P = W in composite rank for fixed determinant

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P = W in composite rank for fixed determinant

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Take a cohomology class on the fixed-determinant Higgs moduli space and work out where it sits in the perverse filtration and in the weight filtration. All you need is its eigenvalue under one operator, h = [e, f]. Then you can explain why that settles P = W in composite rank.

The statement [0:53]

For genus g ≥ 2, degree coprime to n and composite rank n ≥ 4, the paper proves P_k = W_2k = W_2k+1 on the whole rational cohomology. That includes the variant part for Γ = Pic⁰(C)[n], which tautological generators never reach. With the known prime-rank cases, this covers every coprime rank.

The criterion [1:34]

One class e (the integral of the normalized ch₂) raises both filtrations. It is Lefschetz with the same center R on both gradeds: relative hard Lefschetz for P, curious hard Lefschetz for W.
That alone does not pin the filtrations down: two different filtrations can share such a raiser.
One lowering operator f that lowers P by 2 and W by 4, with [[f, e], e] = −2e, forces Gr f to be the standard sl₂ lowering operator [2:34]. Then h = [e, f] acts on grade k by k − r, and it is semisimple. So A_k = ⊕_{j≤k} ker(h − (j − r)) for either filtration, which means P_k = W_2k.

Worked check [3:33]: if h w = 4w and the center is R, then w ∈ P_{R+4} but not P_{R+3}, and w ∈ W_{2R+8} = W_{2R+9}. The doubling acts on the filtration index, never on the eigenvalue.

Where f comes from [4:15]

Let one logarithmic pole move over a finite cover S of the curve and choose a full flag at the pole. The flag correspondence then gives f′ = pr_ π_ ι^* N² ι_* π^* pr^***. N² lowers weight by 4. Integrating over S costs no weight. Nearby cycles over a Higgs line give the perverse shift by 2, and a lattice Riemann–Roch computation, after rescaling, gives the double commutator.

Your rule [5:09]: whenever two filtrations meet on one cohomology, check three things: a shared raiser with the same Lefschetz center, a shared lowerer, and the identity [[f, e], e] = −2e. If all three hold, read both filtrations off the eigenvalues of [e, f].

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