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Let $M$ be a Riemannian manifold, $x$ and $y$ are two points in $M$. Assume that $x$ is not in the cut locus of $y$. Does there exist a neighborhood $U$ of $x$ and a neighborhood $V$ of $y$ such that for every point $u$ in $U$ and for every point $v$ in $V$ we have that $u$ is not in the cut locus of $v$?

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2 Answers

up vote 5 down vote accepted

For a unit tangent vector $u$ with footpoint $p$ let $t(u)$ be the supremum of positive numbers such that the geodesic $t\to \exp_p(tu)$ is minimizing on $[0,t(u)]$. The cut locus at $p$ is the set of points $\exp_{p}(t(u) u)$ of $M$ for which $t_u$ is finite.

A basic result is that $u\to t(u)$ defines a continuous map from the unit tangent bundle to $(0,+\infty]$ where continuity at $+\infty$ is understood in the obvious way. See e.g. Sakai's "Riemannian geometry", Proposition III.4.1.

Now coming to your question fix $x\in M$ and $y=\exp_x(su)$ for some $u=u(x,y)$ and positive number $s$. Suppose $x^\prime$, $y^\prime$ are near $x$, $y$ respectively, and write $y^\prime=\exp_{x^\prime}(s^\prime u^\prime)$. If $y$ is not in the cut locus of $x$, then $t(u)=+\infty$. So $t(v)> s$ on some neighborhood of $u$ in the unit tangent bundle. Since $s^\prime$ and $s$ are almost the same, we conclude that $t(u^\prime)> s^\prime$, i.e. $x^\prime$, $y^\prime$ are not cut points of each other.

In my view Sakai's book is the best source of information about cut and cunjugate loci.

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Thank you very much for your answer, it is very intuitive. –  ProbLe Apr 17 '11 at 16:56
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Yes.

The cut locus for a point $x$ is the closure of the set of points $y$ such that there is more than one minimum length geodesic from $x$ to $y$. Sometimes $y$ may be on the cut locus yet have a unique minimizing geodesic to $x$. This happens when all pairs of minimizing geodesics from $x$ to points $y'$ near $y$ converge to give a single minimizing geodesic from $x$ to $y$ in the limit.

In such a case, $y$ is on the conjugate locus for $x$. This situation can be detected by the first derivative of the exponential map at $Y$, which is singular. An equivalent way to phrase it is that there is a nontrivial Jacobi field along the given geodesic from $x$ to $y$ that is 0 at $x$ and at $y$. (A Jacobi field is the first derivative in the parameter direction of a 1-parameter family of geodesics, that is, an infinitesimal variation of a geodesic). Note that this condition is symmetric in $x$ and $y$.

When $y$ is not a conjugate point of $x$ so that there is no Jacobi field of this sort, then by the implicit function theorem there is a smooth family of geodesics parametrized by $U \times V$ connecting points in a neighborhood $U$ of $x$ to points in a neighborhood $V$ of $y$.

For a complete Riemannian manifold $M$, the set of cut pairs $C(M) \subset M \times M$ is a closed subset invariant under interchanging factors: for any pair of points $(s,t)$ that is a limit of pairs of points $(s', t')$ joined by more than one minimizing geoesic, either there are distinct limits of sequences of minimizing geodesics, giving more than one minimizing geodesic from $s$ to $t$, or $s$ and $t$ are a conjugate pair. This answers your question yes.

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Bill, thank you very much for your detailed answer and for reminding me these basic facts about cut locus and conjugate locus. –  ProbLe Apr 17 '11 at 16:38
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@ProbLe---glad to do it. Knowledge is organic. I enjoy trying to rethink basic facts that I learned some time ago, and attempting to express them in clear terms. Typically when I learned things like this from textbooks and classes, they seemed technical and complicated. Unless I go over and rethink them, my memory gradually decays, making unjustified simplifications and unfeasible shortcuts. I gain something by trying to rethink and express it, trying not to resort too much to citing authorities or my intangible belief system of what is well-known. –  Bill Thurston Apr 17 '11 at 17:44
    
Couldn't have said it better myself! –  drbobmeister Apr 17 '11 at 20:00
    
Bill, it is very kind of you to give me your invaluable comments above, which enlighten me a lot on how to learn mathematics! I am so excited to practice your method just from today! "I gain something by trying to rethink and express it, trying not to resort too much to citing authorities or my intangible belief system of what is well-known." --That's it! –  ProbLe Apr 17 '11 at 22:12
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