Geodesic flows and Killing fields

Question

How well-known is the following result: Let $(M,g)$ be a closed Riemannian manifold and define $D:C^\infty(M,Sym^mT^*M)\to C^\infty(M,Sym^{m+1}T^*M)$, $D\alpha:=\mathbb S(\nabla\alpha)$ where $Sym^m$ denotes the symmetric power of the cotangent bundle, $\mathbb S$ denotes symmetrization and $\nabla$ denotes the covariant derivative on forms induced by the Levi-Civita connection. In the case $m=1$, I can prove $L_{X}g = 2D(X^b)$ for all vector fields $X\in C^\infty(M,TM)$ and $b:TM\to T^*M$ denotes the musical isomorphism. For the general case $m\geq0$, we introduce $\pi_m^*:C^\infty(M,Sym^mT^*M)\to C^\infty(SM)$, $(\pi_m^*f)(v):=f_{\pi(v)}(v,\ldots,v)$, where $SM$ denotes the unit tangent bundle and $\pi$ the footpoint projection. Now let $X\in C^\infty(SM,T(SM))$ be the generator of the geodesic flow.

• Somehow one should be able to show that $X\pi_m^* = \pi_{m+1}^*D$.
• With this, one can somehow deduce that if the geodesic flow is dense (meaning transitive) in $SM$, then there are no non-trivial Killing fields.

Any complete proof or suggestions from the literature on both points would be highly appreciated. - Thanks

Answer 1

If we have a Killing tensor field $K$ of type $(0,d)$, the function $$I:SM\to \mathbb{R}, \ I(v)= K(v,\dots, v) \ \ \ \ \ \ (\ast )$$ is constant along geodesic flow. This is a well-known knowledge and a possible proof is as follows: Let $\gamma$ be an arc length parameterised geodesic. Then, $$\nabla_{\dot \gamma} (K(\dot \gamma,\dots, \dot \gamma))= \nabla K(\dot \gamma, \dots , \dot \gamma). \ \ \ \ (\ast\ast) $$ In the formula above, $\nabla K $ has $d+1$ indices, so I substututed $d+1$ vectors $\dot \gamma$ inside. In deriving $(\ast\ast)$ I used that the geodesic has the property $\nabla_{\dot \gamma}\dot \gamma=0$. Above,
$\dot \gamma$ stays for the velocity vector of the geodesic.

Next, we observe that the Killing equation implies that $$\nabla K(\dot \gamma, \dots \dot \gamma) =0 \ \ \ \ (\ast\ast\ast) .$$

Combining $(\ast\ast) $ and $(\ast\ast\ast)$, we see that the function $I$ given by $(\ast)$ is constant along every geodesic, as its derivative along the velocity vector is zero.

Next, assume that the geodesic flow is transitive. This implies that every continuous function $\tilde I:SM\to \mathbb{R}$ constant along trajectories of the geodesic flow is constant. This implies that the corresponding Killing tensor is trivial.