Hodge Theory for Combinatorial Geometries
Karim Adiprasito, June Huh & Eric Katz · Primary research
Read Definition 1.3 for the two ideals and §9 for the degree map; translate each relation into this six-flat example.
Open the sourceBuild one actual Chow ring
A complete rank-three example you can hold in your hands.
Prerequisites: 11 · Eigenvectors and the sign of a geometry, 18 · The lattice and characteristic polynomial, 22 · Rings: a language for combining constraints
Understand
Objectives
- List the six flats and both types of ring relation.
- Reduce a square and all three mixed degrees.
- Separate the rank-three ring from the U₂,₃ triangle and identify the indefinite restriction.
Start with something you can see
Before using a Hodge signature, build one ring. For the free matroid U₃,₃ on E={1,2,3}, every subset is independent; its nonempty proper flats are the three singletons and three pairs. Name all six before multiplying anything. This is not the triangle edge matroid U₂,₃.
Give the idea a precise name
Put a degree-one symbol x_F on each nonempty proper flat F. Start with ℝ[x₁,x₂,x₃,x₁₂,x₁₃,x₂₃], then quotient by two kinds of relations: products x_Fx_G vanish for incomparable flats, and the sums of x_F over flats containing each fixed element are equal. The rank is 3, so the top degree is 2. A chain product x_i x_ij lies in that top degree 2, and the degree map sends it to deg(x_i x_ij)=1.
α=x₁+x₁₂+x₁₃; β=x₂+x₃+x₂₃; deg(α²,αβ,β²)=(1,2,1)
The six variables are not six independent numbers. Write S₁=x₁+x₁₂+x₁₃, S₂=x₂+x₁₂+x₂₃, S₃=x₃+x₁₃+x₂₃; the linear relations are S₁−S₂=0 and S₁−S₃=0. For example, (S₁−S₂)x₁=0 reduces to x₁²+x₁x₁₃=0, since x₁ is incomparable with x₂ and x₂₃. Hence deg(x₁²)=−1. Multiplying relations by each generator likewise gives square degree −1 for all six. Comparable singleton–pair chains have degree 1.
Work one small world
The exact relations panel sits beside the Hodge lab’s numerical pairing. Each entry describes a rule, not an independent numeric variable.
| Expression | Reduction | Reason |
|---|---|---|
| x₁x₂, x₁x₂₃ | 0 | incomparable flats |
| S₁−S₂, S₁−S₃ | 0 | two linear relations |
| x₁² | −x₁x₁₃ | multiply S₁−S₂ by x₁ |
| deg(x_i x_ij) | 1 | maximal chain |
| deg(α²), deg(αβ), deg(β²) | 1, 2, 1 | three squares −1, two comparable pairs counted twice; two surviving cross-products; symmetric last square |
In αβ only x₁₂x₂ and x₁₃x₃ survive, so its degree is 2. The reduced characteristic polynomial is (q−1)²=q²−2q+1, with absolute coefficients (1,2,1).
Counterexample to positive definiteness: on the two-dimensional restriction span{α,β}, the degree pairing has matrix [[1,2],[2,1]]. Its eigenvectors (1,1) and (1,−1) have eigenvalues 3 and −1; signature (1,1,0). It is indefinite, not positive definite. This restriction is not all of degree one.
THE BRIDGEThese three exact degrees make a small model for the general coefficient identity; they do not prove it for other matroids.
WHERE THIS IDEA STOPSThe degree pairing is indefinite. A restriction with one negative direction is not a positive-definite metric, and U₃,₃ has top degree two.
Experiment
In the Hodge lab set u=α and v=β using p=1,q=0,r=0,s=1. Compare Q(u,u)=1, Q(u,v)=2 and Q(v,v)=1 with the relations panel, then try a nonnegative mixture.
Which exact ring relation turns a square into the negative of a chain product?
Check & explain
What is deg(αβ) for U₃,₃?
What is deg(x₁²) in this ring?
Teach back
Expand α², αβ and β²; mark each vanished incomparable product, then use a linear relation to justify each negative square degree.
- Identify six nonempty proper flats.
- Apply incomparable-product and linear relations separately.
- Give the (1,1,0) signature only for span{α,β}.
Read deeper
Intersection theory of matroids: variations on a theme
Federico Ardila-Mantilla · 2024 · Open exposition
Read Examples 2.3–2.4 for the free three-element matroid computation and Theorem 1.1 for the general degree identity.
Open the source