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I have learnt the Cartan Hadamard theorem,

Let $M$ be a complete Riemannian manifold with nonpositive sectional curvature. Then $\forall x\in M, \exp_x:T_xM\to M$ has no conjugate point.

Then the notes point out completeness assumption is required since for $\mathbb{R}^3\setminus \{0,0,0\}$ with induced metric, the theorem fails.

I don't quite understand the reason. Does it mean $\exp_x$ has some conjugate points for some $x$?

By definition in the notes, if $(d\exp_x)_p$ is singular, then $p$ is called a conjugate point of map $\exp_x$ and $\exp_x(p)$ is called a conjugate point of $x$ along geodesic $\exp_x(tp)$.

I have also learnt, $(d\exp_x)_{p}$ is singular iff there exists a normal Jacobi field $U(t)$ along $\gamma(t)=\exp_x(tp)$ not identically zero such that $U(0)=U(1)=0$.

I think some geometric explanation of conjugate point might be helpful. I saw some online materials say if $p$ and $q$ are conjugate along $\gamma$, one can construct a family of geodesics that start at $p$ and $\underline{almost}$ end at $q$. I don't quite understand the reason and I am not sure whether it's useful to answer my question.

John
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  • The Cartan-Hadamard theorem is considerably stronger than the statement you wrote. It says that if $M$ is complete with nonpositive sectional curvature, then for each $x\in M$, the map $\text{exp}x$ is a _covering map. It has no conjugate points on any manifold with nonpositive curvature, complete or not; but if you don't assume completeness it might not be a covering map (and it won't be globally defined on $T_xM$). – Jack Lee Apr 18 '15 at 18:25
  • @JackLee so for the example I gave, why $exp$ is not a covering map? – John Apr 19 '15 at 04:37
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    For one thing, in your example, given any $x\in \mathbb R^3\setminus{(0,0,0)}$, the map $\exp_x$ not defined on all of $T_xM$. But more to the point, it's not surjective. For example, the point $-x$ is not in the image of $\exp_x$. – Jack Lee Apr 21 '15 at 19:29

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