Show isometry of flow on a compact Riemannian manifold where the vector field is Killing Let $(M,g)$ be a Riemannian manifold, $\nabla$ the Levi-Civita connection of $g$. A vector filed $V$ on $M$ is called a Killing field if for every $p\in M$ and every $X,Y\in T_p M$,
$$
g(\nabla_X V, Y)+g(X,\nabla_Y V)=0
$$
Show that if $(M,g)$ is a compact Riemannian manifold, and $V$ is a Killing field, then the flow $\Psi_t$ of $V$ is an isometry for each $t$. 
Now to get us started,
First we shall show that the rate of change of the metric $g_{\Psi_t(x)}(D_x\Psi_t X, D_x\Psi_t Y)$ with $t$ is zero at $t=0$ for any $X$ and $Y$ in $T_p M$. 
Then use the local group property of the flow.
Any help is appreciated. 
First we claim that
$$
\nabla_{Xg} (X_i,X_j)=X(g(X_i,X_j))-([X,X_i],X_j)-([X,X_j],X_i)=X(g(X_i,X_j)),\dagger
$$
Step 1  Define a connection for differentiating convector fields (1-forms). The derivative $\nabla_Y\omega$ should satisfy
$$
Y(\omega(X))=(\nabla_Y\omega)(X)+\omega(\nabla_Y X).
$$
Hence, 
$$
(\nabla_Y\omega)(X) := Y(\omega(X))-\omega(\nabla_Y X).\qquad (1)
$$
Apply (1), we have
    \begin{align*}
  (\nabla_X g)(X_i, X_j)&=[Y,g](X_i,X_j)\\
 &=Y(g(X_i,X_j))-g(\nabla_Y(X_i,X_j))\\
  &=Y(g(X_i,X_j))-g(\nabla_Y X_i,X_j)-g(X_i,\nabla_Y X_j)
 \end{align*}
 A: To show $\Psi_t$ of $V$ is an isometry for each $t$, that is,
\begin{equation*}
 g_{\Psi_t(x)}(D_x\Psi_t X, D_x\Psi_t Y)=g_x(X,Y)\quad\forall\,\,X,Y\in T_p M
\end{equation*}
Instead, we shall show
\begin{equation*}
 \frac{\partial}{\partial t}( X_i, X_j) =0,
\end{equation*}
which is the same as showing
\begin{equation*}
 \frac{\partial}{\partial t}(D_x\Psi_t X, D_x\Psi_t Y)=\text{const}\quad\dagger
\end{equation*}
Proof:
$V$ is a killing field, then we have for every $p\in M$ and $X,Y\in T_p M$.
\begin{equation*}
 g(\nabla_X V,Y)+g(X,\nabla_Y V)=0
\end{equation*}
Write $Y=V_i=\frac{\partial}{\partial x_i}$,\quad $X= V_j=\frac{\partial}{\partial x_j}$. Since the partial derivative commute in $\mathbb{R}^n$, that is,
\begin{equation*}
 [V_i,V_j]=0\quad\forall\,\,i,j\,\,\text{and}\,\,[V, V_i]=[V_n,V_i]=0
\end{equation*}
It suffices to show that $V$ is a Killing field if and only $L_V g(V_i,V_j)=0$ for all $i,j$. First note that local flows for any given time $t$ always give local diffeomorphisms, since their inverse is provided by the local flow of the vector field $-V$. Thus, for $V$ to be a Killing field near $x$ is equivalent to have the property $(u,v)=((D_x\Psi_t u, D_x\Psi_t v))$ for all $u,v\in T_p M$, and all $p$ near $x$. We claim moreover that for fixed $x_1,\cdots,x_{n-1}$, as $t$ varies the coefficients of $D_x\Psi_t u$ and $D_x\Psi_t v$ in term of the $V_i$ are constant.
Then
\begin{align*}
&\ g(\nabla_{V_j}V,V_i)+g(\nabla_{V_i}V,V_j)\\
= &\ g([V_j,V]-\nabla_V V_j, V_i)+g([V_i,V]-\nabla_V V_i,V_j)\\
= &\ ([V_j,V],V_i)-(\nabla_V V_j,V_i)+([V_i,V],V_j)-(\nabla_V V_i,V_j)\\
= &\ -g([V,V_j],V_i)-g([V,V_i],V_j)-{\color{blue}g((\nabla_V,V_j,V_i)+(\nabla_V V_i,V_j))}\\
= &\ -([V,V_j],V_i)-([V,V_i],V_j)-{\color{blue}g(\nabla_V V_j\cdot V_i+ V_j\nabla_V V_i)}\\
= &\ 
-([V,V_j],V_i)-([V,V_i],V_j)-V(V_j,V_i)\\
= &\
-\frac{\partial}{\partial t}g(V_j,V_i)\\
= &\
-\frac{\partial}{\partial t}g(D_x\Psi_t V_j, D_x\Psi_t V_i)
\end{align*}
The last two lines indicate $\Psi_t$ of $V$ is an isometry of $(M,g)$ for each $t$.
I always feel the minus sigh looks suspicious…
Any correction is appreciated.
