Proof of Killing's Equation

Question

The problem: I am trying to prove that, for a Riemannian manifold $(M,\langle\ , \ \rangle)$, $X \in \Gamma(TM)$ is a Killing field (i.e. one for which $\langle u,v \rangle_{p \in M} = \langle (d \phi^X_{t})_{\phi^{X}(t, p)} (u), (d \phi^X_{t})_{\phi^{X}(t, p)} (v) \rangle_{\phi^{X}(t, p)}$, where $\phi^{X}_{t} = \phi^{X}(t,) $ is the local flow of $X$) if $\forall Y,Z \in \Gamma(TM)$ $$ \langle \nabla_Y X, Z \rangle + \langle \nabla_Z X, Y \rangle = 0, $$ where $\nabla$ is the Levi-Civita connection.

My work so far:

By using the compatibility and torsion-freeness of the connection, it is easy to reduce this assumption to $$ X \langle Y,Z \rangle = \langle [X,Z],Y \rangle + \langle [X,Y],Z \rangle. $$ Then, if we choose $Y,Z$ such that $Y(p) = u$, $Z(p) = v$, and remembering that $X(p) = \frac{d}{dt}\Big|_{t=0} \phi^{X}(t,p)$, our LHS evaluated at $p$ becomes $$ \frac{d}{dt} \Big|_{t=0} \langle Y(\phi^{X}(t, p)), Z(\phi^{X}(t, p)) \rangle_{\phi^{X}(t, p)}. $$ Our RHS at $p$ becomes $$ \langle [X,Z](p),Y(p) \rangle_p + \langle [X,Y](p),Z(p) \rangle_p . $$ At this point I start looking at the definition of the Lie bracket: $$ [X,Y](p) = \frac{d}{dt}\Big|_{t=0} d \phi^X_{-t}(Y(\phi^X(t,p))), $$ so our RHS becomes $$ \langle \frac{d}{dt}\Big|_{t=0} d \phi^X_{-t}(Z(\phi^X(t,p))), Y(p) \rangle_p + \langle \frac{d}{dt}\Big|_{t=0} d \phi^X_{-t}(Y(\phi^X(t,p))) , Z(p) \rangle_p,$$ but I am having a hard time completing the proof. My hunch is that we want to show the RHS at $p$ is zero (I could be wrong here). Any hints or pointers would be very welcome. Apologies if my notation is a bit untidy. There is also probably a much easier way to do this, so any hints away from my solution are also welcome.

Answer 1

Edit (completely new version): This is best formulated in terms of the pullback along diffeomorphisms and the Lie derivative of tensor fields. To write this out, it is better to denote the Riemannian metric by $g$ rather than $\langle\ ,\ \rangle$. Denoting by $\phi^X_t$ the flow of the vector field $X$ up to time $t$, the statement that $\phi^X_t$ is an isometry (as you have written it out) is equivalent to $(\phi^X_t)^*g=g$. Now for a point $x\in M$, $t\mapsto (\phi^X_t)^*g(x)$ is a smooth curve in the finite dimensional vector space $S^2(T_xM)^*$ (defined for small $t$) so you can show that this is constant by proving that its derivative vanishes identically. By the flow property, it follows that $\tfrac{d}{dt}(\phi^X_t)^*g(x)=(\phi^X_t)^*(\tfrac{d}{ds}|_{s=0}(\phi^X_s)^*g)(x)$, and $\tfrac{d}{ds}|_{s=0}(\phi^X_s)^*g=\mathcal L_Xg$, the Lie derivative.

The upshot of this is that $X$ is an infinitesmial isometry if and only if $\mathcal L_Xg=0$. From standard naturality properties of Lie derivatives, you can express $(\mathcal L_Xg)(Y,Z)$ as $$ \mathcal L_X(g(Y,Z))-g(\mathcal L_XY,Z)-g(Y,\mathcal L_XZ)=X\cdot g(Y,Z)-g([X,Y],Z)-g(Y,[X,Z]), $$ which is exactly what you want.

Answer 2

i) If $f$ is flow of $X$, then $$ g(t)=g_{ij}(t) dx^i(t)\otimes dx^j(t) $$

$$ Y(t)= Y^i(t) \partial_i (t) $$

Hence \begin{align*} \frac{d}{dt} g(Y,Z)(t)&= \frac{d}{dt} \{ g_{ij}(t) Y^i(t) Z^j(t) \} \\ &=X (g_{ij}) Y^i Z^j + g_{ij} X(Y^i) Z^j + g_{ij} Y^i X(Z^j)\\&= Xg(Y,Z) \end{align*}

ii) And $$ g(Y,Z)= f(t)^\ast g(t) (Y,Z)= g(t)(df\ Y, df\ Z) $$

iii) Recall definition $$ [X,Y] =\lim\ \frac{1}{t} [ Y(t)- df\ Y(0)] $$

Here $$ -g(Y,Z)(0)= g(t)(Y(t)-df\ Y,df\ Z ) -g(t) (Y(t),df\ Z - Z(t)) -g(t)(Y(t),Z(t )) $$

so that $$ X g(Y,Z)= g( [X,Y],Z) + g(Y, [X,Z]) $$