Killing vector field and critical points In exercise 2, chapter 4, in do Carmo's Riemannian Geometry we have a exercise about Killing vector fields and in the final he says:"by Killing equation in p, $\nabla_{X}X(p) = 0$", where $p$ is a critical point of $f(q) = \langle X,X\rangle(q)$. But the only thing we can obtain from Killing equation is that $2\langle\nabla_{X}X,X\rangle = 0 \Rightarrow \langle\nabla_{X}X,X\rangle = 0 $. And: $df_{p}.v = 2\langle\nabla_{v}X,v\rangle $. What am i ignoring?
Thanks in advance.
 A: Your problem is that you haven't correctly calculated the differential of $f$.
The compatibility with the metric equation is
$$ (d\langle X,Y\rangle) Z = Z\langle X,Y\rangle = \langle\nabla_Z X,Y\rangle + \langle X,\nabla_Z Y\rangle$$
see the definition at wiki.
Taking $X=Y$, and $d_p\langle X,X\rangle = 0$, we get the equation
$0 = 2\langle \nabla_Z X,X\rangle_p,$
or just
$\langle\nabla_Z X,X\rangle_p =0$.
Then the killing equation for $X$ is 
$$\langle\nabla_Y X,Z\rangle + \langle Y,\nabla_Z X\rangle = 0$$
for all $Y$ and $Z$. When $Y=X$, we get
$$\langle \nabla_Z X,X\rangle = - \langle \nabla_X X, Z\rangle.$$
Then together with the first equation, we get
$$\langle \nabla_X X,Z\rangle_p = 0$$
for all $Z\in TM_p$. Taking $Z=(\nabla_X X)_p$, we get $(\nabla_X X)_p = 0$ as desired.
A: Recall the definition of Killing field $X$ : $ \langle
\nabla_YX,Z\rangle+\langle \nabla_ZX,Y\rangle =0$
Hence $Xf=2\langle \nabla_XX,X \rangle =0$. Hence $f$ is const along
flow of $X$.
When $q$ is a critical point of $f$ and $V\in T_qM$ is orthogonal to
$X$ so that $Vf=0$. That is $0=2 \langle
\nabla_VX,X\rangle=-2\langle \nabla_XX,V\rangle$. That is
$\nabla_XX=0$
