Counterexamples to the xz-Conjecture and the Mathieu Conjecture for SU(2)

Let $$\mathcal{I}(h)={\int }_0^1{\int }_{\mathbb{T}}h(x,z)\frac{dz}{2\pi iz}dx(h\in \mathbb{C}[x,z,z^{-1}])\mathrm{.}$$ We give the three-term Laurent polynomial $$f(x,z)=(1-z^{-1})((1-x)+xz)$$ for which $$\mathcal{I}(f^n)=0,\mathcal{I}(z^{-1}{f}^n)=\frac{(-1{)}^{n-1}}{n+1}\mathrm{\ne }0(n\ge 1)\mathrm{.}$$ Since $\mathrm{Sp}(f)=\{-1,0,1\}$, this disproves the $xz$-conjecture already with one interval variable and one torus variable, and it also shows that $\ker \mathcal{I}$ is not a Mathieu–Zhao subspace. Padding gives counterexamples to every mixed case of the $xz$-conjecture. Writing the coordinate functions on $SU(2)$ as $$g=\left(\begin{array}{cc}a& c\\ b& d\end{array}\right),$$ the same example lifts, through the integration formula of Müger and Tuset, to the regular functions $$F=(1+c)(ad+b),G=-c,$$ which satisfy $${\int }_{SU(2)}{F}^{n}dg=0,{\int }_{SU(2)}{F}^{n}Gdg=\frac{(-1{)}^{n-1}}{n+1}\mathrm{\ne }0$$ for every $n\ge 1$. Thus the Mathieu conjecture for $SU(2)$ is false.