# Continuity of retraction on $r : \text{Int }\Omega^c \to B$

I'm reading Milnor's Morse Theory and I have difficulty verifying some claim (which is easy according to Milnor) on page $$88$$, section $$\S 16$$ in the book. Here's the setup for my question. In the end, I only ask about how to show that a certain map between metric spaces is continuous.

Let $$\Omega= \Omega(M;p,q)$$ be the set of piecewise smooth path $$\omega : [0,1]\to M$$ from $$p$$ to $$q$$ in the connected, complete Riemannian manifold $$M$$. This set equipped with metric function $$d : \Omega \times \Omega \to \mathbb{R}$$ defined as $$d(\omega,\omega') = \max_{t \in [0,1] } \rho\big( \omega(t), \omega'(t)\big) + \sqrt{\int_0^1 \Big( \|\dot{\omega}(t)\| - \|\dot{\omega}'(t)\| \Big)^2 dt },$$ where $$\rho : M \times M \to \mathbb{R}$$ is the topological metric of $$M$$ coming from its Riemannian metric.

I already showed that the Energy function $$E : \Omega \to \mathbb{R}$$, $$E(\omega) = \int_0^1 \|\dot{\omega}\|^2 dt$$ is continuous. So for some $$c>0$$ we have open subset $$\text{Int }\Omega^c := E^{-1}([0,c))$$. For a subdivision $$0 = t_0 < t_1 < \cdots < t_k=1$$, let $$\Omega(t_0,\dots,t_k)$$ be a subspace of $$\Omega$$ consisting of paths $$\omega : [0,1] \to M$$ such that segment $$\omega|[t_{i-1},t_i]$$ is a geodesic for each $$i=1,\dots,k$$. Finally we have subspace $$B := \text{Int }\Omega^c \, \cap \, \Omega(t_0,\dots,t_k).$$

It is shown in the text that we can define a map $$r : \text{Int }\Omega^c \to B$$ such that each piecewise-smooth curve $$\omega \in \text{Int }\Omega^c = E^{-1}([0,c))$$ maped to unique broken geodesic $$r(\omega) \in B$$ formed by joining the end points $$\omega(t_{i-1})$$ to $$\omega(t_i)$$ by minimal geodesic. The detail as follows :

for each $$i=1,\dots,k$$, the couple $$\omega(t_{i-1}),\omega(t_i)$$ contained in a neighbourhood $$W_i$$ of a point $$x_i \in M$$ such that $$W_i \times W_i \subset F(U' \times B_{\delta}(0))$$, where $$F : U' \times B_{\delta}(0) \to M \times M$$ is the map $$F(x,v)=(x,\text{exp}_x(v))$$ which map $$U'\times B_{\delta}(0)$$ diffeomorphic onto its image. Therefore the couple $$\omega(t_{i-1}),\omega(t_i)$$ contained in the image $$F(U'\times B_{\delta}(0))$$ which means that there is a unique minimal geodesic from $$\omega(t_{i-1})$$ to $$\omega(t_i)$$. Therefore broken geodesic $$r(\omega)$$ uniquely determined. $$\color{blue}{(\star)}$$

Question : How to show that the map $$r : \text{Int }\Omega^c \to B$$ is continuous? I decided to show this by sequence criteria for continuous function. That is if $$\omega_n \to \omega$$ as $$n \to \infty$$ then $$\gamma_n=r(\omega_n) \to \gamma=r(\omega)$$ as $$n \to \infty$$.

Here is my thought so far: The sequence $$\omega_n \to \omega$$ says that I can make the distance $$d(\omega_n,\omega) = \max_{t \in [0,1] } \rho\big( \omega_n(t), \omega(t)\big) + \sqrt{\int_0^1 \Big( \|\dot{\omega}_n(t)\| - \|\dot{\omega}(t)\| \Big)^2 dt }, \qquad (1)$$ as small as i like by letting $$n$$ large enough. Let $$\epsilon>0$$ be the challenge, i have to show that for $$n$$ large enough, the distance $$d(\gamma_n,\gamma) = \max_{t \in [0,1] } \rho\big( \gamma_n(t), \gamma(t)\big) + \sqrt{\int_0^1 \Big( \|\dot{\gamma}_n(t)\| - \|\dot{\gamma}(t)\| \Big)^2 dt }, \qquad (2)$$ will be small than $$\epsilon$$. I think I can show that $$\max_{t \in [0,1] } \rho\big( \gamma_n(t), \gamma(t)\big)$$ can be made small as I like since I can control the term $$\max_{t \in [0,1] } \rho\big( \omega_n(t), \omega(t)\big)$$. My problem is to control the integral term in $$(2)$$.

I know that the $$\gamma$$'s are broken geodesics, so on each segement $$[t_{i-1},t_i]$$, the integrand $$\Big( \|\dot{\gamma}_n(t)\| - \|\dot{\gamma}(t)\| \Big)^2$$ on the integral term in $$(2)$$, is constant. So if i can show that on each segment that $$\|\dot{\gamma}_n(t)\| \to \|\dot{\gamma}(t)\|$$ as $$n \to \infty$$ for some fix $$t \in [t_{i-1},t_i]$$, then the whole integral goes to zero. To show this, i plan to use the continuity of exponential map $$(q,v) \to \text{exp}(p,v)$$ that define each geodesic segment $$\gamma_n$$ and $$\gamma$$. Am I on the right track ? Any help will be appreciated. Thank you.

Update Here is the detail of my idea in the paragraph above this: lets concentrate on a particular segment $$[t_{i-1},t_i]$$. Since i already showed that $$\max_{t \in [0,1]} \rho\big( \gamma_n(t),\gamma(t) \big) \to 0$$ we have $$\rho\big( \gamma_n(t_{i-1}),\gamma(t_{i-1}) \big) \to 0, \quad \text{and} \quad \rho\big( \gamma_n(t_{i}),\gamma(t_{i}) \big) \to 0.$$ Therefore if $$W_i$$ is the neighbourhood of a point $$x_i \in M$$ such that $$\omega(t_{i-1}) = \gamma(t_{i-1})$$ and $$\omega(t_i) = \gamma(t_i)$$ both contained in $$W_i$$ (as described in $$\color{blue}{(\star)}$$ above), then for $$n$$ large enough the end points $$\gamma_n(t_{i-1})$$ and $$\gamma_n(t_i)$$ also contained in $$W_i$$. Since $$W_i \times W_i \subset F(U'\times B_{\delta}(0))$$ with $$F(x,v):=(x,\text{exp}_x(v))$$ and $$F$$ diffeomorphic to its image, then $$F(\gamma_n(t_{i-1}),v_n) = (\gamma_n(t_{i-1}), \gamma_n(t_i)), \quad \text{and} \quad F(\gamma(t_{i-1}),v) = (\gamma(t_{i-1}), \gamma(t_i))$$ for some tangent vectors $$v_n$$ and $$v$$ at the starting points. But since the curves $$\text{exp}_{\gamma(t_{i-1})}(tv_n)$$ with domain $$[0,1]$$ is just reparametrization of geodesic segment $$\gamma_n|[t_{i-1},t_i]$$, then the initial velocities related by a constant as $$v_n = \lambda \|\dot{\gamma}_n\|$$. Similarly for $$\gamma$$ we have $$v = \lambda \| \dot{\gamma} \|$$. Since $$F$$ diffeomorphism (onto its image) we can write $$(\gamma_n(t_{i-1}),v_n) = F^{-1}(\gamma_n(t_{i-1}), \gamma_n(t_i)), \quad \text{and} \quad (\gamma(t_{i-1}),v) = F^{-1}(\gamma(t_{i-1}), \gamma(t_i)).$$ Now by continuity of $$F^{-1}$$, the convergence $$\gamma_n(t_{i-1}) \to \gamma(t_{i-1})$$ and $$\gamma_n(t_i) \to \gamma(t_i)$$ implies $$v_n = \lambda \|\dot{\gamma}_n\| \to v = \lambda \|\dot{\gamma}\|$$. Therefore $$\|\dot{\gamma}_n\| \to \|\dot{\gamma}\|$$.

I think you are missing only one simple point: As you spotted, in each interval $$[t_{i-1}, t_i]$$, $$\|\gamma_n'\|^2$$ are constants. Moreover, since $$\gamma_n|_{[t_{i-1}, t_i]}$$ is the shortest geodesic joining $$\gamma_n (t_{i-1})$$, $$\gamma_n (t_i)$$,

\begin{align} \rho( \omega_n(t_{i-1}), \omega_n (t_i)) &= \rho (\gamma_n (t_{i-1}), \gamma_n (t_n)) \\ &= \operatorname{length} (\gamma_n |_{[t_{i-1}, t_i]})\\ &= (t_i - t_{i-1}) \| \gamma_n'\|\\ \Rightarrow \|\gamma_n'\| = \frac{\rho( \omega_n(t_{i-1}), \omega_n (t_i))}{t_i - t_{i-1}} \end{align} and similar on $$\gamma$$. Thus

\begin{align} \int_0^1 ( \| \gamma_n'\| - \|\gamma'\|)^2 \mathrm dt &= \sum_{i}\int_{t_{i-1}}^{t_i} ( \| \gamma_n'\| - \|\gamma'\|)^2 \mathrm dt\\ &= \sum_i \frac{\big(\rho ( \omega_n(t_{i-1}), \omega_n (t_i)) - \rho( \omega (t_{i-1}), \omega (t_i)\big)^2}{t_i -t_{i-1}} \end{align}

Now since

$$\max_{t\in [0,1]} \rho (\omega_n(t), \omega(t)) \to 0$$

we have $$\rho ( \omega_n(t_{i-1}), \omega_n (t_i))\to \rho ( \omega(t_{i-1}), \omega (t_i))$$ and so

$$\int_0^1 (\|\gamma_n\| - \|\gamma\|)^2 \to 0.$$

Remark 1 We only need the first term in $$d$$ to conclude the continuity of $$r$$.

Remark 2 The proof that you included at the end is proving a stronger statement. You are trying to show that $$v_n \to v$$, instead of just "$$\|v_n\| \to \|v\|$$". I think one can use the local invertibility of $$F$$ to show that, only that in your argument you write $$v_n = \lambda \|\gamma_n '\|$$, $$v = \lambda \|\gamma'\|$$, which does not make sense as $$v_n, v$$ should be vectors instead of scalars.

• This is definitely much simpler. I was too worried about the integrals, i actually missed this simple approach which actually the formula written by Milnor earlier in the proof. Thank you. But i think you mean $\int (\|\cdot \|- \|\cdot\|)^2 dt$? Thanks for correction by your second remark, i mixed them up. I think what i actually meant was $\|v_n \| = \lambda \|\gamma'_n\|$. – Si Kucing Aug 9 at 19:01
• Yes I will fix it. @SiKucing – Arctic Char Aug 9 at 19:01