# Generic absence of non-trivial first integrals of geodesic flows

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.

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Do you mean apart from the energy integral or do you mean the geodesic flow restricted to the unit tangent bundle? – Robert Bryant Aug 18 '14 at 8:45
@RobertBryant Thanks for pointing out this, I am sorry for less details in my question. I mean the latter, i.e. the geodesic flow restricted to the unit sphere bundle. – Joe Aug 18 '14 at 18:04

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).

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you are very welcome, @Joe – Vladimir S Matveev Aug 19 '14 at 18:26
My earlier (deleted) comment addressed the wrong point because I was careless about reading Vladimir's last sentence. What I should have written is that the last sentence should be completed along the lines of "and the compatibility needed for existence of nontrivial integrals of this system imposes nontrivial differential conditions on the metric $g$." (In fact, the first nontrivial condition on $g$ for such compatibility appears at differential order $\ell = 2k{+}2$, and, modulo diffeomorphism, the metrics with such nontrivial integrals depend on $k$ functions of one variable.) – Robert Bryant Aug 20 '14 at 15:56
Thanks, @Robert. Does the level $\ell=2k +2$ of the first nontrivial condition correspond to dimension two case or is true in arbitrary dimension $\ge 2$? I did not really think about it but it seems that in higher dimensions the overdeterminancy of the PDE system corresponding to the existence of interals is higher that in dimension two which may imply that nontrivial compatibility conditions may start to appear earlier. We saw this phenomenon in the metrization problem: in dimension 4 the first obstruction has $\ell= 3$ and in dimension 2 I believe 4 or 5 -- you know better. – Vladimir S Matveev Aug 22 '14 at 15:57
@VladimirSMatveev: My comment was only about the dimension $2$ case. I agree that the order of obstructing conditions is lower in higher dimensions, and it grows much faster than in dimension $2$. For example, in dimension $n>2$, the existence of a nonvanishing Killing vector field (i.e., $k=1$) implies at least $(n{-}2)$ conditions on the $3$-jet of the metric (and the actual number of third-order conditions probably grows something like $n^4/12$). – Robert Bryant Aug 22 '14 at 17:01

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 $\mathsf{G}_\ell$.

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Thanks a lot @Robert for providing us with an insight view of the problem and where the complication may appear. – Joe Aug 22 '14 at 2:24
@Robert, what you wrote confirms, at least for me, what I wrote in my answer: experts know that generic metric does not have nontrivial Killing tensors but nobody did the efford to publish it. Somebody (may be you? or even me) should write a text for nonexpert explaining it. – Vladimir S Matveev Aug 22 '14 at 16:05
@VladimirSMatveev: I guess you mean that nobody since Darboux has written a 'modern' exposition. Is that true? I'm not actually sure how much further people have gone (at least in the 2-dimensional case) beyond Darboux, Livre VI, Chapitre IV, and he was reporting on results by Lie, Bour, Dini, Bonnet, and Levy (among others) that were already 20 years old at that time (1894). It's true that I haven't seen anything modern that approaches the sometimes amazing results that they got. It's not exactly lost knowledge, but it's not exactly widespread either. – Robert Bryant Aug 22 '14 at 17:15
Thanks, Robert. Suggesting that somebody should write a text about it, I rather thought about the multidimensional case. In dimension two I share your opinion that there is not much left what the classics did not know already -- though of course recently their ideas were fulfilled, particially with the help of computer algebra, by say Kruglikov and also by you with Dunajskii and Eastwood. But in higher dimensions the field is in my understanding completely open – Vladimir S Matveev Aug 23 '14 at 19:09

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

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Thanks @Misha for your answer. Is there any result for the special case that the first integrals are polynomials in the momenta? – Joe Aug 18 '14 at 18:33
The answer of Misha should be of course correct for closed (compact without boundary) manifolds, in the sence that most experts expect that geodesic flow of generic metric is ergodic. In the case of manifolds which are not closed (or at least if the geodesic flow is not complete), the following problem/counterexample appears: if we take the standard flat metric on a disc, any sufficiently small sufficiently smooth perturbation of it is integrable. – Vladimir S Matveev Aug 19 '14 at 6:37