XZ Conjecture for G2: convex-hull criterion from vanishing torus–interval integrals
Prove that for any admissible function f: [0,1]^6 × T^8 → C, if for all positive integers P the integral ∫_{T^8} ∫_{[0,1]^5} (∫_0^{S(ξ_1)} (f(x,ξ,z))^P · ˜J_{G2}(x,ξ) dξ_2) dξ_1 dx_1 … dx_4 · (dz_1/z_1) … (dz_8/z_8) equals 0, then the zero vector does not lie in the convex hull of Sp(f), where ˜J_{G2}(x,ξ)=ξ_1 ξ_2 [ξ_1^2(16(1−ξ_2^2)^3+9(1−ξ_2^2)−24(1−ξ_2^2)^2) − (1−ξ_1^2)(3ξ_2−4ξ_2^2)^2] [ξ_1^2(1−ξ_2^2) − (1−ξ_1^2)ξ_2^2] x_1 x_2 x_3 x_4 and Sp(f) is defined by the nonzero Fourier–Laurent exponents in the torus variables.
References
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 $f:[0,1]{6}\times \mathbb{T}{8}\rightarrow$ be an admissible function in the sense of Definition \ref{def:admissible_function}. If $$\int{\mathbb{T}8}\int_{[0,1]5}\int_0{S(\xi_1)}fP \tilde{J}{G_2}\,d\xi_2 d\xi_1 dx_1\ldots dx_4\frac{dz_1}{z_1}\ldots\frac{dz_8}{z_8} = 0$$ for all $P\in $, where \begin{align} \begin{split} \tilde{J}{G_2}(x_1,\ldots,x_4,\xi_1,\xi_2):=& \xi_1\xi_2\bigg[\xi_12(16(1-\xi_22)3+9(1-\xi_22) -24(1-\xi_22)2)-\ & (1-\xi_12)(3\xi_2-4\xi_22)2\bigg]\big[\xi_12(1-\xi_22)-(1-\xi_12)\xi_22\big]x_1x_2x_3x_4,\end{split} \end{align} then $\vec{0}$ does not lie in the convex hull of $\mathrm{Sp}(f)$.