Let $(M,g)$ be a Riemannian manifold. We equip the tangent bundle $TM$ with the Sasaki metric $g_s$. Assume that $X: M \to TM$ is a vector field on $M$.

We say that $X$ is an orthonormal vector field if $X^* g_s=g$. We say that $X$ is a generalized orthonormal vector field if $X^* g_s=\lambda g$ for some constant $\lambda$. The motivation for such a terminology is that an orthonormal matrix $A\in M_n(\mathbb{R})$ defines a vector field $A:\mathbb{R}^n\to \mathbb{R}^n\times \mathbb{R}^n =T\mathbb{R}^n$ which pulls back the satandard metrics, that is the Sasaki metric associated to the Euclidean metric, to the original Euclidean metric of $M=\mathbb{R}^n$.

Questions: Let $(M,g)$ be a compact Riemannian manifold whose Euler characteristic is zero. Does $M$ admit a non vanishing orthonormal vector field?

The second question:

Let $(M,g)$ be an arbitrary Riemannian manifold with a (generalized) orthonormal vector field $X$ on $M$ with singular set $X$. Is the restriction of the flow of $X$ to $M\setminus S$ a geodesible flow?(A flow whose trajectories are unparametrized geodesics for a Riemannian metric)?

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    $\begingroup$ The first that comes to mind is the observation that $X$ is orthonormal if and only if for any $p \in M$ the differential of $X$ (considered as a map $M \to TM$) at $p$ sends $T_p M$ to the horizontal subspace $H_{(p, X_p)} \subseteq T_{(p, X_p)} (TM)$ (one just needs to unravel the definition of the Sasaki metric). But then it follows that $dX_p \colon T_p M \to H_{(p, X_p)}$ is an isometric isomorphism. $\endgroup$ – Ivan Solonenko Dec 10 '18 at 12:21

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