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.

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    $\begingroup$ 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 $\|v_n \| = \lambda \|\gamma'_n\|$. $\endgroup$ – Si Kucing Aug 9 at 19:01
  • $\begingroup$ Yes I will fix it. @SiKucing $\endgroup$ – Arctic Char Aug 9 at 19:01

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