Generic absence of non-trivial first integrals of geodesic flows

Question

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.

Answer 1

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).

Answer 2

This is an amplification of my comment on Vladimir's answer. It's actually not at all hard to see the generality of the surface metrics that admit a $k$-th degree polynomial first integral of their geodesic flow. Here is a sketch:

The question is local, so we can look at metrics on an open set $U\subset\mathbb{R}^2$. Moreover, modulo diffeomorphism, we can assume that the metric is conformal, i.e., $g = e^{2u}\bigl(dx^2+dy^2\bigr)=e^{2u}dz\circ d\bar{z}$, with cometric $$ \hat g = e^{-2u}\,\frac{\partial}{\partial z}\circ\frac{\partial}{\partial \bar{z}}, $$ which is a function on the symplectic manifold $T^*\mathbb{R}$. Now, a polynomial first integral of degree $k$ is a function $p$ on the tangent bundle of $\mathbb{R}^2$ of the form
$$ p = v_0(x,y)\,dx^k + v_1(x,y)\,dx^{k-1}dy + \cdots + v_k(x,y)\,dy^k. $$ Let $\hat p:T^*\mathbb{R}^2\to\mathbb{R}$ be its $g$-dual, considered as a function on $T^*\mathbb{R}$.

The condition that $p$ be constant on the geodesic flow of $g$ is simply that $\hat g$ and $\hat p$ Poisson commute, i.e., $$ \left\{\hat g, \hat p\right\} = 0. $$

Since the expression $\left\{\hat g, \hat p\right\}$ is polynomial of degree $k{+}1$ in the momenta, this is $k{+}2$ first-order equations for the $k{+}2$ unknowns $u, v_0,\ldots, v_k$. It is not difficult to see that this quasilinear first order system can locally be placed in Cauchy-Kowalewskaya form, so analytic solutions are determined by specifying these $k{+}2$ functions analytically along an appropriately non-characteristic curve. (Not all solutions are real-analytic, however.)

Now, there is still too much symmetry in this formulation, namely the conformal transformations of the complex plane. Generically, one can get rid of this as follows. If one writes $\hat p$ in the form $$ \hat p = h_0(z,\bar z)\ \left(\frac{\partial}{\partial z}\right)^k + h_1(z,\bar z)\ \left(\frac{\partial}{\partial z}\right)^{k-1} \circ \frac{\partial}{\partial {\bar{z}}} + \cdots + h_k(z,\bar z)\ \left(\frac{\partial}{\partial {\bar{z}}}\right)^k $$ where $\overline{h_j} = h_{k-j}$, one finds that the vanishing of the Poisson bracket implies that $h_0$ is actually holomorphic.

Now, one can always assume that $h_0$ is not identically vanishing because, otherwise, one could factor out a number of copies of $\hat g$ from $\hat p$ and so reduce the order of the integral. Now, use a holomorphic transformation to make $h_0\equiv1$. This reduces the number of unknowns by $2$ (as it fixes the real and imaginary parts of $h_0$) and reduces the number of equations by $2$ (because two of the equations are now identities), and the resulting first order system of $k$ equations for $k$ unknowns now has exactly the right generality and can still be put in C-K form locally, showing that the general (analytic) solution depends on $k$ functions of one variable.

All of what I have written was known more than a century ago, and one can find an account of it in Volume 3 of Darboux', Leçons sur la Théorie Générale des Surfaces et les Applications Géométriques du Calcul Infinitésimal.

Possibly, what Vladimir is referring to about the difficulties is something else: Because we now know that there are restrictions on a sufficiently high order jet of a metric in order for it to admit a polynomial geodesic first integral of degree $k$, we might want to know what those conditions are, explicitly in terms of some known invariants. However, this turns out to be extremely complicated once one goes beyond degrees $1$ and $2$.

What one can say is this (which gives a more precise version of the Theorem that Vladimir states in his answer): Consider the space $\mathsf{G}_\ell$ consisting of $\ell$-jets of surface metrics modulo diffeomorphism. This is a (singular) space that is finite dimensional, of dimension $1$ when $\ell=2$ and of dimension $\tfrac12(\ell{-}2)(\ell{+}1)$ when $\ell\ge 3$. (The singular locus of $\mathsf{G}_\ell$ has properly smaller dimension, and is nonempty for $\ell\ge 3$.) Carefully applying the above analysis shows that, when $\ell\ge2k{+}2$, the locus $\mathsf{F}_\ell(k)\subset \mathsf{G}_\ell$ of diffeomorphism classes of $\ell$-jets of metrics that admit a nontrivial polynomial geodesic first integral of degree $k$ or less has codimension $C_\ell(k) = \tfrac12(\ell{-}2k{-}2)(\ell{+}1)+1$ in $\mathsf{G}_\ell$.

Answer 3

This is a famous open problem (ergodicity of geodesic flow for generic metrics). Lohkamp claimed a proof 20 years ago, but nothing appeared.