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Let $M$ be a pseudo-riemannian manifold. Let $p\in M$. Let $\mathcal{D}_p$ be the maximal domain of the exponential map $T_pM\supseteq\mathcal{D}_p\xrightarrow{\exp_p}M$. Define the set of conjugate points of $p$ as $$ \begin{align*} \mathbf{Conj}(p)&\equiv\left\{X\in \mathcal{D}_p\mid T_pM \xrightarrow{d\left(\exp_p\right)_X}T_{e^X}M \text{ is singular}\right\} \\ &=\left\{X\in \mathcal{D}_p\mid \text{there is a non-zero jacobi field between } p \text{ and }e^X\text{ along }\gamma(\lambda)=e^{\lambda X} \text{ that vanishes at } p \text{ and } e^X \right\} \end{align*}$$ Is the lebesgue measure of $\mathbf{Conj}(p)$ zero? In order to define the lebesgue measure one needs to set up a basis in $T_pM$. But whether it is zero is independent of choice of basis.

Rasmus
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  • Since I am viewing $\mathbf{Conj}(p)$ as a subset of $T_pM$ and not $M$, I think $\mathbf{Conj}(p)$ can be uncountable. I imagine that if $p$ has a conjugate point in $q\in M$, then $\exp_p^{-1}({q})$ can be uncountable. – Rasmus Jul 17 '21 at 19:56
  • You really want to define conjugate points as as set of points in the tangent space rather than on the manifold? For example, for a sphere, if $p$ is the north pole then there will be a whole circle of $X$ which correspond to the south pole. Edit: I hadn't seen the above comment. – Pratyush Sarkar Jul 17 '21 at 19:58
  • Yes! @PratyushSarkar. There will be uncountably many circles. And the union of all those circles will have lebesgue measure zero. – Rasmus Jul 17 '21 at 19:59
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    I haven't carefully checked but for the set of conjugate points on the manifold, Sard's theorem might do it. Not sure what happens to the corresponding set of points in the tangent space. – Pratyush Sarkar Jul 17 '21 at 20:04
  • Sard's theorem answers the question for conjugate points on the manfold. – Rasmus Jul 17 '21 at 20:12
  • Sorry, I meant countably in a previous comment – Rasmus Jul 17 '21 at 20:18

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