$C^\infty$, nonvanishing parallel vector field along geodesic, orthogonal to tangent The following question(s) showed up in my admittedly basic undergraduate research in general relativity/cosmology, and I was wondering if anybody could me with it.
Let $(X, g)$ be a $n$-dimensional Riemannian manifold, and $\gamma: S^1 \to X$ a $C^\infty$ embedded closed geodesic.


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*If $n = 2m$, does there exist a $C^\infty$, nonvanishing parallel vector field $U(t)$ along $\gamma(t)$, which is orthogonal to $\gamma'(t)$?

*What can we say if $X$ is not orientable?

 A: We can always find such a vector field when $n=2m$ and $M$ is orientable.
Proof. Let $\gamma:[0,a]\to M$ be the geodesic in question, with $\gamma(0)=\gamma(a)=p$. Let $P_t:T_{p}M\to T_{\gamma(t)}M$ be the parallel transport along this geodesic. From Picard-Lindelöf, it is clear that $P_a\gamma'(0)=\gamma'(a)=\gamma'(0)$. Thus $P_a$ is an isomorphism of $T_pM$ that fixes $\gamma'(0)$. This means that $P_a$ is a rotation (since it preserves lengths and angles) in the subspace of $T_pM$ orthogonal to $\gamma'(0)$, $T_pM^\bot$. Thus, on $T_pM^\bot$, $P_a\in\mathrm{SO}(2m-1)$. We need the following Lemma: If $n$ is odd, any $O\in\mathrm{SO}(n)$ has at least one fixed point. Thus $P_a\lvert T_pM^\bot$ has at least one fixed point, $U$. Then define $U(t)=P_tU$, which is a vector field along $\gamma$ with the desired properties. $\quad\Box$
Proof of the Lemma. Consider $O\in\mathrm{SO}(n)$ as a matrix, it will have unit determinant. Equivalently, the product of its eigenvalues is one. If $O$ has a complex eigenvalue $\lambda$, it will be accompanied by the complex conjugate eigenvalue $\bar\lambda$ by this theorem. If $\lambda$ is real, we have $\lambda=\pm 1$ because $O$ leaves the length of the eigenvector $v$ invariant. But any negative eigenvalue must be accompanied by another negative one because the determinant (product of eigenvalues) is positive. Then, since $n$ is odd, after taking away the negative and complex eigenvalue pairs we are left with at least one positive eigenvalue $+1$. $\quad\Box$
We get $\mathrm{SO}(n-1)$ instead of $\mathrm{O}(n-1)$ because parallel transport preserves orientation. To see this, let $\omega$ be the Riemannian volume form of $M$ and $E_1,\dotsc,E_n$ an oriented orthonormal basis of $T_pM$. Let $c:[0,a]\to M$, $c(0)=p$, be a smooth curve and $P_t$ its parallel transport. Since $P_t$ preserves angles and lengths, $P_tE_1,\dotsc,P_tE_n$ is an orthonormal basis of $T_{c(t)}M$. Consider the function $f(t)=\omega(P_tE_1,\dotsc,P_tE_n)$. Then, by definition, $f(0)=1$. Suppose that $f(a)<0$, which indicates a reversal of orientation at some point along $c$. By Picard-Lindelöf, $f(t)$ is smooth in $t$,  so by the intermediate value theorem there is a  $t^*\in[0,a]$ such that $f(t^*)=0$. But then $P_{t^*}E_1,\dotsc, P_{t^*}E_n$ is not an orthonormal basis of $T_{c(t^*)}M$, a contradiction.
In the odd-dimensional and nonorientable case, the best we can say is that $P_a\lvert T_pM^\bot\in \mathrm{O}(2m)$, which need not have a fixed point at all. In fact, the existence of such a $U(t)$ is equivalent to the existence of a fixed point. In the even-dimensional nonorientable case, we have $P_a\lvert T_pM^\bot\in \mathrm{O}(2m-1)$, which also need not have a fixed point.
A: Let $v^\mu \equiv \frac{d}{dt} (x^\mu \circ \gamma)$, where $x^\mu$ is a chart, and a tangent vector $u^a$, such that $v^a u_a = 0$, $u^a u_a \neq 0$ for certain $t$. Such a vector always exists for $\dim X > 1$. For $\gamma$ a geodesic, we parallel-transport $u^a$, hence $v^a \nabla_a (v^b u_b) = 0$ for every $t$ in the definition of the curve; orthogonality is conserved along the geodesic.
