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Let $(M,g)$ be the (underlying) manifold, $(S,g|)$ be a submanifold. Let $a,b,c \in S$. It's not in general true that $d_M(a,b)\leq d_M(a,c) \implies d_S(a,b)\leq d_S(a,c)$.

QUESTION I:

The above seems to be always locally true if $a,b,c$ are sufficiently close to each other, but not globally. Is it really?If it isn't, could you give a counterexample?

QUESTION II:

1) But is there a sufficient condition one can put on $S$ so that $d_M(a,b)\leq d_M(a,c) \implies d_S(a,b)\leq d_S(a,c)?$

2) How about a necessary and sufficient condition?

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Your condition implies that there is a nondecreasing function, $\phi_a\colon\mathbb R_{\ge0}\to \mathbb R_{\ge0}$ such that $$|a-x|_S=\phi_a(|a-x|_M).$$

One can reformulate it the following way, if you fix a point $a\in S$ then the angle between chord $[ax]_M$, $x\in S$ and the tangent space $T_xS$ depends only on the distance $|a-x|_M$.

This is quite strong global condition.

In particular if $S$ is a hypersurface then any pont is umbilical in the strongest sense ― all its principle curvatures are equal. In the higher codimensions, at each point, the absolute value of the normal curvature vector in all directions has to be the same.

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  • $\begingroup$ Thanks for your answer, but I'm not sure I could agree. Your symbol $|a-x|$ is mine $d(a,x)$ etc. Please see below. I'm not assuming $d_M(a,x_1)=d_M(a,x_2) \implies d_S(a,x_1)=d_S(a,x_2)$. I'm only assuming $d_M(a,x_1)=d_M(a,x_2) \implies d_S(a,x_1)\leq d_S(a,x_2)$, so strict inequality could hold, and hence there may not exist such a map $\phi_a$, as you wrote. $\endgroup$
    – Learning math
    Commented Apr 4, 2016 at 19:45
  • $\begingroup$ I edited the question, and added a question I. Could you give a counterexample in question I? Thanks! $\endgroup$
    – Learning math
    Commented Apr 4, 2016 at 22:56
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    $\begingroup$ @Let'stalkmath Consider the hyperboloid $S=\{z=xy\}$ in $\mathbb R^3$ and $a=0$. Distance from $0$ is preserved along the $x$- and $y$-axis, but distance in $S$ is larger than in $\mathbb R^3$ along all other geodesics emanating from $0$. Use this to find $b$, $c\in S$ yourself. $\endgroup$
    – Sebastian Goette
    Commented Apr 5, 2016 at 6:10
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    $\begingroup$ @SebastianGoette From the first condition it follows that norm of the restriction of differential $|d_x\mathrm{dist}_a|_{T_xS}|$ depends only on $|a-x|_M$ so the condition follows. (Sory, I did not understand what do you mean by "some power of the distance".) $\endgroup$
    – Anton Petrunin
    Commented Apr 5, 2016 at 8:33
  • $\begingroup$ @Anton Of course, sorry ... $\endgroup$
    – Sebastian Goette
    Commented Apr 5, 2016 at 13:27

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