TITLE: A conformal mapping between a Rindler Wedge and a causal diamond. What is the right map and how do I see it does what is expected?
QUESTION [4 upvotes]: I am going through the calculations in arXiv:1312.7856. These involve a conformal map between the Minkowski Rindler Wedge ($\mathcal{R}$), given by
$X^1 \geq 0,X^+\geq 0,X^-\geq 0 \quad$ (with $X^{\pm}=X^1 \pm X^0$),
and the causal diamond ($\mathcal{D}$) given by:
$|x|+|t|\leq R$. 
See the figure below:

The mapping is (supposedly) given by (equation 2.11 in arXiv:1102.0440):
$x^\mu = \frac{X^\mu - X  \cdot X C^\mu} {1-2(X \cdot C) + (X \cdot X) (C \cdot C)} + 2 R^2 C^\mu \qquad$ with $C^\mu = (0,1/2R)$.
This is a special conformal transformation followed by a translation. 
We can check where certain points are mapped to. For instance, we instantly see that $X^\mu = (X^0,X^1)=(0,0)$ maps to $(0,R)$.
However $(0,X^1)$ seems to map to $\left(0, \frac{X^1 - \tfrac{1}{2R}(X^1)^2}{1-\tfrac{X^1}{R} +\tfrac{X^2}{4R^2}} + R \right)$.
We know that it should end up withing $\mathcal{D}$ because our original point was in $\mathcal{R}$. So the first term needs to be negative (or zero) for all $R$. Taking $R=1$, we find that the first term is positive: $(0,1)$ goes to $(0,2+1),2 > R = 1)$, so this lies outside $\mathcal{D}$.
But it should be inside. Can anyone please tell me what is going wrong and how I can see that this map (or another conformal map) does have the desired effect of mapping $R$ to $D$?
P.S.: I have here presented the case for $D=2$ to simplify things, although the mapping should apply for an arbitrary number of dimensions $(1, 1, d-2)$ with $d-2$ 'trivial' dimensions.

REPLY [3 votes]: There's a mistake in the paper. The correct transformation is
$$x^\mu = 2R \frac{X^\mu - b^\mu X^2}{1-2b \cdot X + b^2 X^2} + Rb^\mu,$$
with $b^\mu = (0,-1,\vec{0})$.
I picked the minus sign in the definition of b to make the fact that it's made of a special conformal transformation manifest.
I found it in Rudolf Haag's book Local Quantum Physics in section V.4.2.
EDIT: Also notice that the equation makes no dimensional sense; the coordinates on the right should actually be $\frac{X^\mu}{2R}$.
Absorbing this sign into the parametrising vector $b^\mu$ gives exactly Casini-Huerta-Myers' transformation except for a sign.
[Chronology note: this edit came after OP's comment.]
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