A circle action on a sphere has a pole with trivial normal action

Invariance of $S$ makes $Z{\mathrm{\mid }}_S$ tangent: the curve $t\mapsto {\varphi }_t(q)$ lies in $S$ for $q\in S$. If $f:S^2\to S$ is a smooth identification, the field $x\mapsto d(f^{-1}{)}_{f(x)}{Z}_{f(x)}$ is smooth and tangent to $S^2$; the chain rule makes $df$ invertible, so [F2] gives a point $p\in S$ with $Z(p)=0$. Both $t\mapsto {\varphi }_t(p)$ and the constant curve at $p$ are integral curves with initial value $p$. By [F4] they agree for $\mathrm{\mid }t\mathrm{\mid }<h$ for some $h>0$. For arbitrary $t\in \mathbb{R}$, choose a positive integer $m$ with $\mathrm{\mid }t\mathrm{\mid }\mathrm{/}m<h$; the flow law gives ${\varphi }_t(p)=({\varphi }_{t\mathrm{/}m}{)}^m(p)=p$. Thus the full circle fixes $p$.

Construct a smooth future timelike normal field along all of $S$. For each $q\in S$, the positive-definite restriction of $g_q$ to $T_{q}S$ gives a unique orthogonal projection $P_q:T_{q}M\to T_{q}S$. In a local tangent frame $X_1,X_2$, with $G_{ij}=g(X_i,X_j)$ and inverse matrix $(G^{ij})$, it is $P_q{N}_q={\sum }_{i,j=1}^2{G}^{ij}g(N_q,X_j)X_i$. The matrix is smoothly invertible, so these expressions are smooth; uniqueness of the projection makes them agree on overlaps. Put $u_q=N_q-P_q{N}_q$. Then $u_q\perp T_{q}S$ and $g(u_q,u_q)=g(N_q,N_q)-g(P_q{N}_q,P_q{N}_q)<0$, while $g(u_q,N_q)=g(u_q,u_q)<0$. Thus $u$ is a smooth future timelike normal field along $S$, in the normal-field convention of [F3]. Moreover $T_{q}M=T_{q}S\oplus (T_{q}S{)}^{\perp }$: nondegeneracy of the tangent restriction gives this direct sum, and a vector in the radical of the normal restriction would be orthogonal to all of $T_{q}M$, hence zero. The normal plane has dimension two, contains the timelike vector $u_q$, and has only one negative direction because the ambient metric does. Its signature is therefore $(-,+)$.

Set $W=(T_{p}S{)}^{\perp }$, $n=u_p\mathrm{/}\sqrt{-g(u_p,u_p)}$ and $A_t=d{\varphi }_t{\mathrm{\mid }}_p$. Isometry and invariance of $S$ imply that $A_t$ preserves $W$. The fixed-point property and chain rule give $A_{s+t}=A_s{A}_t$, $A_0=\mathrm{id}$ and $A_{t+2\pi }=A_t$; the coordinate formula makes this a smooth family. Inside the Lorentzian plane $W$, the line $L=W\cap n^{\perp }$ is spacelike: nondegeneracy splits $W=\mathbb{R}n\oplus L$, and the single negative direction is already $n$. Choose a unit $e\in L$ and put $k_{+}=n+e$, $k_{-}=n-e$. For either sign write $A_t{k}_{\pm }={\alpha }_{\pm }(t)n+{\beta }_{\pm }(t)e$. This vector is nonzero and null, so ${\beta }_{\pm }(t{)}^2={\alpha }_{\pm }(t{)}^2$ and ${\alpha }_{\pm }(t)\mathrm{\ne }0$. Since ${\alpha }_{\pm }(0)=1$, [F8] on the interval between $0$ and $t$ gives ${\alpha }_{\pm }(t)>0$. The continuous ratio ${\beta }_{\pm }\mathrm{/}{\alpha }_{\pm }$ takes values in $\{1,-1\}$ and has initial value $\pm 1$, so [F8] makes it identically $\pm 1$. Thus $A_t{k}_{+}=a(t)k_{+}$ and $A_t{k}_{-}=b(t)k_{-}$ for positive continuous scalars $a,b$. Since $g(k_{+},k_{-})=-2$, isometry gives $a(t)b(t)=1$; the representation law gives $a(s+t)=a(s)a(t)$.

By [F7], the positive continuous $2\pi$-periodic function $a$ attains a maximum $B=a(r)>0$ on $[0,2\pi ]$. Periodicity makes $B$ an upper bound on all of $\mathbb{R}$, by reducing any real argument modulo $2\pi$. For every $t$, the identity $a(r+t)=a(r)a(t)$ therefore gives $Ba(t)\le B$, hence $a(t)\le 1$. Applying the same inequality to $-t$ and using $a(t)a(-t)=a(0)=1$ gives $a(t)\ge 1$. Thus $a(t)=b(t)=1$ for every $t$, and the basis $(k_{+},k_{-})$ of $W$ is fixed. Consequently $A_t{\mathrm{\mid }}_W={\mathrm{id}}_W$. This uses only that $2\pi$ is a common period, and covers ineffective and trivial actions without rescaling the generator.

Choose one chart $x$ near the fixed point and write $F(t,y)=x({\varphi }_t(x^{-1}(y)))$ for small $t$ and $y$ near $x(p)$. This is smooth on an open neighborhood of $(0,x(p))$; $F(t,x(p))=x(p)$ makes the coordinate basis at the image independent of $t$. By [F5], $A_t$ is represented by $D_{y}F(t,x(p))$ and $Z^a(y)={\mathrm{\partial }}_t{F}^a(0,y)$. Commuting the mixed derivatives gives ${{\mathrm{\partial }}_t\mid }_{t=0}{D}_{y}F(t,x(p))v=D_{y}Z(x(p))v$. For $v\in W$ the left side vanishes by step 3.1. At $p$ the connection term in [F6] is zero because $Z(p)=0$, so ${\mathrm{\nabla }}_{v}Z=0$ for all $v\in W$.

Finally, if $K=T+\mathrm{\Omega }Z$ vanishes on $S$, evaluate this identity at the pole: $0=K(p)=T(p)+\mathrm{\Omega }Z(p)=T(p)$ by step 1.1. This proves every stated conclusion.