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I'm trying to prove the following result, which is Problem 6-17 in Lee's Introduction to Riemannian Manifolds:

Suppose $(M,g)$ is a compact connected Riemannian manifold. Show that every nontrivial free homotopy class in $M$ is represented by a closed geodesic that has minimal length among all admissible loops in the given free homotopy class.

This problem comes with a hint, which breaks the solution down into the following steps:

  1. Show the given free homotopy class is represented by a geodesic loop;
  2. Show the lengths of such loops have positive greatest lower bound;
  3. Choose a sequence of geodesic loops whose lengths approach this bound, and show that a subsequence converges uniformly to a geodesic loop whose length is equal to this greatest lower bound;
  4. Show the limiting curve is in the given free homotopy class;
  5. Apply the first variation formula to show that the limiting curve is in fact a closed geodesic.

I have most of these steps down except 3 and 5. First of all, if I have such a sequence, finding a universally convergent subsequence seems similar in spirit to Arzelà-Ascoli, but finding the right uniformly bounded and equicontinuous sequence of functions so far has eluded me.

Second of all, in proving 5, the first variation formula says that if $\Gamma_s(t)$ is a variation (i.e. a homotopy) between the limiting curve $\gamma(t)$ and one of the sequence curves $\gamma_n(t)$, then $$ \frac{d}{ds}\bigg|_{s=0} L_g(\Gamma_s) = -\int_a^b \langle V, D_t\dot\gamma\rangle\:dt - \sum_{i=1}^{k-1} \left\langle V(a_i), \Delta_i \dot\gamma\right\rangle + \langle V(b),\dot\gamma(b)\rangle - \langle V(a), \dot\gamma(b)\rangle $$ where $V(t) = \frac{d}{ds}\big|_{s=0} \Gamma_s(t)$ is the variation field along $\gamma$, $(a_0, \ldots, a_k)$ is an admissible partition for $V$, and $\Delta_i \dot\gamma = \dot\gamma(a_i^+) - \dot\gamma(a_i^-)$ is the "jump" in the velocity vector field at $a_i$. In this case, $V(a) = V(b)$, so we need to show $\dot\gamma(a) = \dot\gamma(b)$ and $\Delta_i \dot\gamma = 0$ for all $i$. Since $\Gamma_0$ is a critical point for this variation, this forumula becomes $$ \int_a^b \langle V, D_t\dot\gamma\rangle\:dt = - \sum_{i=1}^{k-1} \left\langle V(a_i), \Delta_i \dot\gamma\right\rangle + \langle V(b),\dot\gamma(b)\rangle - \langle V(a), \dot\gamma(a)\rangle $$ But why does this show the limiting curve $\gamma$ is geodesic? Where in here does it follow that $D_t\dot\gamma \equiv 0$?

Edit 1: I think I can use the Arzelà-Ascoli theorem for $\mathbb R^n$-valued functions of a compact space in a geodesic ball in normal coordinates around a limit point of an appropriate sequence $\{x_n\}$ in $M$; say, for example, $x_n = \gamma_n(0)$, where $\{\gamma_n\}$ is the sequence of geodesic loops.

Edit 2: Using the sequence $\{x_n\}$ coming from $x_n = \gamma_n(0)$, take a convergent subsequence $x_n \to p \in M$, and then let $w_n = \dot\gamma_n(0) \in T_{x_n}M$ for each $n$. Assume after reparametrization that each $\gamma_n$ is unit-speed, so the parallel translations $v_n \in T_pM$ along radial geodesics from $p$ to $x$ are unit tangent vectors. In particular, they too have a convergent subsequence $v_n \to v$. I think the resulting geodesic we want is then just $\gamma(t) = \exp_p(tv)$.

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  • $\begingroup$ The hint says to pick a sequence of geodesic loops. so these should have no jumps but your variation formula contains a list of jumps. I think these should all disappear, so you varation formula simply says critical point if and only if $D_t\dot{\gamma}=0$. $\endgroup$
    – quarague
    Commented Sep 24, 2019 at 8:49
  • $\begingroup$ If the jumps all disappear, then we should get $0 = \langle V(b), \dot\gamma(b) \rangle - \langle V(a), \dot\gamma(a)\rangle$. Why is this equal to $0$? What we really want to show is $\dot\gamma(a) = \dot\gamma(b)$. $\endgroup$
    – D Ford
    Commented Sep 24, 2019 at 15:59
  • $\begingroup$ If it is a geodesic loop there is no jump at the end points either. The end points are only special in your parametrization. $\endgroup$
    – quarague
    Commented Sep 25, 2019 at 7:03

1 Answer 1

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I have an (incomplete) idea for (3), so i'm leaving it here :)

Let $\gamma_i: [0, 1] \to M$ be your geodesics, with $l(\gamma)=l_i$, and $l_i \to l$, the infimum of all distances. We have $\lVert\gamma_i\rVert=l_i$. Now, if $t_1<t_2$, we can bound the (intrinsic) distance between $\gamma_i(t_1)$ and $\gamma_i(t_2)$ using $\gamma_i$ as a path: $$ d(\gamma_i(t_1), \gamma_i(t_2))\leq \int_{t_1}^{t_2} \lVert \gamma_i'(s) \rVert \, ds=(t_2-t_1) l_i. $$ Now, $\sup l_i<\infty$, as $\{l_i\}_{i \in \mathbb{N}}$ is a convergent sequence. This proves that $\{\gamma_i\}_{i \in \mathbb{N}}$ are uniformly Lipschitz functions, and thus equicontinuous.

With that in mind, we can use something like Arzelà-Ascoli: by a diagonal argument and the compactness of $M$, we can extract a subsequence $\{\gamma_{i_j}\}_{j \in \mathbb{N}}$ such that $\gamma_{i_j}(t)$ converges for each rational $t \in [0, 1]$. Equicontinuity (and completeness of $M$) gives uniform convergence, and now we have a uniform limit $\gamma$. In particular, $\gamma: [0, 1] \to M$ is a continuous and closed path.

We just need to bound the lenght, but I'm working on it

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  • $\begingroup$ This is a good argument. I think the length lower bound can come from the fact that the injectivity radius of a compact manifold is always positive, so the length of a geodesic loop is at least twice the injectivity radius $\endgroup$
    – D Ford
    Commented Sep 24, 2019 at 11:17

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