Suppose we have a smooth manifold $M$ and a vector field $X$ on $M$. Then the trajectories of $X$ are pairwise non-intersecting, and further, each trajectory is either injective or periodic. I'm wondering under what conditions there exists a Riemannian metric $g$ on $M$ such that the trajectories of $X$ are all geodesics.

A related problem is solved in Can every curve on a Riemannian manifold be interpreted as a geodesic of a given metric?. Here it is shown that any injective or periodic smooth curve can be realized as a geodesic. This makes me think that this problem should have an affirmative answer for all smooth vector fields.

From the Euler-Lagrange equation, I believe we can reduce this problem as follows: if $\gamma$ is a geodesics of $(M, g)$ (for a hypothetical metric $g$ satisfying the above), then $$\frac{d^2 \gamma^k}{dt^2} + \Gamma_{ij}^k \frac{d\gamma^i}{dt} \frac{d\gamma^j}{dt} = 0 \tag{EL}$$ where $\Gamma_{ij}^k$ denotes the Christoffel symbols. If $\gamma$ is a trajectory of $X$, then $$\frac{d\gamma^\ell}{dt} = X^\ell$$ for all $\ell$ (this is all in local coordinates). Then (EL) becomes $$\frac{dX^k}{dt} + \Gamma_{ij}^k X^i X^j = 0$$ which perhaps makes this easier to solve.

More generally, does there always exist a metric tensor with a given set of Christoffel symbols (assuming they vary smoothly)? Certainly the metric would not be unique, but does existence always hold?

  • 1
    $\begingroup$ the repeated indexes are for sums, so you can't solve for the Christoffel's symbols like that $\endgroup$
    – janmarqz
    May 7, 2021 at 19:05
  • $\begingroup$ Good point haha, I've edited the question accordingly. $\endgroup$
    – Haydn Gwyn
    May 7, 2021 at 19:08
  • $\begingroup$ I found that such vector fields are called parallel, it seems $\endgroup$
    – janmarqz
    May 8, 2021 at 0:46

2 Answers 2


This is not true in general. Note that the linked answer is only local, and making a single curve a geodesic is much simpler than making all integral curves into geodesics simultaneously.

As a counterexample, consider $\mathbb{R}$ with a standard global coordinate $x$ and corresponding vector field $\partial_x$. Define the vector field $V$ by $V(x)=x\partial_x$. The integral curves of $V$ are of the form $t\mapsto Ce^t$ for $C\in\mathbb{R}$. Suppose $g$ is a matric making these curves geodesics. It follows that $g(V,V)$ is constant on $(0,\infty)$ and thus the component $g(\partial_x,\partial_x)$ diverges as $t\to 0^+$, so no such metric can be smoothly extended to all of $\mathbb{R}$.

There is a (limited) local version that works, though: given a point $x\in M$ with $V(x)\neq 0$, there is a neighborhood $U\ni x$ and a local metric $g$ on $U$ such that the integral curves of $V|_U$ are geodesics with respect to $g$. To see this, choose adapted coordinates around $x$ such that $V=\partial/\partial x^0$, and use the Euclidean metric induced by those coordintes.

  • $\begingroup$ Thanks! Is the obstruction here just that the vector field vanishes, or can this fail for nonvanishing fields as well? $\endgroup$
    – Haydn Gwyn
    May 8, 2021 at 1:54
  • $\begingroup$ @HaydnGwyn I'm not sure. The only obstruction for nonvanishing vector fields would have to be global in nature, and there is no obvious way to stitch the local solutions together, or to relate $V$ to the holonomy of the connection. $\endgroup$
    – Kajelad
    May 8, 2021 at 2:42

In Kajelad's answer, she/he mentions that if $V(x)\neq 0$, then one can locally find a metric making the trajectories into geodesics. I wanted to add another answer which indicates why that hypothesis is important. It also provides another obstruction to finding a Riemannian metric having the vector field trajectories as geodesics.

Consider the vector field $V$ on $\mathbb{R}^2$ obtained by rotation around the origin. The trajectories are all circles, except that the origin is a fixed point since $V(0) = 0$.

I claim that there is no Riemannian metric on any neighborhood of $0$ for which all the trajectories in the neighborhood are geodesics.

To see this, first recall that for any point $p$ in a Riemannian manifold, we can always find a totally normal neighborhood $U$ about $p$. In such a neighborhood, given any two points $q_1,q_2\in U$, there is a unique geodesic connecting them which lies entirely in $U$.

As a consequence, a totally normal neighborhood cannot contain a closed geodesic. For if $\gamma:[0,1]\rightarrow U$ is a closed geodesic which is injective except that $\gamma(0) = \gamma(1)$, then the points $\gamma(0)$ and $\gamma(1/2)$ have two minimizing geodesics between them, corresponding to following $\gamma$ in both directions.

Now, for the specific example of $V$ on $\mathbb{R}^2$, any neighborhood of $0$ clearly contains a closed trajectory. Hence, no neighborhood of $0$ can be a totally normal neighborhood with respect to any Riemannian metric, so there must not be a Riemannian metric even locally defined which has all of $V$s trajectories as geodesics.

  • $\begingroup$ Ahh I see; thanks for your answer! It seems, then, like the vector field vanishing is the primary obstruction to this condition holding (at least on $\mathbb R^1$, when the vector field is nonvanishing, the condition definitely holds). Do you know how we might go about constructing a counterexample with a nonvanishing field (say, on $\mathbb R^2$)? $\endgroup$
    – Haydn Gwyn
    May 11, 2021 at 12:12
  • $\begingroup$ One more thing: in both your and Kajelad's answers, it seems like we can remove the obstruction by removing the points of the manifold where the vector field vanishes (the manifold would no longer be simply connected, but c'est la vie). I'd be interested to see if there's an example of a vector field such that even after removing points where it vanishes, the condition does not hold (I'd conjecture that if we have a vector field on $\mathbb R^n$ vanishing at finitely many points, then this removal process will always work). $\endgroup$
    – Haydn Gwyn
    May 11, 2021 at 12:20
  • $\begingroup$ @Haydn: I've thought about it a bit, and I don't know the answer. By the way, removing points can certainly leave a simply connected thing (e.g, $\mathbb{R}^3$ with a point removed is simply connected), but for surfaces you're right. $\endgroup$ May 13, 2021 at 1:59

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