Let us consider the dual space of $n$-simplex tiling. Here the dual space is defined by replacing the $d$-dimensional object $p$ by the co dimension $n-d$ objects at the symmetric center of $p$. It is sort of finding the dual space tiled by the Poincare dual objects.
For example, when we have 2-simplex (triangle) tiling in a 2-dimensional space, the dual space of $2$-simplex tiling is a honeycomb lattice. The reason is that we replace the 1-edge $p$ of 2-simplex (triangle) as the co dimension $2-1=1$ objects at the symmetric center of $p$. The center of the 2-simplex (triangle) becomes the vertices of the dual lattice. Namely, the center of the 2-simplex (triangle) becomes the vertices of the honeycomb lattice. The Poincare dual object is the honeycomb.
question 1. what is the dual space of $3$-simplex tiling in a 3-dimensional Euclidean space? What is the Poincare dual object?
question 2. what is the dual space of $4$-simplex tiling in a 4-dimensional Euclidean space? What is the Poincare dual object?
question 3. what is the dual space of $n$-simplex tiling in a $n$-dimensional Euclidean space? What is the Poincare dual object?
p.s. For example in 3-dimensional space, we may consider this tiling:
For higher dimensions, we may fill the object $n-1$-simplex $\times I^1$ with $n$-simplex first. (e.g. In 3 dimensions, we fill the object $2$-simplex $\times I^1$ (a triangle $\times I^1$ as a elongated triangular pyramid) with $3$-simplex first.) Then we use $n-1$-simplex $\times I^1$ to fill the whole $R^n$ space.
The $n$-simplices need not to exactly the same geometric shape and size, only homeomorphic the same.
Any partial answer is welcome.