Any help with this problem would be appreciated. Thanks

Suppose $(M^3,g)$ is a smooth compact Riemannian manifold with smooth boundary and $\gamma$ is a simple smooth orientable curve in $M$. Does there exist a global smooth function $f:M \to \mathbb{R}$ such that $df(X)|_{\gamma} \neq 0$ along $\gamma$ and $|df|_g \neq 0 $ in $M$. ($X$ denotes the unit tangent vector on $\gamma$.)

in plain words does there exist a smooth function f whose level sets are never tangent to $\gamma$.


2 Answers 2


Edit. Surely $f$ has at least two critical points if $M$ is closed, so we cannot have $|df|_g\ne 0$ everywhere in that case. This arguments fails if we assume that each component of $M$ has a nonempty boundary. To the contrary, now one can assume that $f$ is generic (after a $C^1$-small perturbation, which would not destroy $df(X)|_\gamma>0$ - this definitely works if $\gamma$ is proper, e.g., connects two points on $\partial M$) and hence has only finite many critical points. Because $\dim M=3>2$, we can connect these critical points with the boundary by a family of disjoint paths that do not intersect $\gamma$. Then using an isotopy of $M\cup_{\partial M}\mathrm\partial M\times[0,\varepsilon)$, we can pull out each critical point of $f$ and find the desired function if we started with a function satisfying $df(X)|_\gamma>0$.

Together with Ben's answer, this solves your problem when $\gamma$ is a proper embedding. I am not sure if that condition is necessary. However, whenever you have two sequences $s_i$, $t_i$ with distinct limit points in $\overline{\mathbb R}=\mathbb R\cup\{\pm\infty\}$, such that $\gamma(s_i)$ and $\gamma(t_i)$ converge to the same point in $M$, then such a function $f$ cannot exist.

  • $\begingroup$ Thanks for your answer. I made an error when I was typing the question, and that is to say that the manifold has a boundary and I have edited the question. Surely now $f$ does not necessarily has to have a critical point. $\endgroup$
    – Ali
    Commented Mar 25, 2016 at 21:01
  • $\begingroup$ Also, Your solution seems to take a specific $/gamma$ where as I am asking whether such a function exists for any arbitrary smooth simple curve. $\endgroup$
    – Ali
    Commented Mar 25, 2016 at 21:03
  • $\begingroup$ I just realised that the $C^1$-small perturbation that does not destroy $df(X)_\gamma>0$ might be hard to acchieve if $\gamma$ is not proper, because then $|df(X)|_{\gamma(t)}$ can become arbitrarily small. But in the case where $\gamma$ joins two points on the boundary, I don't see a problem now. $\endgroup$ Commented Mar 26, 2016 at 14:48
  • $\begingroup$ thanks for your detailed remarks. I think now I know how to remove the critical points from the manifold, but still I would like to think more about the existence of a function such that $df(X) >0$ along $\gamma$ when say $\gamma$ is a simple geodesic connecting two points on the boundary so that counter examples such as dense curves are removed. $\endgroup$
    – Ali
    Commented Mar 26, 2016 at 15:10
  • 1
    $\begingroup$ @Ali If $\gamma\colon[a,b]\to M$ is smooth, injective and joins two points on the boundary, then you can find a smooth, injective map $[a-\varepsilon,b+\varepsilon]\times D^2\to M\cup_{\partial M}(\partial M\times[0,\varepsilon))$ (tubular nieghbourhood). On the image, let $f$ be the first coordinate, then extend to a smooth function and continue as above. $\endgroup$ Commented Mar 26, 2016 at 18:07

$f$ will decrease or increase along $\gamma$, so $\gamma$ cannot be closed. But if $\gamma$ is embedded and not closed, then a suitable diffeomorphism will take $\gamma$ into a coordinate ball, so some $f$ exists near $\gamma$. But globally, even with boundary, we might have no function without critical points. Edit: as Sebastian Goette points out, if every component has nonempty boundary, then there are smooth functions without critical points. Still, we need to know more about this $\gamma$.

  • $\begingroup$ If $\gamma$ is dense, as for example on a torus, then clearly there is no such $f$. $\endgroup$
    – Ben McKay
    Commented Mar 25, 2016 at 21:42
  • $\begingroup$ do you think these obstacles could be dealt with by looking at specific curves. E.g $\gamma$ a geodesic connecting two points on the boundary. $\endgroup$
    – Ali
    Commented Mar 25, 2016 at 22:23
  • $\begingroup$ It seems that the condition on $\gamma$ is not purely topological. For example, take $\gamma$ on a 2-dimensional submanifold diffeomorphic to $T^2$, spiralling toward two distinct closed curves. Then there is a function that is constant on each of the limit curves and satisfies $df(X)|_\gamma>0$. It can be extended such that $df$ is nonzero in the normal direction to $T^2$. However, if $\gamma$ converges to a closed limit curve and does not stay in a two-dimensional submanifold close to the limit curves, then the whole limit curve will be critical for any $f$ with $df(X)|_\gamma\ge 0$. $\endgroup$ Commented Mar 26, 2016 at 18:10

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