Given a graph $G = (V, E)$, let the embedding dimension of a graph be the least number $n$ such that there exists a partitioning of $\mathbb{R}^n$ with the contact graph is $G$. Is there any way to bound the embedding dimension?
Suppose a graph is the cartesian product $G \times H$. Is there any way to bound the embedding dimension of the product given the embedding dimensions of $G$ and $H$.
Any connected graph has embedding dimension at most $3$. (Only planar graphs can have embedding dimension $2$, and disconnected graphs aren't the contact graphs of any partition of $\mathbb R^n$ for any $n$, so this describes all embedding dimensions.)
We can begin by taking a spanning tree of $G$, and drawing it as the contact graph of a partition of $\mathbb R^2$, such as in the picture below:
We can think of this as sitting in $\mathbb R^3$, just by repeating this picture indefinitely in the third dimension.
Next, we can add any edge to this spanning tree by adding a narrow tube connecting the regions, and pinching together the regions in between:
In this 2D picture, this makes the in-between regions look disconnected, but since this only happens in a small range of $z$-values, they're still connected elsewhere.
We can do this any number of times (at different "heights" in the third dimension) to add in all the edges we need to build $G$.