I’m interested in whether the complete graph made up of 6 vertices, $K_6$, can be represented as a set of adjacent 3-polytopes. For two polytopes to be adjacent, they should share some nonzero number of facets. Also, the polytopes should be considered solid so that no two polytopes can intersect apart from their shared facets (like building blocks)
So, for example, $K_4$ can be represented as a single tetrahedron, and $K_5$ can be represented as 3 adjacent, non-regular tetrahedra. Is there also some way to represent $K_6$ via 3-polytopes?
I’m pretty sure the answer is no, as I’ve tried and can’t seem to do it, but I don’t know how to prove whether or not this is or isn’t true. Are there any proofs of this, or conversely, does anyone have a valid counter example? If not, what are some tools you think I could use to solve this/where should I look for relevant information and language to better formally pose and answer this question?
UPDATE
The answer I received from Richard works for the problem as I originally stated it, but it doesn't quite satisfy all of the conditions I was going for.
Here is a more formal statement of the problem, starting with some background definitions:
Embedding of $G$ in $\mathbb{R}^n$: Given a graph $G$ with vertex set $verts(G)$ and edge set $edges(G)$, an embedding of $G$ into $\mathbb{R}^n$ is a bijective mapping of each $v \in verts(G)$ to a point $p \in \mathbb{R}^n$, and the bijective mapping of each $e \in edges(G)$ to a line segment between the points corresponding to the vertices of $e$. All the points in a particular embedding of $G$ are given by the set $points(G)$.
Outer Polytope: Given an embedding of $G$ in $\mathbb{R}^n$, the outer polytope $O$ of that embedding is the convex hull of $points(G)$.
Intersecting Polytopes: Given polytopes $A$ and $B$ (which may or may not be convex), each of which are embedded into $\mathbb{R}^n$, $A$ intersects $B$ if the non empty intersection of the space defined by $A$ and $B$ is not a set of facets of both $A$ and $B$. If the non empty intersection of the space defined by $A$ and $B$ is a set of facets of both $A$ and $B$, $A$ and $B$ are adjacent, and do not intersect.
Geometric Embedding of $G$ into $\mathbb{R}^n$: A geometric embedding of a graph $G$ into $\mathbb{R}^n$ is an embedding of $G$ into $\mathbb{R}^n$, where each $p \in points(G)$ is a vertex of at least one polytope in the set $polytopes(G)$ such that
- No two $P \in polytopes(G)$ intersect
- The union of all space defined by each $P \in polytopes(G)$ is equal to the space defined by the outer polytope $O$ (this is a key bit I was missing in my original statement of the problem)
Question: Does $K_6$ have a geometric embedding in $R^3$?
I've created a simulation of a valid geometric embedding and an invalid geometric embedding of $K_5$ to help visualize the problem: https://dideric.is/#/geometric-embeddings.
There is an embedding of $K_6$ in $\mathbb{R}^3$ decomposable into simplices.
Taking the example of $K_5$ in $\mathbb{R}^3$ from your website, the three dimensional convex hull has six faces.
If you orient the graph so that its planar projection has all its vertices on the outside (a convex pentagon), the three dimensional embedding has three upward-facing faces. Place a new vertex in 3-space "above" the planar projection of $K_5$. The new volume in the $3$-D convex hull can be decomposed into three new, adjacent, non-overlapping simplices: each of the three upward-facing faces connected to the new point.
This argument can be extended inductively to any $K_n$ as long as there always exists a planar projection of a decomposable embedding of $K_{n-1}$ that looks like a convex $(n-1)$-gon.