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Given a complete Riemannian manifold $M$ and point $p\in M$, denote $\mathrm{Cut}_p$ the cut locus of $p$ and $\mathrm{Cut}_p^1\subset \mathrm{Cut}_p$ the set of points $q$ which are connected to $p$ by more than one length minimising geodesic. According to a remark in Sakai's Riemannian geometry book (Rmk. 4.9), the latter forms a dense subset - but I don't understand why.

Question: Why is $\mathrm{Cut}_p^1\subset \mathrm{Cut}_p$ dense?

(I use density in this answer on MO. Comments on how to avoid this property to prove regularity of Riemannian distance function are also very welcome.)

Jan Bohr
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  • Does it help to note that the points in $\text{Cut}_p$ that are not in $\text{Cut}^1_p$ must all be points conjugate to $p$? – mjqxxxx Aug 22 '19 at 15:47
  • I had thought about this as well. In other words, if $v\in T_pM$ satisfies $\exp_p(v) \in \mathrm{Cut}_p\backslash \mathrm{Cut}^1_p$, then $d \exp_p$ is singular at $v$. What we need however, is that $\exp_p$ fails to be injective in every neighbourhood of $v$ - but of course this sort of converse of the inverse function theorem fails in general. And I don't know what to do with the information otherwise. – Jan Bohr Aug 22 '19 at 17:03

1 Answers1

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As user Chee kindly pointed out to me in a comment on MO one can use the following result from Klingenberg's Riemannian geometry book (Theorem 2.1.12):

Theorem. On a complete Riemannian manifold $(M,g)$, if $\mathrm{ker}(d\exp_p\vert_v)\neq 0$ for some $(p,v)\in TM$, then $\exp_p$ fails to be injective in every neighbourhood of $v$.

(Thus if injectivity fails infinitesimally, it also fails in every neighbourhood. This is of course false for a general smooth map.)

Now to conclude, take $x \in \mathrm{Cut}_p\backslash \mathrm{Cut}_p^1$, and note that $x=\exp_p(v)$ for some $v\in T_pM$ with $\mathrm{ker}(d\exp_p\vert_v)\neq 0$ (cf. Petersen Lemma 5.78). Take a sequence of opens $U_1\supset U_2\supset\dots\subset T_pM$ with $\bigcap U_n=\{v\}$, then according to the theorem above $\exp_p$ fails to be injective on each $U_n$ and thus $\exists x_n\in \mathrm{Cut}_p^1\cap\exp_p{U_n}$. Clearly $x_n\rightarrow x$ and thus we have proved the density $\mathrm{Cut}^1_p\subset \mathrm{Cut}_p$.

Jan Bohr
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