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

Let $${\mathcal I}(h)=\int_0^1\int_{\mathbb T}h(x,z)\,\frac{dz}{2\pi iz}\,dx \qquad \bigl(h\in{\mathbb C}[x,z,z^{-1}]\bigr).$$ We give the three-term Laurent polynomial $$f(x,z)=(1-z^{-1})\bigl((1-x)+xz\bigr)$$ for which $${\mathcal I}(f^n)=0, \qquad {\mathcal I}(z^{-1}f^n)=\frac{(-1)^{n-1}}{n+1}\neq0 \qquad(n\geq1).$$ Since ${\operatorname{Sp}}(f)=\{-1,0,1\}$, 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 $$g=\begin{pmatrix}a&c\\ b&d\end{pmatrix},$$ the same example lifts, through the integration formula of M\"uger and Tuset, to the regular functions $$F=(1+c)(ad+b),\qquad G=-c,$$ which satisfy $$\int_{SU(2)}F^n\,dg=0, \qquad \int_{SU(2)}F^nG\,dg=\frac{(-1)^{n-1}}{n+1}\neq0$$ for every $n\geq1$. Thus the Mathieu conjecture for $SU(2)$ is false.