Let $E$ be a smooth vector bundle over a manifold $M$ ($\dim M>1$), equipped with a metric. Let $\nabla$ be a metric connection on $E$.
Suppose there exist locally a non-zero section $\sigma \in \Gamma(E)$ which lies in $\ker R(X,Y)$ for all $X,Y \in \Gamma(TM)$. Does $\nabla$ admit a parallel section (locally)?
Note that even if $\| \sigma\|=1$, it is not necessarily true that $\sigma$ parallel. (e.g. if $\nabla$ is flat).
(We have to normalize: the point is that if $\sigma \in \ker R(X,Y)$ so is $f\sigma$ for any function $f$. A parallel section has a constant norm though.)
Clearly, this is a necessary condition:
If $\sigma$ is parallel, then $R(X,Y)\sigma=d_{\nabla}^2\sigma(X,Y)=0$
Consider the reduced holonomy group $Hol^0_p=G$ of $\nabla$ ($p$ is a point in $M$), i.e. the subgroup of the full holonomy group whose elements are defined via parallel transport along null-homotopic loops based at $p$. This is a connected subgroup of the holonomy group. According to Ambrose-Singer theorem (see e.g. Kobayashi-Nomizu's book) endomorphisms of the fiber $E_p$ of the form $Z\mapsto R(X,Y)Z$ span the Lie algebra ${\mathfrak g}$ of $G$. Let $Z$ be a nonzero local section of $E\to M$ near $p$ such that $R(X,Y)Z=0$ for all local sections $X, Y$ of $E\to M$ near $p$. Define the (nonzero) vector $v\in E_p$, $v=Z(p)$. Exponentiating the Lie algebra, we obtain an open neighborhood $U$ of $e\in G$ whose elements fix $v$ (since, by the assumption, $R(X,Y)Z=0$ for all local sections $X, Y$). Let $H$ denote the subgroup of $G$ generated by $U$. Then $H$ is an open subgroup of $G$ which fixes $v$. However, it is also a closed subgroup of $G$ (every open subgroup of a topological group is also closed, this is a nice exercise). Since $G$ is connected, $H=G$. Thus, the entire group $G$ fixes $v$. Now, parallel-translate $v$ in a small (simply-connected) neighborhood $V$ of $$: Since $v$ is fixed by $G$, the parallel transport is independent of the path in $V$. The result is a nonzero local parallel section of $E\to M$.