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The quadratic difference conjecture for symplectic pairs

Let pp and qq be monic polynomials of degree 22, let VV be a vector space, and let (b,u)(b,u) be a symplectic pair on VV. A symplectic (p,q)(p,q)-difference is a symplectic pair satisfying the (p,q)(p,q)-difference condition, while a (p,q)(p,q)-difference in End⁡(V)\operatorname{End}(V) is an endomorphism satisfying the corresponding algebraic condition. Quadratic difference conjecture. For a symplectic pair (b,u)(b,u) to be a symplectic (p,q)(p,q)-difference, it is necessary and sufficient that uu be a (p,q)(p,q)-difference in End⁡(V)\operatorname{End}(V). This would extend the previously established equivalence for (p,q)=(t2,t2)(p,q)=(t^2,t^2) to arbitrary pairs of monic quadratic polynomials; the source gives no resolution, so the conjecture remains open.

Sources & referencesView supporting material

Primary source

Clément de Seguins Pazzis, “The quadratic sum problem for symplectic pairs”, arXiv:2305.19340 (2023).

linear-algebraquadratic-polynomialssymplectic-geometry

Progress summary

Refreshed
Solved

The conjecture is false: a 2023 paper gives an explicit counterexample, so the proposed equivalence does not hold for all quadratic polynomials.

The conjecture asserts an equivalence between a symplectic pair being a symplectic (p,q)(p,q)-difference and its endomorphism component being a (p,q)(p,q)-difference. The 2023 paper The quadratic sum problem for symplectic pairs identifies this as Conjecture 1 and disproves it.

Known results

  • The equivalence was previously established for (p,q)=(t2,t2)(p,q)=(t^2,t^2); the paper shows that this does not extend to arbitrary monic quadratic polynomials.

2023 counterexample

Take p=qp=q irreducible quadratic, let VV have dimension 2n2n over a field with nn odd, choose a symplectic form bb, and set u=0u=0. Then 00 is a (p,p)(p,p)-difference in End⁡(V)\operatorname{End}(V), but (b,0)(b,0) is not a symplectic (p,p)(p,p)-difference: no suitable bb-alternating endomorphism can be annihilated by pp in this dimension. This is a published counterexample to the stated conjecture.

Current status (as of August 2026): The conjecture is resolved negatively; its claimed equivalence fails for some irreducible quadratic p=qp=q, although the special case (p,q)=(t2,t2)(p,q)=(t^2,t^2) remains valid.

Sources

Solutions 1

Counterexample

The paper were this is taken from actually gives a counterexample: see page 6 of https://arxiv.org/pdf/2305.19340

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