Killing Vector Field determined by one point 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?
 A: $X(q)=0,\ \nabla_YX(q)=0\ \forall Y$ 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 $$(dA_t)^{(n)}=0\
(n\geq 2).$$
A: For a Killing field $X$, the formula $\nabla^2X(Y,Z) + R(X,Y)Z = 0$ holds for all $Y$ and $Z$. See here for proofs. With this, define on the bundle $TM\oplus {\rm End}(TM)$ the connection $$\overline{\nabla}_Y(Z,F) = (\nabla_YZ-F(Y), \nabla_YF - R(Y,Z)).$$This is indeed a connection, as it is a connection plus a tensor. Now, if $X$ is Killing, then $(X,\nabla X)$ is a $\overline{\nabla}$-parallel section. Now let $A = \{ p \in M \mid X_p = 0 \mbox{ and } (\nabla X)_p=0\}$ be the zero-set of $X$. By assumption it is non-empty, and it is also closed by continuity. Let's show that it is also open, as follows: let $p \in A$ and consider a path-connected neighborhood $U$ of $p$ in $M$. We will show that $U\subseteq A$. So take $q \in U$ and a curve in $U$ joining $p$ to $q$. Since $(X,\nabla X)$ vanishes at $p$ and it is a parallel section, it vanishes along the entire curve, and in particular at $q$, which shows that $q \in A$. By connectedness $A = M$ and $X=0$, as wanted.
