Topological conditions on manifold from existence of global vector field Does the existence of a (globally-defined, nowhere-vanishing) vector field on a manifold put any topological conditions on the manifold? c.f. the existence of a globally-defined, nowhere-vanishing spinor field implies a reduction in the structure group. e.g. if we take the manifold to be Minkowski space $\mathbb{R}^{3,1}$, does the structure group get reduced by the existence of a global vector field? And if so, to what subgroup is it reduced to?
Update Thinking about it, I guess on a Riemannian manifold with $O(d)$ structure, this gets reduced to $O(d-1)$ by a global vector field. Still, I'd be grateful if anyone could confirm this, and also my second question below still remains.
Further, if the components of this vector field satisfy $\partial_{(\mu} V_{\nu)}=0$, is there any differential condition on the manifold, e.g. a reduction of holonomy or some integrability condition (Ricci-flatness?) or something?
 A: Suppose that $M$ is a smooth connected manifold.
Your question has two sides to it. One is purely topological when $M$ is not given any extra structure.

*

*If $M$ is compact, then $M$ admits a nowhere vanishing vector field if and only if $\chi(M)=0$, where $\chi$ is the Euler characteristic. In one direction, this claim follows from the Poincare-Hopf index formula
$$
\chi(M)=ind(X),
$$
where $X$ is a vector field on $M$ with nondegenerate zeroes, and $ind$ is the index of $X$. The latter is a count of zeroes of $X$ with $\pm$ sign. In particular, if $X$ has no zeroes, then $ind(X)=0$, hence, $\chi(M)=0$.

The opposite direction is also nontrivial, it is proven by starting with a generic vector field $X$ on $M$. Since $ind(X)=0$, all zeroes of $X$ can be grouped in pairs $p_i, q_i, i=1....,k$, where $ind_{p_i}(X)=1, ind_{q_i}(X)=-1$. Connect the points $p_i, q_i$ by simple smooth arcs $\alpha_i$ in $M$ so that the two distinct arcs are disjoint. Lastly, apply a form of the "Whitney trick" to each pair $p_i,q_i$ along the arc $\alpha_i$ "cancelling the two zeroes". See here.


*If $M$ is a noncompact connected manifold, then it admits a nowhere vanishing vectorfield, see here, although I do not like any of the given proofs. The only textbook reference I know is Steenrod, but the result is much more elementary than the general obstruction theory developed there. One day, I will write a self-contained proof. A simple example (which, in many ways suffices for the intuition) is when $M$ admits a vector field with nondegenerate zeroes which all lie on a half-infinite simple proper arc $\alpha$ in $M$, i.e. the image of a smooth proper map $r: [0,\infty)\to M$, where zeroes of $X$ are the images of the nonnegative integers so that $X$ has opposite index at the points
$p_k=r(2k), q_k=r(2k+1)$, $k\in {\mathbb Z}_+$.

Then again group zeroes of $X$ in pairs: $p_k=r(2k), q_k=r(2k+1)$ of opposite index. Now, take arcs $\alpha_k$ in $\alpha$ connecting $p_k, q_k$, where $\alpha_k=r([2k, 2k+1])$. Lastly, apply the "Whitney trick" as above cancelling the pairs of zeroes.
As for the reduction of the structure group $O(d)$ to $O(d-1)$ where $d$ is the dimension of $M$ (I am assuming here that $M$ is given some Riemannian metric, does not matter which one), then, yes, the existence of such $X$ implies such a reduction. The converse is almost true, except the existence of such reduction only implies the existence of a line field. This line field defined a 1-dimensional foliation on $M$. If this foliation is orientable, then taking the unit vector field tangent to the foliation in the "positive direction." In general, such foliation need not be orientable, but this problem can be eliminated by passing to a suitable 2-fold covering space of $M$.


*Suppose, now that $M$ is equipped with an affine connection $\nabla$, for instance, the Levi-Civita connection of a semi-Riemannian metric on $M$. It appears that you are interested in the conditions on $\nabla$ which guarantee the existence of a nonzero parallel vector field $X$ on $M$, i.e. one such that for every tangent vector $v$ in $TM$, $\nabla_v X=0$. This condition is equivalent to the assumption that for every (smooth) path $c$ in $M$, the vector field $X$ is parallel long $c$, i.e. the parallel transport along $c$ preserves $X$. From this, it follows that $X$ is parallel along every loop in $M$. Hence, for every point $p\in M$, the vector $X_p$ is fixed by the holonomy group $Hol_p$ of the connection $\nabla$. In other words, the necessary condition for the existence of such $X$ is that $Hol_p$ has a fixed nonzero vector for some (equivalently, every) $p\in M$. The converse holds as well: if $Hol_p$ has a nonzero fixed vector $v$ then $M$ admits a parallel nonzero vector field. Namely, define $X(p):=v$. Then extend $X$ to points $q\in M$ by parallel transport of $v$ along a path $c$ from $p$ to $q$. Since $v$ is fixed by the holonomy group, this extension is independent of $c$. With a bit more work, one verifies smoothness of such $X$. Again, apart from the orientability issue, such reducibility of the holonomy group is what you asked for.

