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.

  • $\begingroup$ can you do this for $S^2$ and $\gamma$ the equator? $\endgroup$
    – Will Jagy
    Aug 22, 2018 at 1:39

1 Answer 1


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 \begin{equation} \omega_{\gamma (t)}(v_1,...,v_{n-1},\gamma'(t))>0. \end{equation} 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 \begin{equation} \omega_{\gamma (t)}(E_1(t),...,E_{n-1}(t),\gamma '(t))\neq 0 \end{equation} 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, \begin{equation} \omega_{\gamma (t)}(E_1(t),...,E_{n-1}(t),\gamma '(t))>0. \end{equation} 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 \begin{equation} V(1)=A(v)=v, \end{equation}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 \begin{equation} h(s,t)=exp_{\gamma(t)}(sV(t)) \end{equation} is a variation of $\gamma.$ By the same calculation as in the proof of the Synge-Weinstein-Theorem we get \begin{equation} \frac{1}{2}E''(0)=-\int_0^1K(V(t),\gamma'(t))dt<0 \end{equation} and so there is a $s_0$ such that the closed curve $c(t):=h(s_0,t)$ satisfies $l(c)<l(\gamma)$ (details again at the end of the proof of the Synge-Weinstein-Theorem).


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