# Parallel transport along a closed geodesic

It do Carmo, in exercise 9.4, it is claimed that parallel transport along a closed geodesic in an even-dimensional orientable manifold "leaves a vector orthogonal to the geodesic invariant."

So, let $\gamma:[0,a]\to M$ be this geodesic, with $\gamma(0)=\gamma(a)=p$, and let $V\in T_pM$ be orthogonal to $\gamma'(0)$. Let $P_t$ be the parallel transport along the geodesic. From Picard-Lindelöf, it is clear that $P_a\gamma'(0)=\gamma'(a)$. Let $V(t):=P_t V$. I believe the claim is that $V=P_aV$. It is clear that $\langle V(a),\gamma'(0)\rangle=0$, so $V(a)$ lies in the same $\gamma'(0)$-orthogonal subspace ($T_pM^\bot$) as $V$. Let $E_1,\dotsc,E_{n-1}$ be an orthonormal basis of this subspace. Extend these via parallel transport along the geodesic. Then $E_1(a),\dotsc,E_{n-1}(a):=P_aE_1,\dotsc,P_aE_{n-1}$ is also an orthonormal frame of $T_pM^\bot$. Since $P_t$ preserves orientation (I have already proved this), there is an $O\in\mathrm{SO}(n-1)$ such that $E_i(a)=\sum_{j=1}^{n-1}O_i{}^jE_j$.

At this point I get stuck. I see no reason for $O$ to be the identity or what $M$ being even-dimensional has to do with anything. How do I continue?

You've misinterpreted the statement - you to need to show it leaves one such vector invariant, not all such vectors. (You can reasonably easily produce a counterexample to the stronger statement by constructing a flat $R^3$ bundle over $S^1$ with a twist.)
Thus the problem boils down to showing that every $O \in SO(n-1)$ has a fixed point, which is true exactly when $n-1$ is odd. If you're not familiar with this fact, you can establish it by studying the eigenvalues of $O$, remembering that any complex ones must come in conjugate pairs and that their product (the determinant) is positive.