# Closed geodesic on a Riemannian manifold is homotopic to a closed curve of strictly shorter length

This is an exercise from chapter 9 of do Carmo.

Let $M^n$ be an orientable Riemannian manifold with positive sectional curvature and even dimension. Let $\gamma$ be a closed geodesic in $M$, that is, $\gamma$ is an immersion of $S^1$ into $M$ that is geodesic at all of its points. Prove that $\gamma$ is homotopic to a closed curve whose length is strictly less than that of $\gamma$.

What I am really confused about is the hint:

Hint: The parallel transport along the closed curve leaves a vector orthogonal to $\gamma$ invariant (this comes from the orientability of $M$ and the fact that the dimension is even). Therefore there exists a vector field $V(t)$ parallel along the closed curve $\gamma$.

My question are:

1. Why do we need the statement about the parallel transport leaving a vector invariant at all? We know that given any smooth curve $\gamma$ and a vector $v_0$ tangent to $M$ at $\gamma(t_0)$, the parallel transport of $v_0$ exists (given by Proposition 2.6 p. 52 of do Carmo) and that the parallel transport map is an isometry so parallel transport is orthogonal to $\gamma$ at all points.

2. Assuming that I'm wrong and we do actually need the invariance statement, how does this actually follow from the orientability of $M$ and the even dimension? I thought this was from the Synge-Weinstein theorem (page 203) but I cannot see how to apply it.

• can you do this for $S^2$ and $\gamma$ the equator? – Will Jagy Aug 22 '18 at 1:39

I‘m first going to comment on your two questions and then, for completeness, I‘ll give a sketch of proof for the exercise.

1) You are right that if we look at the curve as $$t\to \gamma(e^{2\pi i t})$$ then there is a parallel vector field $$V(t)$$ along $$\gamma$$ with $$V(0)=v$$ for every $$v\in T_pM.$$ However it might be that $$V(1)\neq V(0).$$ But as now we see the curve as a mapping $$\gamma:S^1\to M$$ we also want a parallel VF $$V$$ as a map from $$S^1$$ sucht that $$V(t)\in T_{\gamma(t)}M$$ and therefore the parallel VF should have the same value when starting and when „finishing the whole round“.

If this is not quite clear, then don‘t worry: In my answer below you can still think of the curve as a map $$\gamma:[0,1]\to M,$$ but we will see that there is a parallel vector field with $$V(1)=V(0).$$

2) Let $$\omega$$ be an orientation form for $$M,$$ i.e. a non-vanishing $$n-$$form (where $$n$$ is the dimension of $$M$$). For each $$t\in [0,1]$$ we can define an orientation on the orthogonal complement of $$\gamma‘(t)$$ by declaring $$v_1,...,v_{n-1}$$ to be positively oriented if $$$$\omega_{\gamma (t)}(v_1,...,v_{n-1},\gamma'(t))>0.$$$$ Let $$A:\left(\mathbb{R}\gamma'(0)\right)^{\perp}\to \left(\mathbb{R}\gamma'(0)\right)^{\perp}$$ be the restriction of the parallel transport map along $$\gamma$$ to the orthogonal complement of $$\gamma'(0)=\gamma'(1).$$

$$A$$ is clearly an isometry and also it is orientation preserving. To see the latter just observe that when $$e_1,...,e_{n-1}$$ is a positively oriented ONB of the orthogonal complement of $$\gamma'(0)$$ and if $$E_1(t),...,E_{n-1}(t)$$ denote their parallel transports along $$\gamma,$$ then for all $$t$$ we have $$$$\omega_{\gamma (t)}(E_1(t),...,E_{n-1}(t),\gamma '(t))\neq 0$$$$ because $$E_1(t),...,E_{n-1}(t),\gamma'(t)$$ is a basis for $$T_{\gamma(t)}M$$ and therefore, as $$e_1,...,e_{n-1}$$ is positively oriented, $$$$\omega_{\gamma (t)}(E_1(t),...,E_{n-1}(t),\gamma '(t))>0.$$$$ So $$E_1(1),...,E_{n-1}(1)$$ has the same orientation as $$e_1,...,e_{n-1}.$$ Because of $$E_i(1)=A(e_i),$$ $$A$$ is orientation preserving.

Because $$n$$ is even, Lemma 3.8 in Do Carmo impliess that there is a $$v\in T_{\gamma(0)}M$$ such that $$$$V(1)=A(v)=v,$$$$where $$V(t)$$ denotes the parallel transport of $$v$$ along $$\gamma.$$

Now to the rest of the exercise: Let $$v$$ and $$V(t)$$ such as we obtained it in 2). Then $$$$h(s,t)=exp_{\gamma(t)}(sV(t))$$$$ is a variation of $$\gamma.$$ By the same calculation as in the proof of the Synge-Weinstein-Theorem we get $$$$\frac{1}{2}E''(0)=-\int_0^1K(V(t),\gamma'(t))dt<0$$$$ and so there is a $$s_0$$ such that the closed curve $$c(t):=h(s_0,t)$$ satisfies $$l(c) (details again at the end of the proof of the Synge-Weinstein-Theorem).