Zariski density of integral points on character varieties

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Zariski density of integral points on character varieties

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Recursos

Take a smooth complex curve and fix quasi-unipotent Jordan classes at its punctures. You can then explain why its SL_r-character variety has Zariski-dense integral points over one number field L. You can also trace each step of the proof that produces them.

The statement

Fix quasi-unipotent classes C_1, …, C_s and a number field K that contains their eigenvalues. Then there is one finite extension L/K such that M_r(X)(O_L) ∩ Y(ℂ) is Zariski dense in Y, and so in every component [0:48]. Nonsemisimple classes are allowed, in every rank and every genus. The field is explicit [1:12]:

F = K(√2, ζ_pq), where p and q are distinct primes larger than r;
L = F(ζ_r, u^{1/r}) for all units u of O_F.

Integrality is ambient: the point is integral on the whole character variety, and the exact classes are imposed on the complex representation.

The proof in four moves

Encode each class as a flag [1:38]. Read the Jordan partition by columns. Blocks (2,1) give a flag with pieces of dimension 2 and 1, each with scalar monodromy λ. This turns the class into closed rank bounds. Practice case: blocks (3,2) give columns 2, 2, 1 [2:16].
Lower the rank on disks [2:36]. The diagram density theorem is proved by induction on the largest rank. A top-rank facet with boundary is cut into disks, and each removed band becomes lower-rank strips [2:59]. On each disk, a prefix U of the seam flag extends across the disk because the puncture monodromies are scalars. Put Q = V/U: both have lower rank. The missing extension data in Hom(Q,U) come back through linear choices, and the integral ones are dense [3:27].
Handle closed facets by cutting and twisting [4:06]. A closed facet has no boundary to cut toward. Cut along one nonseparating circle and fix its spectrum α_i = ζ^{i−1}: these values are distinct, and their differences are units. Then Dehn twists (A_1, B_1) ↦ (A_1B_1, B_1) multiply A_1 by a whole torus, so the spectrum can move in every direction. Irreducibility then gives full density [4:43].
Fix determinants, then impose the exact classes last [5:15]. Handle determinants are units, and by Dirichlet's unit theorem one L contains all their r-th roots, so you can twist to determinant one. Only after that do you select Y = q(C) \ q(C_bad) on the character quotient. This order matters because semisimplification can shrink Jordan blocks [5:36].

Try it on a curve of your own: write the flag for each puncture, then name the step that handles each facet.

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