In $\mathbb R^3$ there are 3 natural foliations given by the lines parallel to each axis, which intersect transversally. Let $M^n$ a manifold with $n$ foliations by lines or circles that intersect transversally, is there a way to define a metric such that this fibrations are geodesics?
1 Answer
I'm editing my original answer to add information about the $n=2$ and $n=3$ cases.
The first question to answer is whether such a metric exists locally, and the answer to this is yes for $n=2$, probably (see below) for $n=3$, and no for $n>3$ (in general; the exact answer will depend on the foliations you specify).
In the case $n=2$, the (local) answer is obviously yes, since, locally, any pair of transverse curve foliations is locally equivalent to the standard coordinate foliations, and such metrics obviously exist to make them geodesic. There will be many such local metrics (depending on an arbitrary function of $2$ variables) for a given pair of foliations, but there might be some interesting global obstructions. Here is how one can see the local problem: Locally, any pair of transverse curve foliations on a surface can be described in some coordinate system $(x,y)$ as the curves parallel to the $x$axis and the $y$axis, i.e., as either the curves satisfying $dy=0$ or as the curves satisfying $dx=0$. In this coordinate system, a metric is described by the coefficients $ds^2 = E\ dx^2+2F\ dxdy + G\ dy^2$, where $E$, $G$, and $EGF^2$ are all positive. Set $E = e^2$ and $G = g^2$ for some positive functions $e$ and $g$. The condition that $dy=0$ describe a family of $ds^2$geodesics is equivalent to the condition that $d\xi = 0$ where $\xi = e\ dx +(F/e)\ dy$, while the condition that $dx=0$ describe a family of $ds^2$geodesics is equivalent to the condition that $d\eta = 0$ where $\eta = (F/g)\ dx +g\ dy$. Thus, one can choose $F=F(x,y)$ arbitrarily, and then the equation $d\xi=0$ becomes a (nonlinear) firstorder scalar equation for $e$ (which can easily be solved by the method of characteristics), while the equation $d\eta=0$ becomes a (nonlinear) firstorder scalar equation for $g$ (which can also easily be solved by the method of characteristics). In fact, $e$ will be uniquely specified by its restriction to a single curve in the family $dy=0$, while $g$ will be uniquely specified by its restriction to a single curve in the family $dx=0$. One just has to choose $e$ and $g$ so that the inequality $eg>F$ hold everywhere in the domain. This shows the nature of the local solutions, and it may be useful in understanding possible global obstructions to existence on a surface endowed with two transverse curve foliations.
In the case $n>3$, the generic $n$tuple of transverse curve foliations will not support a metric that all of the leaves (i.e., curves) be geodesic. The reason is this: To specify a curve foliation is to specify a vector field up to multiples, so this requires a choice of $n{}1$ functions of $n$ variables. To specify $n$ such foliations that are transverse requires a choice of $n(n{}1)$ functions of $n$ variables. Of course, two of these might be locally equivalent, and the group of diffeomorphisms (which depends, locally, on $n$ functions of $n$ variables) acts, so that, modulo diffeomorphisms, these structures depend on choosing $n(n{}2)$ functions of $n$ variables. However, metrics in dimension $n$ depend on $\tfrac12 n(n{+}1)$ functions of $n$ variables, and so, modulo diffeomorphism, they depend on $\tfrac12 n(n{}1)$ functions of $n$ variables. Since $n(n{}2)>\tfrac12 n(n{}1)$ when $n>3$, it follows that there must be an $n$tuple of curve foliations that is not, even locally, geodesic for any metric.
There remains the (local) question for $n=3$. The problem is formally determined, being 6 first order equations for 6 unknowns, and one would expect that this would imply that there is such a metric, at least locally in the realanalytic case. This does indeed turn out to be the case. A calculation shows that this first order system can be placed in Cauchy form, and therefore one can get local existence and uniqueness of such metrics for a triple of independent curve foliations that are analytic in some local coordinate system. However, the characteristic variety (which is of degree $6$) turns out, in the generic case, to be the union of three lines and a plane cubic curve. Generically, the cubic curve meets each of the three lines (which are in general position) in three distinct points. Thus, the characteristic variety is singular in the generic case. Consequently, I don't imagine that there are sufficiently strong hyperbolic existence theorems to prove the existence of solutions for smooth data.

$\begingroup$ If I understood, your point is that up to diffeomorphisms you have more transverse curve foliations than metrics, but why does the metric defines one foliation? What I mean is that for example in $\mathbb R^3$ you can take$3$ foliations by lines parallel to the $xy$ that intersect transversally which is not equivalent to taking the lines parallel to each axis, however they are geodesics of the euclidean space $R^3$. On the other hand locally can't one take a diffeomorphism to $\math bb R^n$ so that the foliations correspond to lines in $\mathbb R^n$ and then pullback the metric. $\endgroup$ Jun 1, 2013 at 7:39

1$\begingroup$ @Gerardo: First, given a metric $g$, choosing a (local) foliation by its geodesics involves choosing functions of $n{}1$ variables, not $n$ variables. The $g$geodesics in a $g$convex open set are a manifold $\Lambda$ of dimension $2(n{}1)$, and a geodesic foliation is a submanifold of $\Lambda$ of dimension $n{}1$. Thus, the space "metrics plus $n$tuple of geodesic foliations" still depends only on $\tfrac12 n(n{+}1)$ functions of $n$ variables. Second, when $n>2$, transverse $n$tuples of curve foliations in $\mathbb{R}^n$ have curvature invariants; they are not all locally equivalent. $\endgroup$ Jun 1, 2013 at 11:30