Relation between projection method and valuation map for generating the Rauzy Fractal In my research I am studying the tribonacci substitution $\sigma : \mathcal{A}^* \to \mathcal{A}$
$$
\sigma(0) = 01, \qquad \sigma(1) = 02, \qquad \sigma(2) = 0,
$$
where $\mathcal{A} = \{0,1,2\}$ is the alphabet and $\mathcal{A}^*$ denotes the set of all finite words with letters in $\mathcal{A}$. This substitution rule is related to the polynomial equation $x^3=x^2+x+1$, with real root $\beta = 1.8392...$, called the tribonacci constant. The incidence matrix $M$, the linear map associated to $\sigma$ by abelianisation, is given by
$$
M = \begin{bmatrix}
1 & 1 & 1 \\ 
1 & 0 & 0 \\ 
0 & 1 & 0 
\end{bmatrix}.
$$
The matrix $M$ is Pisot, what means that there is only one eigenvalue with $|\lambda|>1$ and the corresponding eigenspace has dimension 1. In this case this eigenvalue is the tribonacci constant $\beta>1$ with eigenvector $\vec{v} = (\beta^2,\beta,1)^T$, which spans the unstable (expanding) eigenspace $E^u$. The other two eigenvalues of $M$ obey $|\lambda|<1$ and span the stable (contracting) eigenspace $E^s$. The fixed point of $\sigma$ is called the tribonacci word and is given by
$$w = 010201001020101020100102010...$$
Now let $\vec{e_i}$, with $i=0,1,2$ be the canonical basis of $\mathbb{R}^3$, one defines a stair of points in $\mathbb{R}^3$ by
$$
\vec{x_m} = \sum^m_{j=1} \vec{e_{w_j}},
$$
where $w_j$ is the $j$th symbol in $w$ and $m \geq 0$. From $w$, a so-called Rauzy fractal can be constructed in two ways: 1) project $\vec{x}_m$ along $\vec{v}$ onto $E^s$ for all $m \geq 0$, or 2) use the valuation map.
Method 1) is the common way to construct Rauzy fractals according to literature [2,3]. Method 2) is dealt with in this question and in this paper [1]. I have been able to generate a Rauzy fractal method 2) that is perfectly self-similar in the way that the tiles are perfect copies of each other, up to a scaling factor: My figure of the Rauzy fractal with valuation map
When I use method 1), I do get a Rauzy fractal, but it is skewed. By this I mean that it is still a Rauzy fractal that tiles the plane, but its three domains are not simply scaled versions of each other, but are rather related by a more general affine transformation: My figure of a Rauzy fractal by projecting along unstable eigenspace onto stable eigenspace
The problem is, the "correct" Rauzy fractal, with up to scaling identical domains as obtained from method 2), is depicted in literature and claimed to be obtained from method 1).
My question is now: does method 1) indeed produce a skewed version of the "correct" Rauzy fractal, or am I doing something wrong?
[1] Sirvent, V. F., & Wang, Y. (2002). Self-affine tiling via substitution dynamical systems and Rauzy fractals. Pacific journal of mathematics, 206(2), 465-485.
[2] Fogg, N. P., Berthé, V., Ferenczi, S., Mauduit, C., & Siegel, A. (Eds.). (2002). Substitutions in dynamics, arithmetics and combinatorics. Berlin, Heidelberg: Springer Berlin Heidelberg.
[3] Arnoux, P., & Ito, S. (2001). Pisot substitutions and Rauzy fractals. Bulletin of the Belgian Mathematical Society-Simon Stevin, 8(2), 181-207.
 A: Your pictures are exactly correct. The difference between the two images arises because one is an oblique projection and the other is an orthogonal projection. Thus, the pictures are quite similar in appearance but somewhat distorted.
You've described Rauzy's original construction perfectly: We project the staircase along the real eigenvector of $M$ and onto the two dimensional stable subspace spanned by the real and imaginary parts of a complex eigenvector. Let us emphasize the fact that the stable space need not be orthogonal to the real eigenvector and, in fact, is not in this case. The projection is oblique.
Next, the valuation map can also be thought of as a projection. This time though, we project onto the one-dimensional space spanned by one of the complex eigenvectors of $M^T$. Of course, $M$ and $M^T$ have the same eigenvalues but different eigenvectors. Furthermore, if $\vec{v}$ is an eigenvector of $M$ and $\vec{v}_T$ is an eigenvector of $M^T$ corresponding to different eigenvalues, then they are orthogonal. Thus, this projection is an orthogonal projection.
