# Dense subset of cut locus

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.)

• 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 $p$? Commented Aug 22, 2019 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. Commented Aug 22, 2019 at 17:03

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 $$v$$.
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 $$\mathrm{Cut}^1_p\subset \mathrm{Cut}_p$$.