Killing Vector Field determined by one point

Question

I am trying to prove that if $X$ is a Killing vector field on a connected Riemannian manifold $(M,g)$ (i.e. $\mathfrak L_X g = 0$), then $X$ is determined by $X_p$ and $\nabla X|_p$ for any point $p \in M$. It suffices to prove that if $X_p = 0$ and $\nabla X|_p = 0$ then $X = 0$. Clearly, the condition implies that $X$ is zero along the flow line of $X$ through $p$. However, I am having trouble finding a way to show that $X$ must be 0 on all of $M$. Does anyone have any suggestions?

Answer 1

$X(q)=0,\ \nabla_YX(q)=0\ \forall Y$ Then $X\equiv 0$

Solution : Step (1) : $$[X,Y](q) = \nabla_0Y - \nabla_YX = 0$$ so that $$ 0 = [X,Y] = \lim \frac{1}{t} [d\varphi_{-t} Y_0 - Y_0 ] = [\frac{d}{dt} d\varphi_{-t} ]Y_0 $$ Since $Y_0$ is arbitrary, so $ \frac{d}{dt} d\varphi_{-t} =0$.

Step (2) : Since $X(p)=0$ so $\varphi_t$ sends a geodesic sphere onto itself. So $\varphi_t$ defines $A_t: T_qM\rightarrow T_qM$ such that $$ A_t(cv)=cA_t(v),\ \varphi_t\circ {\rm exp}\ (v) ={\rm exp}\ (A_t v) $$ Here $$|A_t(v)| = |v|\ (\ast)$$ where $|\cdot |$ is canonical. Note that $$ d\varphi_t (v) = dA_t (v),\ v\in T_qM $$ so that $d\varphi_t = dA_t$. From $\ast$ $(A_t(sw),A_t(sw)) = s^2|w|^2$ so that $$ | dA_t(w)| = |w|\ (\ast\ast) $$ Hence $\frac{d}{dt} dA_t=0$. So $$d A_t = dA_0 + \frac{1}{2}t^2 (dA_t)'' + ... $$ so that $(\ast\ast)$ imlies $$(dA_t)^{(n)}=0\ (n\geq 2).$$