Let $M$ be a connected complete Riemannian manifold, $p\in M$ and $\mathrm{Cu}(p)$ denotes the cut locus of a point. This is a standard result that $M\setminus\mathrm{Cu}(p)$ deforms to $p$.

Now if $L$ is a compact submanifold of $M$ and $\mathrm{C}u(L)$ denotes the cut locus of $L$ then is it true that $M\setminus \mathrm{Cu}(L)$ deforms to $L$. I checked some of the examples and it worked. But I am unable to prove this fact. Any reference or proof will be appreciated.


  • We say $q\in \mathrm{Cu}(L)$ if any distance minimal geodesic joining $L$ to $q$ is no longer distance minimal beyond $q$.
  • By deformation, I mean to find $$H:M\setminus \mathrm{Cu}(L)\times [0,1]\to M\setminus \mathrm{Cu}(L)$$ such that $H(x,0)= x,~ H(q,1)\in L$ and $H(q,t)=q$ (if $q\in L$).


  • 1
    $\begingroup$ Can you include the definition of $Cu(L)$ in the post? Thanks. $\endgroup$ Commented Aug 25, 2020 at 16:50
  • $\begingroup$ You should also define what you mean by "deforms to $L$." $\endgroup$ Commented Aug 25, 2020 at 17:16
  • $\begingroup$ Your definition of "deform" is what one usually calls strong deformation retraction/retract. $\endgroup$ Commented Aug 25, 2020 at 19:00
  • $\begingroup$ Yeah, that is what I mean. $\endgroup$
    – XYZABC
    Commented Aug 26, 2020 at 0:31
  • $\begingroup$ I guess if you take a $\varepsilon$ disk bundle of $N$ and then take the exponential map that is precisely the set $M-\mathrm{Cu}(N)$. Then you can deform this to $N$. Probably this should work. $\endgroup$ Commented Aug 27, 2020 at 11:49

1 Answer 1


I am assuming the submanifold has no boundary. In this case it is true that $M\setminus cut(L)$ deformation retracts to $L$. I'm not sure what happens if there is boundary.

Let $\nu L$ denote the normal bundle of $L$. For each $v\in \nu L$, let $t_v\in [0,\infty]$ denote the time where the geodesic $\exp(tv)$ stops minimizing distance to $L$.

We define $U\subseteq \nu L$ by $U = \{tv \in \nu L: \|v\| = 1\text{ and } t<t_v\}$.

Then, according to this paper, $\exp|_{U}:U\rightarrow M\setminus Cut(L)$ is a diffeomorphism. Just to make notation a bit nicer, I'll use $\rho$ to denote $\exp|_{U}$.

Now, define $H:M\setminus C(L)\times [0,1]\rightarrow M\setminus C(L)$ by $H(m,t) = \exp((1-t)\rho^{-1}(m))$.

This is a composition of continuous functions, so is continuous. Further, $H(m,0) = \exp(\rho^{-1}(m)) = m$. Further, $H(m,1) = \exp(0\cdot \rho^{-1}(m)) \in L$. Lastly, for any $\ell\in L$, $\rho^{-1}(\ell)$ is in the zero section of $\nu L$, so $\exp((1-t) \rho^{-1}(\ell)) = \ell$, independent of $t$.

  • $\begingroup$ Thanks, it's a nice answer. $\endgroup$
    – XYZABC
    Commented Aug 28, 2020 at 9:10
  • $\begingroup$ @XYZABC: I'm glad you like it. That said, I'd hold of on accepting it - maybe someone will treat the case where $L$ is a manifold with boundary? $\endgroup$ Commented Aug 28, 2020 at 12:48

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