# Are two regularity properties on hypergraphs equivalent?

Let $$H=\left( E_0 ,E_1 ,E_2 , \ldots , E_d \right)$$ be a $$d$$-dimensional full-hyper graph/complex. That is to say, if for some $$i\in \left[d \right]$$ the hyper-edge $$e_j \in E_i$$ than for any $$i-1$$-dimensional $$e_k \subset e_j$$: $$e_k \in E_{i-1}$$. The vertices are $$E_0$$, the hyper-edges between $$2$$ vertices are $$E_1$$, between $$3$$ are $$E_2$$ and so on.

I bring forward 2 definitions of regularity, and I'd like to show their equivalence:

• There exists $$r_1,\ldots , r_d$$ such that for each $$1\leq i \leq d$$ each vertex $$v\in E_0$$ has exactly $$r_i$$ hyper-edges from $$E_i$$ to which it belongs.
• There exists $$r_1,\ldots , r_d$$ such that for each $$1\leq i \leq d$$ each $$e_j \in E_{i-1}$$ has exactly $$r_i$$ hyper-edges from $$E_i$$ that contain it.

I claim that the definitions are equivalent, obviously with different values of $$r_i$$'s.

My attempted proof for the case of $$d=3$$:

$$\Leftarrow$$

Each vertex $$v$$ has $$r_1$$ $$2$$-edges, each edge has $$r_2$$ $$3$$-edges. Each $$3$$-edge, by definition, uses exactly a distinct pair of $$v$$'s $$2$$-edges. Therefore, each $$3$$-edge is counted twice - one per each $$2$$-edge of $$v$$. This makes the total number of $$3$$-edges exactly $$\frac{r_1\cdot r_2}{2}$$.

Having a bit trouble for the other direction. Could not disprove it, and I believe it holds, but not every pair of edges that share a common vertex must yield a $$3$$-edge.

Also, I'd like to try to generalize this for any $$d$$, but first, to understand the other direction for $$d=3$$.

Are the 2 definitions for regularity equivalent? I have 2 questions:

• For the case of d=3?
• For any d?

(What I provided was a proof in one direction. As for the other direction, I am not sure it holds.)

• "Each vertex $v$ has $r_1$ $2$-edges". Do you mean each vertex $v$ has $r_1$ $1$-edges? Jan 17, 2023 at 8:45
• Because of the comment above, do you mean "For the case of d=2?"? Jan 17, 2023 at 8:46
• Is there any problem you are not satisfied with Discrete lizard's answer? I would like to encourage you to upvote and accept it. (This comment will be deleted upon feedback.) Jun 10, 2023 at 3:16

Consider the following full hypergraph $$H$$ (or abstract simplicial complex) of dimension $$2$$, represented as a simplicial complex (nodes are in black, $$2$$-edges in red, $$3$$-edges are turqoise triangles):
(or, algebraically: $$E_0 = \{v_1,v_2,v_3,v_4,v_5, v_6\}$$, $$E_1=\{(v_1,v_2), (v_2,v_3), (v_3,v_4), (v_4,v_5), (v_5,v_6), (v_1,v_6), (v_1,v_4), (v_2,v_6), (v_3,v_5)\}$$, $$E_2 =\{(v_1,v_2,v_6),(v_3,v_4,v_5)\}$$ )
$$H$$ is regular according to the first definition, since each vertex belongs to 3 $$2$$-edges and 1 triangle. However, $$H$$ is not regular according tot the second definition, because e.g. $$(v_1,v_2)$$ is contained in 1 triangle while $$(v_2,v_3)$$ is contained in 0 triangles.
This means the "other direction" does not hold for dimensions $$2$$ and higher, so the two properties are not equivalent. (Of course, the two definitions are trivially equivalent for dimension $$1$$)
As for whether the second property implies the first, your proof (which seems to be for dimension $$2$$) looks correct to me. The step of going from $$2$$-edges to $$3$$-edges can be generalized into $$i$$-edges to $$(i+1)$$-edges, after the proof follows by induction. So, it seems the second property is stronger than the first.