advanced34 min studyLesson 629 in Mathematics

Extremal hypergraphs and codegrees

Extend edge and degree counting to uniform set systems and diagnose why high local codegree can coexist with a global matching obstruction.

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What you’ll understand

  • Define uniformity, link neighborhoods, and minimum codegree with explicit subset sizes.
  • Derive the codegree-sum identity by double counting incidences.
  • Verify a parity obstruction to a perfect matching without overstating a general threshold.

An ordinary edge connects two vertices. A k-uniform hypergraph instead has edges that are k-element subsets of a finite vertex set. We use simple set systems: edges have distinct vertices and no repeated copies. For k at least three, familiar graph intuition becomes less decisive because higher-order correlations can hide from pairwise summaries. Codegrees measure local extension opportunities, while global divisibility can still prevent those opportunities from fitting together into a spanning structure.

Count extensions of a fixed vertex subset

For an s-element set S with s smaller than k, its s-degree is the number of edges containing S. When s equals k minus one, this is the codegree: the number of vertices that can extend S to an edge. The minimum is taken over all such subsets, not only those already lying in some edge. A missing extension therefore contributes a zero codegree. In a three-uniform hypergraph, codegree belongs to a pair, whereas vertex degree counts edges containing one vertex.

The link of a vertex v is the (k−1)-uniform hypergraph whose edges are obtained by removing v from edges containing it. More generally, links translate extension questions into lower-uniformity set systems. Summing all (k−1)-codegrees counts each original edge exactly k times, once for each vertex that can be removed. This gives an immediate lower bound on the number of edges from a minimum-codegree assumption, but it does not describe how those edges overlap.

d(S)=∣{v∈V∖S:S∪{v}∈E}∣,∑S∈(Vk−1)d(S)=k∣E∣d(S)=|\{v\in V\setminus S:S\cup\{v\}\in E\}|,\qquad \sum_{S\in\binom V{k-1}}d(S)=k|E|
Here |S|=k−1, and edges are k-element sets. Therefore δₖ₋₁·C(n,k−1)≤k|E| for the minimum codegree δₖ₋₁.↔ Scroll wide equations horizontally.

An extremal density does not transfer unchanged from graphs

The hypergraph extremal number ex_k(n,F) is the largest edge count in an n-vertex k-uniform hypergraph avoiding an ordinary copy of fixed F. The natural edge density divides by the number of k-subsets. Complete multipartite constructions still supply examples, but the graph theorem that chromatic number determines asymptotic extremal density has no automatic direct replacement for arbitrary hypergraphs. One must specify the forbidden uniform set system and prove the relevant claim rather than importing an exponent or threshold by analogy.

High vertex degree and high codegree are also different hypotheses. Many edges through every vertex can coexist with a particular pair having no extension. Conversely, a positive minimum codegree gives extension opportunities for every pair in a three-uniform system, but choosing edges greedily can consume vertices needed later. Perfect matching questions require disjoint edges covering every vertex, including the divisibility requirement that k divides the vertex count. Local abundance cannot repair a failed arithmetic condition.

δk−1(H)(nk−1)≤ke(H),a perfect matching requires k∣n\delta_{k-1}(\mathcal H)\binom n{k-1}\le k e(\mathcal H),\qquad \text{a perfect matching requires }k\mid n
The first condition is a counting necessity. The divisibility condition is another necessity; neither alone is a sufficient perfect-matching theorem.↔ Scroll wide equations horizontally.

A parity barrier survives local extension opportunities

Partition the vertices into A and B and include exactly those three-element edges meeting A in an even number of vertices. Every permitted edge uses either zero or two vertices of A. Any disjoint union of permitted edges therefore covers an even number of A vertices. If A has odd size, no perfect matching can exist, even when the total order is divisible by three. This is a global obstruction visible in arithmetic modulo two.

The codegrees depend on pair type. A pair inside A must be extended by a B vertex; a mixed pair must be extended by a different A vertex; a pair inside B must be extended by another B vertex. Provided both parts are large enough, each type has many extensions. This construction does not disprove a correctly stated sufficiently high-codegree matching theorem. It shows why such theorems must account for sharp thresholds and structural barriers, and why an unspecified phrase such as “dense enough” cannot stand in for a hypothesis.

A twelve-vertex parity barrier

Take a three-uniform hypergraph with |A|=5 and |B|=7, including exactly the triples with an even intersection with A. Compute its edge count and minimum pair-codegree, then rule out a perfect matching.

  1. Edges either use three B vertices or two A vertices and one B vertex. Their number is C(7,3)+C(5,2)·7=35+70=105.

  2. A pair inside A has seven extensions in B. A mixed pair has four other A choices. A pair inside B has five other B choices. Thus the minimum pair-codegree is four.

  3. A perfect matching would use four disjoint triples because there are twelve vertices. Every selected triple covers an even number of A vertices, so four such triples cover an even number. They cannot cover the five A vertices.

The result

The hypergraph has 105 edges and minimum codegree four, yet no perfect matching. The obstruction is parity rather than a lack of individual pair extensions.

Explore the model

Interactive lab

Interactive lab · modular

Track the parity invariant of selected edges

Pause, reset, then set modulus m=2 and additive step a=0. Predict period one and residue zero after every addition. Each allowed edge contributes either zero or two A vertices, both zero modulo two. The target of covering five A vertices has residue one and is unreachable by such contributions. The model tracks this necessary arithmetic invariant; it neither chooses disjoint hyperedges nor proves any general matching threshold.

Track the parity invariant of selected edgesAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA residue invariant can forbid a global matching even when many individual extensions are available.

Check your understanding

In the twelve-vertex parity example, what is the minimum pair-codegree?

Try it yourself

For a simple three-uniform hypergraph on ten vertices with minimum pair-codegree at least four, derive an edge-count lower bound. Does this imply a perfect matching?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the codegree-sum identity and then check whether three divides ten.

Compare with the solution

There are C(10,2)=45 pairs. Their total codegree is at least 4·45=180, and the identity says this equals three times the edge count. Therefore at least 60 edges are required. A perfect matching is impossible because each edge covers three vertices and ten is not divisible by three. The edge bound is a necessary consequence of the assumption, not a sufficiency statement for a spanning matching.

Take it with you

  • Minimum codegree quantifies every (k−1)-set's extension count, including zero cases.
  • Summed codegrees count each k-edge exactly k times.
  • Global divisibility and parity barriers can survive substantial local density.

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