A variation of domino tiling problem with fusions I know several specific variations of the domino tiling problem has been determined to be decidable or undecidable, such as the seed domino problem. I have a variation which  I have not been able to find, but as I am not familiar with this subject, I thought someone here might know better.
Given a Wang tile $T=(T_N,T_E,T_S,T_W)$ with colors $\mathcal{C}$, each entry corresponds to the color in the north, east, south or west sides. Given two Wang tiles $T^{(1)},T^{(2)}$ that  agree on opposite sides, their fusion would be gluing them on the side on which they agree. For example, if $T^{(1)}_N=T^{(2)}_S$, their fusion along South-North side is a new Wang tile
$$T^{(1)} \oplus_{SN} T^{(2)}=\Big( T^{(2)}_N, (T^{(2)}_E,T^{(1)}_E), T^{(1)}_S, (T^{(2)}_W,T^{(1)}_W) \Big),$$
over a different set of colors obviously. We can similarly define $T^{(1)}\oplus_{WE} T^{(2)}$ when they satisfy $T^{(1)}_E=T^{(2)}_W$. If we take a Wang tile set $\mathcal{T}$ with colors $\mathcal{C}$, we can obtain new Wang tiles sets
$$ \mathcal{T_1} = \big\{ T \oplus_{SN} T' : T,T' \in \mathcal{T}, T_N=T'_S  \big\} $$
and
$$ \mathcal{T_2} = \big\{ T \oplus_{WE} T' : T,T' \in \mathcal{T}_1, T_E=T'_W  \big\}. $$
$\mathcal{T}_2$ will again be a collection of Wang tiles over the new color set $\mathcal{C}\times \mathcal{C}$. I can consider two related questions:

*

*Does $\mathcal{T}_2$ tile the plane periodically?

*Does $\mathcal{T}_1$ tile the plane periodically?

Is it known whether question (1) being un\decidable imply question (2) is un\decidable? I am assuming that $\mathcal{T}_1$ tiles the plane.
This seems like a natural question to me, but I really have a bad intuition on the subject, so it may be obviously resolvable. I would be glad to find appropriate references, or even solutions.
 A: Dynamically, the set of tilings $X = X_{\mathcal{T}} \subset \mathcal{T}^{\mathbb{Z}^2}$ by a set of Wang tiles $\mathcal{T}$ is a compact dynamical system under the action of $\mathbb{Z}^2$, with the product topology on $\mathcal{T}^{\mathbb{Z}^2}$.
If $H \leq \mathbb{Z}^2$ is a finite-index subgroup, then $H$ is isomorphic to $\mathbb{Z}^2$, and it is well-known that $(X, H)$ is itself topologically conjugate (in bijection through a shift-commuting homeomorphic) with a set of tilings. Your $\mathcal{T}_1$ realizes this construction for the subgroup $H = \mathbb{Z} \times 2\mathbb{Z}$, i.e. we have
$(X_{\mathcal{T}_1}, \mathbb{Z}^2) \cong (X_{\mathcal{T}}, H)$
where $H$ is identified in the obvious way with $\mathbb{Z}^2$. In other words, as far as dynamical properties are concerned, we can just think of $X_{\mathcal{T}_1}$ as the same space $X$ with a different action.
Now, a totally periodic point is a point which is fixed by a finite-index subgroup $K$ of the acting group. If $x \in X$ is fixed by $K$, then it is also fixed by the finite index subgroup $K \cap H$, thus total periodicity is equivalent whether or not we consider the $H$-action or the $\mathbb{Z}^2$-action.
(This is all completely general, and works for any acting group.)
A: Each tiling with tiles from $\mathcal{T}$ gives rise to exactly two tilings with tiles from $\mathcal{T}_1$ and exactly two tilings with tiles from $\mathcal{T}_2$, given by taking the two choices of fusing neighbouring pairs (either $2n\longleftrightarrow 2n+1$ or $2n-1 \longleftrightarrow 2n$). This process also preserves periodicity.
Up to translation, this correspondence is also reversible and again preserves periodicity. So $\mathcal{T}$ admits a periodic tiling if and only if $\mathcal{T}_1$ admits a periodic tiling if and only if $\mathcal{T}_2$ admits a periodic tiling.
The same argument works for non-periodic tilings of course.
