The Hodge bridge: signature → inequality

First recall what the preceding ring permits: products of degree-one classes, a top-degree map, and relations that make the products computable. The three U₃,₃ degrees (1,2,1) illustrate one inequality, 2²≥1×1. They do not establish a theorem for all matroids. Which arrows in the argument require a theorem rather than another multiplication?

For a finite loopless matroid M of rank r+1, the Chow ring A*(M) has top degree r. Its reduced characteristic polynomial is χ̄_M(q)=χ_M(q)/(q−1)=Σₖ₌₀ʳ(−1)ᵏμᵏqʳ⁻ᵏ. Here α and β are the degree-one classes defined by sums over flats containing or avoiding one element. Adiprasito–Huh–Katz, Theorem 1.4, proves Poincaré duality, hard Lefschetz and Hodge–Riemann for this ring with an ample class arising from a strictly submodular function; Proposition 9.5 gives the coefficient identity. α and β lie on the nef boundary, reached by limits of ample classes.

For 1≤k≤r−1, the three entries of Qₖ on α,β are Qₖ(α,α)=μᵏ⁻¹, Qₖ(α,β)=μᵏ, Qₖ(β,β)=μᵏ⁺¹. The mixed Hodge–Riemann input and its nef limit give at most one positive direction for this pairing, with a positive class available for the elementary argument. If Q(u,u)>0, write v=cu+w with Q(u,w)=0. The complement has Q(w,w)≤0, hence Q(u,v)²≥Q(u,u)Q(v,v). Applying this to α and β gives (μᵏ)²≥μᵏ⁻¹μᵏ⁺¹. The coefficient identity, mixed signature and passage to the nef boundary are theorem-level inputs; the last inequality is algebra.

Set u=α and v=β in the Hodge lab. Compare the displayed reversed inequality with the negative direction α−β; the sliders restrict u and v to nonnegative combinations, so compute that negative vector on paper.