The local twisted Gan–Gross–Prasad conjecture for unitary groups

Let FF be a pp-adic field, let EE and KK be quadratic field extensions of FF, and let VV be an nn-dimensional skew-hermitian space relative to E/FE/F. Set VK=VFKV_K=V\otimes_F K, G=ResK/FU(VK)G=\operatorname{Res}_{K/F}U(V_K), and HV=U(V)H_V=U(V). For a nontrivial additive character ψ\psi of FF and a conjugate-symplectic character μ\mu of E×E^{\times}, let ωV,ψ,μ\omega_{V,\psi,\mu} be the Weil representation and define

mV(π)=dimHomHV(π,ωV,ψ,μ).m_V(\pi)=\dim \operatorname{Hom}_{H_V}(\pi,\omega_{V,\psi,\mu}).

Let L=KFEL=K\otimes_F E, let MM be a generic LL-parameter for GG with associated LL-packet ΠMIrr(G)\Pi_M\subset \operatorname{Irr}(G), and write AM=iIZ/2ZaiA_M=\prod_{i\in I}\mathbb{Z}/2\mathbb{Z}\cdot a_i. The local twisted Gan–Gross–Prasad conjecture. (1) For every irreducible representation π\pi of G(F)G(F), mV(π)1m_V(\pi)\leq 1. (2) Summing over the two skew-hermitian spaces VV over EE of dimension nn and over πΠM\pi\in\Pi_M, one has

VπΠMmV(π)=1.\underset{V}{\sum}\sum_{\pi\in \Pi_M}m_V(\pi)=1.

(3) If V0V_0 is the unique space giving a nonzero contribution, then, for eE0×e\in E_0^{\times} with E=F(e)E=F(e),

μ(detV0)=ε(1/2,AsL/E(M)×μ1,ψE)det(AsL/E(M))(e)ωK/F(e2)n(n1)/2.\mu(\det V_0)=\varepsilon(1/2,\operatorname{As}_{L/E}(M)\times\mu^{-1},\psi_E)\cdot\det(\operatorname{As}_{L/E}(M))(e)\cdot\omega_{K/F}(e^2)^{n(n-1)/2}.

(4) The unique πΠM\pi\in\Pi_M contributing nontrivially corresponds to the character satisfying

χ(ai)=ε(1/2,[As(Mi)+As(M)+As(M/Mi)]μ1,ψE,e),\chi(a_i)=\varepsilon(1/2,[\operatorname{As}(M_i)+\operatorname{As}(M)+\operatorname{As}(M/M_i)]\cdot\mu^{-1},\psi_{E,e}),

where ψE,e=ψ(TrE/F(ex))\psi_{E,e}=\psi(\operatorname{Tr}_{E/F}(ex)).

Sources & referencesView supporting material

Primary source

Nhat Hoang Le, “The local twisted Gan-Gross-Prasad conjecture for U(V_K)/U(V)”, arXiv:2511.01301 (2025).

Additional references

2 papers in this index state this conjecture (2025). The statement above is taken from the most recent of them; the others are arXiv:2504.21002.

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.