This is an exercise from my General Relativity course, so Einstein summation is implied.
On a Riemannian manifold with metric $g$, show that if for a vector field $X^{\mu}$ and a geodesic with tangent $T^{\nu}$ the quantity $g(T,X)\equiv T_\mu X^\mu$ is conserved along the geodesic, then $X$ satisfies Killing's equation: $$ \nabla_{\mu}X_{\nu} + \nabla_{\nu}X_\mu = 0 $$
I can show the converse, however I am struggling to show how $T_\mu X^\mu$ being conserved along the geodesic implies that $X^\mu$ is a Killing field.
Here is how far I got:
Starting from the initial assumption $T^\mu\nabla_\mu(T^\nu X_\nu) = 0$, we immediately get: $$ 0 = T^\mu(\nabla_\nu T^\nu)X_\nu + T^\mu T^\nu(\nabla_\nu X_\mu) $$
The first term vanishes due to the geodesic equation, and we can expand the second term into
$$ 0 = T^\mu T^\nu(\nabla_\nu X_\mu + \nabla_\mu X_\nu) - T^\mu T^\nu(\nabla_\mu X_\nu) $$
Where the last term must vanish as a direct consequence of the previous equation. However, I am not sure how (if even) $T^\mu T^\nu(\nabla_\nu X_\mu + \nabla_\mu X_\nu) = 0$ implies that $\nabla_{\mu}X_{\nu} + \nabla_{\nu}X_\mu = 0$. My best hope is to rewrite it as $$ T^\mu T^\nu \nabla_\mu X_\nu = - T^\mu T^\nu \nabla_\nu X_\mu $$ and appeal to some argument about antisymmetry, but the presence of the $T^\mu T^\nu$ term bothers me. Am I even going about this the correct way? Would there be a better way to approach this?
You need to assume that for all geodesics $\gamma$, $g(\dot{\gamma},X)$ is constant along $\gamma$. If this is true for all then by taking arbitrary $T$, your equation implies the vanishing of $\nabla_{\mu}X_{\nu}+\nabla_{\nu}X_{\mu}$ (in general if $a_{\mu\nu}$ is symmetric in its two indices, and $a_{\mu\nu}T^{\mu}T^{\nu}=0$ for all $T$’s, then $a_{\mu\nu}=0$).
Essentially the same (but index-free way) to phrase the argument is that for any $p\in M$ and any tangent vector $\xi_p\in T_pM$, fix a geodesic $\gamma$ with $\gamma(p)=0$ and $\dot{\gamma}(0)=\xi_p$. Then, $g(\dot{\gamma},X)$ is constant along $\gamma$, so differentiating and evaluating at $0$ gives $g(\xi_p,\nabla_{\xi_p}X)=0$. In other words, on the tangent space $T_pM$, the bilinear functional $f(\xi_p,\eta_p):=g(\xi_p,\nabla_{\eta_p}X)$ is such that for all $\xi_p$, $f(\xi_p,\xi_p)=0$, and so it is skew-symmetric: $f(\xi_p,\eta_p)=-f(\eta_p,\xi_p)$, i.e $g(\xi_p,\nabla_{\eta_p}X)=-g(\eta_p,\nabla_{\xi_p}X)$, i.e the operator $\xi_p\mapsto \nabla_{\xi_p}X$ on $T_pM$ is skew-adjoint relative to $g$, which is precisely equivalent to $X$ being a Killing vector field.
But, notice that we have to assume $g(\dot{\gamma},X)$ is constant for all geodesics. It is not enough to assume it for just one geodesic, because this condition doesn’t tell us anything about how $X$ behaves away from the geodesic.