# Geodesic distance from point to manifold

This is question 1 of chapter 9 from Manfredo do Carmo's Riemmanian Geometry.

$M$ is a complete Riemmanian manifold and $N\subset M$ a closed submanifold. $p_0\in M$ and $p_0\notin N$. Let $d(p_0,N)$ denote the distante of $p_0$ to $N$. Show that there exists some $q_0\in N$ such that $d(p_0,q_0)=d(p_0,N)$ and the minimizing geodesic connecting $p_0$ and $q_0$ is orthogonal to $N$ in $q_0$.

I solved the first part of the problem. How can I prove the orthogonality property?

## 2 Answers

Intuitively, if the minimizing geodesic does not hit orthogonally, then a geodesics-with-corner is shorter, contradicting minimality. To see this, consider the following picture:

The submanifold $$N$$ is on the left, $$c$$ is the geodesic which connects $$p_0$$ (which is too far away to be in view) to $$q_0\in N$$. Of course, in real life the geodesics isn't "straight" nor is $$N$$ (whatever that even means for an abstract Riemannian manifold), but if you zoom in enough, everything looks closer and closer to Euclidean.

The point is that by breaking your geodesic at the dotted line, you make a broken geodesic between $$p_0$$ and $$q_1$$ which is shorter than $$c$$. But a broken geodesic is never minimizing, so the actual distance from $$p_0$$ to $$q_1$$ is shorter than this broken geodesic, so is, in particular, shorter than $$c$$. It follows that $$q_1$$ is strictly closer to $$p_0$$ than $$q_0$$ is, contradicting minimality of $$q_0$$.

The thing which makes this all rigorous is the first variation formula for energy (top of page 195 in my book).

Edit I think I have another approach that avoids this issue we've been discussing in the comments. As before, $$\gamma$$ is a minimizing geodesic between $$p_0$$ and $$q_0$$, where $$q_0\in N$$ is a point minimizing the distance between $$N$$ and $$p_0$$.

Choose a radius $$r$$ such that $$B(2r,q_0)$$ is a totally normal neighborhood of $$q_0$$. Pick any point $$p\in B(r,q_0)$$ which also lies on $$\gamma$$. Let $$t$$ be the smallest radius for which the distance sphere $$S_t$$ around $$p$$ intersects $$N$$. I now claim two things.

1. $$q_0$$ is in $$S_t\cap N$$ and

2. For any $$q\in S_t\cap N$$, we have $$T_q N\subseteq T_q S_t$$ and so, in particular, a vector normal to $$S_t$$ at $$q$$ is necessarily normal to $$N$$ at $$q$$.

Believing these two claims for a moment, by the Gauss lemmma (pg 69-70), we know that given any $$q\in S_t\cap N$$, the unique minimizing geodesic connecting $$p$$ to $$q$$ hits $$S_t$$ orthogonally, and hence hits $$T_q N$$ orthogonally. Further, if $$\alpha$$ is the unique minimizing geodesic from $$p$$ to $$q_0$$, then $$\alpha$$ is a portion of $$\gamma$$ (by uniqueness), showing $$\gamma$$ hits $$N$$ orthogonally.

We now prove claim 1.

If $$d(p,q_0)< t$$, then $$S_{d(p,q_0)}$$ hits $$N$$ with a smaller radius, contradicting minimality. If $$d(p,q_0)>t$$, then the broken geodesic beginning as $$\gamma$$ from $$p_0$$ to $$p$$, then from $$p$$ to a point of $$N\cap S_t$$ has length smaller than $$\gamma$$, contradicting the choice of $$q_0$$. So, we have $$d(p,q_0) = t$$, so $$q_0 \in S_t\cap N$$.

We now prove claim 2.

Assume it's false and choose $$v\in T_{q_1}N$$ which is not in $$T_{q_1}S_t$$. By replacing $$v$$ by $$-v$$, we may assume that $$v$$ points "inside" $$S_t$$. (More specifically, the ball of radius $$t$$ around $$p$$ is a manifold with boundary $$S_t$$, so we have a notion of inside and outside). Then by the Gauss lemma, the geodesic in $$N$$ in the direction of $$v$$ gets closer to $$p$$, contradicting the fact that $$S_t$$ is the smallest sphere intersecting $$N$$.

• I was trying to do this: Let $\gamma :[0,a]\rightarrow M$ be the geodesic joining the points $p_0$ and $q_0$. Let $v\in T_{q_0}N$ and let $\phi :(-\epsilon,\epsilon)\rightarrow M$ be the geodesic with $\phi(0)=q_0$ and $\phi'(0)=v$. Consider a variation $f:(-\epsilon,\epsilon)\times [0,a]\rightarrow M$ of $\gamma$ such that $f(s,0)=p_0$ and $f(s,a)=\phi(s)$. Now its just remain to prove that the energy of this variation satisfy $E'(0)=0$. How can i prove it? Nov 20, 2012 at 18:53
• (I deleted my old comment, sorry): First, you're trying to prove $E'(0)<0$ so that you get shorter geodesics. Second, I think you should adjust your approach a bit. First, the problem is that even if $\gamma$ gets shorter in the direction of $v$, it might not hit $N$. If $N$ is totally geodesic, there is no problem because $\phi(\epsilon,\epsilon)\subseteq N$ in that case, but in general we could have $\phi(t)\in N$ iff $t = 0$. I'm going to be busy for the next while, but later I'll think about it and update my answer with more details - it's trickier than I was thinking. Nov 20, 2012 at 20:41
• In fact Jason, if the exercise is right, my approach just prove that the two things are equivalentes, i.e. $E'(0)=0$ if and only if the property of orthogonality is right. Hence if we assume that the problem is right, $E'(0)$ must be zero. Nov 20, 2012 at 23:02
• I think we're saying the same thing - I'm suggesting a proof by contradiction. Assume $\gamma$ is not perpendicular and show that $E'(0)<0$ Nov 20, 2012 at 23:41
• First, see unc.edu/~sawon/MATH782/solutions7.pdf problem 2. The issue I was running into is handled(ish) there. Namely, the variation need not end in $N$. I think the hand wavy stuff about "first order" and "second order" can be made rigorous by the first variation formula ;-), but I also said that earlier and it ended up hairier than I was thinking. Nov 21, 2012 at 3:28

I think i have a answer too. Please verify if it is correct.

Let $$\gamma: [0,a]\rightarrow M$$ be the geodesic joining the points $$p_0$$ and $$q_0$$. Let $$v\in T_{q_0}N$$ and $$\phi: (-\epsilon,\epsilon)\rightarrow N$$ be a differentiable curve such that $$\phi(0)=q_0$$ and $$\phi'(0)=v$$ (note that the image of $$\phi$$ is contained in N).

Consider a variation $$f:(-\epsilon,\epsilon)\times [0,a]\rightarrow M$$ such that $$f(s,0)=p_0$$ and $$f(s,a)=\phi(s)$$. Let $$E$$ be the energy associated with $$f$$. By the first variation of energy formula, we have that $$\frac{E'(0)}{2}=\langle v,\gamma'(a)\rangle$$

Now, because $$\phi$$ is contained in $$N$$, we can conclude that $$E'(0)=0$$. This is true because $$\gamma$$ is the curve that minimizes the distance between $$p_0$$ and $$N$$ which implies that the curves $$f(s,t)$$ for fixed $$s$$ have more energy/length than $$\gamma$$.

• I think your answer is fine (much better than my own!) I'll delete my own answer in a second. The only thing I would change is that I'd pick a particular $\phi$, because for some $\phi$, the image of $\phi$ might not be contained in $N$. Nov 24, 2012 at 15:41
• Please dont delete your answer @JasonDeVito. Im gonna give you the bounty (if no better answer come). Nov 24, 2012 at 15:43
• Ok, I'll leave it up. Nov 24, 2012 at 15:45
• This is very nice, however I have a minor technical question: Why is it true that there exists a smooth variation $f$, such that $f(s,a)=\phi(s) \, \,\forall s$? The "classic" way of showing existence of smooth variations for any given smooth variational field $V(t) \in T_{\gamma(t)} M$, is to define something like $f(s,t)=exp_{\gamma(t)}(sV(t))$. In our case, the naive approach (given some $v \in T_{q_0}N$) would be to define a variational field by $V(0)=0$ (so all curves start from $p_0$) and $V(a)=v$; Now there are many (irrelevant) ways to proceed; Jun 21, 2016 at 20:04
• for instance one could parallel transport $V(a)=v$ along $\gamma$ (in the inverse direction) and multiply it by $t^2$,so it will decrease smoothly to zero. My point is that this way of construction will not assure us that $f(s,a) \in N$ at all "times" $s$. This will only be true if $S$ was a totally geodesic submanifold, so geodesics in $S$ will be geodesics in $M$. (Note that the $exp_{\gamma(a)}=exp_{q_0}$ is the exponential map of $M$, not of $S$ at $q_0$). Jun 21, 2016 at 20:04