XZ Conjecture for Sp(N): convex-hull criterion from vanishing nested integrals
Prove that for any admissible function f: [0,1]^{N^2} × T^{N(N+1)} → C, if for all positive integers P the integral ∫_{[0,1]^{N(N−1)}} ∫_{T^{N(N+1)}} (∫_0^1 ∫_0^{ξ_N} … ∫_0^{ξ_2} (f(x,ξ,z))^P · ˜J_{Sp(N)}(x,ξ) dξ_1 … dξ_N) · (dz_1/z_1) … (dz_{N(N+1)}/z_{N(N+1)}) dx_1 … dx_{N(N−1)} equals 0, then the zero vector does not lie in the convex hull of Sp(f), where ˜J_{Sp(N)}(x,ξ)=˜J_{SU(N)}(x_1,…,x_{N(N−1)/2}) · (∏_{j=1}^N ξ_j) · (∏_{j>k}[ξ_j^2(1−ξ_k^2)−(1−ξ_j^2)ξ_k]) · ˜J_{SU(N)}(x_{N(N−1)/2+1},…,x_{N(N−1)}) and Sp(f) is defined by the nonzero Fourier–Laurent exponents in the torus variables.
References
In , we used the decomposition of $SU(N)$ to obtain results about $Sp(N)$ and $G_2$. Using Lemma \ref{lemma:Euler_Angles_SU(N)revisited} and the techniques given in this paper instead of Lemma 2.5 in , we arrive at the following two conjectures and theorems: Let $f:[0,1]{N2}\times \mathbb{T}{N(N+1)}\rightarrow$ be an admissible function in the sense of Definition \ref{def:admissible_function}. If $$\int{[0,1]{N(N-1)}\int_{\mathbb{T}{N(N+1)}\int_01\int_0{\xi_N}\ldots\int_0{\xi_2}fP \tilde{J}{Sp(N)} \,\,d\xi_1\ldots d\xi_N\frac{dz_1}{z_1}\ldots \frac{dz{N(N+1)}{z_{N(N+1)}dx_1\ldots dx_{N(N-1)} = 0$$ for all $P\in $, where \begin{align*} \tilde{J}{Sp(N)}(x_1,\ldots,x{N(N-1)},\xi_1,\ldots,\xi_N):=\,&\tilde{J}{SU(N)}(x_1,\ldots x\frac{N(N-1)}{2})\left(\prod_{j=1}N\xi_j\right)\cdot\ &\left(\prod_{j>k}\left(\xi_j2(1-\xi_k2)-(1-\xi_j2)\xi_k\right)\right)\cdot\ &\tilde{J}{SU(N)}(x{\frac{N(N-1)}{2}+1},\ldots,x_{N(N-1)}), \end{align*} then $\vec{0}$ does not lie in the convex hull of $\mathrm{Sp}(f)$.