Is it true that given a smooth manifold M (with or without boundary), a "generic" metric g on M does not possess any non-trivial (non-constant) first integral for the geodesic flow induced by g on the unit sphere bundle?


What if we restrict ourselves to considering the first integrals of the geodesic flow which is a polynomial in the momenta? I did some search, it seems that for the case that the first integrals are linear in momenta, then a generic metric does not possess such first integral. This somehow related to the fact that a generic metric does not possess Killing vector fields. What about for higher order (Killing tensors)? 


Thanks a lot.



Generic absence of non-trivial first integrals of geodesic flows
In the case the integrals are polynomial in momenta, a generic metric does not poses those (except trivial integrals such as the energy and polynomial functions of the energy). This is a local statement.


The formal theorem, which also explains the notion ``generic'' in this case, could be formulated here as following. We call a metric $k$-rigid if it does not admit an nontrivial integral which is polynomial in momenta of degree $k$. 


Theorem. Any ($C^\infty$) metric can be locally $C^\infty$ perturbed, arbitrary small, such that the obtained metric has a small neigborhood in $C^\infty$ topology such that all metrics in this neighborhood are $k$-rigid. 


I did not see this theorem or its proof in the literature, may be because its proof is both relatively long and still obvious to experts. It follows from the observation that the system of PDE on the coefficients of the polynomial integral that corresponds to the condition that there exists an integral of degree $k$ is overdetermined and of finite type. 


Edited a year after the initial answer was posted: Together with Boris Kruglikov, I wrote the proof of the statement above, see <http://xxx.lanl.gov/abs/1510.01493> . In fact, we used slightly different trick from the one explained above and also from the one in the comments/answer of Robert Bryant, which allowed to prove absence of a nontrivial integral of a generic in the $C^2$ (and not $C^\infty$ as in the answer above) topology metric. Actually, a request to write the proof came from Gabriel Paternain, who need the result for his paper <http://arxiv.org/abs/1511.04547> (joint with H. Zhou). 


