SO(2) orbit with two slice intersections and a stabilizer at the origin

The set $SU(2)$ is closed and bounded in the finite-dimensional space $M_2(\mathbb{C})$, hence compact; the matrices $\mathrm{diag}(i,-i)$ and $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ lie in $SU(2)$ and do not commute. The projection $P\to \{\star \}$ with right multiplication is a smooth principal bundle, $VP=TP$, and $H=0$ is a smooth right-invariant complement. For $Z\in \mathfrak{s}\mathfrak{u}(2)$, the defining one-parameter subgroup $t\mapsto {\exp }_G(tZ)$ has initial velocity $Z$, so differentiating $q{\exp }_G(tZ)$ gives the fundamental vector $Z_P(q)=qZ$ and ${\omega }_q(Z_P(q))=Z$; also $\ker \omega =0=H$ and ${\omega }_{qh}(Vh)=h^{-1}{\omega }_q(V)h$. Thus $\omega$ is a principal connection. The specified $s_0$ and $U$ satisfy the slice-data requirement. The regulator group is independent of the ambient full gauge group, as required by the Definition. The rotations satisfy $((x,y)\cdot R_{\theta })\cdot R_{\varphi }=(x,y)\cdot R_{\theta +\varphi }$, so the displayed formula is a smooth right action. The zero set of $\chi$ is exactly $S$, and at every point of $S$, $d\chi (\delta x,\delta y)=\delta y$ has kernel $T_{s}S=\{(\delta x,0)\}$.

Differentiating the action at $\theta =0$ gives $R_{(x,y)}(1)=(y,-x)$. Hence at $s=(r,0)$, $M_s(1)=-r$ and ${\mathrm{\Delta }}_{\mathrm{F}\mathrm{P}}(s)=\mathrm{\mid }r\mathrm{\mid }$, since both the angle density and the standard Lie-algebra density have unit coefficient. For $r>0$, the differential of $\mathrm{\Psi }$ at $(e,s)$, with source coordinates $(\theta ,r)$ and target coordinates $(y_{\mathrm{t}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{t}},z)=(y,x)$, has determinant of absolute value $r$, as the theorem's density formula predicts.

For $x_0=(r,0)$, $F_{x_0}(R_{\theta })=-r\sin \theta$, so its zeros on the circle are exactly $0$ and $\pi$ modulo $2\pi$. Their slice points are $(r,0)$ and $(-r,0)$, which are distinct. At the two zeros the orbit derivatives in the $y$-direction are respectively $-r$ and $r$, nonzero for $r>0$; thus both intersections are transverse.

The density factor along the orbit is ${\mathrm{\Delta }}_{\mathrm{F}\mathrm{P}}(x_0\cdot R_{\theta })=r\mathrm{\mid }\cos \theta \mathrm{\mid }$, which is the absolute derivative $\mathrm{\mid }d{F}_{x_0}\mathrm{/}d\theta \mathrm{\mid }$. For $Cϵ<r$, the inverse image of $\mathrm{supp}{\eta }_{ϵ}$ lies in two disjoint arcs about $0$ and $\pi$; on each arc, $F_{x_0}$ is a diffeomorphism onto $(-Cϵ,Cϵ)$, which contains $\mathrm{supp}{\eta }_{ϵ}$. Changing variables $z=F_{x_0}(R_{\theta })$ on each arc makes its integral $\int {\eta }_{ϵ}(z)dz=1$. Therefore the sum is $2$, the two distinct transverse intersections.

At the origin the orbit map is zero, so $M_{(0,0)}=0$ and the FP Jacobian is zero. Also every group element fixes the origin, hence $F_{(0,0)}\equiv 0$ and the zero set is the whole circle, not a finite set of transverse roots. This verifies exactly why the multiplicity theorem does not apply at a point with this stabilizer.