the hard Lefschetz theorem
/ LEFF-shets /
On a compact Kähler manifold there is a natural operation that wedges a cohomology class with the Kähler form, raising the degree by two. Iterating it walks you up the cohomology in steps of two. The hard Lefschetz theorem says this walk is as symmetric and rigid as possible: wedging k times sets up a perfect mirror between cohomology in degree n-k and cohomology in degree n+k. It is a deep, non-formal constraint forcing the Betti numbers of a compact Kähler manifold to rise to the middle dimension and fall symmetrically afterward.
Precisely, let M be a compact Kähler manifold of complex dimension n, and let L be the Lefschetz operator that sends a cohomology class alpha to [omega] wedge alpha, where [omega] is the Kähler class. The hard Lefschetz theorem states that for each k from 0 to n, the iterated map L^k from H^{n-k}(M) to H^{n+k}(M) is an isomorphism. Two immediate consequences follow. First, the Betti numbers are unimodal and symmetric about the middle: b_0 <= b_2 <= ... up to the middle, with b_{n-k} = b_{n+k}. Second, cohomology has a primitive (Lefschetz) decomposition: every class decomposes uniquely into pieces L^j times a primitive class (one killed by enough applications of the adjoint Lambda), an sl(2)-representation-theoretic structure on the total cohomology.
Why it matters: hard Lefschetz is a strong topological obstruction. A compact Kähler manifold whose Betti numbers are not unimodal-symmetric simply cannot exist, which rules out vast families of symplectic or complex manifolds from admitting any Kähler metric. The proof again flows from the Kähler identities via the sl(2)-action generated by L, Lambda, and the degree-counting operator. A caution: the name 'hard' distinguishes it from the easier Lefschetz hyperplane theorem (which is about cohomology of a hyperplane section); and although the statement is purely topological, its truth genuinely requires the Kähler hypothesis — it can fail on compact complex non-Kähler manifolds.
On CP^2 (complex dimension 2) the Betti numbers are b_0 = b_2 = b_4 = 1 and b_1 = b_3 = 0. Hard Lefschetz with k=1 gives an isomorphism L from H^1 to H^3 (both zero, consistent) and with the cup structure forces b_0 = b_4 and b_2 in the middle — the symmetric 1, 0, 1, 0, 1 profile, exactly as the theorem predicts.
Hard Lefschetz forces Betti numbers symmetric about the middle: CP^2 gives 1, 0, 1, 0, 1.
Do not confuse hard Lefschetz (an isomorphism L^k on cohomology of a compact Kähler manifold) with the Lefschetz hyperplane theorem (comparing a variety with its hyperplane section). Hard Lefschetz genuinely needs the Kähler hypothesis and can fail without it.