# $M\setminus \mathrm{Cut}(L)$ deforms to $L$

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.

Edit

• 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 $$q\in L$$).

Thanks!

• Can you include the definition of $Cu(L)$ in the post? Thanks. Commented Aug 25, 2020 at 16:50
• You should also define what you mean by "deforms to $L$." Commented Aug 25, 2020 at 17:16
• Your definition of "deform" is what one usually calls strong deformation retraction/retract. Commented Aug 25, 2020 at 19:00
• Yeah, that is what I mean. Commented Aug 26, 2020 at 0:31
• 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. Commented Aug 27, 2020 at 11:49

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.

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 $$t$$.

• Thanks, it's a nice answer. Commented Aug 28, 2020 at 9:10
• @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? Commented Aug 28, 2020 at 12:48