Let $(M,g)$ be a smooth Riemannian manifold, and let $S \subseteq M$ be a compact submanifold.

Assume that for each $p \in M$, there exist a unique closest point on $S$, i.e a unique point $\tilde s(p) \in S$ such that $d(p,\tilde s (p))=d_S(p)$.

It is easy to see the map $\tilde s:M \to S$ is continuous.

Is it differentiable? (at which points)? If not, are there directional derivatives everywhere?

Does anything change if we assume every point has a unique minimizig geodesic to $S$? or that $M$ is complete? or both?

Edit: As shown in the example given by Willie Wong, when both conditions do not hold, $\tilde s$ does not have to be differentiable.


Consider the following:

Take $M$ the plane with the standard (flat) metric, with the origin and the ray $[0,\infty) \times \{2\}$ removed.

Let $S$ be the unit circle centered at the origin. Clearly for every point in $M$ there exists a unique point on $S$ that is closest to it:

  1. When $(x,y)\in M$ is such that either $x < 0$ or $y < 2$, then the closest point is $(x,y) / (x^2 + y^2)$.
  2. On the other hand, when $(x,y)\in M$ is such that $x \geq 0$ and $y > 2$, then the closest point is $(0,1)\in S$.

Clearly this mapping from $M \to S$ given by the proximal point is not differentiable on the subset $\{(0,y): y > 2\}$.

Note however that this manifold does not satisfy the stronger version of the question where there "exists a minimizing geodesic", since the exponential map from $S$ is not surjective.

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    $\begingroup$ Thanks, this is a nice example. I am wondering what is the situation if we restrict $(M,g)$ to be complete. (Of course, this does not imply that every point has a unique minimizig geodesic to $S$). $\endgroup$ – Asaf Shachar Jun 13 '16 at 4:07
  • $\begingroup$ @AsafShachar: it is a lot more delicate! Try, for example, to run an abstract argument to prove that your map is differentiable on $M = \mathbb{T}^2$ and $S = \mathbb{T}$. There are some obstructions to running the inverse function theorem argument. One way I see requires identifying the conjugate points, showing it is a manifold, and using it to run some sort of implicit function theorem argument. And we also have to worry now about focal points since the argument breaks. $\endgroup$ – Willie Wong Jun 15 '16 at 15:02

REVISED VERSION: My original answer was at best a mess. Here is what I think is a much shorter and cleaner version:

Assume that $M$ is an open Riemannian manifold and $S\subset M$ a submanifold such that there is a unique minimal geodesic joining each $x \in M$ to $\tilde{s}(x) \in S$. Since any minimizing geodesic must be normal to $S$, there is an open subset $\Omega$ of the normal bundle of $S$ such that the exponential map $\exp: \Omega \rightarrow M$ is a smooth bijective map. The argument given by Willie below shows that there are no focal points with respect to $S$ in $M$. This implies that the exponential map is a diffeomorphism. If $\pi: N_*S \rightarrow S$ is the bundle projection, then $\tilde{s} = \pi\circ\exp^{-1}$, which is a smooth submersion.

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    $\begingroup$ Willie, thanks for improving the answer. You're welcome to add more, if you think of anything else. As for it not being a focal point, it is because you can extend the geodesic a little bit (since $M$ is open) and it remains the unique minimizing geodesic. $\endgroup$ – Deane Yang Jun 13 '16 at 17:22
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    $\begingroup$ As an aside, this answer is really about the exponential map from $S$ to $M$ and its differential. I always found discussions about the differential of the exponential map to be very confusing. I prefer to recast everything in terms of Jacobi fields. Although there is no use of curvature here, the application of Sturm-Liouville theory to Jacobi equation (essentially $J'' + KJ = 0$) is, I think, the easiest way to understand the role of curvature on the behavior of geodesics and the volume of geodesic balls and tubular neighborhoods of a submanifold. $\endgroup$ – Deane Yang Jun 13 '16 at 18:20
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    $\begingroup$ It's known that a geodesic cannot be minimizing past a conjugate point (where some nontrivial Jacobi field vanishes). I glanced at a proof, and it appears to require the second variation of the length function. I believe the argument should extend to saying that a minimizing geodesic from a submanifold cannot go beyond the focal point. This would, I believe, complete the full argument. $\endgroup$ – Deane Yang Jun 14 '16 at 23:50
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    $\begingroup$ To prove that the exponential map is a diffeomorphism requires there to be no focal point in $M$. This is equivalent to saying that if a set of Jacobi fields orthogonal to the geodesic are linearly independent at any point along the geodesic, then they are linearly independent on the entire geodesic. $\endgroup$ – Deane Yang Jun 14 '16 at 23:54
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    $\begingroup$ @DeaneYang: Let $\Omega(S,x)$ denote the set of curves from $S$ to $x$. Suppose $\gamma$ is a length minimizing geodesic in $\Omega(S,x)$. Then the second variation formula for the energy is $$ H_\gamma(V,W) = \int \langle V', W'\rangle - R(V,\gamma', W,\gamma') \mathrm{d}\tau - \langle \gamma'(0), II(V_0, W_0)\rangle $$ So the proof that minimizing geodesics cannot go past a conjugate point goes through pretty much line by line. (1) Suppose $\gamma$ as a focal point before $x$. Let $V$ be such that before the focal point it is a non-trivial Jacobi field vanishing at $x$... $\endgroup$ – Willie Wong Jun 15 '16 at 14:41

When $M=\mathbb{R}^n$ and $S$ is at least $C^2$, your map $\tilde{s}$ is $C^1$ except at points $p$ where $d_S(p)={1\over\kappa(\tilde{s}(p))}$, where $\kappa$ is the largest principal curvature of $S$ at $\tilde{s}(p)$. The proof boils down to the inverse function theorem, and I don't think is much harder in the general Riemannian case (After giving it some thought, it seems that the proof in the Euclidean case does not apply readily to the Riemannian case).

Most references are concerned with the smoothness of the distance function $d_S$, but the smoothness of $\tilde{s}$ comes as a byproduct and is usually buried in the proof. See for example http://www.ams.org/tran/2007-359-12/S0002-9947-07-04260-2/

  • $\begingroup$ Thanks. Can you please elaborate on why $\tilde s$ is not differentiable at the points you mentioned? $\endgroup$ – Asaf Shachar Jun 12 '16 at 20:20
  • $\begingroup$ @AsafShachar: I think it is meant that $\tilde{s}$ is $C^1$ except possibly at points $p$ where ... There are obviously points that satisfy the condition $d_s(p) = 1 / \kappa(\tilde{s}(p))$ where $\tilde{s}$ is still $C^1$; for example, points on the convex side of the surface $S$. $\endgroup$ – Willie Wong Jun 12 '16 at 22:58
  • $\begingroup$ OK, is it easy to see then why in all the other points $\tilde s$ is indeed differentiable? $\endgroup$ – Asaf Shachar Jun 13 '16 at 2:31
  • $\begingroup$ A high powered way to do this is to use what Deane wrote + my comment thereof. Roughly speaking the condition $d_S(p) = 1/\kappa$ is implied by "being a focal point". An alternative way to see what was written here is to note that the normal exponential map on $S$ is a diffeomorphism to its image when restricted to $U$ in the normal bundle of $S$ where for $s\in S$ we have the section $U_s \subset B(s; 1 / \kappa)$ where $\kappa$ is the largest principal curvature at $s$. $\endgroup$ – Willie Wong Jun 13 '16 at 13:34
  • $\begingroup$ The final statement in the previous comment can be found in most textbooks on differential geometry of curves and surfaces, and boils down to some multivariable calculus. $\endgroup$ – Willie Wong Jun 13 '16 at 13:36

The following paper discusses precisely the relation of smoothness and closest point property:


See also Unexpected regularity of the distance from a $C^2$ submanifold

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    $\begingroup$ The first paper (presumably the second post also) discusses regularity of the distance function, whereas the original question here is about the regularity of the unique closest point in a sub-manifold, if the latter is defined. $\endgroup$ – John Jiang Jun 12 '16 at 18:35
  • $\begingroup$ The paper alzo discusses the regularity of the projection to the submanifold, which is, when defined, the unique closest point. $\endgroup$ – Raziel Jun 12 '16 at 23:13
  • $\begingroup$ OK I guess the lemma is somewhat related. But the neighborhood U cannot be any that satisfies the unique closest point property. $\endgroup$ – John Jiang Jun 13 '16 at 3:13


This is an attempt to prove rigorously that there exists an open subset $\Omega$ of the normal bundle to $S$, such that $exp:\Omega \to M$ is a bijection. This proof seems to work only when $S$ is compact.

Reminder-we assume the following:

1) For every point in $M$, there exist a unique closest point in $S$

2) For every point in $M$, there exist a unique minimizing geodesic from it to $S$

Define $\Omega=\{(s,v)\in NS|\, \, exp_s(tv) \, \text{ is the unique minimizing geodesic from } s \in S \text{ to } exp_s(v) \}$

Assumption (2) (together with the fact minimizing geodesics are orthogonal to the submanifold) implies that $exp: \Omega \to M$ is a bijection. (This is the only candidate).

Assume by contradiction $\Omega$ is not open, then there exists a point $(s,v)$ in $\Omega$, and a sequence $(s_n,v_n)$ in $NS\setminus\Omega$ converging to $(s,v)$. By the definition of $\Omega$, there exists a sequnce $(\til s_n,\til v_n) \in NS$ s.t $exp_{\til s_n}\til v_n=exp_{s_n}v_n$, $|\til v_n| < |v_n|$ (the norms are taken w.r.t the metric at the different base points). since $S$ is compact, $NS$ is also compact, so we can assume $(\til s_n,\til v_n)$ is converging to some $(\til s,\til v) \in NS$. Then, continuity of $exp$ forces $exp_{\til s}\til v=exp_sv$, and continuity of the metric forces $|\til v| \le |v| $.

The only thing we can conclude is that $(s,v)=(\til s,\til v)$....

Actually I think $\Omega$ must be closed: If $\Omega \ni (s_n,v_n) \to (s,v)$ then $exp_{s_n}v_n \to exp_sv$. It is not hard to see that if there is a geodesic from $S$ to $exp_sv$ whose length is smaller than $|v|$, then the $v_n$ won't be the optimal lengths of geodesics from $S$ to $exp_{s_n}v_n$, for $n$ sufficiently large.

I am trying to fill in some details based on Deane's original answer (now changed) and Willie's comments, for the case when $M$ is an open Riemannian manifold and $S\subset M$ a submanifold such that there is a unique minimal geodesic joining each $x \in M$ to $\tilde{s}(x) \in S$.

Given $v \in T_xM$, Let $\alpha(\tau)$ be a path in $M$, $\alpha(0)=x,\dot \alpha(0) =v$. Given $\alpha$ there exists a unique 1-parameter family of of constant speed minimal geodesics along $\alpha$, i.e:

$f: (t,\tau) \in [0,1] \times (-\delta,\delta) \rightarrow M$ such that: $$f(0,0) = x, f(1,0) = \tilde{s}(x), \,f(t,0) \text{ is the unique minimizing geodesic from } x \text{ to } \tilde s(x),$$ $$ f(0,\tau)=\alpha(\tau), \, f(t,\tau) \text{ is the unique minimizing geodesic from } \alpha(\tau) \text{ to } \tilde s(\alpha(\tau))$$

Notice, that the associated Jacobi field is $J(t)=\frac{\partial f}{\partial \tau} (t,0)$, and that $\tilde s(\alpha(\tau))=f(1,\tau)$, hence $$\frac{d}{d\tau}|_{s=0}\tilde s(\alpha(\tau))=\frac{\partial f}{\partial \tau} (1,0)=J(1)$$

The Jacobi field $J(t)$ satisfies $J(0) = \dot \alpha(0)=v$. We would like to show that $J(1)$ is in fact the directional derivative of $\tilde{s}$ in the direction $v$. A-priori, $J$ might depend on the chosen path $\alpha$ which represents $v \in T_xM$. (Note that $J(1)$ depends on $f(1,\tau)=\tilde s(\alpha(\tau))$).

In order to do so we will show $\frac{DJ}{Dt}(0)$ is determined only by $v$. (Since a Jacobi field $J(t)$ is uniquely determined by $J(0),\frac{DJ}{Dt}(0)$ this finishes the proof of independence).

We can represent the family $f$ via the exponential map as follows: $$f(t,\tau)=exp_{\alpha(\tau)}\big(tv(\tau)\big) $$

$$\frac{DJ}{Dt}(t)=\frac{D}{Dt}\frac{\partial f}{\partial \tau} (t,0)=\frac{D}{D\tau}\frac{\partial f}{\partial t} (t,0)$$

Note that, $$\frac{D}{D\tau}\frac{\partial f}{\partial t} (t,\tau)=\frac{D}{D\tau}d(exp_{\alpha(\tau)})_{tv(\tau)}(v(\tau)),$$


$$ \frac{DJ}{Dt}(0)= \frac{D}{D\tau}d(exp_{\alpha(\tau)})_0(v(\tau))|_{\tau=0}=\frac{D}{D\tau}(v(\tau))|_{\tau=0}=\frac{Dv}{D\tau}(0)$$

Here I am stuck. How can we show $\frac{Dv}{D\tau}(0)$ depends only on $v$ (and not on $\alpha$)?

Other points that need clarification:

1) Demonstrate linearity in $v$:

For a function to be differentiable, the directional derivatives needs to be linear in their arguments. Of course, since Jacobi equation is linear, and we know $J_v(0)=v$ (which is linear), then if we could show $\frac{DJ_v}{Dt}(0)$ depends linarly on $v$ we were done.

2) Finally, linear behaviour of the directional derivatives does not in general imply differentiability.

  • $\begingroup$ I'll take a look at your answer and might try to rewrite my answer with more details. $\endgroup$ – Deane Yang Jun 15 '16 at 2:39
  • $\begingroup$ A few comments. First en.wikipedia.org/wiki/Jacobi_field#Solving_the_Jacobi_equation may help. For the argument to work you probably need $v$ to be orthogonal to the minimizing geodesic to ensure $J(1)$ is parallel to $S$. (Think about the case where $\alpha$ is equal to the minimizing geodesic.) $\endgroup$ – Willie Wong Jun 15 '16 at 13:35
  • $\begingroup$ The problem with $D_t J_v$ is real, however, because you have to choose it so that the perturbed geodesic is the minimizing geodesic from the nearby point to $S$. Deane mentioned Sturm-Liouville in his comments: a slight problem is that the endpoint "1" is not always "1"; you have a bit of a free boundary issue there due to what I wrote in my previous comment. These can all be sorted out using some sort of inverse function theorem argument. But if you are going that way you might as well just directly prove that for every $x\in M$ and $\tilde{s}(x)\in S$, that the normal exponential map.. $\endgroup$ – Willie Wong Jun 15 '16 at 13:40
  • $\begingroup$ from $S$, based at $\tilde{s}(x)$ with appropriate velocity, has surjective differential, and use the local inverse function theorem. Since now the "constrained" surface $S$ is on the side of the initial data,, and not the final data, you avoid the free boundary problem. And now at the initial data level you have the right degree of freedom: $v \in T_{\tilde{s}} S$ has $\dim(S)$, and $J'(0)$ has $\dim(M) - \dim(S)$ after using the constraint that your Jacobi field corresponds to a family of geodesics orthogonal to $S$. (You can pick the components of $J'(0)$ orthogonal to $S$ freely, but... $\endgroup$ – Willie Wong Jun 15 '16 at 13:46
  • $\begingroup$ ... the tangential components are determined by $J(0)$ (which is tangent) by the shape operator of $S$). $\endgroup$ – Willie Wong Jun 15 '16 at 13:48

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