112), we have

$\displaystyle\tilde{v}^{(S)}_{k}(X)$
$\displaystyle=\int{d^{2}p\over(2\pi)^{2}}\theta(-p^{2})\sqrt{\pi}z^{d\over
2}J_{\nu}(pz)e^{ip\cdot x}A_{kp}^{(S)}$ (176)
$\displaystyle=\sqrt{\pi}z^{d\over
2}{\pi\epsilon(\omega)\over\sqrt{2\sinh\pi|\omega|}}e^{-i{\theta}(k^{+},k^{-})}\left(e^{\pi\omega\over
2}I_{+}-e^{-{\pi\omega\over 2}}I_{-}\right)$

where

$\displaystyle
I_{+}(z,x^{\mu})=\int{d^{2}p\over(2\pi)^{2}}\theta(-p^{2}){\theta}(p^{0})J_{\nu}(pz)e^{ip\cdot
x}(p^{+})^{-ik^{+}-\frac{1}{2}}(p^{-})^{ik^{-}-\frac{1}{2}},$ (177)
$\displaystyle
I_{-}(z,x^{\mu})=\int{d^{2}p\over(2\pi)^{2}}\theta(-p^{2}){\theta}(-p^{0})J_{\nu}(pz)e^{ip\cdot
x}(-p^{+})^{-ik^{+}-\frac{1}{2}}(-p^{-})^{ik^{-}-\frac{1}{2}}=I_{+}(z,-x^{\mu})\
.$ (178)

We can change integration variables in $I_{+}$ as follows

$\displaystyle I_{+}$ $\displaystyle={1\over
2\pi^{2}}\int_{0}^{\infty}dp^{+}dp^{-}\,J_{\nu}(2\sqrt{p^{+}p^{-}}z)e^{-ip^{+}x^{-}-ip^{-}x^{+}}(p^{+})^{-ik^{+}-\frac{1}{2}}(p^{-})^{ik^{-}-\frac{1}{2}}$
(179) $\displaystyle={1\over\pi^{2}}\int_{0}^{\infty}{pdpdr\over
r}\,J_{\nu}(2pz)p^{-ik-1}r^{-{i\omega}}e^{-ip(rx^{-}+{1\over r}x^{+})}$

where $p=\sqrt{p^{+}p^{-}}$ and $r=\sqrt{p^{+}\over p^{-}}$. If
$x\in\hat{\bar{S}}$, i.e. $x^{+}<0,x^{-}>0$, we can rotate the $r$-contour to
the negative imaginary axis , i.e.

$\displaystyle
I_{+}(z,x^{\mu})={1\over\pi^{2}}\int_{0}^{\infty}dp\,J_{\nu}(2pz)p^{-ik}\int_{0}^{-i\infty}{dr\over
r}\,r^{-{i\omega}}e^{-ip(rx^{-}+{1\over r}x^{+})}$ (180)
$\displaystyle={e^{-{\pi\omega\over
2}}\over\pi^{2}}\int_{0}^{\infty}dp\,J_{\nu}(2pz)p^{-ik}\int_{0}^{\infty}{dr\over
r}\,r^{-{i\omega}}e^{-p(rx^{-}-{1\over r}x^{+})}\equiv e^{-{\pi\omega\over
2}}I\ .$ (181)

Similarly we can rotate the $r-$contour to the positive imaginary axis for
$I_{+}(z,-x^{\mu})$,

$I_{+}(z,-x^{\mu})={e^{{\pi\omega\over
2}}\over\pi^{2}}\int_{0}^{\infty}dp\,J_{\nu}(2pz)p^{-ik}\int_{0}^{\infty}{dr\over
r}\,r^{-{i\omega}}e^{-p(rx^{-}-{1\over r}x^{+})}=e^{\pi\omega\over 2}I\ .$
(182)

We thus conclude that

$e^{\pi\omega\over 2}I_{+}-e^{-{\pi\omega\over 2}}I_{-}=0,\quad
x\in\hat{\bar{S}}\quad\Rightarrow\quad\tilde{v}^{(S)}_{k}(X)=0\ .$ (183)

If $x\in\hat{S}$ we have instead

$I_{+}=e^{{\pi\omega\over 2}}I,\quad I_{-}=e^{-{\pi\omega\over 2}}I,$ (184)

which gives

$\tilde{v}^{(S)}_{k}(X)={z^{d\over 2}\pi^{3\over
2}\epsilon(\omega)\over\sqrt{2\sinh\pi|\omega|}}e^{-i{\theta}(k^{+},k^{-})}2\sinh\pi\omega\,I=\pi^{3\over
2}z^{d\over 2}\sqrt{2\sinh\pi|\omega|}e^{-i{\theta}(k^{+},k^{-})}I\ .$ (185)

Similarly, if $x^{+},x^{-}>0$ or $x^{+},x^{-}<0,$ one has $v^{(S)}_{k}(X)\neq
0$ and $v^{(\bar{S})}_{k}(X)\neq 0.$ Thus, only for $x\in\hat{\bar{S}}$ does
$v^{(S)}_{k}(X)$ vanish identically, demonstrating that
${\mathfrak{c}}_{\hat{\bar{S}}}=\\{X=(z,x^{\mu}),~{}x\in\hat{\bar{S}}\\}$ is
the algebraic entanglement wedge, ${\mathfrak{b}}_{\bar{S}},$ for the half-
space $\bar{S}$.

The region ${\mathfrak{b}}_{\bar{S}}={\mathfrak{c}}_{\hat{\bar{S}}}$ found as
the algebraic entanglement wedge for the half-space ${\bar{S}}:x^{1}<0$ (on
the $x^{0}=0$ slice) exactly coincides with the entanglement wedge that one
expects from the RT prescription for the boundary spatial subregion $\bar{S}$.
In particular, ${\mathfrak{b}}_{\bar{S}}$ is the bulk domain of dependence of
the homology hypersurface $E_{\bar{S}}:\\{(z,x^{\mu}),~{}x^{0}=0,x^{1}<0\\}$
whose boundary is $\partial E_{\bar{S}}={{\gamma}}_{\bar{S}}\cup{\bar{S}}$
with ${{\gamma}}_{\bar{S}}=\\{(z,x^{\mu}),~{}x^{0}=0,x^{1}=0\\}$ where
$\bar{S}$ is viewed as a curve on the conformal boundary ($z=0$) of Poincaré
AdS${}_{3}.$ Defining the RT surface as the (bulk part of the) boundary of the
bulk spatial subregion whose domain of dependence is the algebraic
entanglement wedge, we find the RT surface to be given by
${{\gamma}}_{\bar{S}}.$ This exactly agrees with the RT surface one obtains
using the usual extremal surface prescription.

##  Appendix D RT surface without entropy: interval and its complement

In this appendix we show that the algebra
${\mathcal{X}}_{I}={\mathcal{Y}}_{\hat{I}}$ of a diamond-shaped boundary
region is dual to the expected bulk region given by the domain of dependence
of the homology hypersurface between a time-slice of the boundary diamond and
the RT surface. We also show that the algebra, ${\mathcal{X}}_{O},$ of the
union, $O=O_{L}\cup O_{R},$ of two finitely separated half-spaces,
$O_{L}:x^{1}<-a$ and $O_{R}:x^{1}>a,$ describes the bulk region expected from
the RT prescription. In particular, we will explicitly see that the
entanglement wedges are ‘superadditive,’ i.e.
${\mathfrak{b}}_{O}\supset{\mathfrak{b}}_{O_{L}}\vee{\mathfrak{b}}_{O_{R}}.$

From our discussion of appendix B.3 we see that the algebra
${\mathcal{X}}_{I}$ is generated by oscillators $\\{a^{(I)}_{k}\\}$ and the
algebra ${\mathcal{X}}_{O}$ is generated by oscillators $\\{a^{(O)}_{k}\\}.$
Thus, any operator that can be written in terms of only
$\\{a^{(I)}_{k}\\}~{}(\\{a^{(O)}_{k}\\})$ clearly lies in
${\mathcal{X}}_{I}~{}({\mathcal{X}}_{O}).$

We consider a bulk scalar field $\phi$, which has the expansion

$\displaystyle\phi(X)=\int{d^{2}p\over(2\pi)^{2}}v_{p}(X)b_{p}=\int{d^{2}k\over(2\pi)^{2}}\left(\tilde{v}^{(I)}_{k}(X)a^{(I)}_{k}+\tilde{v}^{(O)}_{k}(X)a^{(O)}_{k}\right)\
,$ (186) $\displaystyle v_{p}(X)=\theta(-p^{2})\sqrt{\pi}z^{d\over
2}J_{{{\Delta}}-1}(pz)e^{ip\cdot x},\qquad p=\sqrt{-p^{2}},$ (187)

as in (67). The expansion in terms of interval/complement oscillators is
obtained by using the basis change from interval/complement to Minkowski
oscillators as in (135). The expansion coefficients are then

$\displaystyle\tilde{v}^{(I)}_{k}(X)=\int{d^{2}p\over(2\pi)^{2}}v_{p}(X)P_{kp}^{(I)},\qquad\tilde{v}^{(O)}_{k}(X)=\int{d^{2}p\over(2\pi)^{2}}v_{p}(X)P_{kp}^{(O)}\
.$ (188)

We may then derive an explicit expression for a bulk operator at any point $X$
on the Poincaré patch in terms of the interval/complement basis. Using (153),
we find the coefficient of the interval oscillator to be

$\displaystyle\tilde{v}^{(I)}_{k}(X)$
$\displaystyle={N_{k}z(2a)^{{{\Delta}}}\pi^{2}\over
2^{{{\Delta}}-2}{{\Gamma}}({{\Delta}})\sinh\pi\omega}\left(I_{+}(k)-I_{-}(k)\right),\qquad
I_{\pm}(k)=\int_{0}^{\infty}{dp^{+}~{}dp^{-}\over(2\pi)^{2}}~{}j_{k}(\pm
p^{+},\pm p^{-}),$ (189) $\displaystyle j_{k}(p^{+},p^{-})$
$\displaystyle=(4p^{+}p^{-})^{{{\Delta}}-1\over
2}{}_{1}F_{1}\left(\bar{q}_{+};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left(\bar{q}_{-};{{\Delta}};2iap^{-}\right)e^{-i(x^{-}+a)p^{+}-i(x^{+}+a)p^{-}}J_{{{\Delta}}-1}(2z\sqrt{p^{+}p^{-}})\
.$

From the form of $P^{(O)}_{kp}$ in (153) we find that $\tilde{v}^{(O)}_{k}(X)$
can easily be expressed in terms of the same quantities as
$\tilde{v}^{(I)}_{k}(X).$ Explicitly we have

$\tilde{v}^{(O)}_{k}(X)={N_{k}z(2a)^{{{\Delta}}}\pi^{2}\over
2^{{{\Delta}}-2}{{\Gamma}}({{\Delta}})\sinh\pi\omega}\left(e^{\pi\omega}I_{+}(-k)-e^{-\pi\omega}I_{-}(-k)\right)\
.$ (190)

An explicit calculation of $I_{\pm}(k)$ (see appendix D.1 for details) gives

$\displaystyle I_{+}(k)$
$\displaystyle={2^{\nu}\over(2\pi)^{2}}z^{\nu}(i(x^{-}+a-i\epsilon))^{-q_{+}}(i(x^{-}-a-i\epsilon))^{-\bar{q}_{+}}{{\Gamma}}({{\Delta}})\left[i\left({(x^{+}+a)(x^{-}+a-i\epsilon)-z^{2}\over
x^{-}+a-i\epsilon}\right)\right]^{-i\omega}$ (191)
$\displaystyle\cdot\left[i\left({(x^{+}-a)(x^{-}+a-i\epsilon)-z^{2}\over
x^{-}+a-i\epsilon}\right)\right]^{-\bar{q}_{-}}\left[i\left({(x^{+}+a)(x^{-}-a-i\epsilon)-z^{2}\over
x^{-}-a-i\epsilon}\right)\right]^{-\bar{q}_{+}}$
$\displaystyle\cdot{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};{{\Delta}};{4a^{2}z^{2}\over((x^{+}-a)(x^{-}+a-i\epsilon)-z^{2})((x^{+}+a)(x^{-}-a-i\epsilon)-z^{2})}\right)\
,$

with $\epsilon>0$ needed to ensure convergence of the integrals. In all
factors except the hypergeometric function we can safely take $\epsilon$ to
zero. The argument of the hypergeometric function to first order in $\epsilon$
is

$\displaystyle{4a^{2}z^{2}\over((x^{+}-a)(x^{-}+a)-z^{2})((x^{+}+a)(x^{-}-a)-z^{2})}-{8a^{2}z^{2}((a^{2}-(x^{+})^{2})x^{-}+z^{2}x^{+})\over(a^{4}-a^{2}((x^{-})^{2}+(x^{+})^{2}-2z^{2})+(z^{2}-x^{-}x^{+})^{2})^{2}}~{}i\epsilon+O(\epsilon^{2})$
(192)
$\displaystyle={4a^{2}z^{2}\over((x-a)^{2}+z^{2}-t^{2})((x+a)^{2}+z^{2}-t^{2})}-((a^{2}-(x^{+})^{2})x^{-}+z^{2}x^{+})i\epsilon\
,$

where we absorb positive definite factors into $\epsilon.$ When the real part
of the expression above is in $(1,\infty),$ we are on the branch cut of the
hypergeometric function and thus must use the $i\epsilon$ prescription to
understand from which direction the branch cut is approached.

A similar calculation (see appendix D.1 for details) yields

$\displaystyle I_{-}(k)$
$\displaystyle={2^{\nu}\over(2\pi)^{2}}z^{\nu}(-i(x^{-}+a+i\epsilon))^{-q_{+}}(-i(x^{-}-a+i\epsilon))^{-\bar{q}_{+}}{{\Gamma}}({{\Delta}})\left[-i\left({(x^{+}+a)(x^{-}+a+i\epsilon)-z^{2}\over
x^{-}+a+i\epsilon}\right)\right]^{-i\omega}$ (193)
$\displaystyle\cdot\left[-i\left({(x^{+}-a)(x^{-}+a+i\epsilon)-z^{2}\over
x^{-}+a+i\epsilon}\right)\right]^{-\bar{q}_{-}}\left[-i\left({(x^{+}+a)(x^{-}-a+i\epsilon)-z^{2}\over
x^{-}-a+i\epsilon}\right)\right]^{-\bar{q}_{+}}$
$\displaystyle\cdot{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};{{\Delta}};{4a^{2}z^{2}\over((x^{+}-a)(x^{-}+a+i\epsilon)-z^{2})((x^{+}+a)(x^{-}-a+i\epsilon)-z^{2})}\right)\
.$

One finds that, due to the complex powers and branch cuts, (191) and (193)
take on distinct functional forms depending on the location of the bulk point
$X=(t,x,z).$ In particular, we find that the Poincaré patch is split into
twelve distinct regions. They are defined as follows

$\displaystyle{\mathfrak{r}}_{1}$
$\displaystyle=\\{(x^{0},x^{1},z)~{}|~{}(x^{1})^{2}+z^{2}>(a+|x^{0}|)^{2}\\}$
(194) $\displaystyle{\mathfrak{r}}_{2}$
$\displaystyle=\\{(x^{0},x^{1},z)~{}|~{}(x^{1})^{2}+z^{2}<(a-|x^{0}|)^{2},~{}|x^{0}|<a\\}$
$\displaystyle{\mathfrak{r}}_{3}$
$\displaystyle=\\{(x^{0},x^{1},z)~{}|~{}(x^{0})^{2}>(\sqrt{(x^{1})^{2}+z^{2}}+a)^{2},~{}x^{0}>a\\}$
$\displaystyle{\mathfrak{r}}_{4}$
$\displaystyle=\\{(x^{0},x^{1},z)~{}|~{}(x^{0})^{2}>(x^{1}+a)^{2}+z^{2},~{}(x^{0})^{2}<(x^{1}-a)^{2}+z^{2},~{}x^{0}>0\\}$
$\displaystyle{\mathfrak{r}}_{5}$
$\displaystyle=\\{(x^{0},x^{1},z)~{}|~{}(x^{0})^{2}<(x^{1}+a)^{2}+z^{2},~{}(x^{0})^{2}>(x^{1}-a)^{2}+z^{2},~{}x^{0}>0\\}$
$\displaystyle{\mathfrak{r}}_{6}$
$\displaystyle=\\{(x^{0},x^{1},z)~{}|~{}(x^{0})^{2}>(\sqrt{(x^{1})^{2}+z^{2}}+a)^{2},~{}x^{0}<-a\\}$
$\displaystyle{\mathfrak{r}}_{7}$
$\displaystyle=\\{(x^{0},x^{1},z)~{}|~{}(x^{0})^{2}>(x^{1}+a)^{2}+z^{2},~{}(x^{0})^{2}<(x^{1}-a)^{2}+z^{2},~{}x^{0}<0\\}$
$\displaystyle{\mathfrak{r}}_{8}$
$\displaystyle=\\{(x^{0},x^{1},z)~{}|~{}(x^{0})^{2}<(x^{1}+a)^{2}+z^{2},~{}(x^{0})^{2}>(x^{1}-a)^{2}+z^{2},~{}x^{0}<0\\}$
$\displaystyle{\mathfrak{r}}_{9}$
$\displaystyle=\\{(x^{0},x^{1},z)~{}|~{}(x^{0})^{2}<(x^{1}+a)^{2}+z^{2},~{}(x^{0})^{2}<(x^{1}-a)^{2}+z^{2},~{}(x^{0})^{2}>(\sqrt{(x^{1})^{2}+z^{2}}-a)^{2},~{}x^{0}>0\\}$
$\displaystyle{\mathfrak{r}}_{10}$
$\displaystyle=\\{(x^{0},x^{1},z)~{}|~{}(x^{0})^{2}<(x^{1}+a)^{2}+z^{2},~{}(x^{0})^{2}<(x^{1}-a)^{2}+z^{2},~{}(x^{0})^{2}>(\sqrt{(x^{1})^{2}+z^{2}}-a)^{2},~{}x^{0}<0\\}$
$\displaystyle{\mathfrak{r}}_{11}$
$\displaystyle=\\{(x^{0},x^{1},z)~{}|~{}(x^{0})^{2}>(x^{1}+a)^{2}+z^{2},~{}(x^{0})^{2}>(x^{1}-a)^{2}+z^{2},~{}(x^{0})^{2}<(\sqrt{(x^{1})^{2}+z^{2}}+a)^{2},~{}x^{0}<-a\\}$
$\displaystyle{\mathfrak{r}}_{12}$
$\displaystyle=\\{(x^{0},x^{1},z)~{}|~{}(x^{0})^{2}>(x^{1}+a)^{2}+z^{2},~{}(x^{0})^{2}>(x^{1}-a)^{2}+z^{2},~{}(x^{0})^{2}<(\sqrt{(x^{1})^{2}+z^{2}}+a)^{2},~{}x^{0}>a\\}\
.$

The region ${\mathfrak{r}}_{1}$ is exactly the entanglement wedge of $O$ that
one expects from the RT prescription. Similarly, ${\mathfrak{r}}_{2}$ is the
expected entanglement wedge of $I.$

Defining

$\displaystyle L_{1}$ $\displaystyle\equiv(x^{+}+a)(x^{-}-a)-z^{2}$ (195)
$\displaystyle L_{2}$ $\displaystyle\equiv(x^{+}-a)(x^{-}+a)-z^{2}$
$\displaystyle L_{3}$ $\displaystyle\equiv(x^{+}+a)(x^{-}+a)-z^{2}\ ,$

the expressions for $\tilde{v}^{(I)}_{k}(X)$ and $\tilde{v}^{(O)}_{k}(X)$ can
be written in a (somewhat) compact manner.

The explicit results are

$\displaystyle\tilde{v}^{(I)}_{k}(X)$
$\displaystyle=N_{k}(2a)^{{{\Delta}}}z^{{{\Delta}}}$ (196)
$\displaystyle\cdot\begin{cases}0,&X\in{\mathfrak{r}}_{1}\\\
\left(-L_{1}\right)^{-\bar{q}_{+}}\left(-L_{2}\right)^{-\bar{q}_{-}}L_{3}^{-i\omega}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{2}\\\
-i{\sin\pi{{\Delta}}\over\sinh\pi\omega}L_{1}^{-\bar{q}_{+}}L_{2}^{-\bar{q}_{-}}L_{3}^{-i\omega}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{3}\\\ -i{\sin\pi
q_{-}\over\sinh\pi\omega}L_{1}^{-\bar{q}_{+}}\left(-L_{2}\right)^{-\bar{q}_{-}}L_{3}^{-i\omega}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{4}\\\ -i{\sin\pi
q_{+}\over\sinh\pi\omega}\left(-L_{1}\right)^{-\bar{q}_{+}}L_{2}^{-\bar{q}_{-}}L_{3}^{-i\omega}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{5}\\\
i{\sin\pi{{\Delta}}\over\sinh\pi\omega}L_{1}^{-\bar{q}_{+}}L_{2}^{-\bar{q}_{-}}L_{3}^{-i\omega}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{6}\\\
i{\sin\pi\bar{q}_{+}\over\sinh\pi\omega}L_{1}^{-\bar{q}_{+}}\left(-L_{2}\right)^{-\bar{q}_{-}}\left(-L_{3}\right)^{-i\omega}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{7}\\\
i{\sin\pi\bar{q}_{-}\over\sinh\pi\omega}\left(-L_{1}\right)^{-\bar{q}_{+}}L_{2}^{-\bar{q}_{-}}\left(-L_{3}\right)^{-i\omega}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{8}\\\
{{{\Gamma}}({{\Delta}}){{\Gamma}}(i\omega)\over{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})}\left(-L_{1}\right)^{-\bar{q}_{+}}\left(-L_{2}\right)^{-\bar{q}_{-}}L_{3}^{-i\omega}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};1-i\omega;1-{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{9}\\\
{{{\Gamma}}({{\Delta}}){{\Gamma}}(-i\omega)\over{{\Gamma}}(\bar{q}_{+}){{\Gamma}}(\bar{q}_{-})}\left(-L_{1}\right)^{-\bar{q}_{+}}\left(-L_{2}\right)^{-\bar{q}_{-}}\left(-L_{3}\right)^{-i\omega}\left({4a^{2}z^{2}\over
L_{1}L_{2}}-1\right)^{i\omega}{}_{2}F_{1}\left(q_{+},q_{-};1+i\omega;1-{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{10}\\\
{i\over\sinh\pi\omega}L_{1}^{-\bar{q}_{+}}L_{2}^{-\bar{q}_{-}}\left(-L_{3}\right)^{-i\omega}\left(\sin\pi({{\Delta}}-i\omega){{{\Gamma}}({{\Delta}}){{\Gamma}}(i\omega)\over{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};1-i\omega;1-{4a^{2}z^{2}\over
L_{1}L_{2}}\right)\right.&{}\textstyle\\\
\left.+\sin\pi{{\Delta}}{{{\Gamma}}({{\Delta}}){{\Gamma}}(-i\omega)\over{{\Gamma}}(\bar{q}_{+}){{\Gamma}}(\bar{q}_{-})}\left({4a^{2}z^{2}\over
L_{1}L_{2}}-1\right)^{i\omega}{}_{2}F_{1}\left(q_{+},q_{-};1+i\omega;1-{4a^{2}z^{2}\over
L_{1}L_{2}}\right)\right),&X\in{\mathfrak{r}}_{11}\\\
{-i\over\sinh\pi\omega}L_{1}^{-\bar{q}_{+}}L_{2}^{-\bar{q}_{-}}L_{3}^{-i\omega}\left(\sin\pi{{\Delta}}{{{\Gamma}}({{\Delta}}){{\Gamma}}(i\omega)\over{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};1-i\omega;1-{4a^{2}z^{2}\over
L_{1}L_{2}}\right)\right.&{}\textstyle\\\
\left.+\sin\pi({{\Delta}}+i\omega){{{\Gamma}}({{\Delta}}){{\Gamma}}(-i\omega)\over{{\Gamma}}(\bar{q}_{+}){{\Gamma}}(\bar{q}_{-})}\left({4a^{2}z^{2}\over
L_{1}L_{2}}-1\right)^{i\omega}{}_{2}F_{1}\left(q_{+},q_{-};1+i\omega;1-{4a^{2}z^{2}\over
L_{1}L_{2}}\right)\right),&X\in{\mathfrak{r}}_{12}\\\ \end{cases}$

$\displaystyle\tilde{v}^{(O)}_{k}(X)$
$\displaystyle=N_{k}(2a)^{{{\Delta}}}z^{{{\Delta}}}$ (197)
$\displaystyle\cdot\begin{cases}\left(-L_{1}\right)^{-q_{+}}\left(-L_{2}\right)^{-q_{-}}\left(-L_{3}\right)^{i\omega}{}_{2}F_{1}\left(q_{+},q_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{1}\\\ 0,&X\in{\mathfrak{r}}_{2}\\\
-i{\sin\pi({{\Delta}}+i\omega)\over\sinh\pi\omega}L_{1}^{-q_{+}}L_{2}^{-q_{-}}L_{3}^{i\omega}{}_{2}F_{1}\left(q_{+},q_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{3}\\\ -i{\sin\pi
q_{+}\over\sinh\pi\omega}L_{1}^{-q_{+}}\left(-L_{2}\right)^{-q_{-}}L_{3}^{i\omega}{}_{2}F_{1}\left(q_{+},q_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{4}\\\ -i{\sin\pi
q_{-}\over\sinh\pi\omega}\left(-L_{1}\right)^{-q_{+}}L_{2}^{-q_{-}}L_{3}^{i\omega}{}_{2}F_{1}\left(q_{+},q_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{5}\\\
i{\sin\pi({{\Delta}}-i\omega)\over\sinh\pi\omega}L_{1}^{-q_{+}}L_{2}^{-q_{-}}L_{3}^{i\omega}{}_{2}F_{1}\left(q_{+},q_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{6}\\\
i{\sin\pi\bar{q}_{-}\over\sinh\pi\omega}L_{1}^{-q_{+}}\left(-L_{2}\right)^{-q_{-}}\left(-L_{3}\right)^{i\omega}{}_{2}F_{1}\left(q_{+},q_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{7}\\\
i{\sin\pi\bar{q}_{+}\over\sinh\pi\omega}\left(-L_{1}\right)^{-q_{+}}L_{2}^{-q_{-}}\left(-L_{3}\right)^{i\omega}{}_{2}F_{1}\left(q_{+},q_{-};{{\Delta}};{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{8}\\\
{{{\Gamma}}({{\Delta}}){{\Gamma}}(i\omega)\over{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})}\left(-L_{1}\right)^{-q_{+}}\left(-L_{2}\right)^{-q_{-}}L_{3}^{i\omega}\left({4a^{2}z^{2}\over
L_{1}L_{2}}-1\right)^{-i\omega}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};1-i\omega;1-{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{9}\\\
{{{\Gamma}}({{\Delta}}){{\Gamma}}(-i\omega)\over{{\Gamma}}(\bar{q}_{+}){{\Gamma}}(\bar{q}_{-})}\left(-L_{1}\right)^{-q_{+}}\left(-L_{2}\right)^{-q_{-}}\left(-L_{3}\right)^{i\omega}{}_{2}F_{1}\left(q_{+},q_{-};1+i\omega;1-{4a^{2}z^{2}\over
L_{1}L_{2}}\right),&X\in{\mathfrak{r}}_{10}\\\
{i\over\sinh\pi\omega}L_{1}^{-q_{+}}L_{2}^{-q_{-}}\left(-L_{3}\right)^{i\omega}\left(\sin\pi{{\Delta}}{{{\Gamma}}({{\Delta}}){{\Gamma}}(-i\omega)\over{{\Gamma}}(\bar{q}_{+}){{\Gamma}}(\bar{q}_{-})}{}_{2}F_{1}\left(q_{+},q_{-};1+i\omega;1-{4a^{2}z^{2}\over
L_{1}L_{2}}\right)\right.\\\
\left.+\sin\pi({{\Delta}}-i\omega){{{\Gamma}}({{\Delta}}){{\Gamma}}(i\omega)\over{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})}\left({4a^{2}z^{2}\over
L_{1}L_{2}}-1\right)^{-i\omega}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};1-i\omega;1-{4a^{2}z^{2}\over
L_{1}L_{2}}\right)\right),&X\in{\mathfrak{r}}_{11}\\\
{-i\over\sinh\pi\omega}L_{1}^{-q_{+}}L_{2}^{-q_{-}}\left(-L_{3}\right)^{i\omega}\left(\sin\pi({{\Delta}}+i\omega){{{\Gamma}}({{\Delta}}){{\Gamma}}(-i\omega)\over{{\Gamma}}(\bar{q}_{+}){{\Gamma}}(\bar{q}_{-})}{}_{2}F_{1}\left(q_{+},q_{-};1+i\omega;1-{4a^{2}z^{2}\over
L_{1}L_{2}}\right)\right.\\\
\left.+\sin\pi{{\Delta}}{{{\Gamma}}({{\Delta}}){{\Gamma}}(i\omega)\over{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})}\left({4a^{2}z^{2}\over
L_{1}L_{2}}-1\right)^{-i\omega}{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};1-i\omega;1-{4a^{2}z^{2}\over
L_{1}L_{2}}\right)\right),&X\in{\mathfrak{r}}_{12}\ .\end{cases}$

Notice that we have $\tilde{v}^{(O)}_{k}(X)=0$ when $X\in{\mathfrak{r}}_{2}.$
Thus local bulk field operators in ${\mathfrak{r}}_{2}$ are supported only on
$\\{a^{(I)}_{k}\\}$ oscillators and thus are elements of ${\mathcal{X}}_{I}$.
Thus the algebraic entanglement wedge of $I$ is
${\mathfrak{b}}_{I}={\mathfrak{r}}_{2}.$ This region is the bulk domain of
dependence of the hypersurface
$E_{I}=\\{(z,x^{\mu}),~{}x^{0}=0,0<z<\sqrt{a^{2}-(x^{1})^{2}}\\}$ whose
boundary is $\partial E_{I}={{\gamma}}_{I}\cup I.$ With the RT surface defined
as the (bulk part of the) boundary of this hypersurface whose domain of
dependence is ${\mathfrak{b}}_{I},$ we find the RT surface to be
${{\gamma}}_{I}=\\{(z,x^{\mu}),~{}x^{0}=0,z=\sqrt{a^{2}-(x^{1})^{2}},|x^{1}|<a\\}.$
This is exactly as expected from the RT prescription.

When $X\in{\mathfrak{r}}_{1},$ we have $\tilde{v}^{(I)}_{k}=0.$ This shows
that local bulk field operators located in ${\mathfrak{r}}_{1}$ are supported
solely on the $a^{(O)}_{k}$ oscillators and thus are elements of
${\mathcal{X}}_{O}.$ The algebraic entanglement wedge of $O$ is
${\mathfrak{b}}_{O}={\mathfrak{r}}_{1}.$ Looking at the definition of
${\mathfrak{r}}_{1}$ from (194) we see that it is the bulk domain of
dependence of the hypersurface
$E_{O}=\\{(z,x^{\mu})~{}|~{}x^{0}=0,(x^{1})^{2}+z^{2}>a^{2}\\}.$ Defining the
RT surface as the (bulk part of the) boundary, $\partial
E_{O}={{\gamma}}_{O}\cup O,$ of this hypersurface whose domain of dependence
is ${\mathfrak{b}}_{O},$ we find the RT surface to be292929Note that
${{\gamma}}_{O}={{\gamma}}_{I}$ in this case since ${\mathcal{X}}_{O}$ and
${\mathcal{X}}_{I}$ are commutants.
${{\gamma}}_{O}=\\{(z,x^{\mu}),~{}x^{0}=0,z=\sqrt{a^{2}-(x^{1})^{2}},|x^{1}|<a\\}.$
Again, this is exactly as expected from the RT prescription.

An important point is that we never used the RT prescription. Instead we
simply asked which bulk operators can be described by ${\mathcal{X}}_{O}$ and
we find that the region ${\mathfrak{r}}_{1}$ emerges as a result of the
calculation whose only inputs are global reconstruction and the basis change
from Minkowski oscillators to interval/complement oscillators.

###  D.1 Calculation of $I_{\pm}(k)$

We now explicitly calculate the integrals $I_{\pm}(k),$ defined in (189),
which are relevant to the computation of the $a^{(I)}_{k}/a^{(O)}_{k}$ mode
expansion of local bulk operators. For convenience, we copy (189) below

$\displaystyle I_{\pm}(k)$
$\displaystyle=\int_{0}^{\infty}{dp^{+}~{}dp^{-}\over(2\pi)^{2}}~{}j_{k}(\pm
p^{+},\pm p^{-}),$ (198) $\displaystyle j_{k}(p^{+},p^{-})$
$\displaystyle=(4p^{+}p^{-})^{{{\Delta}}-1\over
2}{}_{1}F_{1}\left(\bar{q}_{+};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left(\bar{q}_{-};{{\Delta}};2iap^{-}\right)e^{-i(x^{-}+a)p^{+}-i(x^{+}+a)p^{-}}J_{\nu}(2z\sqrt{p^{+}p^{-}})\
,$

where we define $\nu={{\Delta}}-1$ to simplify the expressions below.

Noting that

$J_{\nu}(2\sqrt{p^{+}p^{-}}z)={(2\sqrt{p^{+}p^{-}}z)^{\nu}\over(i2\sqrt{p^{+}p^{-}}z)^{\nu}}I_{\nu}(i2\sqrt{p^{+}p^{-}}z)=e^{-{i\pi\nu\over
2}}I_{\nu}(2i\sqrt{p^{+}p^{-}}z)\ ,$ (199)

we can apply the following formula PBMvol4

$\int_{0}^{\infty}dx~{}e^{-px}x^{B-1\over
2}I_{B-1}(\sigma\sqrt{x}){}_{1}F_{1}\left(A;B;\omega
x\right)=\left(\sigma\over
2\right)^{B-1}p^{A-B}(p-\omega)^{-A}e^{\sigma^{2}\over
4p}{}_{1}F_{1}\left(A;B;{\sigma^{2}\omega\over 4p(p-\omega)}\right)$ (200)

to (198). (200) applies for

$\mathop{\rm Re}(B)>0,~{}\mathop{\rm Re}(p)>0,~{}\mathop{\rm Re}(p-\omega)>0\
.$ (201)

We will perform the $p^{+}$ integral of (198) using (200). We can then perform
the $p^{-}$ integral using the formula PBMvol4

$\displaystyle\int_{0}^{\infty}dx{}$ $\displaystyle
e^{-px}x^{B-1}{}_{1}F_{1}\left(A^{\prime};B;\sigma
x\right){}_{1}F_{1}\left(A;B;\omega x\right)$ (202)
$\displaystyle={{\Gamma}}(B)p^{A+A^{\prime}-B}(p-\sigma)^{-A^{\prime}}(p-\omega)^{-A}{}_{2}F_{1}\left(A,A^{\prime};B;{\sigma\omega\over(p-\sigma)(p-\omega)}\right)\
,$

which applies for

$\mathop{\rm Re}(B)>0,~{}\mathop{\rm Re}(p)>0,~{}\mathop{\rm
Re}(p-\sigma)>0,~{}\mathop{\rm Re}(p-\omega)>0,~{}\mathop{\rm
Re}(p-\sigma-\omega)>0\ .$ (203)

In the computation of $I_{+},$ in order to apply (200) we need to regulate by
sending $x^{-}\to x^{-}-i\epsilon,$ for infinitesimal $\epsilon>0.$ In
particular, recalling that we have $\nu={{\Delta}}-1,$ the parameters in (200)
for this application are

$\displaystyle p$ $\displaystyle=i(x^{-}+a)\to_{x^{-}\to
x^{-}-i\epsilon}i(x^{-}+a)+\epsilon$ (204) $\displaystyle A$
$\displaystyle=\bar{q}_{+},~{}~{}B={{\Delta}}$ $\displaystyle\sigma$
$\displaystyle=2iz\sqrt{p^{-}},~{}~{}\omega=2ia\ .$

With these parameters and using (200) we obtain

$\displaystyle I_{+}(k)$ $\displaystyle={2^{\nu}e^{-{i\pi\nu\over
2}}\over(2\pi)^{2}}\int_{0}^{\infty}dp^{-}(p^{-})^{\nu\over
2}e^{-i(x^{+}+a)p^{-}}{}_{1}F_{1}\left(\bar{q}_{-};{{\Delta}};2iap^{-}\right)$
(205)
$\displaystyle\cdot\left(iz\sqrt{p^{-}}\right)^{\nu}(i(x^{-}+a-i\epsilon))^{-q_{+}}(i(x^{-}-a-i\epsilon))^{-\bar{q}_{+}}e^{{iz^{2}\over
x^{-}+a-i\epsilon}p^{-}}{}_{1}F_{1}\left(\bar{q}_{+};{{\Delta}};{2iaz^{2}\over(x^{-}+a-i\epsilon)(x^{-}-a-i\epsilon)}p^{-}\right)$
$\displaystyle={2^{\nu}e^{-{i\pi\nu\over
2}}\over(2\pi)^{2}}(iz)^{\nu}(i(x^{-}+a-i\epsilon))^{-q_{+}}(i(x^{-}-a-i\epsilon))^{-\bar{q}_{+}}$
$\displaystyle\cdot\int_{0}^{\infty}dp^{-}(p^{-})^{\nu}e^{-i(x^{+}+a)p^{-}}e^{{iz^{2}\over
x^{-}+a-i\epsilon}p^{-}}{}_{1}F_{1}\left(\bar{q}_{-};{{\Delta}};2iap^{-}\right){}_{1}F_{1}\left(\bar{q}_{+};{{\Delta}};{2iaz^{2}\over(x^{-}+a-i\epsilon)(x^{-}-a-i\epsilon)}p^{-}\right)\
.$

With the $p^{+}$ integral performed, we may now apply (202) to the $p^{-}$
integral. From the final line above, we see that the parameters are

$\displaystyle p$ $\displaystyle=i(x^{+}+a)-{iz^{2}\over x^{-}+a-i\epsilon}$
(206) $\displaystyle A^{\prime}$
$\displaystyle=\bar{q}_{-},~{}~{}B={{\Delta}},~{}~{}A=\bar{q}_{+}$
$\displaystyle\sigma$
$\displaystyle=2ia,~{}~{}\omega={2iaz^{2}\over(x^{-}+a-i\epsilon)(x^{-}-a-i\epsilon)}\
.$

With these parameter values we have

$\displaystyle\mathop{\rm Re}(B)$ $\displaystyle={{\Delta}}>0,~{}\mathop{\rm
Re}(p)=\mathop{\rm
Re}(p-\sigma)={z^{2}\over(x^{-}+a)^{2}+\epsilon^{2}}\epsilon>0$ (207)
$\displaystyle\mathop{\rm Re}(p-\omega)$ $\displaystyle=\mathop{\rm
Re}(p-\sigma-\omega)={z^{2}\over(x^{-}-a)^{2}+\epsilon^{2}}\epsilon>0\ ,$

so (202) applies without any need for further regulation.

Using (202) we then obtain

$\displaystyle I_{+}(k)$
$\displaystyle={2^{\nu}\over(2\pi)^{2}}z^{\nu}(i(x^{-}+a-i\epsilon))^{-q_{+}}(i(x^{-}-a-i\epsilon))^{-\bar{q}_{+}}{{\Gamma}}({{\Delta}})\left[i\left({(x^{+}+a)(x^{-}+a-i\epsilon)-z^{2}\over
x^{-}+a-i\epsilon}\right)\right]^{-i\omega}$ (208)
$\displaystyle\cdot\left[i\left({(x^{+}-a)(x^{-}+a-i\epsilon)-z^{2}\over
x^{-}+a-i\epsilon}\right)\right]^{-\bar{q}_{-}}\left[i\left({(x^{+}+a)(x^{-}-a-i\epsilon)-z^{2}\over
x^{-}-a-i\epsilon}\right)\right]^{-\bar{q}_{+}}$
$\displaystyle\cdot{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};{{\Delta}};{4a^{2}z^{2}\over((x^{+}-a)(x^{-}+a-i\epsilon)-z^{2})((x^{+}+a)(x^{-}-a-i\epsilon)-z^{2})}\right)\
,$

as quoted in (191).

Similarly, in the computation of $I_{-},$ in order to apply (200) to the
$p^{+}$ integral, we now need to regulate by sending $x^{-}\to
x^{-}+i\epsilon,$ for infinitesimal $\epsilon>0.$ For this application the
parameters in (200) are

$\displaystyle p$ $\displaystyle=-i(x^{-}+a)\to_{x^{-}\to
x^{-}+i\epsilon}-i(x^{-}+a)+\epsilon$ (209) $\displaystyle A$
$\displaystyle=\bar{q}_{+},~{}~{}B={{\Delta}}$ $\displaystyle\sigma$
$\displaystyle=2iz\sqrt{p^{-}},~{}~{}\omega=-2ia\ .$

With these parameters and using (200) we obtain

$\displaystyle I_{-}(k)={2^{\nu}e^{-{i\pi\nu\over
2}}\over(2\pi)^{2}}\int_{0}^{\infty}dp^{-}(p^{-})^{\nu\over
2}e^{i(x^{+}+a)p^{-}}{}_{1}F_{1}\left(\bar{q}_{-};{{\Delta}};-2iap^{-}\right)$
(210)
$\displaystyle\cdot\left(iz\sqrt{p^{-}}\right)^{\nu}(-i(x^{-}+a+i\epsilon))^{-q_{+}}(-i(x^{-}-a+i\epsilon))^{-\bar{q}_{+}}e^{-{iz^{2}\over
x^{-}+a+i\epsilon}p^{-}}{}_{1}F_{1}\left(\bar{q}_{+};{{\Delta}};{-2iaz^{2}\over(x^{-}+a+i\epsilon)(x^{-}-a+i\epsilon)}p^{-}\right)$
$\displaystyle={2^{\nu}e^{-{i\pi\nu\over
2}}\over(2\pi)^{2}}(iz)^{\nu}(-i(x^{-}+a+i\epsilon))^{-q_{+}}(-i(x^{-}-a+i\epsilon))^{-\bar{q}_{+}}$
$\displaystyle\cdot\int_{0}^{\infty}dp^{-}(p^{-})^{\nu}e^{i(x^{+}+a)p^{-}}e^{-{iz^{2}\over
x^{-}+a+i\epsilon}p^{-}}{}_{1}F_{1}\left(\bar{q}_{-};{{\Delta}};-2iap^{-}\right){}_{1}F_{1}\left(\bar{q}_{+};{{\Delta}};{-2iaz^{2}\over(x^{-}+a+i\epsilon)(x^{-}-a+i\epsilon)}p^{-}\right)\
.$

With the $p^{+}$ integral performed, we may now apply (202) to the $p^{-}$
integral. The parameters are

$\displaystyle p$ $\displaystyle=-i(x^{+}+a)+{iz^{2}\over x^{-}+a+i\epsilon}$
(211) $\displaystyle A^{\prime}$
$\displaystyle=\bar{q}_{-},~{}~{}B={{\Delta}},~{}~{}A=\bar{q}_{+}$
$\displaystyle\sigma$
$\displaystyle=-2ia,~{}~{}\omega={-2iaz^{2}\over(x^{-}+a+i\epsilon)(x^{-}-a+i\epsilon)}\
.$

With these parameter values we have

$\displaystyle\mathop{\rm Re}(B)$ $\displaystyle={{\Delta}}>0,~{}\mathop{\rm
Re}(p)=\mathop{\rm
Re}(p-\sigma)={z^{2}\over(x^{-}+a)^{2}+\epsilon^{2}}\epsilon>0$ (212)
$\displaystyle\mathop{\rm Re}(p-\omega)$ $\displaystyle=\mathop{\rm
Re}(p-\sigma-\omega)={z^{2}\over(x^{-}-a)^{2}+\epsilon^{2}}\epsilon>0\ ,$

so (202) applies without any further regulation.

Using (202) we then obtain

$\displaystyle I_{-}(k)$
$\displaystyle={2^{\nu}\over(2\pi)^{2}}z^{\nu}(-i(x^{-}+a+i\epsilon))^{-q_{+}}(-i(x^{-}-a+i\epsilon))^{-\bar{q}_{+}}{{\Gamma}}({{\Delta}})\left[-i\left({(x^{+}+a)(x^{-}+a+i\epsilon)-z^{2}\over
x^{-}+a+i\epsilon}\right)\right]^{-i\omega}$ (213)
$\displaystyle\cdot\left[-i\left({(x^{+}-a)(x^{-}+a+i\epsilon)-z^{2}\over
x^{-}+a+i\epsilon}\right)\right]^{-\bar{q}_{-}}\left[-i\left({(x^{+}+a)(x^{-}-a+i\epsilon)-z^{2}\over
x^{-}-a+i\epsilon}\right)\right]^{-\bar{q}_{+}}$
$\displaystyle\cdot{}_{2}F_{1}\left(\bar{q}_{+},\bar{q}_{-};{{\Delta}};{4a^{2}z^{2}\over((x^{+}-a)(x^{-}+a+i\epsilon)-z^{2})((x^{+}+a)(x^{-}-a+i\epsilon)-z^{2})}\right)\
,$

as quoted in (193).

##  Appendix E Showing the additivity and intersection anomalies

In this appendix we use the results of appendix D to demonstrate the second
relation of (72) and relation (73). The first relation of (72) is discussed in
appendix F.

###  E.1 Showing the additivity anomaly

The additivity anomaly we studied in section IV.4 is the relation

${\mathcal{X}}_{R_{+}}\vee{\mathcal{X}}_{R_{-}}\subset{\mathcal{X}}_{R_{U}}\
.$ (214)

From the definitions of ${\mathcal{X}}_{R_{\pm}},~{}{\mathcal{X}}_{R_{U}}$ it
is clear that
${\mathcal{X}}_{R_{+}}\vee{\mathcal{X}}_{R_{-}}\subseteq{\mathcal{X}}_{R_{U}}.$
It therefore remains to show that this is a proper inclusion. Consider the
operator $\phi(x^{0}=0,x^{1}=0,z)$ with $\sqrt{a^{2}-b^{2}}<z<a+b.$ From (197)
(with half-width $a\to a+b$) we see that, since $z<a+b,$ $\phi(0,0,z)$ can be
interpreted as a boundary operator with support only on $\\{a^{R_{U}}_{k}\\}$
(and no support on $\\{a^{\bar{R_{U}}}_{k}\\}$). Thus,
$\phi(0,0,z)\in{\mathcal{X}}_{R_{U}}.$

We will now show that $\phi(0,0,z)$ is not in
${\mathcal{X}}_{R_{+}}\vee{\mathcal{X}}_{R_{-}},$ establishing the proper
inclusion. In fact,
$\phi(0,0,z)\in\left({\mathcal{X}}_{R_{-}}\cup{\mathcal{X}}_{R_{+}}\right)^{\prime}.$
To see this, consider (196)–(197) translated to $R_{\pm}$ which is centered at
$t=0,~{}x=\pm b$ rather than the origin. The overlap of $\phi(0,0,z)$ with the
corresponding oscillators is then given by (196)–(197) with $x\to x\mp d$.
Since $z>\sqrt{a^{2}-b^{2}}$ we see that $(0,0,z)$ lies in the
${\mathfrak{r}}_{1}$ region with respect to both $R_{+}$ and $R_{-}.$ Thus,
$\phi(0,0,z)$ has no overlap with $\\{a^{(R_{\pm})}_{k}\\}$ and lies in the
commutant of both ${\mathcal{X}}_{R_{+}}$ and ${\mathcal{X}}_{R_{-}}.$ Thus,
$\phi(0,0,z)$ is a non-zero (non-central) operator in
$\left({\mathcal{X}}_{R_{+}}\cup{\mathcal{X}}_{R_{-}}\right)^{\prime}$ and
thus it does not lie in
$\left({\mathcal{X}}_{R_{+}}\cup{\mathcal{X}}_{R_{-}}\right)^{\prime\prime}={\mathcal{X}}_{R_{+}}\vee{\mathcal{X}}_{R_{-}}.$
It does however lie in ${\mathcal{X}}_{R_{U}}$ establishing the proper
inclusion (214).

###  E.2 Showing the intersection anomaly

The intersection anomaly (73) of section IV.4 reads

${\mathcal{X}}_{R_{I}}\subset{\mathcal{X}}_{R_{-}}\land{\mathcal{X}}_{R_{+}}\
.$ (215)

Clearly ${\mathcal{X}}_{R_{\pm}}\supset{\mathcal{X}}_{R_{I}}$ so the inclusion

${\mathcal{X}}_{R_{I}}\subseteq{\mathcal{X}}_{R_{-}}\land{\mathcal{X}}_{R_{+}}$
(216)

is immediate. It then remains to show that there are operators in
${\mathcal{X}}_{R_{-}}\land{\mathcal{X}}_{R_{+}}$ that are not in
${\mathcal{X}}_{R_{I}}.$ Consider the operator $\phi(x^{0}=0,x^{1}=0,z)$ with
$a-b<z<\sqrt{a^{2}-b^{2}}.$

This operator has localized mode expansions in $R_{\pm}$ described by
(196)–(197) with $x\to x\mp b.$ Since $z<\sqrt{a^{2}-b^{2}},$ we see that
$(0,0,z)$ is in ${\mathfrak{r}}_{2}$ with respect to both $R_{+}$ and $R_{-}$
and thus $\phi(0,0,z)$ has no overlap with $\\{a^{\bar{R}_{\pm}}_{k}\\}.$ Thus
$\phi(0,0,z)$ can be expanded in terms of either the $\\{a^{(R_{-})}_{k}\\}$
or $\\{a^{(R_{+})}_{k}\\}$ and is therefore part of both
${\mathcal{X}}_{R_{-}}$ and ${\mathcal{X}}_{R_{+}}.$ i.e.
$\phi(0,0,z)\in{\mathcal{X}}_{R_{-}}\land{\mathcal{X}}_{R_{+}}.$

We may also express $\phi(0,0,z)$ in terms of
$\\{a^{R_{I}}_{k},~{}a^{\bar{R}_{I}}_{k}\\}.$ The explicit description is
given by (196)–(197) with half-width $a\to a-b.$ Since $z>a-b,$ $(0,0,z)$ is
in ${\mathfrak{r}}_{1}$ with respect to the $R_{I}$ decomposition. Thus
$\phi(0,0,z)$ can be expressed in terms of the $\\{a^{\bar{R}_{I}}_{k}\\}$
only and cannot be expressed in terms of the $\\{a^{R_{I}}_{k}\\}.$ Thus,
$\phi(0,0,z)\notin{\mathcal{X}}_{R_{I}}.$ This, along with the discussion of
the previous paragraph, then establishes that $\phi(0,0,z)$ is an operator in
${\mathcal{X}}_{R_{-}}\land{\mathcal{X}}_{R_{+}}$ but not in
${\mathcal{X}}_{R_{I}}$ which shows that we have the proper inclusion (215).

##  Appendix F Boundary Geometric Superadditivity

We consider a Minkowski boundary in the vacuum state. For the overlapping
spatial subregions $R_{-}:x^{1}\in(-b-a,-b+a)$ and $R_{+}:x^{1}\in(b-a,b+a)$
with $a>b>0$ we have boundary causal completions

${\hat{R}}_{\pm}=\\{(x^{0},x^{1})~{}|~{}|x^{0}+(x^{1}\mp
b)|<a,~{}|x^{0}-(x^{1}\mp b)|<a\\}\ .$ (217)

These regions overlap in a small diamond of half-width $a-b$ centered at the
origin. The union, $R_{U}=R_{+}\cup R_{-},$ has causal completion

${\hat{R}}_{U}=\\{(x^{0},x^{1})~{}|~{}|x^{0}+x^{1}|<a+b,~{}|x^{0}-x^{1}|<a+b\\}\
,$ (218)

and the intersection $R_{I}=R_{+}\cap R_{-},$ has causal completion

${\hat{R}}_{I}=\\{(x^{0},x^{1})~{}|~{}|x^{0}+x^{1}|<a-b,~{}|x^{0}-x^{1}|<a-b\\}\
.$ (219)

We are interested in understanding the relations among
${\mathcal{X}}_{R_{-}}\lor{\mathcal{X}}_{R_{+}},~{}{\mathcal{X}}_{R_{-}}\land{\mathcal{X}}_{R_{+}},{\mathcal{X}}_{R_{U}},~{}{\mathcal{X}}_{R_{I}}.$
In particular, we will see that
${\mathcal{X}}_{R_{-}}\lor{\mathcal{X}}_{R_{+}}={{\mathcal{M}}}_{\hat{R_{-}}}\lor{{\mathcal{M}}}_{\hat{R_{+}}}\supset{{\mathcal{M}}}_{\hat{R_{-}}\cup\hat{R_{+}}},$
i.e. the addition of the algebras in $\hat{R_{-}}$ and $\hat{R_{+}}$ leads to
a larger algebra then expected from the boundary spacetime union of regions.

###  F.1 Bulk dual of ${\mathcal{X}}_{R_{-}}\lor{\mathcal{X}}_{R_{+}}$

We first construct the bulk dual of the algebra
${\mathcal{X}}_{R_{-}}\lor{\mathcal{X}}_{R_{+}}={{\mathcal{M}}}_{\hat{R}_{+}}\lor{{\mathcal{M}}}_{\hat{R}_{-}}$.
Bulk additivity implies that this algebra is dual to the causal completion of
the union of causal domains of the two boundary diamonds,
$\left({\mathfrak{c}}_{\hat{R}_{+}}\cup{\mathfrak{c}}_{\hat{R}_{-}}\right)^{\prime\prime}$.
We construct this causal completion directly in the bulk using the bulk causal
structure.

The causal domains of the boundary diamonds can be explicitly described on
Poincaré AdS3 by

${\mathfrak{c}}_{{\hat{R}}_{\pm}}=\\{(x^{0},x^{1},z)~{}|~{}(x^{1}\mp
b)^{2}+z^{2}<(a-|x^{0}|)^{2},~{}|x^{0}|<a\\}\ .$ (220)

The union ${\mathfrak{c}}_{{\hat{R}}_{-}}\cup{\mathfrak{c}}_{{\hat{R}}_{+}}$
is not a causally complete bulk subregion. However, its causal complement is
causally complete and thus the domain of dependence of a subregion $\bar{H}$
of a bulk Cauchy slice (in this case, the $t=0$ slice). The spatial subregion
$\bar{H}$ is described by

${\bar{H}}=\\{(x^{0},x^{1},z)~{}|~{}x^{0}=0,(x^{1}-b)^{2}+z^{2}>a^{2}\\}\cap\\{(x^{0},x^{1},z)~{}|~{}x^{0}=0,(x^{1}+b)^{2}+z^{2}>a^{2}\\}\
.$ (221)

The causal complement is then
$\left({\mathfrak{c}}_{{\hat{R}}_{-}}\cup{\mathfrak{c}}_{{\hat{R}}_{+}}\right)^{\prime}={\bar{H}}^{\prime\prime}.$
We are interested in the causal completion of
${\mathfrak{c}}_{{\hat{R}}_{-}}\cup{\mathfrak{c}}_{{\hat{R}}_{+}}$ which we
can clearly see is ${\bar{H}}^{\prime}.$

Since ${\bar{H}}^{\prime}$ is causally complete in the bulk, it is the domain
of dependence of the subregion $H,$ the complement of ${\bar{H}}$ on the
$x^{0}=0$ bulk Cauchy slice. Thus we have established that
$\left({\mathfrak{c}}_{{\hat{R}}_{-}}\cup{\mathfrak{c}}_{{\hat{R}}_{+}}\right)^{\prime\prime}=H^{\prime\prime}.$
If $H$ had a smooth boundary, the boundary of $H^{\prime\prime}$ would be
generated by orthogonal null geodesics sent from the boundary of $H.$ However,
in this case the boundary of $H$ has a kink at the point
$(x^{0}=0,x^{1}=0,z=\sqrt{a^{2}-b^{2}}),$ and thus there is not a well-defined
normal at that point. For this reason we construct $H^{\prime\prime}$ as
${\bar{H}}^{\prime}$ rather than from null geodesics.

Clearly $H^{\prime\prime}$ has both $x^{0}-$ reflection and $x^{1}-$reflection
symmetry about the $x^{0}=0$ and $x^{1}=0$ axes, respectively. For this reason
we only need to consider the part of $H^{\prime\prime}$ with
$x^{0}>0,~{}x^{1}>0$ and the rest can be obtained by reflections. We first
exclude from $H^{\prime\prime}$ all points connected by pure
$x^{0}-$translation to ${\bar{H}}$. Notice that any point with
$0<x^{1}<a+b,z>\sqrt{a^{2}-(x^{1}-b)^{2}}$ is clearly timelike separated (by a
constant $(x^{1},z)$ curve) from ${\bar{H}}$ and thus doesn’t lie in
$H^{\prime\prime}.$ Thus there are no points with $z>a$ in $H^{\prime\prime}.$
For fixed $z<a,$ there can also not be any points with
$x^{1}-b>\sqrt{a^{2}-z^{2}}.$ Similarly, when $\sqrt{a^{2}-b^{2}}<z<a,$ there
cannot be any points with $0<x^{1}<b-\sqrt{a^{2}-z^{2}}.$

Thus, from considerations of pure $x^{0}-$translations alone, we conclude that
points in $H^{\prime\prime}$ (with $x^{1}>0$) must satisfy $z<a$ and
furthermore, for $\sqrt{a^{2}-b^{2}}<z$ they must have
$|x^{1}-b|>\sqrt{a^{2}-z^{2}}$, while for $0<z<\sqrt{a^{2}-b^{2}}$ we must
have $x^{1}-b>\sqrt{a^{2}-z^{2}}.$

Since Poincaré AdS is conformally flat (i.e. null curves are the same as in
Minkowski space), a point $(x^{0},x^{1},z)$ is only in
$H^{\prime\prime}={\bar{H}}^{\prime}$ if for all
$(\bar{x^{1}},{\bar{z}})\in{\bar{H}}$ we have

$(x^{0})^{2}<(x^{1}-\bar{x^{1}})^{2}+(z-\bar{z})^{2}\ .$ (222)

Notice that the ‘closest’ points to $H^{\prime\prime}$ in ${\bar{H}}$ are
precisely those which lie on its boundary with $H.$ For $\bar{x^{1}}>0,$ the
boundary is the curve $\bar{z}=\sqrt{a^{2}-(\bar{x^{1}}-b)^{2}}.$ Thus, for
fixed $(x^{1},z)$ that could potentially be associated to a point in
$H^{\prime\prime},$ the minimal value of the RHS of (222) over all
$(\bar{x^{1}},\bar{z})\in\bar{H}$ will be obtained at a point on the boundary
curve for which the RHS of (222) reads

$F(\bar{x^{1}};x^{1},z)=(x^{1}-\bar{x^{1}})^{2}+\left(z-\sqrt{a^{2}-(\bar{x^{1}}-b)^{2}}\right)^{2}\
.$ (223)

For fixed (allowed) $(x^{1},z)$, $F$ is minimized over all
$\bar{x^{1}}\in(0,b+a)$ at $\bar{x^{1}}_{*}$ with

$\bar{x^{1}}_{*}=\begin{cases}b+{a(x^{1}-b)\over\sqrt{(x^{1}-b)^{2}+z^{2}}},&(\sqrt{a^{2}-b^{2}}<z<a)\lor\left(z<\sqrt{a^{2}-b^{2}}\land
x^{1}>b\left(1-\sqrt{z^{2}\over a^{2}-b^{2}}\right)\right)\\\
0,&z<\sqrt{a^{2}-b^{2}}\land 0<x^{1}<b\left(1-\sqrt{z^{2}\over
a^{2}-b^{2}}\right).\end{cases}$ (224)

We therefore obtain that the points $(x^{0},x^{1},z)\in H^{\prime\prime}$ with
$x^{1}>0$ must satisfy $|x^{0}|<t_{\max}(x^{1},z)$ where

$t_{\max}(x^{1},z)=\begin{cases}\sqrt{(x^{1})^{2}+(z-\sqrt{a^{2}-b^{2}})^{2}},&z<\sqrt{a^{2}-b^{2}}\land
0<x^{1}<b\left(1-\sqrt{z^{2}\over a^{2}-b^{2}}\right)\\\
a-\sqrt{(x^{1}-b)^{2}+z^{2}},&(\sqrt{a^{2}-b^{2}}<z<a)\lor\left(z<\sqrt{a^{2}-b^{2}}\land
x^{1}>b\left(1-\sqrt{z^{2}\over a^{2}-b^{2}}\right)\right),\\\ \end{cases}$
(225)

and recall that we also must have $|x^{1}-d|<\sqrt{a^{2}-z^{2}}.$

Using the $x^{0}-$ and $x^{1}-$reflection symmetries we obtain the complete
characterization of the causal completion of the union of the two causal
domains of the boundary diamonds, $H^{\prime\prime}.$ It is the set of bulk
points $(t,x^{1},z)$ such that $z<a$ and $|x^{0}|<t_{\max}(x^{1},z)$ with

$t_{\max}(x^{1},z<\sqrt{a^{2}-b^{2}})=\begin{cases}a-\sqrt{(x^{1}+b)^{2}+z^{2}},&x^{1}\in\left(-b-\sqrt{a^{2}-z^{2}},-b\cdot\left(\sqrt{a^{2}-b^{2}}-z\over\sqrt{a^{2}-b^{2}}\right)\right)\\\
\sqrt{(x^{1})^{2}+(\sqrt{a^{2}-b^{2}}-z)^{2}},&x^{1}\in\left(-b\cdot\left(\sqrt{a^{2}-b^{2}}-z\over\sqrt{a^{2}-b^{2}}\right),b\cdot\left(\sqrt{a^{2}-b^{2}}-z\over\sqrt{a^{2}-b^{2}}\right)\right)\\\
a-\sqrt{(x^{1}-b)^{2}+z^{2}},&x^{1}\in\left(b\cdot\left(\sqrt{a^{2}-b^{2}}-z\over\sqrt{a^{2}-b^{2}}\right),b+\sqrt{a^{2}-z^{2}}\right)\\\
\end{cases}$ (226)
$t_{\max}(x^{1},\sqrt{a^{2}-b^{2}}<z<a)=\begin{cases}a-\sqrt{(x^{1}+b)^{2}+z^{2}},&x^{1}\in\left(-b-\sqrt{a^{2}-z^{2}},-b+\sqrt{a^{2}-z^{2}}\right)\\\
0,&x^{1}\in\left(-b+\sqrt{a^{2}-z^{2}},b-\sqrt{a^{2}-z^{2}}\right)\\\
a-\sqrt{(x^{1}-b)^{2}+z^{2}},&x^{1}\in\left(b-\sqrt{a^{2}-z^{2}},b+\sqrt{a^{2}-z^{2}}\right)\\\
\end{cases}$ (227)

Two constant $z$ cross-sections of $H^{\prime\prime}$ are shown in figure 28.
The intersection of $H^{\prime\prime}$ with the boundary is shown in figure
29. We will call this boundary subregion
$(\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}$ and sometimes use the shorthand
$R_{e}$. Notice that
$(\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}\supset\hat{R}_{+}\cup\hat{R}_{-},$
i.e. the causal completion of the union of the two causal domains contains
additional near-boundary points. Thus, bulk additivity suggests that
${{\mathcal{M}}}_{\hat{R}_{+}\cup\hat{R}_{-}}\subset\left({{\mathcal{M}}}_{\hat{R}_{+}\cup\hat{R}_{-}}\right)^{\prime\prime},$
i.e. ${{\mathcal{M}}}_{\hat{R}_{+}\cup\hat{R}_{-}}$ is not a von Neumann
algebra. One may then ask, is
$\left({{\mathcal{M}}}_{\hat{R}_{+}\cup\hat{R}_{-}}\right)^{\prime\prime}$ the
same as
$\left({{\mathcal{M}}}_{(\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}}\right)^{\prime\prime}$?
In the next subsection, we give a bulk argument that the answer is yes by
identifying
$\left({{\mathcal{M}}}_{(\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}}\right)^{\prime\prime}$
with the algebra of bulk operators in its causal domain, which we will see is
exactly equal to $H^{\prime\prime}.$ Moreover,
$(\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}$ is the largest boundary subregion
whose causal domain is equal to $H^{\prime\prime}.$

(a)

(b)

Figure 28: Characterizing $H^{\prime\prime},$ for $a=1,~{}b=0.5$. (a)
($z=0.5$) $z<\sqrt{a^{2}-b^{2}}$ cross-section of $H^{\prime\prime}.$ The
points between the blue and red curves are in $H^{\prime\prime}$ but not in
the union of causal domains. (b) ($z=0.9$) $z>\sqrt{a^{2}-b^{2}}$ cross-
section of $H^{\prime\prime}$.  Figure 29: The near-boundary points of the
bulk causal completion of the union of causal domains of two overlapping
boundary diamonds. The points between the black and blue curves lie in $R_{e}$
but not in $\hat{R}_{-}\cup\hat{R}_{+}.$

One final interesting point is that the intersection of $H^{\prime\prime}$
with the boundary is not a causally complete boundary subregion. In
particular, we have
$(\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}\subset\widehat{(R_{+}\cup R_{-})}.$
Thus bulk causal completion is not equivalent to boundary causal completion.
This was already clear from the time-band algebras but the signature in this
case is more interesting. In this case, bulk causal completion gives some
extra near-boundary points that are not in the original boundary subregion of
interest; however, it does not give all the extra near-boundary points of the
boundary causal completion of the original boundary subregion.

###  F.2 Causal wedge of $(\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}$

We saw in the previous subsection that
$\left({\mathfrak{c}}_{{\hat{R}}_{-}}\cup{\mathfrak{c}}_{{\hat{R}}_{+}}\right)^{\prime\prime}$
contains extra near-boundary points that are not in
$\hat{R}_{+}\cup\hat{R}_{-}.$ In fact the intersection of
$\left({\mathfrak{c}}_{{\hat{R}}_{-}}\cup{\mathfrak{c}}_{{\hat{R}}_{+}}\right)^{\prime\prime}$
with the boundary is the larger region
$(\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}$ shown in figure 29. We can
explicitly describe this subregion as the region of the $(x^{0},x^{1})$ plane
with $|x^{0}|<t_{\max}(x^{1}),$ where

$t_{\max}(x^{1})=\begin{cases}a+(x^{1}+b),&x^{1}\in\left(-b-a,-b\right)\\\
\sqrt{(x^{1})^{2}+a^{2}-b^{2}},&x^{1}\in\left(-b,b\right)\\\
a-(x^{1}-b),&x^{1}\in\left(b,b+a\right)\\\ \end{cases}.$ (228)

We wish to understand the causal wedge of this boundary subregion
${\mathfrak{a}}_{(\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}}=J^{+}\left((\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}\right)\cap
J^{-}\left((\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}\right).$ By
$x^{0}-$reflection symmetry about $x^{0}=0$ we need only study its bulk causal
past as its bulk causal future is simply the $x^{0}-$reflection of that bulk
subregion. In particular, we will study the bulk points $(x^{0},x^{1},z)$ in
the boundary of the causal past with $x^{0}>0.$ For simplicity we will
suppress the bulk subregion name and write $J^{-}$ rather than
$J^{-}\left((\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}\right)$ and $R_{e}$
rather than $(\hat{R}_{+}\cup\hat{R}_{-})_{\text{ext}}$.

A point $(x^{0},x^{1},z)$ is only in $J^{-}$ if
$x^{0}<a~{}(=\max_{x^{1}}(t_{\max}(x^{1})))$ and

$(x^{0}-t^{\prime})^{2}>(x^{1}-x^{\prime})^{2}+z^{2}\ ,$ (229)

for some $(t^{\prime},x^{\prime})\in R_{e}.$ The boundary of $J^{-}$ is given
by finding the maximum value of $x^{0}$ for a fixed $(x^{1},z)$ so that (229)
can still be satisfied. Clearly the maximal value of $x^{0}$ will be obtained
by taking $t^{\prime}$ as large as possible, i.e.
$t^{\prime}=t_{\max}(x^{\prime}).$ To be in the past of $R_{e}$ it is
necessary that $x^{0}<t^{\prime},$ thus, (229) can be stated in terms of
$x^{0}$ as

$x^{0}<t_{\max}(x^{\prime})-\sqrt{(x^{1}-x^{\prime})^{2}+z^{2}}\ .$ (230)

The maximum of the RHS is obtained at $x^{\prime*}$ where, for
$z<\sqrt{a^{2}-b^{2}}$

$x^{\prime*}=\begin{cases}-b,&x^{1}\in\left(-\infty,-b\left(1-\sqrt{z^{2}\over
a^{2}-b^{2}}\right)\right)\\\
x^{1}{\sqrt{a^{2}-b^{2}}\over\sqrt{a^{2}-b^{2}}-z},&x^{1}\in\left(-b\left(1-\sqrt{z^{2}\over
a^{2}-b^{2}}\right),b\left(1-\sqrt{z^{2}\over a^{2}-b^{2}}\right)\right)\\\
b,&x^{1}\in\left(b\left(1-\sqrt{z^{2}\over
a^{2}-b^{2}}\right),\infty\right)\\\ \end{cases},$ (231)

and for $z>\sqrt{a^{2}-b^{2}}$

$x^{\prime*}=\begin{cases}-b,&x^{1}\in\left(-\infty,0\right)\\\
b,&x^{1}\in\left(0,\infty\right)\\\ \end{cases}.$ (232)

Plugging $x^{\prime*}$ into (230) we obtain the maximal value of $x^{0}$ for
any fixed $(x^{1},z),$ so that $(x^{0},x^{1},z)$ is in the causal past of
$R_{e}.$ For the intermediate values of $x^{1}$ in (231) we have

$\displaystyle t_{\max}(x^{\prime*})-\sqrt{(x^{1}-x^{\prime*})^{2}+z^{2}}$
(233)
$\displaystyle=\sqrt{\left(x^{1}{\sqrt{a^{2}-b^{2}}\over\sqrt{a^{2}-b^{2}}-z}\right)^{2}+a^{2}-b^{2}}-\sqrt{\left(x^{1}-x^{1}{\sqrt{a^{2}-b^{2}}\over\sqrt{a^{2}-b^{2}}-z}\right)^{2}+z^{2}}$
$\displaystyle=\sqrt{(\sqrt{a^{2}-b^{2}}-z)^{2}+(x^{1})^{2}}\ .$

For the other values of $x^{1}$ the evaluation of
$t_{\max}(x^{\prime*})-\sqrt{(x^{1}-x^{\prime*})^{2}+z^{2}}$ is trivial, using
$t_{\max}(\pm b)=a.$ We therefore obtain the maximum value of $x^{0}$ for any
fixed $(x^{1},z)$ to be in the past of $R_{e}$

$t_{\max}^{\text{(past)}}(x^{1},z<\sqrt{a^{2}-b^{2}})=\begin{cases}a-\sqrt{(x^{1}+b)^{2}+z^{2}},&x^{1}\in\left(-\infty,-b\left(1-\sqrt{z^{2}\over
a^{2}-b^{2}}\right)\right)\\\
\sqrt{(x^{1})^{2}+(\sqrt{a^{2}-b^{2}}-z)^{2}},&x^{1}\in\left(-b\left(1-\sqrt{z^{2}\over
a^{2}-b^{2}}\right),b\left(1-\sqrt{z^{2}\over a^{2}-b^{2}}\right)\right)\\\
a-\sqrt{(x^{1}-b)^{2}+z^{2}},&x^{1}\in\left(b\left(1-\sqrt{z^{2}\over
a^{2}-b^{2}}\right),\infty\right)\\\ \end{cases}$ (234)
$t_{\max}^{\text{(past)}}(x^{1},z>\sqrt{a^{2}-b^{2}})=\begin{cases}a-\sqrt{(x^{1}+b)^{2}+z^{2}},&x^{1}\in\left(-\infty,0\right)\\\
a-\sqrt{(x^{1}-b)^{2}+z^{2}},&x^{1}\in\left(0,\infty\right)\\\ \end{cases}.$
(235)

However, for $(x^{0},x^{1},z)$ to be in the causal wedge of $R_{e}$ it must
also be in the bulk causal future of $R_{e},$ not only in its bulk causal
past. This places a minimum on the possible value of $x^{0}$ for fixed
$(x^{1},z).$ We thus only have $(x^{0},x^{1},z)$ in the causal wedge of
$R_{e}$ if

$t^{\text{(future)}}_{\min}(x^{1},z)<x^{0}<t_{\max}^{\text{(past)}}(x^{1},z)\
.$ (236)

By time reflection symmetry

$t^{\text{(future)}}_{\min}(x^{1},z)=-t_{\max}^{\text{(past)}}(x^{1},z)\ .$
(237)

Clearly, no points with spatial coordinates $(x^{1},z)$ can be in the causal
wedge of $R_{e}$ when $t_{\max}^{\text{(past)}}(x^{1},z)<0.$ This prevents any
points with $z>a,$ or with $z<a$ but $x^{1}<-b-\sqrt{a^{2}-z^{2}}$ or
$x^{1}>b+\sqrt{a^{2}-z^{2}}$ from lying in the causal wedge of $R_{e}.$ The
final result is then that the causal wedge of $R_{e}$ is the set of points
$(x^{0},x^{1},z)$ with $|x^{0}|<t_{\max}(x^{1},z)$ where

$t_{\max}(x^{1},z<\sqrt{a^{2}-b^{2}})=\begin{cases}a-\sqrt{(x^{1}+b)^{2}+z^{2}},&x^{1}\in\left(-b-\sqrt{a^{2}-z^{2}},-b\left(1-\sqrt{z^{2}\over
a^{2}-b^{2}}\right)\right)\\\
\sqrt{(x^{1})^{2}+(\sqrt{a^{2}-b^{2}}-z)^{2}},&x^{1}\in\left(-b\left(1-\sqrt{z^{2}\over
a^{2}-b^{2}}\right),b\left(1-\sqrt{z^{2}\over a^{2}-b^{2}}\right)\right)\\\
a-\sqrt{(x^{1}-b)^{2}+z^{2}},&x^{1}\in\left(b\left(1-\sqrt{z^{2}\over
a^{2}-b^{2}}\right),b+\sqrt{a^{2}-z^{2}}\right)\\\ \end{cases}$ (238)
$t_{\max}(x^{1},\sqrt{a^{2}-b^{2}}<z<a)=\begin{cases}a-\sqrt{(x^{1}+b)^{2}+z^{2}},&x^{1}\in\left(-b-\sqrt{a^{2}-z^{2}},-b+\sqrt{a^{2}-z^{2}}\right)\\\
a-\sqrt{(x^{1}-b)^{2}+z^{2}},&x^{1}\in\left(b-\sqrt{a^{2}-z^{2}},b+\sqrt{a^{2}-z^{2}}\right)\\\
\end{cases}.$ (239)

This then shows that
${\mathfrak{a}}_{R_{e}}=\left({\mathfrak{c}}_{\hat{R}_{+}}\cup{\mathfrak{c}}_{\hat{R}_{-}}\right)^{\prime\prime},$
i.e. the causal wedge of the extended boundary subregion $R_{e}$ is equal to
the bulk causal completion of the union of causal wedges for the two boundary
diamonds. In particular, this establishes that ${\mathfrak{a}}_{R_{e}}$ is
itself causally complete (i.e.
${\mathfrak{a}}_{R_{e}}={\mathfrak{c}}_{R_{e}}\equiv({\mathfrak{a}}_{R_{e}})^{\prime\prime}$)
and thus expectations from the bulk field theory imply that it should have an
associated von Neumann algebra which should be identified with
${{\mathcal{M}}}_{R_{e}}=\left({{\mathcal{M}}}_{R_{e}}\right)^{\prime\prime}.$
Identifying ${{\mathcal{M}}}_{\hat{R}_{+}}\lor{{\mathcal{M}}}_{\hat{R}_{-}}$
(which is by construction a vN algebra) with the bulk causal completion of the
union of causal wedges and the equivalence of this bulk subregion with the
causal wedge of $R_{e}$ then suggests

${{\mathcal{M}}}_{\hat{R}_{+}}\lor{{\mathcal{M}}}_{\hat{R}_{-}}={{\mathcal{M}}}_{R_{e}}\
.$ (240)

Since $R_{e}$ properly contains $\hat{R}_{+}\cup\hat{R}_{-},$ this is a
phenomenon of superaddivity, i.e. the addition of algebras associated to two
spacetime subregions is equivalent to the algebra of a region that is larger
than the geometric union of the original spacetime subregions.

###  F.3 A boundary only approach to superadditivity

Here we discuss a boundary approach to finding the double commutant. In
particular, we would like to show that the operator ${{\mathcal{O}}}(x)$ lies
in ${{\mathcal{M}}}_{\hat{R}_{-}}\vee{{\mathcal{M}}}_{\hat{R}_{+}}$ whenever
$x\in R_{e}.$ We first characterize the commutant
$\left({{\mathcal{M}}}_{\hat{R}_{-}}\cup{{\mathcal{M}}}_{\hat{R}_{+}}\right)^{\prime}.$

Consider an operator ${{\mathcal{O}}}_{g}$ supported solely on oscillators
$a^{({\bar{R}_{-}})}_{k},$ i.e. in the commutant of
${{\mathcal{M}}}_{\hat{R_{-}}}.$ Such an operator has a mode expansion

${{\mathcal{O}}}_{g}=\int{d^{2}k\over(2\pi)^{2}}g(k)a^{({\bar{R}_{-}})}_{k}\
.$ (241)

Using the basis change to Minkowski oscillators, we can equivalently write

${{\mathcal{O}}}_{g}=\int{dp^{+}dp^{-}\over(2\pi)^{2}}\theta(p^{+}p^{-})K_{p}(g)b_{p}\
,$ (242)

where

$K_{p}(g)=\int{d^{2}k\over(2\pi)^{2}}R^{(\bar{R}_{-})}_{pk}g(k)\ .$ (243)

If we wish ${{\mathcal{O}}}_{g}$ to also lie in the commutant of
${{\mathcal{M}}}_{\hat{R}_{+}},$ then we must demand that its overlap with
$a^{R_{+}}_{k}$ vanishes. Using the basis change from $R_{+}$ interval
oscillators to Minkowski oscillators, this condition is

$\int{dp^{+}dp^{-}\over(2\pi)^{2}}\theta(p^{+}p^{-})K_{p}(g)P^{(R_{+})}_{k^{\prime}p}=0,~{}\forall
k^{\prime}\ .$ (244)

This is a stringent condition on $K_{p}(g)$ (and therefore $g$).

For a spatial interval of width $2a$ centered at at a point $x_{0},$ the basis
change from interval to Minkowski oscillators can be explicitly computed to be

$\displaystyle P^{(R_{x_{0}})}_{kp}$
$\displaystyle=e^{ip^{+}x_{0}^{-}+ip^{-}x_{0}^{+}}{\epsilon(\omega)\epsilon(p^{0})\theta(p^{+}p^{-})\over\sqrt{2\sinh\pi|\omega|}}C\kappa^{\nu}e^{-ip^{0}a}{}_{1}F_{1}\left({\bar{q}_{+}};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left({\bar{q}_{-}};{{\Delta}};2iap^{-}\right)$
(245) $\displaystyle P^{(\bar{R}_{x_{0}})}_{kp}$
$\displaystyle=e^{ip^{+}x_{0}^{-}+ip^{-}x_{0}^{+}}{\epsilon(\omega)\epsilon(p^{0})\theta(p^{+}p^{-})e^{\pi\epsilon(p^{0})\omega}\over\sqrt{2\sinh\pi|\omega|}}C^{*}\kappa^{\nu}e^{-ip^{0}a}{}_{1}F_{1}\left({q_{+}};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left({q_{-}};{{\Delta}};2iap^{-}\right)\
.$

The inverse transformation from Minkowski to interval oscillators is

$\displaystyle R^{(R_{x_{0}})}_{pk}$
$\displaystyle=e^{-ip^{+}x_{0}^{-}-ip^{-}x_{0}^{+}}{\theta(p^{+}p^{-})\over\sqrt{2\sinh\pi|\omega|}}C^{*}\kappa^{\nu}e^{-ip^{0}a}{}_{1}F_{1}\left({\bar{q}_{+}};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left({\bar{q}_{-}};{{\Delta}};2iap^{-}\right)$
(246) $\displaystyle R^{(\bar{R}_{x_{0}})}_{pk}$
$\displaystyle=e^{-ip^{+}x_{0}^{-}-ip^{-}x_{0}^{+}}{\theta(p^{+}p^{-})e^{\pi\epsilon(p^{0})\omega}\over\sqrt{2\sinh\pi|\omega|}}C\kappa^{\nu}e^{-ip^{0}a}{}_{1}F_{1}\left({q_{+}};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left({q_{-}};{{\Delta}};2iap^{-}\right)\
.$

The explicit formula for $K_{p}(g)$ is then

$\displaystyle K_{p}(g)$
$\displaystyle=\theta(p^{+}p^{-}){\pi\kappa^{\nu}\over
2^{\nu}}(2a)^{{\Delta}}e^{-ip^{+}(a+b)-ip^{-}(a-b)}$ (247)
$\displaystyle\cdot\int{d^{2}k\over(2\pi)^{2}}{e^{\pi\epsilon(p^{0})\omega}\over\sqrt{2\sinh\pi|\omega|}}{|{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})|\over{{\Gamma}}({{\Delta}})^{2}}b^{ik^{+}}c^{-ik^{-}}{}_{1}F_{1}\left({q_{+}};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left({q_{-}};{{\Delta}};2iap^{-}\right)g(k)\
.$

The condition for ${{\mathcal{O}}}_{g}$ to also lie in the commutant of
${{\mathcal{M}}}_{R_{+}}$ is explicitly

$\displaystyle 0$
$\displaystyle={\pi^{2}(2a)^{2{{\Delta}}}\epsilon(\omega^{\prime})b^{ik^{\prime+}}c^{-ik^{\prime-}}\over\sqrt{2\sinh\pi|\omega^{\prime}|}}{|{{\Gamma}}(q^{\prime}_{+}){{\Gamma}}(q^{\prime}_{-})|\over{{\Gamma}}({{\Delta}})^{4}}$
(248)
$\displaystyle\cdot\int{dp^{+}dp^{-}\over(2\pi)^{2}}(p^{+}p^{-})^{{{\Delta}}-1}e^{-2ip^{+}(a+b)-2ip^{-}(a-b)}\epsilon(p^{0})\theta(p^{+}p^{-}){}_{1}F_{1}\left({\bar{q}^{\prime}_{+}};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left({\bar{q}^{\prime}_{-}};{{\Delta}};2iap^{-}\right)$
$\displaystyle\cdot\int{d^{2}k\over(2\pi)^{2}}{e^{\pi\epsilon(p^{0})\omega}\over\sqrt{2\sinh\pi|\omega|}}|{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})|b^{ik^{+}}c^{-ik^{-}}{}_{1}F_{1}\left({q_{+}};{{\Delta}};2iap^{+}\right){}_{1}F_{1}\left({q_{-}};{{\Delta}};2iap^{-}\right)g(k)\
.$

The $p^{+},~{}p^{-}$ integrals can be done explicitly leaving a $k-$integral
condition on $g(k).$ The condition for ${{\mathcal{O}}}_{g}$ to lie in
$\left({{\mathcal{M}}}_{\hat{R}_{-}}\cup{{\mathcal{M}}}_{\hat{R}_{+}}\right)^{\prime}$
is then

$\displaystyle 0$
$\displaystyle=\int{d^{2}k\over(2\pi)^{2}}g(k){|{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})|\over\sqrt{2\sinh\pi|\omega|}}b^{ik^{+}}c^{-ik^{-}}\left({a+b\over
b}\right)^{i(k^{+}-k^{\prime+})}\left({a-b\over
b}\right)^{i(k^{-}-k^{\prime-})}$ (249)
$\displaystyle\cdot\left(e^{\pi(k^{+}+k^{\prime-})}F\left(q_{+},{\bar{q}^{\prime}_{+}};{{\Delta}};{a^{2}\over
b^{2}}+i\epsilon\right)F\left(q_{-},{\bar{q}^{\prime}_{-}};{{\Delta}};{a^{2}\over
b^{2}}-i\epsilon\right)\right.$
$\displaystyle\left.-e^{-\pi(k^{+}+k^{\prime-})}F\left(q_{+},{\bar{q}^{\prime}_{+}};{{\Delta}};{a^{2}\over
b^{2}}-i\epsilon\right)F\left(q_{-},{\bar{q}^{\prime}_{-}};{{\Delta}};{a^{2}\over
b^{2}}+i\epsilon\right)\right)\ .$

Computing the commutator of ${{\mathcal{O}}}_{g}$ with ${{\mathcal{O}}}(x)$
one obtains

$[{{\mathcal{O}}}(x),{{\mathcal{O}}}_{g}]=\int{dp^{+}dp^{-}\over(2\pi)^{2}}u_{p}(x)K_{-p}(g)\epsilon(p^{0})\
.$ (250)

This can be explicitly written as

$\displaystyle\int{dp^{+}dp^{-}\over(2\pi)^{2}}{\sqrt{\pi}\over{{\Gamma}}({{\Delta}})}e^{-ip^{+}x^{-}-ip^{-}x^{+}}\theta(p^{+}p^{-})\pi(p^{+}p^{-})^{{{\Delta}}-1}(2a)^{{\Delta}}e^{+ip^{+}(a+b)+ip^{-}(a-b)}$
(251)
$\displaystyle\cdot\int{d^{2}k\over(2\pi)^{2}}{e^{-\pi\epsilon(p^{0})\omega}\over\sqrt{2\sinh\pi|\omega|}}{|{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})|\over{{\Gamma}}({{\Delta}})^{2}}b^{ik^{+}}c^{-ik^{-}}{}_{1}F_{1}\left({q_{+}};{{\Delta}};-2iap^{+}\right){}_{1}F_{1}\left({q_{-}};{{\Delta}};-2iap^{-}\right)g(k)\
.$

Again the $p^{+},~{}p^{-}$ integrals can be explicitly performed to yield

$\displaystyle{\pi\sqrt{\pi}(2a)^{{\Delta}}\over{{\Gamma}}({{\Delta}})(2\pi)^{2}}\int{d^{2}k\over(2\pi)^{2}}\left(e^{\pi(k^{+}+k^{-})}e^{{i\pi\over
2}\epsilon(x^{-}-b-a){\bar{q}_{+}}}e^{{i\pi\over
2}\epsilon(x^{-}-b+a){q_{+}}}e^{{i\pi\over
2}\epsilon(x^{+}+b-a){\bar{q}_{-}}}e^{{i\pi\over
2}\epsilon(x^{+}+b+a){q_{+}}}\right.$ (252)
$\displaystyle\left.-e^{-\pi(k^{+}+k^{-})}e^{-{i\pi\over
2}\epsilon(x^{-}-b-a){\bar{q}_{+}}}e^{-{i\pi\over
2}\epsilon(x^{-}-b+a){q_{+}}}e^{-{i\pi\over
2}\epsilon(x^{+}+b-a){\bar{q}_{-}}}e^{-{i\pi\over
2}\epsilon(x^{+}+b+a){q_{+}}}\right)$
$\displaystyle{|{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})|\over\sqrt{2\sinh\pi|\omega|}}b^{ik^{+}}c^{-ik^{-}}|x^{-}-b-a|^{-\bar{q}_{+}}|x^{-}-b+a|^{-q_{+}}|x^{+}+b-a|^{-\bar{q}_{-}}|x^{+}+b+a|^{-q_{-}}g(k)\
.$

The case we are most interested in is when $(x^{+},x^{-})$ is in the “$F_{R}$”
region with respect to $R_{-}$ as this is a region that contains points in
$R_{e}$ that are not in $\hat{R}_{-}\cup\hat{R}_{+}.$ In this case,
$-a<x^{-}-b<a$ and $x^{+}+b>a>-a.$ The commutator expression then simplifies
to

$\displaystyle{\pi\sqrt{\pi}(2a)^{{\Delta}}\over{{\Gamma}}({{\Delta}})(2\pi)^{2}}\int{d^{2}k\over(2\pi)^{2}}2i\sin\pi{\bar{q}_{-}}$
(253)
$\displaystyle{|{{\Gamma}}(q_{+}){{\Gamma}}(q_{-})|\over\sqrt{2\sinh\pi|\omega|}}b^{ik^{+}}c^{-ik^{-}}|x^{-}-b-a|^{-\bar{q}_{+}}|x^{-}-b+a|^{-q_{+}}|x^{+}+b-a|^{-\bar{q}_{-}}|x^{+}+b+a|^{-q_{-}}g(k)\
.$

We expect that (253) must vanish whenever we have
$x^{+}x^{-}<a^{2}-b^{2},x^{+}>a-b,~{}x^{-}>a-b$ and $g$ satisfies (249) but we
have not been able to demonstrate this explicitly.

####  F.3.1 Double commutant is not the causal completion

One can easily show, using the global mode expansion, that the bulk operator
$\phi(t=0,x=0,z=\sqrt{a^{2}-b^{2}+\epsilon})$ fails to commute with a boundary
operator ${{\mathcal{O}}}(t^{\prime},x^{\prime})$ only when
$(t^{\prime}-x^{\prime})(t^{\prime}+x^{\prime})>a^{2}-b^{2}+\epsilon,$ for
$\epsilon>0.$ This is exactly what we expect from bulk causality. We have
previously shown that this bulk operator, when written in terms of boundary
GFF modes, is supported only on $a^{(\bar{R}_{-})}_{k}$ modes in the
$\\{a^{(R_{-})}_{k},~{}a^{(\bar{R}_{-})}_{k}\\}$ basis while it only has
support on $a^{(\bar{R}_{+})}_{k}$ modes when written in the
$\\{a^{(R_{+})}_{k},~{}a^{(\bar{R}_{+})}_{k}\\}$ basis. Thus, such an operator
clearly lies in
$\left({{\mathcal{M}}}_{\hat{R}_{-}}\cup{{\mathcal{M}}}_{\hat{R}_{+}}\right)^{\prime}.$
It does not commute with boundary operators with
$(t^{\prime}-x^{\prime})(t^{\prime}+x^{\prime})>a^{2}-b^{2}+\epsilon,$ thus,
such boundary operators cannot be in
$\left({{\mathcal{M}}}_{\hat{R}_{-}}\cup{{\mathcal{M}}}_{\hat{R}_{+}}\right)^{\prime\prime}.$

In particular, for $x^{\prime}=0$, the operators with
$t^{\prime}>\sqrt{a^{2}-b^{2}}$ are not in the double commutant. This is
interesting since this includes operators with
$\sqrt{a^{2}-b^{2}}<t^{\prime}<a+b$ which are in the causal completion of the
boundary subregion.

##  Appendix G Proof of geodesic convexity

We wish to show that the complement, $D_{w}=\overline{C_{w}},$ of the bulk
dual of a time-reflection symmetric time-band on a bulk Cauchy slice is
geodesically convex. This means that for any two points $p_{1},~{}p_{2}\in
D_{w},$ there is a unique minimizing geodesic, ${{\gamma}}_{12},$ on the $t=0$
bulk slice that connects them and this geodesic lies entirely in $D_{w}.$ Let
$(x_{1},z_{1}),~{}(x_{2},z_{2})$ be the coordinates of $p_{1},~{}p_{2}$,
respectively. Without loss of generality we take $x_{2}>x_{1}.$ Since the
induced geometry on the $t=0$ bulk slice is simply that of the Poincaré half-
plane, there is a unique geodesic connecting these two points given by

${{\gamma}}_{12}=\\{(x-x_{c})^{2}+z^{2}=w_{c}^{2}~{}|~{}x_{1}<x<x_{2},~{}z>0\\},$
(254)

with

$x_{c}={x_{1}^{2}+z_{1}^{2}-(x_{2}^{2}+z_{2}^{2})\over 2(x_{1}-x_{2})},\qquad
w_{c}={\sqrt{\left((x_{1}-x_{2})^{2}+(z_{1}-z_{2})^{2}\right)\left((x_{1}-x_{2})^{2}+(z_{1}+z_{2})^{2}\right)}\over
2|x_{1}-x_{2}|}\ .$ (255)

An explicit computation shows that, for $\bar{x}\in(x_{1},x_{2}),$ the point
$(\bar{x},\bar{z})$ only lies on ${{\gamma}}_{12}$ if

$\bar{z}^{2}=z_{1}^{2}\left({x_{2}-\bar{x}\over
x_{2}-x_{1}}\right)+z_{2}^{2}\left({\bar{x}-x_{1}\over
x_{2}-x_{1}}\right)+(\bar{x}-x_{1})(x_{2}-\bar{x})\ .$ (256)

Since $p_{1},~{}p_{2}$ are in $D_{w}$ we have $z_{i}>z_{w}(x_{i})$ or more
explicitly

$z_{i}^{2}>\max_{x_{0}\in I}\left(w(x_{0})^{2}-(x_{i}-x_{0})^{2}\right)\ ,$
(257)

for $i=1,2.$ Since $x_{1}<\bar{x}<x_{2}$ we can use (257) in (256) to conclude
that for $(\bar{x},\bar{z})\in{{\gamma}}_{12},$ we have

$\bar{z}^{2}>\left(\max_{x_{0}\in
I}\left(w(x_{0})^{2}-(x_{1}-x_{0})^{2}\right)\right)\left({x_{2}-\bar{x}\over
x_{2}-x_{1}}\right)+\left(\max_{x_{0}^{\prime}\in
I}\left(w(x_{0}^{\prime})^{2}-(x_{2}-x_{0}^{\prime})^{2}\right)\right)\left({\bar{x}-x_{1}\over
x_{2}-x_{1}}\right)+(\bar{x}-x_{1})(x_{2}-\bar{x})\ .$ (258)

The geodesic ${{\gamma}}_{12}$ remains in $D_{w}$ so long as for each
$\bar{x}\in(x_{1},x_{2})$ we have $\bar{z}>z_{w}(\bar{x}).$ At the coordinate
$x=\bar{x},$ the curve (88) has303030Here we assume $\bar{x}$ is such that
$\max_{x_{0}\in I}\left(w(x_{0})^{2}-(\bar{x}-x_{0})^{2}\right)>0,$ otherwise
we have $z_{w}(\bar{x})=0$ by definition and clearly since ${{\gamma}}_{12}$
is a geodesic in the Poincaré half-plane we always have $\bar{z}>0,$ so at
such points ${{\gamma}}_{12}$ is obviously in $D_{w}.$

$z_{w}(\bar{x})^{2}=\max_{x_{0}\in
I}\left(w(x_{0})^{2}-(\bar{x}-x_{0})^{2}\right)=w(x_{0}^{*})^{2}-(\bar{x}-x_{0}^{*})^{2}\
,$ (259)

where $x_{0}^{*}$ is the value of $x_{0}\in I$ where the maximum is obtained.
In (258), each of the maxima is taken over $I$ and the result multiplies a
positive number. Thus, the right-hand-side of (258) must be larger than the
same expression evaluated at $x_{0}\to x_{0}^{*}$ and $x_{0}^{\prime}\to
x_{0}^{*}.$ In particular we have

$\bar{z}^{2}>\left(w(x_{0}^{*})^{2}-(x_{1}-x_{0}^{*})^{2}\right)\left({x_{2}-\bar{x}\over
x_{2}-x_{1}}\right)+\left(w(x_{0}^{*})^{2}-(x_{2}-x_{0}^{*})^{2}\right)\left({\bar{x}-x_{1}\over
x_{2}-x_{1}}\right)+(\bar{x}-x_{1})(x_{2}-\bar{x})\ .$ (260)

The right-hand-side of (260) can be explicitly computed to be
$w(x_{0}^{*})-(\bar{x}-x_{0}^{*})^{2}=z_{w}(\bar{x}).$ So, (260) implies that
for any $(\bar{x},\bar{z})\in{{\gamma}}_{12}$ we have
$\bar{z}^{2}>z_{w}(\bar{x})^{2}$ and thus $(\bar{x},\bar{z})\in D_{w},$
completing the proof that

${{\gamma}}_{12}\subset D_{w}\ ,$ (261)

and thus showing that $D_{w}$ is geodesically convex.

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