Three-dimensional triangulations with fixed number of vertices

Question

My question is the following:

Are there triangulations of $S^3$ which (a) are non-degenerate, (b) have four vertices, and (c) have no edges of degree two?

A side question:

If one represents this triangulation as a graph, and draws it's dual graph, is it true that it is a planar graph?

(At the moment I'm unable to construct a planar graph starting from an edge degree 6 or 8 with the above requirements and taking into account the comments , that all degrees are even, as I end up with growing number of tetrahedra.)

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Here are the definitions.

We use labeled triangulations of $S^3$. We allow adjacent tetrahedra to share multiple faces. The degree of a vertex (or an edge) is the number of tetrahedra connected to it, counted with multiplicity.

We only allow non-degenerate triangulations. That is, we require that every edge is embedded (meets two vertices). Note that this implies that all triangles and tetrahedra are embedded.

However, we do not require our triangulations to be simplicial (where a cell is determined by its vertices). In particular, we allow many edges to share the same two endpoints as vertices. In this case we distinguish the edges by their labels. The same applies to triangles and tetrahedra.

There are various moves that can change the configuration to another. To be concrete, imagine two adjacent tetrahedra defined by vertices $\{a, b, c, d\}$ sharing four faces. Note that all edges have degree two. Let's call this our "initial configuration". We can add additional tetrahedra to the initial configuration (maintaining the topology) in various ways.

0-2 face move: Choose a triangle (between two tetrahedra $A$ and $B$) and glue in a pair of tetrahedra ($C$ and $D$) sharing three faces, (and a new vertex: $e$), and their remaining face is glued between let's say the triangle $(a b c)$ and its copy $(a,b,c)$, defining the set now as: $\{a b c d\}$, $\{a b c d\}$, $\{a b c e\}$, $\{a b c e\}$. This is like inflating a triangle into a sphere and putting a vertex in the middle. The same vertices can define unique triangles and edges. If we would give labels to triangle $(a b c)$ or the other $(a b c)$ they would differ due to their adjacency relation. This move increases the number of vertices.

1-4 move: Choose a tetrahedron. Add a vertex in the middle of the tetrahedron, connecting it to all the four vertices, splitting the interior into 4 tetrahedra.

2-3 move Choose a face meeting distinct tetrahedra. Add an edge to connect the distant vertices of two adjacent tetrahedra, exchanging a triangular face with an edge. This turns two tetrahedra into three.

0-2 edge move: Choose two adjacent triangles $(a b c)$, $(a b d)$, and open them up, inserting two tetrahedra $\{a b c d\}$ and $\{a b c d\}$, by creating a copy of $(a b c)$ and $(a b d)$ in the process. The move adds two tetrahedra, a new edge with degree 2, and splits the original [a b] edge into be two with smaller or equal degree than their original. (0-2 edge move)

One could devise other type of moves, adding extra edges, adding/ removing extra vertices, to change the triangulation into another one. (Note that all moves can be obtained from the bistellar flips: the 2-3 and 1-4 moves.)

Answer 1

Here are some observations that might be useful: not an answer, but too long for a comment.

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The Euler characteristic of $S^3$ is zero (this holds for any compact three-manifold without boundary). So we have $0 = V - E + F - T$.

By hypothesis we have $V = 4$. Thus $4 = E - F + T$. In any triangulation (of a three-manifold) every tetrahedron gives us four faces, and every face meets two tetrahedra. We deduce that $F = 2T$. Thus we have $4 = E - T$.

Let $E_k$ be the number of edges with degree $k$. So $E = \sum E_k$. From the definition of degree we have $6T = \sum (k \cdot E_k)$. All edges have positive degree. (So $E_0 = 0$.) By hypothesis we have that $E_2 = 0$. Non-degeneracy gives us the following.

Claim: All edge degrees are even.

Proof: Suppose that $e$ is an edge of degree $k$. Suppose that the endpoints of $e$ are the vertices $v_0$ and $v_1$. Suppose that $t_0, t_1, \ldots, t_{k-1}$ are the tetrahedra (in order) meeting $e$. Thus the vertices of the edges of $t_i$ not meeting $e$ are $v_2$ and $v_3$. These alternate as we walk around $e$. Thus $k$ is even. $\square$

Note that $$ 4 = E - T = \sum \left(1 - \frac{k}{6}\right)E_k $$ Thus there are at least some edges with degree four. (This can also be seen by noting that triangulations of $S^2$ must have some vertices of degree less than six.) Finally, since $6T = \sum (k \cdot E_k) \geq 4E$ we have that $3T \geq 2E$. Thus $4 = E - T \leq E - \frac{2}{3}E = \frac{1}{3}E$ and so $12 \leq E$. Similarly, $8 \leq T$.

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Inspired by the equalities $V = 4$, $T = 8$, and $E = 12$, we recall that the boundary of the cross-polytope is a triangulation of $S^3$ having $8$ vertices, $24$ edges, $32$ faces, and $16$ tetrahedra. There is a natural (orientation-preserving) involution on the boundary of cross-polytope induced by the antipodal map on $\mathbb{R}^4$. The quotient is a triangulation of the real projective space with $4$ vertices, $12$ edges, $16$ faces, and $8$ tetrahedra.

That is, there is a triangulation of some three-manifold obeying all of the desired combinatorial hypotheses. Thus to prove that there is no such triangulation of $S^3$ we would actually need to use some further property of $S^3$ (such as the fact that it has trival $H_1$ or trivial $\pi_1$.)

Answer 2

This is going to be a long one. I believe I now have a triangulation of the $3$-sphere with $28$ tetrahedra, $56$ triangles, $32$ edges and $4$ vertices, obeying all of the rules.

Our triangulation will be built from $7$ torii, nested one inside the other, as in this picture that I stole from Martin Fernandez Cufre:

Number the torii as $T_1$, $T_2$, ..., $T_7$. I'll refer to $T_1$ as the "innermost torus" and $T_7$ as the ``outermost". So, for each $T_i$, I can talk about the "interior of $T_i$" and the "exterior of $T_i$", each of which is a $3$-fold with boundary $T_i$. The torii $T_i$ and $T_{i+1}$ are identified along vertices and along certain edges, but their triangles are all distinct.

Each $T_i$ has $4$ vertices, $12$ edges and $8$ triangles, and they are all isomorphic as abstract simplicial complexes. However, I want to explain not only what the $T_i$ look like as abstract triangulations of the torii, but how they are nested inside each other. In the diagrams below, I have unwrapped each of the tori unto a square; glue the sides of the square together to get the torus back. The horizontal sides of the square will become contractible in the interior of $T_i$, and the vertical sides will become contractible in the exterior of $T_i$. The same letter always indicates the same vertex.

I'll now describe the regions between the torii.

The interior of $T_1$:

In the figure below I have redrawn $T_1$ with arrowheads on some of the edges and numbers inside the triangles. The two edges with single arrow heads are glued together (in the direction shown by the arrowhead), as are two edges with double arrow heads. The triangles numbered $1$ form the faces of a tetrahedron, as do the triangles numbered $2$.

This three dimensional figure thus has boundary $T_1$, and the horizontal paths around $T_1$ become contractible in this figure.

The region between $T_1$ and $T_2$

In the figure below, I have redrawn $T_1$ and $T_2$ with numbers in the triangles. For each number, the four triangles with that number form the faces of a tetrahedron. Glue these together, identifying vertices with the same name and the corresponding edges, to make a three dimensional figure with two boundaries, 0ne which is $T_1$ and the other of which is $T_2$. We have drawn an edge dashed if it is in only one of the two torii, and solid if it is in both.

This one is relatively easy to visualize, because we can overlay the diagrams on each other. Note that two edges are now completely surrounded: The edge in $T_1$ with the single arrowheads is surrounded by tetrahedra $(1,2,4,6)$, and the edge with double arrow heads is surrounded by tetrahedra $(1,2,3,5)$.

The region between $T_2$ and $T_3$

This one is hard to visualize. Here are $T_2$ and $T_3$ labeled with the same rules as before: the numbers on the triangles are the forces of the tetrahedra; solid edges are in both $T_2$ and $T_3$; dashed edges are in just one.

The challenge is to see that the solid edges form isomorphic graphs on the two tori. In the picture below, we have redrawn the figures above with new fundamental domains:

Hopefully, it is now clear that we can glue in tetrahedra between these two surfaces.

Between $T_4$ and $T_3$ we do the same thing as between $T_1$ and $T_2$.

Between $T_{8-i}$ and $T_{8-(i+1)}$, we do the same thing as between $T_i$ and $T_{i+1}$, but reflected over the northeast/southwest diagonal line. Similarly, to fill the exterior of $T_8$, we use the pattern that we used inside $T_1$, reflected over that same diagonal.

I'll start the process of drawing the dual graph and leave the rest to you. Let $v_i$ be the vertex dual to the $i$-th tetrahedron. Comparing the two ways that we have numbered the triangles of $T_1$, we see that each of $\{ v_1, v_2 \}$ borders each of $\{ v_3, v_4, v_5, v_6 \}$. Comparing the two ways that we have numbered the triangles of $T_2$, we see that $(v_3, v_7, v_6, v_{10}, v_5, v_9, v_4, v_8)$ forms an octagon (in that order). Hopefully, that's enough to get you going. I don't think it is planar.

I leave it to you to track the edges and check that each one is in at least $4$ tetrahedra. :)